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Proofs

(An Alternative to Discrete ) G. Viglino Ramapo College of New Jersey January 2018

CONTENTS

Chapter 1 A LOGICAL BEGINNING 1.1 1 1.2 Quantifiers 14 1.3 Methods of 23 1.4 of 33 1.5 The Fundamental of 43 Chapter 2 A TOUCH OF 2.1 Basic Definitions and Examples 53 2.2 Functions 63 2.3 Infinite Counting 77 2.4 Equivalence Relations 88 2.5 What is a Set? 99 Chapter 3 A TOUCH OF ANALYSIS 3.1 The System 111 3.2 123 3.3 Metric Structure of  135 3.4 Continuity 147 Chapter 4 A TOUCH OF 4.1 Metric 159 4.2 Topological Spaces 171 4.3 Continuous Functions and 185 4.4 Product and Quotient Spaces 196 Chapter 5 A TOUCH OF THEORY 5.1 and Examples 207 5.2 Elementary Properties of Groups 220 5.3 Subgroups 228 5.4 Automorphisms and 237 Appendix A: Solutions to CYU-boxes Appendix B: Answers to Selected Exercises PREFACE The underlying goal of this text is to help you develop an appreciation for the essential roles and definitions play throughout mathematics. We offer a number of paths toward that goal: a path (Chapter 1); a path (Chapter 2); an Analysis path (Chapter 3); a Topology path (Chapter 4); and an Algebra path (Chapter 5). You can embark on any of the last three independent paths upon comple- tion of the first two:

pter 3 Cha alysis h of An . A Touc y Ch 2 or A A To apter er he uch 4 pt T to of T a et u C opolo h S ch gy C of ha h of p uc A te Chapter 1 To b r A st 5 .A Logical Beginning. ra c t A lg eb ra

A one-semester course may include one, two, or even all three of the last three chapters, in total or in part, depending on the preparation of the students, and the ambition of the instructor. For our part, we have made every effort to assist you in the journey you are about to take. We did our very best to write a readable book, without compromising mathematical integrity. Along the way, you will encounter numerous Check Your Understanding boxes designed to challenge your understanding of each newly-introduced concept. Detailed solutions to each of the Check Your Understanding problems appear in Appendix A, but you should only turn to that appendix after making a valiant effort to solve the given prob- lem on your own, or with others. In the words of Desecrates: We never understand a thing so well, and make it our own, when we learn it from another, as when we have discovered it for ourselves. There are also plenty of exercises at the end of each section, covering the gamut of diffi- culty, providing you with ample opportunities to further hone your mathematical skills. Exercises that play a role in future developments are marked with a sharp (#). I wish to thank my colleague, Professor Marion Berger, for her numerous comments and suggestions throughout the development of the text. Her generous participation is deeply appreciated. I am also grateful to Professor Maxim Goldberg-Rugalev for his invaluable input.

1.1 Propositions 1

1 CHAPTER 1 A Logical Beginning Mathematics strives, as best it can, to transpose our physical into a non-physical form—a thought-universe, as it were, where defini- tions are the physical objects, and where the reasoning process rules. At some , one has to ask whether we humans are creating mathemat- ics or whether, through some kind of wonderful mental telescope, we are capable of discovering a meager portion of the mathematical uni- verse which, in all of its glory, may very well be totally independent of us the spectators. Whichever; but one thing appears to be clear: the nature of mathematics, as we observe it or create it, appears to be inspired by what we perceive to be intuitive logic. §1. PROPOSITIONS We begin by admitting that “True” and “” are con- cepts, and that the word “sentence” is also undefined. Nonetheless: DEFINITION 1.1 A (or ) is a sentence PROPOSITION that is True or False, but not both. Fine, we have a definition, but its meaning rests on undefined words! We feel your frustration. Still, mathematics may very well be the closest thing to perfection we have. So, let’s swallow our pride and proceed. We can all agree that 53+8= is a True proposition, and that 53+9= is a False proposition. But how about this sentence: THE SUM OF ANY TWO ODD IS EVEN. First off, in order to understand the sentence one has to know all of its words, including: sum, , odd, and even. Only then can one hope to determine if the sentence is a proposition; and, if it is, to establish whether it is True or False. All in good time. For now, let us consider two other sentences: HE IS AN AMERICAN CITIZEN. This sentence cannot be evaluated to be True or False without first knowing exactly who “He” is. It is a variable proposition (more formally called a predicate), involving the variable “He.” Once He is specified, then we have a proposition which is either True or False. THE TWO OF US ARE RELATED. Even if we know precisely who the “two of us” are, we still may not have a proposition. Why not? Because the word “related” is just too vague. 2 Chapter 1 A Logical Beginning

COMPOUND PROPOSITIONS The two numbers 3 and 5 can be put together in several ways to arrive at other numbers: 35+ , 35– , 35 , and so on. Similarly, two propo- sitions, p and q, can be put together to arrive at other propositions, called compound propositions. One way, is to “and-them:” In table form: DEFINITION 1.2 Let p, q be propositions. The conjunction of p q pq p and q p and q (or simply: p and q), written pq , T T T pq is that proposition which is True if both p and T F F q are True, and is False otherwise. F T F For example, the proposition: F F F 75 and 35+8= is True. (since both 75 and 35+8= are True) The proposition: 75 and 35+9= is False. (since 75 and 35+9= are not both True) Another way to put two propositions together is to “or-them:” In form: DEFINITION 1.3 Let p, q be propositions. The disjunction of p p q pq and q (or simply: p or q), written pq , is T T T p or q T F T pq that proposition which is False when both p F T T and q are False, and is True otherwise. F F F For example, the proposition: 75 or 35+8= is True, as is the Note that p or q is True when both p and q are True. As such, it proposition 75 or 35+9= . The proposition 75 or 35+9= is said to be the inclusive-or. The exclusive-or is typically used in is False, since neither 75 nor 35+9= is True. conversation, as in: Do you want coffee or tea (one or the other, but not both) CHECK YOUR UNDERSTANDING 1.1 Let p be the proposition 75= , and q be the proposition 35+8= . (a) True (b) False (a) Is pq True or False? (b) Is pq True or False? The above and/or operators, which act on two given propositions, are said to be binary operators. The following operator is a unary operator in that it deals with only one given proposition: In truth table form: DEFINITION 1.4 Let p be a proposition. The of p, p  p NOT P read “not p,” and written  p , is that proposi- T F tion which is False if p is True, and True if p F T is False. For example, the proposition: ~7 5 is True, since the proposition 75 is False. The proposition: ~3+8 5= is False, since the proposition 35+8= is True. 1.1 Propositions 3

CHECK YOUR UNDERSTANDING 1.2 Indicate if the proposition  p is True of False, given that (a) p is the proposition 53 . (a) False (b) False (b) p is the proposition  q , where q is the proposition 35= . The negation operator “~” takes precedence (is performed prior) to both the “ ” and the “ ” operators. For example, the proposition:  35  35+9= is True. [since  35 is True] As is the case with algebraic expressions, the order of operations in a logical expression can be overridden by means of parentheses. Brackets”[ ]”play the same role as parentheses, and are For example, the proposition: used to bracket an expres- sion which itself contains  53  35+8= is False parentheses. [since 53  35+8= is True ] CHECK YOUR UNDERSTANDING 1.3 Assume that p and q are True propositions, and that s is a False prop- osition. Determine if the given compound proposition is True or False. (a)  pq (b)  pq (c)  sq (d)  sq

True: d, f, g, h, j (e)  pq (f)  pq (g)  sq (h)  sq False: a, b, c, e, i, k (i) ps  qs (j) ps  qs (k)  ps  qs

CONDITIONAL STATEMENT Consider the sentence: IF IT IS MONDAY, THEN JOHN WILL CALL HIS MOTHER. For it to become a proposition, we must attribute logical values to it (True or False); and so we shall: DEFINITION 1.5 Let p, q be propositions. The conditional of q if p then q by p, written pq , and read if p then q, is False if p is True and q is False, and is True In truth table form: pq otherwise. p q pq p is called the and q is T T T called the conclusion of pq . T F F F T T You probably feel comfortable with the first two rows in the adjacent F F T truth table but maybe not so much with the last two. Why is it that for p False, the proposition pq is specified to be True for every proposi- tion q, whether q is True or False? 4 Chapter 1 A Logical Beginning

In an attempt to put your mind a bit at ease, let’s reconsider the condi- tional proposition: IF IT IS MONDAY, THEN JOHN WILL CALL HIS MOTHER. Suppose it is not Monday. Then John may or may not call his mother. So, if p is False (it is not Monday), then pq should not be assigned a value of False. Yes, but can’t the same “” be given to the assertion that if p is False, then pq should not be assigned a value of True? It could, but that option would lead us up an intuitively illogical path (see Exercise 85). Does the statement: IF IT IS MONDAY, THEN I WILL GO TO SCHOOL allow you to go to school on Tuesday? Sure, but the following state- ment does not: I WILL GO TO SCHOOL IF AND ONLY IF IT IS MONDAY Or, if you prefer: IT IS MONDAY IF AND ONLY IF I GO TO SCHOOL Formally:

p q pq DEFINITION 1.6 Let p, q be propositions. The biconditional of T T T p if and only if q p and q, written pq , and read p if and T F F only if q (abbreviated “p iff q), is True if p F T F p iff q and q have the same truth values and is False F F T pq if they have different truth values.

ORDER OF OPERATIONS First “ ”, then “ ,  ”, and finally “ , ” Use parentheses to establish precedence between  ,  and between  , 

LOGICAL IMPLICATION AND EQUIVALENCE A is a proposition which is necessarily always True; as is the case with the proposition pq  p  q : p q pq pq  p pq  p  q The third column follows from Definition 1.5; the fourth from TT T T T Definition 1.2, and the fifth from Definition 1.5 TF F F T FT T F T FF T F T

Definition 1.5 1.1 Propositions 5

We use the pq , read p implies q, to mean that if the prop- Subjected to the logical impli- osition p is True, than q must also be True; equivalently, that pq is cation “ ” the proposition: If pigs can fly then Donald a tautology: Duck is the president of the United States For given propositions p and q, the only thing that is True. can keep pq from being a tautology is if p is It is a well-known that True and q is False (Definition 1.5). most ignore flying pigs. Typically, they In particular, as we have just observed: pq  p  q . start with a True proposition in the hope of being able to estab- lishing the of another. Note that pq is not a proposition for it does not assume values of True or False. It is an assertion (sometimes called a meta-proposition).

It is only logical: If p or q is True and p is EXAMPLE 1.1 (a) Show that pq  ~ p  q . False, then q must be True. (b) Show that pq  qs  ps .

SOLUTION: (a) We verify that pq  ~ p  q is a tautology: The Truth-Values in the fifth column are a conse- p q pq  p pq  ~ p pq  ~ p  q quence of the Truth-Values in columns 3 and 4 (Defini- T TTF F T tion 1.2). The tautology (column 6) follows from T FTF F T Definition 1.5. F TTT T T F FFT F T

Definition 1.5

(b) We verify that pq  qs  ps is a tautology:

p q s pq qs pq  qs ps pq  qs  ps TTT T T TT T TTF T F FF T TFT F T FT T TFF F T FF T FTT T T TT T FTF T F FT T FFT T T TT T FFF T T TT T

CHECK YOUR UNDERSTANDING 1.4

Verify: Answer: See page A-1. pq  ~q  ~p 6 Chapter 1 A Logical Beginning

DEFINITION 1.7 Two propositions p and q are logically equiv- LOGICALLY alent, written pq , if pq and qp . EQUIVALENT In other words: pq if the proposition pq is a tautology; or, equivalently: p is True if and only if q is True (Definition 1.6).

Augustus DeMorgan THEOREM 1.1 (a) ~pq  ~p  ~q (1806-1871). DEMORGAN’S LAWS (b) ~pq  ~p  ~q PROOF: (a)

The driving force in the table p q  p  q pq pq ~p ~q was to arrive at the truth-values of the compound propositions: TTFF T F F pq and ~p ~q TFFT F T T FTTF F T T FFTT F T T same Observing that the values of  pq and ~p ~q coincide, we conclude that ~pq  ~p  ~q . (b) ~pq  ~p  ~q : p q  p  q pq  pq ~p ~q TTFF T F F TFFT T F F FTTF T F F FFTT F T T same EXAMPLE 1.2 Use DeMorgan’s laws to negate the given prop- osition: (a) Mary is a math major and lives on campus. (b) Bill will go running or swimming. (c) 5  x  19 . SOLUTION: (a)  Mary is a math major  Mary lives on campus  ~ Mary is a math major  ~ Mary lives on campus : Mary is not a math major or she does not live on campus (Either Mary is not a math major or she does not live on campus.) 1.1 Propositions 7

(b)  Bill will go running  Bill will go swimming  ~ Bill will go running  ~ Bill will go swimming : Bill will not go running and he will not go swimming (Bill will neither go running nor swimming) (c) The negation of the proposition 5  x  19 : 5  x AND x  19 is the proposition 5  x OR x  19 equivalently: 5  x OR x  19

CHECK YOUR UNDERSTANDING 1.5 Use DeMorgan’s laws to negate the given proposition: (a) Mary is going to a movie or she is going shopping. (b) Bill weighs more than 200 pounds and is less than 6 feet tall. Answer: See page A-2. (c)x  0 or x  –5 . There are several distributive properties involving propositions. Here is one of them:

THEOREM 1.2 pqs   pq  ps

In general, a truth table PROOF: The eight possible value-combinations of p, q, and s, are involving n propositions will listed in the first three columns of the table below. The remaining col- n contain 2 rows. That being umns speak for themselves: the case, they quickly become “unmanageably tall”: A four-proposition-table 4 calls for 2 = 16 rows, p q s qs pqs  pq ps pq  ps while a five-proposi- tion-table already has 32 TTT T T TT T rows. TTF F T TT T TFT F T TT T TFF F T TT T FTT T T TT T FTF F F TF F FFT F F FT F FFF F F FF F same

EXAMPLE 1.3 Are p  q   ~ p and pq  ~ p logi- cally equivalent? 8 Chapter 1 A Logical Beginning

SOLUTION: No: p q pq  p pq  ~p q  ~p pq  ~p TT T F FFF TF F F TTT FT F T T T F FF F T T T F

not the same truth values, so not logically equivalent

CHECK YOUR UNDERSTANDING 1.6 Verify: (a) ~~p p (b) pq  ~p  q Answer: See page A-2. (c) ~pq  ~~p q (d) pq   pq  qp

To say that pq is Not True is to say that p is True and q is False. To put it another way:

THEOREM 1.3 NEGATION OF A CONDITIONAL PROPOSITION: ~pq  p  ~q

PROOF: We offer two proofs for your consideration: Truth Table Method: Using Established Results CYU 1.6(b) p q pq ~pq ~q p  ~q proof aside, we ~pq  ~~pq suggest that this result TT T F F F is intuitively appealing. 1.1:  ~~p ~q After all, to say that it is TF F T T T not true that p implies CYU 1.6(a)  p  ~q q, is to say that p is true FT T F F F while q is false. FF T F T F

While the above truth table method is the more straight-forward approach, the development on the right is a better representation of what you will be doing in subsequent math courses. Indeed, truth be told: Once you leave this course, you may never again encounter a truth table.

CONTRAPOSITIVE Are the following two statements saying the same thing? IF IT IS MONDAY, THEN I WILL GO TO SCHOOL. IF I DO NOT GO TO SCHOOL, THEN IT IS NOT MONDAY. Yes: This is a very useful result. It tells you that: THEOREM 1.4 pq  ~q  ~p as pq goes so does ~q  ~p 1.1 Propositions 9

PROOF: p q pq ~p ~q ~q  ~p TT TFF T TF FFT F FT TTF T FF TTT T same

We note that the proposition ~q  ~p is said to be the contrapositive of the proposition pq .

Getting a bit ahead of the game, we point out that by virtue of 1.4, one can establish the validity of the proposition: If n2 is even then n is even by verifying that: If n is odd, then n2 is not even

CHECK YOUR UNDERSTANDING 1.7

Prove that pq  ~s  s  ~p  ~q using: Answer: See page A-3. (a) A truth table. (b) Previously established results.

CONVERSE AND INVERSE OF A CONDITIONAL STATEMENT

DEFINITION 1.8 The converse of pq is qp . The inverse of pq is ~p ~q . For example: Proposition pq If it is Monday, then I will go to school. Converse qp If I go to school, then it is Monday. Inverse ~p ~q If it is not Monday, then I will not go to school. It is important to note that neither the converse nor the inverse of a conditional proposition is logically equivalent to the given proposition: p q pq qp  p  q ~p  ~q TT T T F F T TF F T F T T FT T F T F F FF T T T T T

not the same not the same 10 Chapter 1 A Logical Beginning

Truth table aside, consider the True statement: If n is divisible by 4, then n is even. Its converse is False: If n is even, then n is divisible by 4. (6 is even but is not divisible by 4) Its inverse is also False: If n is not divisible by 4, then n is not even. (6 is not divisible by 4 but is even)

So, the inverse is the con- CHECK YOUR UNDERSTANDING 1.8 trapositive of the converse. Prove that the converse and the inverse of pq are logically equiv- Answer: See page A-3. alent.

SET NOTATION A bit of set notation is needed for the development of the remainder of The order in which ele- this chapter. Roughly speaking, a set is a collection of objects, or ele- ments appear in a set is of ments. A “small” set can be specified by simply listing all of its elements no consequence. The sets 123 and 231 inside brackets, as is done with the set A below: for example, are consid- A = –251199  ered to be one and the same, or equal. The above so-called roster method can also be used to denote a set whose elements follow a discernible pattern, as is done with the set O of odd positive integers below: O = 13579 The descriptive method can also be used to define a set. In this method, a statement or condition is used to specify the elements of the set, as is done with the set O below: Oxx=  is an odd positive integer

Read: O is the set of all x such that x is an odd positive integer For a given set A, xA , is read: x is an of A (or x is con- tained in A), and xA is read: x is not an element of A. For example: 5 – 251199  while 9 – 2 5 11  99

Finally, we note that throughout the text:  denotes the set of real numbers. + denotes the set of positive real numbers. Z denotes the set of integers: Z = –23 –1 –  0123 .

Z+ denotes the set of positive integers: Z+ = 1234 . Q denotes the set of rational numbers (“fractions”). Q+ , the set of positive rational numbers. 1.1 Propositions 11

EXERCISES

Exercises 1-10. State whether the given proposition is True or False. 1. 10 is an even number. 2. 15 is an even number. 3. 10 or 15 is an even number. 4. 10 and 15 are even numbers.

5.32 or (57 or 9 is odd) 6.32 and (57 or 9 is odd)

7.32 or (57 and 9 is odd) 8. (32 and 57 ) or 9 is odd

9. (32 or 57 ) and 9 is odd 10. (32 and 57 ) or 9 is not odd

Exercises 11-31. Assume that p and q are True propositions, and that r and s are False propositions. Determine if the given compound proposition is True or False. 11.ps 12.~ps 13. ~p  ~s

14.rs 15.~rs 16. ~r  ~s

17.rs 18.~rs 19. ~r  ~s

20.ps  rs 21.ps  rs 22. pq  ~r  ~s

23.pq  rs 24.pq  rs 25. p  ~r  qs

26.~pq  ~rs 27.~pq  rs 28.  pq  ps

29.pq  ~rs 30.~pr  ~ps 31. ~pq  ~r  ~s 

Exercises 32-39. Determine if the given statement is a tautology. 32.p  ~p 33. p  ~p  q

34.p  ~p  q 35. p  ~p  q

36.~pq  p  ~q 37. pq  ~p  p  ~q

38.pq  s  pqs  39. pq  s  ~pqs 

Exercises 40-43. Proceed as in Example 1.1 to establish the given implication.

40.pq  p 41. pq  p ~q

42. pq  ~p  q  q 43. p  ~p  q  pq

Exercises 44-59. Determine if the given pair of statements are logically equivalent. 44.ppq  , p 45.ppq  , q

46.p  ~q   p , p 47.p  ~q   p , q

12 Chapter 1 A Logical Beginning

48.~p  ~q  , q  ~p 49. ~p  ~q   ~p  ~q  ,  q

50.~p  ~pq , ~p  ~q 51.~p  ~pq , ~p  ~q

52.~pq , p  ~q 53.~pq , ~q  ~p

54.pq  ps , pqs  55.pq  ps , pq  s

56.pr  qs , pq  s 57.ps  qs , pqs 

58.pq  qs , ps 59.pq  qs , ps

Exercises 60-62. Establish the given logical equivalence. 60. Commutative Laws: 61. Associative Laws: (a) pq  qp (a) pq  r  pqr  (b) pq  qp (b) pq  r  pqr 

(a) pqr   pq  pr 62. Distributive Laws: (b) pqr   pq  pr Already proved (Theorem 1.2)

Exercises 63-65. Indicate True or False. For any proposition p and any tautology t: 63.pt  p 64.pt  p 65. pt  t

66. . A contradiction is a proposition which is False for all possible values of the statements from which it is constructed; as is the case with p  ~p : p ~p p  ~p T FF F TF (a) Verify that ~pq  p  ~q is a contradiction. (b) Verify that pq  ~p  p  ~q is a contradiction. (c) Indicate True or False. For any proposition p and any contradiction c: (i) pc  p (ii) pc  c (iii) pc  p (d) Indicate True or False. For any tautology t and any contradiction c: (i) tc is a tautology. (ii) tc is a tautology. (iii) tc is a contradiction. (iv) tc is a contradiction. (e) Indicate True or False: ~tc  p is a contradiction for any tautology t, contradiction c, and proposition p. Exercises 67-74. Negate the given statement. 67. Joe is a math major and Mary is a biology major. 68. Joe is a math major or Mary is a biology major. 1.1 Propositions 13

69. Joe is neither a math major nor a biology major. 70. Mary does not live in the dorms but does eat lunch in the cafeteria. 71.3x + 5 = y and both x and y are integers. 72. x is divisible by both 2 and 3, but is not divisible by 7. 73. x is not divisible by either 2 or 3, but is divisible by 7. 74. x is greater than y and z, but is less than yz– . Exercises 75-83. Formulate the contrapositive, converse, and inverse of the given conditional statement. 75. If it rains, then I will stay home. 76. If it does not rain, then I will go to the game. 77. If Nina feels better, then she will either go to the library or go shopping. 78. If either Jared or Tommy gets paid, then the two of them will go to the concert.

79. If XZ= then MN .

80. If XZ then M is a solution of the given equation.

81. If XZ then M or N is a solution of the given equation.

82. If XZ then M and N are solutions of the given equation.

83. If XZ then the given equation has no solution. 84. Exclusive-Or. Let p and q be propositions. Define, via a Truth Table, the exclusive-or propo- sition: p or q but not both p and q. 85. In defense of Definition 1.5. (a) Prove that If we chose to define: p q pq Then pqqp   T T T (which runs contrary to T F F our logical instincts) F T F F F T (b) Prove that If we chose to define: p q pq Then pq  ~q  ~p T T T (which runs contrary to T F F our logical instincts) F T T F F F (c) What if we chose to define: p q pq Anything “bad” happens? T T T T F F F T F F F F

14 Chapter 1 A Logical Beginning

1

Chances are you already know most of the material §2. QUANTIFIERS of this section. Not neces- sarily because you for- In the previous section, we considered the variable proposition: He is mally saw it before, but an American citizen. As it stands, the sentence is not a (determined) because it is, well, “logi- cal.” You know, for exam- proposition, for its depends on the variable “He.” Here is ple, that to disprove the another variable proposition, px : claim that every dog has fleas one needs to exhibit px: x + 515 a specific dog which has no fleas. On the other Once we substitute a number for the variable x, we arrive at a proposi- hand, finding a dog with tion which can be evaluated to be True or False. For example: fleas will not prove that all dogs have fleas. p7 is False, since 7+ 5 15 while p25 is True, since 25+ 5 15 Now consider the sentences: For all real numbers x, x + 515 (*) For some real numbers x, x + 515 (**) Are they propositions? Yes: (*) is a False proposition since it fails to hold for the number 7 (among other numbers), while (**) is a True proposition since it holds for the number 25 (among others). In both cases a variable x is present, but in each case it is quantified by a condition: “For all” in (*), and “For some” in (**). Lets begin by focusing on the “For all” sit- uation:

THE UNIVERSAL :  The symbol  , called the universal quantifier, is read “for all.” The proposition x   x + 515 is False, We remind you that  since it does not hold for every real number x. denotes the set of real num- bers, and that x   is read: The proposition x   x2  0 is true, since it x is an element of  ; which does hold for every real number x. is to say, x is a real number. In general: DEFINITION 1.9 Let px be a variable proposition, and let X denote the set of values which the variable x UNIVERSAL can assume (called the domain of x). PROPOSITION If px is True xX , then the universal proposition xXpx   is True. If px is false for at least one xX , then the universal proposition xXpx   is False (any such px is said to be a coun- terexample). The above is a long definition, but it’s really quite straightforward. Consider the following examples. 1.2 Quantifiers 15

Indicate if the given proposition is True or False. We remind you that Z denotes EXAMPLE 1.4 the set of all integers, and that Justify your claim. + Z denotes the set of all posi- + tive integers. (a) nZ  2nn + The expression nm  Z in (c) is shorthand for: (b) nZ  2nn + + nZ and mZ (c) nm  Z+ nmnm+ 

+ SOLUTION: (a) We show nZ  2nn is True: For any nZ + : 2nnnn==+  + 0 n (b) We show that the proposition nZ  2nn is False by exhibit- ing a specific couterexample: For n = –5 , 2n ==25– –10 – 5 Any negative integer whatsoever can be used to show that the given proposition is False. (c) The proposition nm  Z+ nmnm+  is also False. We need to display a counterexample — a SPECIFIC pair of integers, n and m, for which nmnm+  . Here is one such pair: n = 1 and m = 2 : nm+12==+3 while nm ==12 2

CHECK YOUR UNDERSTANDING 1.9 Indicate if the given proposition is True or False. If False, exhibit a counterexample. (a) All months have at least thirty days. (b) Every month contains (at least) three Sundays. (c) nm  Z+ nm+  Z+ . (b) and (d) are True (a) and (d) are False (d) nZ + and m  Zn + m Z+

THE EXISTENTIAL QUANTIFIER:  The symbol “ ,” called the existential quantifier, is read “there exists.” The proposition nZ + such that n + 3 = 5 is True, since 2  Z+ and 23+5= . The proposition nZ + such that n + 5 = 3 is False, for there does not exist a positive integer which when added to 5 yields 3. We point out that the symbol “ ” is read “such that.” For example: nZ +  n + 5 = 3 translates to: nZ + such that n + 5 = 3 and: x  Xpx  translates to x  X such that px . 16 Chapter 1 A Logical Beginning

In general: DEFINITION 1.10 Let px be a variable proposition with The proposition: domain X. x  Xpx  EXISTENTIAL If px is True for at least one xX , then the can be further abbreviated: PROPOSITION x  Xpx  existential proposition x  Xpx  is True. If px is false for every xX , then the exis- tential proposition x  Xpx  is False.

EXAMPLE 1.5 Indicate if the given proposition is True or False. (a) nZ  2nn – (a) x    2x = x

(b) nZ +  2n = n SOLUTION: (a) nZ  2nn – is True: for n = –5 , 2n ==–10 5 –n . Any negative integer whatsoever can be used to establish the validity of this proposition (b) x    2x = x is True: for x = 0 , 2x ===20 0 x . Only the number 0 can be used to establish the validity of this proposition. One is all we needed! (c) nZ +  2n = n is False: for any positive integer n, 2nn .

CHECK YOUR UNDERSTANDING 1.10 Indicate if the given proposition is True or False. If True, verify. (a) There exists a month with more than thirty days. (b) There exists a week with more than seven days. (c) nm  Z+  nm+ = 100 (a), (c), and (d) are True. (b) is False. (d) nm  Z+  nm = nm+

PROPOSITIONS CONTAINING MULTIPLE QUANTIFIERS When a statement contains more than one quantifier, the left-most quantifier takes precedence. For example, the proposition: nZmZnm     + = 5 is True, since: For every integer n, there does exists an integer m such that nm+5= ; namely, m = 5 – n .

On the other hand, the proposition: n  ZmZnm    + = 5 is False, since: For any given integer n we can find some m for which nm+  5 . One such m is 6 – n : n + 6 – n  5 . 1.2 Quantifiers 17

The four mathematical sen- EXAMPLE 1.6 Indicate if the given proposition is True or False. tences of this example are + + nice and compact. Their (a) nm  Z sZ  ns = m. may call for a + less compact consideration. (b) nm  Z s    ns = m. Please try to arrive at each + answer before looking at (c) nZ   mZ nm . our solution. (d) nZ   mZ nm . SOLUTION: (a) Here is what the proposition “nm  Z+sZ +  ns = m ”is saying: For all positive integers, n and m, there exists a positive integer, s, such that ns= m. Is it True? No: For n = 2 and m = 1 , there does not exist a posi- tive integer s such that ns= m ; which is to say, there is no positive integer s such that 2s = 1

(b) nm  Z+ s    ns = m is True: + m m For given nm  Z let s = ----   . Then: ns== n  ---- m . n n

(c) Here is what the proposition nZ   mZ + nm is saying: There is an integer that is smaller than every positive integer. We had to exhibit a par- ticular n, and went with This proposition is True: n = –7 . Any non-positive The integer n = –7 is smaller than every positive integer. integer would do as well. (d) The proposition nZ   mZnm   is False: For any given nZ , the integer mn= – 1 is less than n.

CHECK YOUR UNDERSTANDING 1.11 Indicate if the given proposition is True or False. Justify your claim. (a) nm  Z+ sZ +  snm . (b) sZ + nm  Z+  snm . (c) x    nZ + xn  x. (a), and (d) are True. (b) and (c) are False (d) nZ +  mZ + nm = n .

18 Chapter 1 A Logical Beginning

NEGATION OF QUANTIFIED PROPOSITIONS The negation of the proposition: All dogs have fleas. is NOT: No dog has fleas. It IS the proposition: At least one dog does not have fleas. In general: The negation of the universal proposition: xXpx   is the existential proposition: x  X  ~px

EXAMPLE 1.7 Negate the given proposition. (a) All things go bang in the night. (b) x  Xx is this and x is that. (c) x  Xx is this or x is that.

SOLUTION: (a) The negation of: All things go bang in the night. is: Some thing does not go bang in the night. (b) The negation of: x  X , x is this and x is that. is:xX such that x is not this or x is not that. (possibly neither) (c) The negation of: x  X , x is this or x is that. is: xX such that x is not this and x is not that.

CHECK YOUR UNDERSTANDING 1.12 Write an existential negation of the given universal proposition. (a) All college students study hard. (b) Everyone takes a bath at least once a week. (c) x  Xpx  qx Answer: See page A-4. (d) x  Xpx  qx The negation of the proposition: Some music rocks. Is the proposition: No music rocks (or: all music does not rock) In general: The negation of the existential proposition: x  Xpx  is the universal proposition: xX  ~px 1.2 Quantifiers 19

EXAMPLE 1.8 Negate the given proposition. (a) There is a perfect person. (b) x  X such that x is this and x is that. (c) x  X such that x is this or x is that. SOLUTION: (a) The negation of: There is a perfect person. is: Nobody is perfect. (b) The negation of: x  X such that x is this and x is that. is: xX , x is not this or x is not that. (possibly neither) (c) The negation of: x  X such that x is this or x is that. is:xX , x is not this and x is not that.

CHECK YOUR UNDERSTANDING 1.13 Write a universal negation of the given existential proposition. (a) There are days when I don’t want to get up. (b) x  Xpx  qx Answer: See page A-4. (c) x  Xpx  qx The negation of the universal/existential proposition: In every day there is a beautiful moment. Is the existential/universal proposition: Some day contains no beautiful moment. In general:

The notation pxy is used The negation of the universal/existential proposition: to indicate that the truth x  Xy   Ypxy  value of the proposition, p, is a of two vari- is the existential/universal proposition: ables, x and y. xX   yY  ~pxy

EXAMPLE 1.9 Negate the given proposition. (a) To every problem there is a solution. (b) x   nZ +  nx SOLUTION: (a) The negation of: To every problem there is a solution. is: There is a problem that has no solution. (b) The negation of: x   nZ +  nx is: x    nZ + nx 20 Chapter 1 A Logical Beginning

CHECK YOUR UNDERSTANDING 1.14 Write an existential/universal negation of the given universal/exis- tential proposition. (a) For every xX there exists a yY such that y blips at x.

Answer: See page A-4. (b) x   nZ +  x = 2n The negation of the existential/universal proposition: There is a girl at school who can jump higher than every boy. Is the universal/existential proposition: For any girl in school, there is a boy who can jump as high or higher than that girl. Or: No girl in school can jump higher than every boy. In general: The negation of the existential/universal proposition: x  XyYp    xy is the universal/existential proposition: x  Xy   Y ~pxy

EXAMPLE 1.10 Negate the given proposition. (a) Some cheetahs run faster than every gazelle. (b) x    nZ + 0  xn  100

(c) x    nZ + xn  0 or xn  100 SOLUTION: (a) The negation of: Some cheetahs run faster than every gazelle. is: For any cheetah there is a gazelle that runs as fast or faster than that cheetah. (b) The negation of: x    nZ + 0  xn  100 is:x   n  Z+  xn  0 or xn  100 .

(c) The negation of: x    nZ + xn  0 or xn  100 is:x   n  Z+  0  xn  100 .1

CHECK YOUR UNDERSTANDING 1.15 Write a universal/existential negation of the given existential/univer- sal proposition. (a) There is a motorcycle that gets better mileage than any car. Answer: See page A-4. (b) nZ   mZ +nm 1.2 Quantifiers 21

EXERCISES

Exercises 1-51. Indicate if the given statement is True or False. If you indicate False, then justify your claim by means of a specific counterexample. To illustrate: ab : ab+ 2 = a2 + b2 is False: 32+ 2 ==25 while 32 +1322 . 1.nZ  –nn + 1 2. r   –rr 3.n  Znn –  + 1 4. r   –rr

5. nm  Znm  n+ m 6. nm  Z+ nm Z

7. n m  ZnmZ  + 8. nm  Z: nm or mn 9.ab  ab+ = ab+ 10. ab  ab+  ab+

11. ab   ab+ = ab+ 12. nm  Z+ nm+ = nm+ 13. nm  ZsZns    + = m 14. nm  ZsZ   +  ns+ = m 15.xy  nZxn  + = y 16. xy  r    xr+ = y

17. nZ : n2 = n 18. nZ +: n 2 = n 19. xy  r    xr = y 20. nZ + ab  Z+  an+ = b 21.nZ  ab  Z+  an+ = b 22. nZ + ab  Zan + = b

23. nZ + n  2n 24. nZn   2n

25.n  Z+  n  2n and n  2n 26. n  Z+  n  2n or n  2n

27.nZ +r    5n = n 28. r  nZ +  5n = r 29.nZnn   2 30. nZ +  nn 2 31. nZ +  nn 2 32. xy    x = xy+ 33. n  Z+ nn 2 34. x y  s    xs+ = xy+ 35.xy    xy+ 2  2xy 36. x y  : xy+ 2  2xy

37. nZ   mZ + nm n 38. nZ +  mZnm  

39.n  Zm   Z+  nm n 40. n  Z+ m  Znmn  41. nZ   mZ + nm 42. nm  Z  s Zn + m s 43.n  Zm   Zn = 2m 44. a   b   a = 2b 45.nm  Z+  sZnms  +  46. nm  Z+  sZ + nms+  22 Chapter 1 A Logical Beginning

47.nZ and x    mZ + xmn+  48. nZ + and x    mZ + nmx+  49.nZ + and x    mZ + nmx+  50. nZ +  ms  Zn + m n+ s 51.nZ +  ms  Z+ nm = ns 52. nZ   ms  Z+ nm = ns Exercises 53-64. Write an existential negation of the given universal proposition. 53. All roads lead to Rome. 54. Every cloud has a silver lining. 55. Every time it rains, it rains pennies from 56. All good things must come to an end. heaven. 57.nZna   58. x   xa 59.nZna   and nb 60. x   xa or xb 61.xXpx   qx sx 62.xXpx   qx sx 63.xXpx   qx sx 64. xXpx   qx sx

Exercises 65-74. Write a universal negation of the given existential proposition. 65. There is a for everything. 66. There is no room for error. 67.n  Zna  68. n  Zna  69.n  Zna  and nb 70. x   xa or xb 71.x  Xpx  qx sx 72. x  Xpx  qx sx 73.x  Xpx  qx sx 74. x  Xpx  qx sx

Exercises 75-84. Write an existential/universal negation of the given universal/existential propo- sition. 75. Everybody loves somebody. 76. All dogs go to heaven. 77. Every day contains a special moment. 78. Every good deed has a reward. 79.xXyYxy     + = 0 80. xXyYxy      81.ab  S  mn  Za + b= mn 82. ab  S  x X a+ b= x 83. xXab     Ya ==xb+ or b ax+ 84. xX and yYzZz     ==x or zy

Exercises 85-94. Write a universal/existential negation of the given existential/universal proposi- tion. 85. Some operas are longer than every symphony. 86. There is a solution to every problem. 87. Someone is greater than everyone else. 88. At some point in time, everything will be fine. 89.x  Xy   Yx + y= 0 90. x  Xy   Yx  y 91.a bS   m n  Za + b= mn 92. a bS   x Xa + b= x 93. x  Xa   b  Ya ==xb+ or b ax+ 94. x  X and yY   z Zz ==x or zy 1.3 Methods of Proof 23

1

§3. METHODS OF PROOF THROUGHOUT THIS SECTION WE WILL BE DEALING EXCLUSIVELY WITH INTEGERS. (A TOUCH OF )

DIRECT PROOF

The direct method of proving pq is to assume that p is True and then apply mathematical reasoning to deduce that q is True.

Let us accept the fact that: DEFINITION 1.11 nZ is even if kZn  = 2k . n is odd if and only EVEN AND ODD if it is not even. nZ is odd if kZn  = 2k + 1 .

For example, n = 14 is even since: k 14= 2 7 While n = –23 is odd since: k –21223 = – + 1

EXAMPLE 1.11 Prove that the sum of any two even integers is A proof is like a journey, even. which begins at a given point (the hypothesis), and SOLUTION: The hypothesis and conclusion of the journey estab- ends at a given point (the conclusion). You must be lishes a frame within which you are to paint a logical path:. mindful of both the begin- ning and the end of the jour- n and m are even Let n and m be even integers. ney. For if you know where By Definition 1.11, k hZ such that: you want to go (the conclu- the frame sion), but do not know n ==2k and m 2h where you are (the hypoth- a logical esis), then chances are that path Then: you aren’t going to reach nm+2==k +22h kh+ your destination. More- over, if you know where you are, but do not know nm+ is even Thus: nm+ is even (Definition 1.11). where you are going, then you probably won’t get Note how Definition 1.11 is used in both directions in the above proof. It was used there; and even if you do, in one direction to accommodate the given that n and m are even inte- you won’t know it. gers, and was then used in the other direction to conclude that nm+ is even.

CHECK YOUR UNDERSTANDING 1.16 (a) Prove that the sum of any two odd integers is even. (b) Formulate a concerning the sum of an even integer Answer: See page A-4. with an odd integer. Establish the validity of your conjecture. 24 Chapter 1 A Logical Beginning

EXAMPLE 1.12 Prove that 2mn+ is odd if and only if n is odd. SOLUTION: We need to establish validity in both directions; namely: (1) If 2mn+ is odd, then n is odd. (2) If n is odd, then 2mn+ is odd. Lets do it, beginning with (1): If 2mn+ is odd, then 2mn+2= k + 1 for some k. Solving for n we have: n ==2k +212– m km– + 1 . Conclusion: n is odd. Now for (2): If n is odd, then n = 2k + 1 for some k. Consequently: 2mn+2==m ++2k 1 2mk+ + 1 . Conclusion: 2mn+ is odd. Sometimes, one may be able establish both directions of an “if and only if” proposition simultaneously; as we are able to do here: Definition 1.11 2mn+ is odd 2mn+2= k + 1  n = 2k –12m +  n = 2km– + 1  n is odd

Definition 1.11

CHECK YOUR UNDERSTANDING 1.17 Answer: See page A-4. Prove that 2mn+ is even if and only if n is even.

CONTRAPOSITIVE PROOF pq  ~q  ~p Theorem 1.4, page 8 (see margin) tells us that if you can show that ~q  ~p , then pq will follow. Formality aside: Suppose you know that ~q  ~p . Can p be True and q False? No, for if p were True and q False, then ~p would be False and ~q True — contradicting ~q ~p . Another argument: Suppose (i): ~q  ~p . Can it be that (ii): p  ~q ? No, for if so: p  ~q  ~p Tisk! (ii) (i) 1.3 Methods of Proof 25

EXAMPLE 1.13 Show that for all integers n: (a) n is even if and only if n2 is even. (b) n is odd if and only if n2 is odd.

PROOF: (a) We use a direct argument to show that n even  n2 even : n even  n = 2k (for some k   n2 ==4k2 22k2 2  n is even A proof which estab- We use a contrapositive argument (margin) and show that: lished the validity of n2 even  n even pq by showing that ~q ~p is said to be a Specifically, we show that: n not evenn not even  n2 not even i.e : Contrapositive Proof. n odd  n2 odd :

n odd  n = 2k + 1 (for some k 

 n2 = 2k + 1 2 = 4k2 ++4k 1 = 22k2 + 2k + 1  n2 odd

(b) n is odd  n is not even Part (a):  n2 is not even  n2 is odd

CHECK YOUR UNDERSTANDING 1.18 Prove that: (a) 3n is odd if and only if n is odd . Answer: See page A-5. (b) n3 is odd if and only if n is odd .

This method of proof is called: proof by contradiction or, if you prefer Latin: reductio ad One can establish that a proposition p is True by demonstrating that absurdum. the assumption that p is False leads, via a logical argument, to a False conclusion. Invoking the “logical commandment” that from Truth only Truth can follow, one can then conclude that the assumption that p is False must itself be False, and that therefore p has to be True. 26 Chapter 1 A Logical Beginning

Whatever you can do by the contrapositive method can also be done by the contradiction method. To illustrate, in Figure 1.1(b) we recall the contrapositive proof of n2 even  n even given in Example 1.13(a), and offer a proof by contradiction in Figure 1.1(b). As you can see there is precious little difference between the two methods. In (b), rather than showing that n not even  n2 not even , we start off accepting he given condition that n2 is even, and then go on to show that the assumption that n is not even contradicts that given condition.

n2 is even  n is even Contrapositive Proof: We show n odd  n2 odd : Proof by Contradiction: 2 n odd  n = 2k + 1 (for some k  Suppose n is even (given condition), and assume that n is odd, say  n2 = 2k + 1 2 n = 2k + 1 for some k. Then: n2 = 4k2 ++4k 1 = 4k2 ++4k 1 = 22k + 2 + 1  n2 is odd = 22k2 + 2k + 1  n2 odd — contradicting our stated assumption that n2 is even. (a) (b) Figure 1.1 We have exhibited three methods of proof: Contrapositive Proof Proof by Contradiction In attempting to prove something, which method should you use? No general answer can be provided, other than if one does not work then try another. We do note, however, that a direct proof is considered to be more aesthetically pleasing than either a contrapositive proof or a proof by contradiction, but a direct proof may not always be a viable option.

EXAMPLE 1.14 Prove that if 3n + 2 is even, then n is even.

SOLUTION: Let’s try to prove that 3n + 2 even  n even using a direct approach: 3n +22 ==k  3n 2k – 2 — Now what? If you try something like n = ------2k – 2- , where does it take you 3 (nowhere, other than to an expression that may be outside the realm of integers)? Okay the direct approach does not appear to work, should we try the contrapositive approach or the proof by contradiction approach? That is basically a matter of taste, and so we offer both methods for your consideration: 1.3 Methods of Proof 27

3n + 2 even  n even Contrapositive Proof by Contradiction: Proof:n not even 3n + 2 not even Let 3n + 2 be even (given condition) i.e: n odd 3n + 2 odd n odd  n = 2k + 1 (for some k) Assume that n is odd, say n = 2k + 1 .  3n + 2 = 32k + 1 + 2 Then: 3n +322 = k + 1 + 2 = 6k + 5 ==6k +65 k ++41 = 6k ++41 = 23k + 2 + 1 = 23k + 2 + 1  3n + 2 odd — contradicting the stated condition that  3n + 2 odd 3n + 2 is even.

CHECK YOUR UNDERSTANDING 1.19

Answer: See page A-5. Prove that 3n + 2 is odd if and only if n is odd.

Note that while -----15- is the DEFINITION 1.12 We say that a nonzero integer a divides an 5 DIVISIBILITY integer b, written ab , if bak= for some number 3. 515 is not a integer k. number; it is the state- ment that there exists an In the event that ab , we say that b is divis- integer k (in this case 3) such that 15= 5k . ible by a and that b is a multiple of a. Here are two statements for your consideration: (a) If ab and bc , then ac . (b) If abc , then ab or ac . Let’s Challenge (a) with some specific integers: Challenge 1: 3 divides 9 and 9 divides 18; does 3 divide 18? Yes. Challenge 2: 2 divides 4 and 4 divides 24; does 2 divide 24? Yes. Challenge 3: 5 divides 55 and 55 divides 110; does 5 divide 110? Yes. The above “Yeses” may certainly suggest that (a) does indeed hold for all abc  Z — suggest, yes, but NOT PROVE. A proof is provided in Theorem 1.5(a), below. Statement (b): If abc , then ab or ac is False. A counterexample: 623 but 62 and 63 .

TYPICALLY: A GENERAL ARGUMENT is needed to establish the validity of a statement. A (specific) COUNTEREXAMPLE is needed to establish that a statement is False. 28 Chapter 1 A Logical Beginning

THEOREM 1.5 Let b and c be nonzero integers. Then: (a) If ab and bc , then ac . (b) If ab and ac , then ab+ c . (c) If ab , then abc for every c.

PROOF: (a) If ab and bc , then, by Definition 1.12: bak== and cbh for some h and k . Consequently: cbhak==hakh = =at (where tkh= ). It follow, from Definition 1.12, that ac . Note how Definition 1.12 is used in both directions in the above proof. (b) If ab and ac , then bah= and cak= for some h and k. Consequently: bc+ ==ah+ ak ah+ k =at (where thk= + ). It follows that ab+ c . (c) If ab , then bak= for some k. Consequently, for any c: bc=== ak cakc at (where tkc= ). It follows that abc .

EXAMPLE 1.15 Prove or give a counterexample. (a) If ab+ c , then ab or ac . (b) If ab and ab+ c , then ac .

SOLUTION: (a) Unless you are fairly convinced that a given statement is True, you may want to start off by challenging it: Challenge 1.48+ 16 , and 4 certainly divides 8 or 16 (in fact, it divides both 8 and 16). Inconclusive. Challenge 2.43+ 1 , and 4 divides neither 3 nor 1 — a counterexample! The statement is False. (b) Challenging the statement “If ab and ab+ c , then ac ” will not yield a counterexample. It can’t, since the statement is True. To establish its validity, a general argument is called for. One proof: Since ab , there exists h such that: (1) bah= . ab and ab+ c Since ab+ c , there exists k such that: (2) bc+ = ak . (Now we have to go ahead and show that cat= for some t) From (2): cakb= – . From (1): cakah==– ak– h . ac Since cat= (where tkh= – ): ac. 1.3 Methods of Proof 29

An alternate proof: Observing that: cbc==+ – b bc+ + –1 b we employ Theorem 1.6(b) and (c) to conclude that ac . (Recall that we are given that ab+ c and that ab )

CHECK YOUR UNDERSTANDING 1.20 (a) Prove: (a-i) If an and am then an+ m . (a-ii) If na then nca for every cZ . (b) Prove or give a counterexample: (b-i) If ab or ac, then ab+ c . (b-ii) If a and b are even and if ab+ c , then c must be even. (b-iii) If ab+ cd+ , then there exist k such that: Answer: See page A-6. ak+++ bk c d = 0.

As you worked your way through this section, you must have observed how: DEFINITIONS RULE! They are the physical objects in the mathematical universe. 30 Chapter 1 A Logical Beginning

EXERCISES

Exercises 1-25. Establish the validity of the given statement. 1. 0 is even and 1 is odd. 2. The product of any two even integers is even. 3. The product of any two even integers is divisible by 4. 4. The sum of any two odd integers is even.

5. If ab then a2 b2 .

6.3 n if and only if 3 n + 9 .

7. If ab , and ac then abncm+ for every n and m.

8. If ac , and bd then ab cd .

9. If 5n – 7 is even then n is odd. 10.11n – 7 is even if and only if n is odd.

11. If 4 n2 – m2 then either n and m are both even or they are both odd.

12.n2 ++3n 5 is odd for all n. 13.3n + 1 is even if and only if n is odd.

14.n3 is even if and only if n is even.

15. If n4 is even then so is 3n.

16.3n3 is even if and only if 5n2 is even.

17.5n – 11 is even if and only if 3n – 11 is even. 18. If 3n + 5m is even, then either n and m are both even or they are both odd.

19. The of every odd integer is of the form 4n + 1 for some nZ .

20. If 4 n2 – 1 then 2 n – 1 .

# 21. Let baqr= + . Prove that if ca and cb , then cr . 22. If bc is not a multiple of a, then neither b nor c can be a multiple of a. 23. If bc+ is a multiple of a, and b is a multiple of a, then c is a multiple of a.

24. If a and b are odd positive integers and if ca+ b , then c is even.

25. If a is odd and b is even and if ca+ b , then c is odd. 1.3 Methods of Proof 31

Exercises 25-33. Disprove the given statement. . 26. The sum of any two even integers is divisible by 4. 27. The sum of any three odd integers is divisible by 3. 28. The product of any two even integers is divisible by 6.

29. If 4 n2 – 1 then 4 n – 1 .

30. If 2mn+ is odd then both m and n are odd. 31. If 2mn+ is even then both m and n are even.

32. If ab or ac then ab+ c .

33. If ab and ba then ab= .

PROVE OR GIVE A COUNTEREXAMPLE

34. If n is odd and m is even then nm– is odd. 35. If nm+ is odd then neither n nor m can be even. 36. If nm+ is even then neither n nor m can be odd. 37. If nm+ is odd then n or m must be odd. 38. If nm+ is even then n or m must be even.

39. If n + 1 is even then so is n3 – 1 .

40. If n + 1 is even then so is n2 – 1 .

41. If n + 1 is even then so is n3 + 1 .

42. If n + 1 is even then so is n2 + 1 .

43. If n + 1 is odd then so is n3 – 1 .

44. If n + 1 is odd then so is n2 – 1 .

45. If n + 1 is odd then so is n3 + 1 .

46. If n + 1 is odd then so is n2 + 1 .

47. If a is even and b is odd then a2 + 2b is even.

48. If a is even and b is odd then a2 + 3b is odd.

49. If a is odd then so is a2 + 2a .

50. If a is odd then so is a2 + 3a . 32 Chapter 1 A Logical Beginning

51. If a is even and b is odd then a + 2 2 + b – 1 2 is even. 52. If 9 n + 3 then 3 n

53. If a + 2 2 + b – 1 2 is even then a or b has to be even. 54. If abbc  and ca, then abc==.

55. If abbc  and ca , then ab= or ac= or bc= . 1.4 Principle of Mathematical Induction 33

§4. Principle of Mathematical Induction This section introduces a most powerful mathematical tool, the Prin- A form of the Principle of Math- ciple of Mathematical Induction (PMI). Here is how it works: ematical Induction is actually one of Peano’s axioms, which PMI serve to define the positive inte- Let Pn denote a proposition that is either true or false, depend- gers. ing on the value of the integer n. [ (1858-1932).] If: I.P1 is True. And if, from the assumption that: II. Pk is True one can show that: III.Pk+ 1 is also True. then the proposition Pn is valid for all integers n  1 Step II of the induction procedure may strike you as being a bit strange. After all, if one can assume that the proposition is valid at nk= , why not just assume that it is valid at nk= + 1 and save a step! Well, you can assume whatever you want in Step II, but if the proposition is not valid for all n you simply are not going to be able to demonstrate, in Step III, that the proposition holds at the next value of n. It’s sort of like the domino theory. Just imagine that the propositions P1  P2  P3 Pk Pk+ 1  are lined up, as if they were an of dominoes:

P(1 P(2 P P P(6) P(7) P(8) P(9) P(10) ) ) (3) (4) P(5) ......

If you knock over the first domino (Step I), and if when a domino falls (Step II) it knocks down the next one (Step III), then all of the domi- noes will surely fall. But if the falling kth domino fails to knock over the next one, then all the dominoes need not fall. The Principle of Mathemati- cal Induction might have been To illustrate how the process works, we ask you to consider the sum better labeled the Principle of of the first n odd integers, for n = 1 through n = 5 : Mathematical Deduction, for Sum of the first n odd integers Sum n Sum is used to formulate a hypothesis or con- 111 1 jecture, while deductive rea- 1 + 3 4 2 4 soning is used to rigorously 1 + 3 + 5 9 3 9 establish whether or not the 1 + 3 + 5 + 7 16 4 16 conjecture is valid. 1 + 3 + 5 + 7 + 9 25 5 25 6 ? Figure 1.2 34 Chapter 1 A Logical Beginning

Looking at the pattern of the table on the right in Figure 1.1, you can probably anticipate that the sum of the first 6 odd integers will turn out to be 62 = 36 , which is indeed the case. Indeed, the pattern suggests that: The sum of the first n odd integers is n2 Using the Principle of Mathematical Induction, we now establish the validity of the above conjecture: Let Pn be the proposition that the sum of the first n odd integers equals n2 . I. Since the sum of the first 1 odd integers is 12 , P1 is true. The sum of the first 3 odd II. Assume Pk is true; that is: integers is: 135+++ +2k – 1 = k2 135++ 2  3 – 1 The sum of the first 4 odd see margin integers is: III. We show that Pk+ 1 is true, thereby completing the proof: 1357+++ 2  4 – 1 the sum of the first k + 1 odd integers Suggesting that the sum of the first k odd integers is: 135+++ +2k – 1 + 2k + 1 ==k2 + 2k + 1 k + 1 2 13++ +2k – 1 (see Exercise 1). induction hypothesis: Step II

EXAMPLE 1.16 Use the Principle of Mathematical Induction to establish the following formula for the sum of the first n integers: nn+ 1 123+++ +n = ------ 2

SOLUTION: Let Pn be the proposition: nn+ 1 123+++ +n = ------(*) 2 11+ 1 I.P1 is true: 1 =------ Check! 2 kk+ 1 II. Assume Pk is true: 123+++ +k = ------2 III. We are to show that Pk+ 1 is true; which is to say, that (*) holds when nk= + 1 : k + 1 k + 1 + 1 k + 1 k + 2 123+++ ++kk+ 1 ==------------------- 2 2 Let’s do it: 123+++ ++kk+ 1 = 123+++ +k + k + 1 kk+ 1 induction hypothesis: = ------+ k + 1 2 ==------kk + 1------+ 2k------+ 1------k + 1 ------k + 2 2 2 1.4 Principle of Mathematical Induction 35

CHECK YOUR UNDERSTANDING 1.21 (a) Use the formula for the sum of the first n odd integers, along with that for the sum of the first n integers, to derive a formula for the sum of the first n even integers. (b)Use the Principle of Mathematical Induction directly to establish Answer: See page A-6. the formula you obtained in (a).

The “domino effect” of the Principle of Mathematical Induction need not start by knocking down the first domino P1 . Consider the fol- lowing example where domino P0 is the first to fall. EXAMPLE 1.17 Use the Principle of Mathematical Induction to establish the n  2n for all n  0 .

SOLUTION: Let Pn be the proposition n  2n . I.P0 is true: 02 0 , since 20 = 1 . II. Assume Pk is true: k  2k . III. We show Pk+ 1 is true; namely that k + 12 k + 1 : k +221  2k + 12 k + 2k ==k 2k + 1 II 12 k

EXAMPLE 1.18 Use the Principle of Mathematical Induction to show that 45n – 1 for all integers n  0 . SOLUTION: Let Pn be the proposition 45n – 1 .

I.P0 is true: 450 – 1 , since 50 –111 ==–0 . II. Assume Pk is true: 4 5k – 1 .

III. We show Pk+ 1 is true; namely, that 45k + 1 – 1 : What motivated us to write –1 in the form 5k + 1 –551 ==k –551 k – 5 + 4 (see margin) – 5 + 4 ? Necessity did: We had to do something = 55k – 1 + 4 to get “5k – 1 ” into the The desired conclusion now follows from CYU 1.17, page 24: picture (see II). Clever, to be sure; but such CYU 1.20(a), page 29: 45k – 1  455k – 1 and then: a clever move stems from stubbornly focusing on CYU 1.20(b): 455k – 1 and 4 4 4 5 5k – 1 + 4 what is given and on what has to be established. EXAMPLE 1.19 Use the Principle of Mathematical Induction to show that 322n – 1 for all integers n  1 . 36 Chapter 1 A Logical Beginning

SOLUTION: Let Pn be the proposition: 322n – 1 .

I.P1 is true: 221 –31 = Check! II. Assume Pk is true: 3 22k – 1 . III. We show that Pk+ 1 is true; which is to say: 322k + 1 – 1 : 22k + 1 –21 = 2k + 2 – 1 anm+ = anam: = 22  22k – 1 = 42 2k – 1 wanting to get a 3 into the picture: = 31+ 22k – 1 regrouping: = 32 2k + 22k – 1 Clearly 3 divides 32 2k , and, by the induction hypothesis, 322k – 1 . The desired result now follows from CYU 1.18(a), page 25.

CHECK YOUR UNDERSTANDING 1.22 Use the Principle of Mathematical Induction to show that Answer: See page A-7. 6 n3 + 5n for all integers n  1 .

Recall that:. EXAMPLE 1.20 Use the Principle of Mathematical Induction to n!12=  n show that n!  n2 for all integers n  4 .

SOLUTION: Let Pn be the proposition n!  n2 :

I.P4 is true: 4!1234== 24 42 . II. Assume Pk is true: k!  k2 (for k  4 ) III. We show Pk+ 1 is true; namely, that k + 1 !  k + 1 2 : k + 1 ! = k!k + 1  k2k + 1 II Now what? Well, if we can show that k2k + 1  k + 1 2 , then we will be done. Let’s do it: Since k  4 , k  2 , and therefore k2 = kk  2kk + 1 Multiplying both sides by the positive number k + 1 : k2k + 1  k + 1 2 .

CHECK YOUR UNDERSTANDING 1.23 Use the Principle of Mathematical Induction to show that Answer: See page A-7. 2n  n + 2 ! for all integers n  0 . 1.4 Principle of Mathematical Induction 37

Our next application of the Principle of Mathematical Induction Edouard Lucas formalized involves the following Tower of Hanoi puzzle: the puzzle in 1883, basing Start with a number of washers of differing sizes on spindle A, it on the following legend: In a temple at Benares, as is depicted below: there are 64 golden disks mounted on one of three diamond needles. At the beginning of the world, all the disks were stacked on the first needle. The priests attending the temple have the sacred A B C obligation to move all the disks to the last nee- The objective of the game is to transfer the arrangement cur- dle without ever placing a larger disk on top of a rently on spindle A to one of the other two spindles. The rules smaller one. The priests are that you may only move one washer at a time, without ever work day and night at this task. If and when placing a larger one on top of a smaller one. they finally complete the job, the world will end. EXAMPLE 1.21 Show that the tower of Hanoi game is winna- ble for any number n of washers.

SOLUTION: If spindle A contains one washer, then simply move that washer to spindle B to win the game (Step I). Assume that the game can be won if spindle A contains k washers (Step II —the induction hypothesis). We now show that the game can be won if spindle A contains k + 1 washers (Step III): Just imagine that the largest bottom washer is part of k + 1 wahsers the of spindle A. With this sleight of hand, we are looking at a situation consisting of k washers on a modified spindle A (see margin). By the induction . . k washers hypothesis, we can move those k washers onto spindle { } B. We now take the only washer remaining on spindle new base A (the largest of the original k + 1 washers), and move it to spindle C, and then think of it as being part of the base of that spindle. Applying the induction hypothesis one more time, we move the k washers from spindle B onto the modified spindle C, thereby winning the game.

CHECK YOUR UNDERSTANDING 1.24 (Challenging) Use the Principle of Mathematical Induction to show that any n lines, no two of which are parallel and no three of which pass through a n2 ++n 2 Answer: See page A-7. common point, will separate the into ------regions. 2 38 Chapter 1 A Logical Beginning

TWO ALTERNATE FORMS OF THE PRINCIPLE OF MATHEMATICAL INDUCTION We complete this section by introducing two equivalent forms of the Principle of Mathematical Induction — equivalent in that any one of API is often called the them can be used to establish the remaining two. Strong Principle of Induc- One version, which we will call the Alternate Principle of Induction tion. A bit of a misnomer, (API), is displayed in Figure 1.3(b). As you can see, the only difference since it is, in fact, equiva- between PMI and API surfaces in (*) and (**). Specifically, the propo- lent to PMI. sition “Pk True” in (a) is replaced, in (b), with the proposition “Pm True for all integers m up to and including k”. Let Pn denote a proposition that is either true or false, depending on the value of the integer n. PMI API If P1 is True, and if: If P1 is True, and if (*) Pk True  Pk+ 1 True (**): Pm True for 1 mk  Pk+ 1 True then Pn is True for all integers n  1 then Pn is True for all integers n  1 (a) (b) Figure 1.3 We establish the equivalence of PMI and API by showing that (*) holds if and only (**) holds. Clearly, if (*) holds then (**) must also hold. As for the other way around: Assume that (**) holds and that (*) does not. (we will arrive at a contradiction)

If (*) does not hold, then there must exist some k0 for which

Pk0 is True and Pk0 + 1 is False. Since Pk0 + 1 is False,

and since (**) holds, we know that Pk1 is False for some 1 k1 k0 . But we are assuming that Pk0 is True. Hence

Pk1 is False for some 1  k1  k0 .

Repeating the above procedure with k1 playing the role of

k0 we arrive at Pk2 is False for some 1  k2  k1 . Continuing in this fashion we shall, after at most

k0 – 1 steps, be forced to conclude that P1 is False — contradicting the assumption that P1 is True.

EXAMPLE 1.22 Use API to show that for any given integer n  12 there exist integers a  0 b  0 such that n = 3a + 7b .

SOLUTION: I. Claim holds for n = 12 : 12= 3 4 + 70 1.4 Principle of Mathematical Induction 39

II. Assume claim holds for all m such that 12 mk . III. To show that the claim holds for nk= + 1 we first show, directly, that it does indeed hold if k +131 = or if k +141 = : 13== 3 2 +30727 1 and 14  +  Now consider any k + 115 . If k + 115 , then 12  k + 1 – 3  k . Appealing to the induction hypothesis, we choose a  0 b  0 such that: k + 1 –33 = a + 7b It follows that k +31 = a + 1 + 7b , and the proof is complete.

Here is another important property which turns out to be equivalent to the Principle of Mathematical Induction:

THE WELL-ORDERING PRINCIPLE FOR Z+ Note that of Z need not have first elements. A Every nonempty of Z+ has a smallest (or least, or first) element. case in point  –42024 –  We show that the Alternate Principle of Mathematical Induction Nor does the bounded set: implies the Well-Ordering Principle: x   5 x 9 Let S be a NONEMPTY subset of Z+ . contain a smallest element (note that 5 is not in the If 1  S , then it is certainly the smallest element in S, and we are done. above set). Assume 1  S , and suppose that S does not have a smallest element (we will arrive at a contradiction): Let Pn be the proposition that nS for nZ + . Since, 1  S , P1 is True. Suppose that Pm is True for all 1 mk, can Pk+ 1 be False? No: To say that Pk+ 1 is False is to say that k + 1  S . But that would make k + 1 the smallest element in S, since none of its prede- cessors are in S. This cannot be, since S was assumed not to have a smallest element. Since P1 is True (1  S ) and since the validity of Pm for all 1 mk implies the validity of Pk+ 1 , Pn must be True for all nZ + ; which is the same as saying that no element of Z+ is in S — contradicting the assump- tion that S is NONEMPTY. CHECK YOUR UNDERSTANDING 1.25

Answer: See page A-8. Show that the Well-Ordering Principle implies the Principle of Math- ematical Induction. 40 Chapter 1 A Logical Beginning

EXERCISES

Exercises 1-33. Establish the validity of the given statement.

1. The nth odd integer is 2n – 1 .

3n2 – n 2. For every integer n  1 , 147+++ +3n – 2 = ------. 2

n 2n – 1 2n + 1 3. For every integer n  1 , 12 ++++32 52  2n – 1 2 = ------------------. 3

4. For every integer n  1 , 12 ++++22 32  n2 = nn------+ 1------2n------+ 1 . 6

44n – 1 5. For every integer n  1 , 44++++2 43  4n = ------------. 3

1 1 1 1 1 6. For every integer n  1 , --- +++------ +----- = 1– ----- . 2 4 8 2n 2n

1 1 1 1 n 7. For every integer n  1 , ------++++------ ------= ------. 12 23 34 nn+ 1 n + 1

1 1 1 n 8. For every integer n  1 , ------+++------ ------= ------. 23 34 n + 1 n + 2 2n + 4

1 1 1 1 9. For every integer n  1 , 1 + --- 1 + --- 1 + --- 1 + --- = n + 1 . 1 2 3 n

1 – xn + 1 10. For every integer n  1 and any real number x  1 , x0 ++++x1 x2  xn = ------. 1 – x

n a1 – rn + 1 11. For every integer n  1 , and any real number r  1 , ari = ------.  1 – r i = 0

12. For every integer n  0 : 524n + 2 + 1 .

13. For every integer n  1 : 943n – 1 .

14. For every integer n  1 : 35n – 2n .

15. For every integer n  1 : 732n – 2n . 1.4 Principle of Mathematical Induction 41

16. For every integer n  1 : 6 n3 + 5n .

17. For every integer n  1 , 52n + 7 is divisible by 8.

18. For every integer n  1 , 33n + 1 + 2n + 1 is divisible by 5.

19. For every integer n  1 , 4n + 1 + 52n – 1 is divisible by 21.

20. For every integer n  1 , 32n + 2 –98n – is divisible by 64.

21. For every integer n  0 , 2n  n .

22. For every integer n  5 , 2n – 4  n .

23. For every integer n  5 , 2n  n2 .

24. For every integer n  4 , 3n  2n + 10 .

2n ! 25. For every integer n  1 , ------is an odd integer. 2nn!

26. For every integer n  4 , 2nn ! . 27. ( Dependent) Show that the sum of n differentiable functions is again differentiable. d 28. (Calculus Dependent) Show that for every integer n  1 , xn = nxn – 1 . dx Suggestion: Use the product Theorem: If f and g are differentiable functions, then so is fg d d d differentiable, and fxgx = fx gx+ gx fx . dx dx dx

1 29. Let a1 = 1 and an + 1 = 3 – ----- . Show that an + 1  an . an

1 30. Let a1 = 2 and an + 1 = ------. Show that an + 1  an . 3 – an

1 1 1 31. For every integer n  1 , 1 ++++------ ------ 2n +11 – . 2 3 n

32. For any positive number x, 1 + x n  1 + nx for every n  1 .

33. For every integer n  8 , there exist integers a  0 b  0 such that n = 3a + 5b . 42 Chapter 1 A Logical Beginning

# 34. Let m be any nonnegative integer. Use the Well-Ordering Principle to show that every non- empty subset of the set nZnm  – contains a smallest element.

# 35. Use the Principle of Mathematical Induction to show that there are n! different ways of ordering n objects, where n!123=  n . 36. What is wrong with the following “Proof” that any two positive integers are equal: Let Pn be the proposition: If a and b are any two positive integers such that maxab = n , then ab= . I.P1 is true: If maxab = 1 , then both a and b must equal 1. II. Assume Pk is true: If maxab = k , then ab= . III. We show Pk+ 1 is true: If maxab = k + 1 then maxa – 1 b – 1 = k . · By II, a – 1 ==b – 1  a b . 1.5 The Division and Beyond 43

§5. The Division Algorithm and Beyond ALL LETTERS IN THIS SECTION WILL BE UNDERSTOOD TO REPRESENT INTEGERS. In elementary school you learned how to divide one integer into another to arrive at a quotient and a remainder, and could then check 5 q your answer (see margin). That checking process reveals an important 3 17 a d 15 result: 2 r THEOREM 1.6 For any given aZ and dZ + , there exist Check: 17= 3 5+ 2 THE DIVISION unique integers q and r, with 0  r  d , such that: adqr= + ALGORITHM ad= qr+ PROOF: We begin by establishing the existence of q and r such that: Here is a “convincing argu- ment” for your consideration: adqr= + with 0  rd Mark off multiples of d on the Consider the set: number : SadnnZ= –  and adn–  0 (*) | | | | | -2d -d 0 d 2d We first show that S is not empty: Case 1. If adq= , then let If a  0 , then aad= –  00 , and therefore aS . r = 0 . [0 is playing the role of n in (*)] Case 2. If a is not a multiple If a  0 , then ada–  0 , and therefore adaS–  . of d, then let dq be such that [a is playing the role of n in (*) and remember that dZ + ] dq a d + 1 q . We then have ad= qr+ , where: Since S is a nonempty subset of 0  Z+ , it has a least element d (Exercise 34, page 42); let’s call it r. Since r is in S, there exists q  Z r such that: r = ad– q (**) .a dq dq+ d To complete the existence part of the proof, we show that r  d . In either case 0  rd . Assume, to the contrary, that r  d . From: r – d ==ad– q – d ad– q + 1 (**) we see that r – d is of the form adn– (with n = q + 1 ). Moreover, our assumption that r  d implies that r – d  0 . It follows that r – dS , contradicting the minimality of r. To establish uniqueness, assume that: adqr= + with 0  rd and adq=  + r with 0  r  d This is a common mathe- matical theme: [We will show that qq=  and rr=  (see margin)] To establish that some- thing is unique, consider Since r  0 and r  d (or –r  –d ): rr–   0 – r  0 – d = –d . two such “somethings” Since rd and r  0 (or –r  0 ): rr–   dr–   d – 0 = d and then go on to show that the two “some- Thus: –d  rr–   d , or rr–   d things” are, in fact, one and the same. From dq+ r = dq + r we have: rr–  = dq – q (a multiple of d) But if rr–   d and if rr–  is a multiple of d, then rr–0 = (or rr=  ). Returning to dq+ r = dq + r we now have: dq+ r = dq + r  dq= dq  dq– q ==0  q q d  0 44 Chapter 1 A Logical Beginning

. EXAMPLE 1.23 Show that for any odd integer n, 8 n2 – 1 .

SOLUTION: There are, at times, more than one way to stroke a cat: Using Induction Using the Division Algorithm We show that the proposition: We know that for any n there exists q such that: 82m + 1 2 – 1 n ==2q or n 2q + 1 (*) holds for all m  0 (thereby covering all n ==3q or n 3q +31 or n =q + 2 (**) odd integers n). n ====4q or n 4q +41 or n q +42 or n q + 3 I. Valid at m = 0 : 20 + 1 2 –01 = . II. Assume valid at mk= ; that is: While (*) and (**) may not lead us to a fruitful conclu- sion, the bottom line does. Specifically: 2 2 2k + 1 –81 = t or4k +84k = t For any n: for some integer t. n ====4q or n 4q +41 or n q +42 or n q + 3 III. We are to establish validity at If n is odd, then there are but the two possibilities: mk= + 1 ; that is, that: n ==4q +41 or n q + 3 2 2k + 1 + 1 –81 = s We now show that, in either case 8 n2 – 1 . for some integer s. Let’s do it: If n = 4q + 1 , then: 2k + 1 + 1 2 – 1 n2 –41 ==q + 1 2 –161 q2 ++8q 11– = 8k = 2k + 3 2 – 1 with k = 2q2 + q 2 = 4k ++12k 8 If n = 4q + 3 , then: = 4k2 + 4k + 8k + 8 n2 –41 ==q + 3 2 –161 q2 ++24q 91– = 8h ===8t +88k + 1 tk++1 8s with h = 2q2 ++3q 1 II

CHECK YOUR UNDERSTANDING 1.26

Prove that for any integer n, n2 = 3q or n2 = 3q + 1 for some inte- Answer: See page A-8. ger q.

DEFINITION 1.13 For given a and b not both zero, the greatest GREATEST COMMON common divisor of a and b, denoted by DIVISOR gcd(a,b), is the largest positive integer that divides both a and b. If the magnitude of either a or b is “small,” then one can easily find their greatest common divisor. Consider, for example, the two numbers 245 and 21. The only divisors of 21 are 1, 3, 7 and 21. Since 21 does not divide 245, it cannot be the greatest common divisor of 245 and 21. The next contender is 7, and 7 does divide 245 245= 7 35 . Conse- quently: gcd245 21 = 7 . 1.5 The Division Algorithm and Beyond 45

The following example illustrating a procedure that can be used to find the greatest common divisor of any two numbers — a procedure that will surface within the proof of Theorem 1.7 below.

EXAMPLE 1.24 Determine gcd4942 1680

SOLUTION: Employing the Division Algorithm:

Divide 1680 into 4942 to arrive at: 4942= 2 1680+ 1582 (1)

r1

Divide r1 = 1582 into 1680:1680= 1 1582+ 98 (2)

r1 r2

Divide r2 = 98 into r1 = 1582 :1582= 16 98+ 14 (3)

r1 r2 r3

Divide r3 = 14 into r2 = 98 :98= 7 14+ 0 (4) Figure 1.4 Looking at equation (4), we see that 14 divides 98. Moving up to equation (3) we can conclude that 14 must also divide 1582 [see The- orem 1.5(b), page 28]. Equation (2) then tells us that 14 divides 1680, and moving up one more time we find that 14 divides 4942. At this point, we know that 14 divides both 1680 and 4942. To see that 14 is, in fact, the greatest common divisor of 1680 and Since each resulting remain- 4942, consider any divisor d of 1680 and 4942. From equation (1), der is strictly smaller than its predecessor, the algorithm rewritten in the form 1582= 4942– 2 1680 , we see that d must must eventually terminate, as also divide 1582. From Equation (2), rewritten as it did in step (4), with a zero remainder. 1680–98 1582 = , we see that d must also divide 98. Equation (3) then tells us that d divides 14. Having observed that any divisor of 1680 and 4942 also divides 14, we conclude that gcd1680 4942 = 14 . Working the algorithm of Figure 1.4 in reverse, we show that the greatest common divisor of 4942 and 1680, namely 14, can be expressed as a multiple of 4942 plus a multiple of 1680: From (3):14 = 1582– 16 98 From (2): = 1582– 16 1680– 1582 From (1): = 4942– 2 1680 – 16 1680–  4942– 2 1680 regrouping: = 17  4942 + –50 1680

In general:

In the above illustration: THEOREM 1.7 If a and b are not both 0, then there exist s and a ==4942 b 1680 t such that: s ==17 and t –50 gcd a b = sa+ tb 46 Chapter 1 A Logical Beginning

PROOF: Let Gx=  0 xmanb= + for some m and n Assume, without loss of generality that a  0 . Since both a and –a are of the form ma+ nb : a = 1a + 0b while –1a = – a + 0b ; and since either a or –a is positive: G   . That being the case the Well Ordering Principle (page 39) assures us that G has a smallest element gsatb= + . We show that ggcdab=  by showing that (1): g divides both a and b, and that (2): every divisor of a and b also divides g. (1) Applying the Division Algorithm we have: aqgr= + with 0  rg . (*) (**) Substituting gsatb= + in (*) brings us to: aqsatb= + + r r = 1 – qs aqtb– Since r is of the form ma+ nb with rg , it cannot be in G, and must therefore be 0 [see (**)]. Consequently aqg= , and ga. The same argument can be used to show that gb . (2) If da and db , then, by Theorem 1.5(b) and (c), page 28: dg .

CHECK YOUR UNDERSTANDING 1.27 (a) Determine gcd1870 5605 . (b) Find integers s and t such that gcd1870 5605 = sa+ tb . (a) 5 (b) s ==3 t –1 (c) Show that for any a and b not both zero: (c) See page A-8 gcd a b = gcd a b .

DEFINITION 1.14 Two integers a and b, not both zero, are rela- RELATIVELY PRIME tively prime if: gcd a b = 1 For example: Since gcd15 8 = 1 , 15 and 8 are relatively prime. Since gcd15 9 = 31 , 15 and 9 are not relatively prime.

THEOREM 1.8 Two integers, a and b, are relatively prime if and only if there exist st  Z such that 1 = sa+ tb

PROOF: To say that a and b are relatively prime is to say that gcd a b = 1 . The existence of integers s and t such that 1 = sa+ tb follows from Theorem 1.7. For the converse, assume that there exist integers s and t such that 1 = sa+ tb. Since gcd a b divides both a and b, it divides 1 [The- orem 1.6(b) and (c), page 28]; and, being positive, must equal 1 [CYU 1.20(a-1), page 29]. 1.5 The Division Algorithm and Beyond 47

THEOREM 1.9 Let abc  Z . If abc , and if gcd a b = 1 , then ac.

PROOF: Let s and t be such that: 1 = sa+ tb Multiplying both sides of the above equation by c: csactbc= + Clearly asac. Moreover, since abc: atbc. The result now follows from Theorem 1.5(b), page 28. CHECK YOUR UNDERSTANDING 1.28 Let abc  Z . Show that if abc and ab , then a and c can not be Answer: See page A-8. relatively prime.

PRIME NUMBERS Chances are that you are already familiar with the important concept of a prime number; but just in case:

DEFINITION 1.15 An integer p  1 is prime if 1 and p are its PRIME only divisors.

For example: 2, 5, 7, and 11 are all prime, while 9 and 25 are not. Moreover, since any even number is divisible by 2, no even number So, 2 is the oddest prime (sorry). greater than 2 is prime.

THEOREM 1.10 If p is prime and if pab , then pa or pb .

PROOF: If pa , we are done. We complete the proof by showing that if pa, then pb: Since the greatest common divisor of p and a divides p, it is either 1 or p. As it must also divide a, and since we are assuming pa , it must be that gcd p a = 1 . The result now follows from Theorem 1.9.

CHECK YOUR UNDERSTANDING 1.29 Let p be prime. Use the Principle of Mathematical Induction to show Answer: See page A-8.  that if pa1a2 an , then pai for some 1 in . 48 Chapter 1 A Logical Beginning

The following result is important enough to be called the Fundamen- tal Theorem of Arithmetic. THEOREM 1.11 Every integer n greater than 1 can be expressed uniquely (up to order) as a product of primes.

PROOF: We use API of page 38 (starting at n = 2 ) to establish the existence part of the theorem: I. Being prime, 2 itself is already expressed as a product of primes. II. Suppose a prime factorization exists for all m with 2 mk . III.We complete the proof by showing that k + 1 can be expressed as a product of primes: If k + 1 is prime, then we are done. If k + 1 is not prime, then k + 1 = ab , with 2 abk . By our induction hypothesis, both a and b can be expressed as a product of primes. But then, so can k + 1 = ab . For uniqueness, consider the set: SnZ=  + n has two different prime decompositions Assume that S   (we will arrive at a contradiction). The Well-Ordering Principle of page 39 assures us that S has a smallest element, let’s call it m. Being in S, m has two distinct prime factorizations, say:

m ==p1p2ps q1q2qt

Since p1 p1p2ps and since p1p2ps = q1q2qt we have

p1 q1q2qt . By CYU 1.28, p1 qj for some 1 jt .

Without loss of generality, let us assume that p1 q1 . Since q1 is prime, its only divisors are 1 and itself. It follows, since p1  1 , that p1 = q1 . Consequently:

p1p2ps = q1q2qt  p1p2ps = p1q2qt

 p1p2ps – p1q2qt = 0

 p1p2ps – q2qt = 0

p1  0:  p2ps – q2qt = 0

 p2ps = q2qt two distinct prime decompositions for an integer smaller than m — contradicting the minimality on m in S 1.5 The Division Algorithm and Beyond 49

A pairs of prime number, such THEOREM 1.12 There are infinitely many primes. as 35 , 57 , and 11 13 , that differ by 2 are said to be PROOF: Assume that there are but a finite number of primes, say twin primes. Whether or not there exist infinitely many Sp= 1p2  pn , and consider the number: twin primes remains an open question. mp= 1p2pn + 1 Since mS , it is not prime. By Theorem 1.11, some prime must

divide m. Let us assume, without loss of generality, that p1 m . Since

p1 divides both m and p1p2pn : p1 mp– 1p2pn [Theorem

1.5(b), page 28]. A contradiction, since mp–11p2pn = .

CHECK YOUR UNDERSTANDING 1.30 Answer: See page A-9. Leta and b be relatively prime. Prove that if an and bn , then ab n . 50 Chapter 1 A Logical Beginning

EXERCISES

Exercises 1-3. For given a and d, determine integers q and r, with 0  r  d , such that ad= qr+ . 1.a ==0 d 1 2.a ==–5 d 133 3. a ==–133 d 5 Exercises 4-6. Find the greatest common divisor of a and b, and determine integers s and t such that gcd a b = sa + tb 4.a ==120 b 880 5.a ==–5 d 133 6. a ==–133 d 5

7. Prove that if 3 does not divide n, then n = 3k + 1 or n = 3k + 2 for some kZ .

8. Let n be such that 3 n2 – 1 . Show that 3 n .

9. Show that if n is not divisible by 3, then n2 = 3m + 1 for some integer m.

10. Show that an odd prime p divides 2n if and only if p divides n.

11. Prove that if a = 6n + 5 for some n, then a = 3m + 2 for some m.

12. Show that 2 n4 – 3 if and only if 4 n2 + 3 .

13. Prove that any two consecutive odd positive integers are relatively prime.

14. Let a and b not both be zero. Prove that there exist integers s and t such that nsatb= + if and only if n is a multiple of gcd a b .

15. Prove that the only three consecutive odd numbers that are prime are 3, 5, and 7.

16. Show that a prime p divides n2 if and only if p divides n.

17. Prove that every odd prime p is of the form 4n + 1 or of the form 4n + 3 for some n.

18. Prove that every prime p  3 is of the form 6n + 1 or of the form 6n + 5 for some n.

19. Prove that every prime p  5 is of the form 10n + 1 , 10n + 3 , 10n + 7 , or 10n + 9 for some n.

20. Prove that a prime p divides n2 – 1 if and only if pn– 1 or pn+ 1 .

21. Prove that every prime of the form 3n + 1 is also of the form 6k + 1 .

22. Prove that if n is a positive integer of the form 3k + 2 , then n has a prime factor of this form as well.

23. Prove that if the integer n  1 satisfies the property that if nab , then na or nb for every pair of integers a and b, then n is prime.

24. Prove that n  1 is prime if and only if n is not divisible by any prime p with pn . 1.5 The Division Algorithm and Beyond 51

PROVE OR GIVE A COUNTEREXAMPLE

25. There exists an integer n such that n2 = 3m – 1 for some m. 26. If a = 3m + 2 for some m, then a = 6n + 5 for some n. 27. If m and n are odd integers, then either mn+ or mn– is divisible by 4. 28. For any a, and b not both 0, there exist a unique pair of integers s and t such that gcd a b = sa + tb . 29. For every n, 34n – 1 . 30. For every nZ + , 34n + 1 . 52 Chapter 1 A Logical Beginning 2.1 Basic Defintions 53

2 CHAPTER 2 A Touch of Set Theory Modern mathematics rests on set theory. Some of that foundation is introduced in the first four sections of the chapter, and a glimpse into the axiomatic construction of the theory is offered in Section 5.

A bit of set notation has already been introduced §1. BASIC DEFINITIONS in the previous chapter. (See page 10) As you might expect, one defines two sets to be equal if each element of either set is also an element of the other:

DEFINITION 2.1 Two sets A and B are equal, written AB= if: We remind you that the SET xA  xB and xB  xA symbol  is read “is an element of.” or: xA  xB Note that when it comes to set notation, order is of no consequence. For example: 123 = 213 . Moreover, to avoid an unpleas- antness such as 1123 = 123 we stipulate that elements cannot be listed more than once within the set notation.

DEFINITION 2.2 Let A and B be sets. A is said to be a subset SUBSET of B, written AB if every element in A is B AND also an element in B, i.e: xA  xB . A PROPER SUBSET A is said to be a proper subset of B, written AB , if A is a subset of B and AB . For example: 123  123 12  123  and  1 2  123 Just as you can add or multiply numbers to obtain other numbers, so then can sets be combined to obtain other sets: DEFINITION 2.3 Let A and B be sets. INTERSECTION AND The intersection of A and B, written AB , OF SETS is the set consisting of the elements common to both A and B. That is: If someone asks you if you AB = xx A and xB want tea or coffee, you are read such that being offered one or the other, but not both: the The union of A and B, written AB , is the exclusive-or is being used. set consisting of the elements that are in A or In mathematics and sci- ence, however, the inclu- in B (see margin). That is: sive-or is generally used. In particular, to say that x AB = xx A or xB is in A or B, allows for x to be both in A and in B. 54 Chapter 2 A Touch of Set Theory

The adjacent visual repre- sentations of sets are called A A Venn diagrams. B B [John Venn (1834-1923)]. AB = xx A and xB AB = xx A or xB (a) (b) Figure 2.1 For example, if A = 35911 and B = 125611 , then: AB ==35911  125611 511 and: AB ==35911  125611 35911126  . We note that the set that contains no elements is called the (or the null set), and is denoted by  . I A B DEFINITION 2.4 Two sets A and B are disjoint if AB= . DISJOINT SETS In particular, since 123 45 = : 123 and 45 are disjoint.

DEFINITION 2.5 Let A and B be sets. A minus B, denoted by A B A MINUS B AB– , is the set of elements in A that are not AB– in B: AB– = xx A and xB

For example, if A = 12561011  and B = 35911 , As it is with numbers, set then: subtraction is not a com- AB–== 12561011  –35911  12610 mutative : AB– and: need not equal BA– . BA–== 35911 –3912561011  

When dealing with sets, one usually has a in mind — a set that encompasses all elements under current consideration. The let- ter U is typically used to denote the universal set, and it generally takes on a rectangular form in Venn diagrams (see margin). U Let A be a subset of the universal set U. The c DEFINITION 2.6 A c complement of A in U, written A , is the set OF A SET of elements in U that are not contained in A: A Ac = x xU and xA (More simply: xx A , if U is understood) Note that:Ac = UA– For example, if U = 12 3 4 5 6 789 , then: c 13578 = 2469 2.1 Basic Defintions 55

CHECK YOUR UNDERSTANDING 2.1 (a) Let U = 1234567 , A = 135 , B = 2347 , and C = 345 . Determine: (i) AB c (ii) AB c  AB c (iii) AB– c  C (iv) ABC–  c (v) xU xy= + 2 yB (a-1) 6 (a-ii) 156 (a-iii) 34 (a-iv) 23467 (b) Indicate True or False: (a-v) 456 (i) 2  12 (ii) 212  (iii) 2  12  True: (b-i), (b-ii), (b-iii), (b-iv), (b-vii) False: (b-v), (b-vi) (iv)   12 (v)   12 (vi)   (vii)   Here is a link between the difference and complement concepts:

THEOREM 2.1 AB– = AB c

PROOF: xAB –  xA and xB xA and xB c xAB  c We can also use a Membership Table (an approach reminiscent of Truth Tables) to establish the above result. As you can see, we labeled the first two columns of Row 1 in Figure 2.2: A and B. Other involved sets, built from those two sets, are acknowledged in the remaining three columns of Row 1. Now, for any given x in the universal set, there are four possibilities: xA and xB And so we positioned a 1 in the first two columns of Row 1 to indicate containment. xA and xB And so we positioned a 0 in the second column of Row 3 to indicate non-containment. xA and xB 0 and then 1 appear in the first two columns of Row 4. xA and xB 0 and 0 appear in the first two columns of Row 5. Since 1 appears under both A and B in Row 2 of Figure 2.2, a 0 appears in the third column of that row. Why? Because: xA and xB  xAB – 1 1 0 Why does a 1 appear in the third column of Row 3? Because: xA and xB  xAB – 1 0 1 You get the developing pattern, no? But just in case, let’s finish up with Row 3: The Bc column: xB  xB c 0 1 The AB c column: xA and xB c  xAB  c 1 1 1 56 Chapter 2 A Touch of Set Theory

You can turn this mem- Set Row 1: ABAB– Bc AB c bership table into a truth table by replacing the Row 2: 11 0 0 0 sets A and B with the propositions Row 3: 10 1 1 1 xA and xB Row 4: 01 0 1 0 respectively. Row 5: 00 0 0 0

xAB –  xAB  c Figure 2.2

EXAMPLE 2.1 Use a Membership Table to establish: (a) AB  AB c = A (b) AB–  AC–  ABC– 

SOLUTION: (a)

ABAB Bc AB c AB  AB c 11 1 0 0 1 10 0 1 1 1 01 0 1 0 0 00 0 0 0 0

xA  x  AB  AB c (b) ABCAB– AC– AB–  AC– BC ABC–  111 0 0 010 110 0 1 001 101 1 0 001 011 0 0 010 100 1 1 101 010 0 0 000 001 0 0 000 000 0 0 000

if xAB –  AC– then xA –  BC

EXAMPLE 2.2 Prove or give a counterexample. (a) AB c  C = CAB–  (b) AB c  AB

c SOLUTION: (a) True: xAB   CxC  and x  AB  xC –  AB 2.1 Basic Defintions 57

(b) A glance at the adjacent figure suggests that, c A AB

in general, AB c is not a subset of AB . AB B  Suggestion aside, a specific counterexample is needed to nail down the claim. Here is one: For U = 12 , A = 1 , B = 1 : AB c ==2 while AB 1

CHECK YOUR UNDERSTANDING 2.2 Prove or give a counterexample: (a) ABC–  = AB–  AC– Answer: See page A-9. (b) AB c c  B = Ac  B The two DeMorgan’s laws in the margin were established on page 6 Let p and q be proposi- (see margin). Here are the “set-versions” of those laws: tions. Then: (a) ~pq  ~p  ~q THEOREM 2.2 For any sets A and B: (b) ~pq  ~p  ~q DEMORGAN’S (a) AB c = Ac  Bc LAWS (b) AB c = Ac  Bc

PROOF: (a) In accordance with Definition 2.1, we are to show that: xAB  c  xA c  Bc and that xA c  Bc  AB c . Let’s do them both at the same time: Definition 2.5 Definition 2.3 Definition 2.5 xAB  c  xA   B   xA or xB  xA c or xB c

Definition 2.3:  xA c  Bc

(b) We choose to use a Membership Table to establish AB c = Ac  Bc :

ABAc Bc Ac  Bc AB AB c 1100 0 1 0 1001 0 1 0 0110 0 1 0 0011 1 0 1

xA c  Bc  xAB  c 58 Chapter 2 A Touch of Set Theory

CHECK YOUR UNDERSTANDING 2.3 (a) Use a Membership Table to prove Theorem 2.1(a) (b) Prove Theorem 2.1(b) by a method similar to that used in the Answer: See page A-10. proof of Theorem 2.1(a).

The union of n sets S n = S S  S can be denoted by in i = 1 1 2 n S  S   S or by S , and the union of the sets S  by 1 2 n  i i i = 1 i = 1 S or S . For any set A, S , denotes a collection of sets  i  i    A i = 1 iZ + indexed by A, and just as Si can be used to denote the union of the n  sets in the collection S iZ + , so then does S denote the union Similarly  Si ,  Si , i iZ +   i = 1 iZ +   A S of the sets in the collection S . For example, Sr denotes the and   represent the   A    A r   intersection of n sets, sets union of a collection of sets indexed by the set of real numbers  . indexed by the positive integers, and sets indexed We are now in a position to address the most general form of DeMor- by the set A, respectively. gan’s Laws:

THEOREM 2.3 For any collection S   A of sets: DEMORGAN’S c c LAWS c c (a) S = S (b) S = S       A   A   A   A

c PROOF: (a) xS   xS    The containment table   A   A approach cannot be used to establish this  xS  for every   A result. Why not? c  xS  for every   A c  xS     A

As for (b): CHECK YOUR UNDERSTANDING 2.4 Answer: See page A-10. Establish Theorem 2.2(b) 2.1 Basic Defintions 59

INTERVAL NOTATION Certain subsets of  , called intervals, warrant special recognition. As is indicated below, brackets are used to indicate endpoint-inclu- sions, while parenthesis denote endpoint-exclusions. Interval Notation Geometrical Representation All real numbers strictly 15 = x 1 x 5 | | | | | | | | | | between 1 and 5 (not including ( ) -3 -2 -1 0 1 2 3 4 5 6 1 or 5) excluding 1 and 5 All real numbers between 1 and 15 = x 1 x 5 5, including both 1 and 5. | | | | [| | | | ] | | -3 -2 -1 0 1 2 3 4 5 6 including 1 and 5 All real numbers between 1 and 15  = x 1  x  5 5, including 1 but not 5. | | | | [| | | | ) | | -3 -2 -1 0 1 2 3 4 5 6 including 1 and excluding 5 All real numbers between 1 and 1 5  = x 1  x  5 5, including 5 but not 1. | | | | (| | | | ] | | -3 -2 -1 excluding 1 and including 5 0 1 2 3 4 5 6 All real numbers greater than 1. 1 = xx 1 | | | | (| | | | | | -3 -2 -1 the symbol 0 1 2 3 4 5 6 All real numbers greater than or 1  = xx 1 equal to 1. | | | | [| | | | | | -3 -2 -1 0 1 2 3 4 5 6 All real numbers strictly less – 5  = xx 5 than 5. | | | | | | | | ) | | -3 -2 -1 0 1 2 3 4 5 6 All real numbers less than or – 5  = xx 5 equal to 5. | | | | | | | | ] | | -3 -2 -1 0 1 2 3 4 5 6 The set of all real numbers. – = x – x  | | | | | | | | | | -3 -2 -1 0 1 2 3 4 5 6 Figure 2.3 In general, for ab : ab = rRarb  is said to be a closed interval. ab = rRarb  is said to be an open interval. ab  = rRarb   and ab  = rRar   b are said to be half-open (or half-closed) intervals. In addition: a  and – b are said to be open. a  and – b  are said to be half-open (or half-closed). = –   is said to be both closed and open (or clopen).

CHECK YOUR UNDERSTANDING 2.5 Simplify: (a) –22  05 (b) –13 c  5  (a) 02  (b) 5  (c) –22   35 (c) –20 –21 35 60 Chapter 2 A Touch of Set Theory

EXERCISES

Exercises 1-19. For U = 123 , O = 135 , E = 246 , A = 5nn U , B = 3nn U , C = 123 15 , D = 24610 , and F = 11 12 13 14 , determine: 1.OE 2.OE 3.AB 4. AB 5.BC 6.BC 7.CD 8. CD 9.Oc  Ec 10. Oc  A 11. CO 12. OA c 13.CD  F 14.CDF  15. CF  D 16. CF c  F 17. Bc  C  DO 18. OE c  AB c 19. OE c  Cc

Exercises 20-23. Determine the set of all subsets of the given set A. 20.A = 1 21.A = 12 22.A = 123 23. A = 

24. Establish the following set identities (all capital letters represent subsets of a universal set U): A   = A Domination Laws AU = U AU = A A  = AA = A Complementation Ac c = A

Commutative AB = BA Associative Laws ABC  = AB  C Laws AB = BA ABC  = AB  C Distributive Laws ABC  = AB  AC ABC  = AB  AC Exercises 25-51. Prove that: 25. AB–  Bc 26. Ac  B c = AB c 27. ABA– = 28.AC– CB = 29. AB– – C  AC– 30. AB  AB c = A 31. AAB–  = AB– 32.AB– c = BA c 33. UAB=   AB c 34.Ac  Ac – B c = A 35. Ac  Bc – A c = A 36. ABA – = AB 37.AC– – BC– = AB– – C 38. AC–  BC– = AB – C 39.AC–  BC– = AB – C 40. AB– – AC– = ACB – 41.AB–  AC– = ABC–  42. AB–  AC– = ABC–  43.BA–  CA– = BC – A 44. AB – C = AC–  BC– 45.AB– c  BA– c = AB c 46. AB  Cc = AB – C 2.1 Basic Definitions 61

47.AB– – BC– = AB– 48. AB – CA– = ABC – 49. AB  AC  ABC  50. AB c c  B = Ac  B 51. ABC Ac BC Bc Cc = U

Exercises 52-59. Give a counterexample to show that each of the following statements is False. 52. AB– = BA– 53.AB– c = Bc – Ac 54. AB c = Ac  Bc 55.AB c = Ac  Bc 56.AB c = Ac  Bc 57. AB c = Ac  Bc 58. AB – C = ABC – 59. AB c  Cc = Ac  BC c

60. For n an integer distinct from 0, let An = aZa is divisible by n . Determine the set: c c (a) A5  A10 (b) A5  A10 (c) A9  A3 (d) A9  A3

61. Prove that for any given set A,   A and AA .

62. Prove that if AB and BC , then AC .

63. Prove that if AB , BC , and CA then ABC== .

64. Prove that for any sets A and B of a universal set U: AB– = AB c .

65. Prove that AB if and only if Bc  Ac .

66. Prove that AAB=   AB– and AB AB– = . [So, AB  AB– is a representation of A as a ]

67. Prove that AB = ABA – and that ABA– = . (So, ABA – is a representation of AB as a disjoint union.)

68. Prove that AB if and only if AC  BC for every CU .

69. Prove that AB if and only if AC  BC for every CU . 70. Prove that the three statements (i) AB (ii) AB = A (iii) AB = B are equivalent by sowing that i ii iii i . # 71. Let S   A be any collection of sets. Prove that for any set X:   (a) S  X = S  X (b) S  X = S  X       A   A   A   A 62 Chapter 2 A Touch of Set Theory

PROVE OR GIVE A COUNTEREXAMPLE

72. If AB   and BC   , then AC   .

73. If AB= or BC= , then AC= .

74. If ABC  and CBA  , then AC= .

75. If AB = AC , then AC= .

76. If AB = AC , then AC= .

77. If AB = AB , then either A =  or B =  .

78.ABC– – = AB– – C .

79.AB– – C = AC– – BC– .

80.AB– = BA– if and only if AB= .

81. If ABC  and BCD  , then AC  BD .

82.AB  AC = ABC  . 83. If no element of a set A is contained in a set B, then A cannot be a subset of B. 84. Two sets A and B are equal if and only if the set of all subsets of A is equal to the set of all subsets of B.

85. = . 2.2 Functions 63

2

§2. FUNCTIONS You’ve dealt with functions in one form or another before, but have All “objects” in mathematics you ever been exposed to a definition? If so, it probably started off with are sets, and functions are no something like: exceptions. The function f A function is a rule...... 2 given by fx = x , is the Or, if you prefer, a rule is a function...... subset f = xx 2 x   You are now too sophisticated to accept this sort of “circular definition.” of the plane. Pictorially: Alright then, have it your way: DEFINITION 2.7 For given sets X and Y, we define the Carte- CARTESIAN sian Product of X with Y, denoted by XY , PRODUCt to be the set of ordered pairs: A function such as XY = xy xX and yY f = xx 2 x   is often simply denoted by fx= x2 . Still, in spite of DEFINITION 2.8 A function f from a set X to a set Y is a subset their dominance throughout FUNCTION fXY  such that for every xX there mathematics and the , functions that can be described exists a unique yY with xy  f , and we in terms of algebraic expres- write: yfx= . sions are truly exceptional. Scribble a in the plane DOMAIN The set X is said to be the domain of f, and for which no vertical line cuts RANGE yYxy   f for some xX the curve in more than one point and you have yourself a is said to be the range of f. function. But what is the “rule” While the domain of f is all of X, for the set g below? the range of f need not be all of Y. g Moreover, for AX and BY : gx. OF AX fA= faaA is called the image of INVERSE IMAGE OF A under f, and f –1B = xXfx  B .x BY is called the inverse image of B.

Note that the set S below, is not a function: The notation f: XY is used to denote a function from a set X to a set Y. Moreover, as is customary, we write yfx=  to indicate that . xy  f . . S The function f: XY is also called a mapping from X to Y. In the . event that yfx=  , we will say that f maps x to y. Finally, we note that x Why not? the symbols Df and Rf are used to denote the domain and range of f, respectively. Since functions are sets, the definition of equality is already at hand. Two functions f and g are equal if the set f is equal to the set g. In other words, if:

Df = Dg and fx= gx for every xD f = Dg .

When not specified, the expressed in terms of a variable x is understood to consist of all values of x for which the expression can be evaluated. 64 Chapter 2 A Touch of Set Theory

CHECK YOUR UNDERSTANDING 2.6 For X = 0123 and Y = 23456 determine if the sub- set f of XY is a function f: XY . If so, determine its range. (a) f = 02 12 23 26 (b) f = 02 12 22 32 (c) f = 02 12 23 (a) No (b) Range: 2 (c) No (d) No (d) f = 22 30 41 53 60

COMPOSITION OF FUNCTIONS Consider the schematic representation of the functions f: XY and g: YT in Figure 2.4, along with a third function gf: XT . f . g fx .gfx x.

X Y T gf Figure 2.4 As is suggested in the above figure, the function gf: XT is given by: gf x = gfx first apply f and then apply g Formally: DEFINITION 2.9 Let f: XY and g: YT be such that COMPOSITION the range of f is contained in the domain of g. The composite function gf: XT is given by: gf x = gfx In set notation: gfxt=  yY   xy  f and yt  g

EXAMPLE 2.3 For fx= x2 + 1 and gx= 2x – 5 find:

(a) gf 3 (b) fg 3 (c) gf x (d) fg x 2.2 Functions 65

SOLUTION: 2 (a) gf 3 ==gf3 g3 + 1 ===g10 2105 – 15 2 (b) fg 3 ==f g3 f23 – 5 ===f1 1 +21 2 2 2 (c) gf x ==gfx gx+ 1 =2x + 1 –25 =x – 3 2 2 (d) fg x ==fgx f2x – 5 =2x – 5 +41 =x –2620x +

CHECK YOUR UNDERSTANDING 2.7 3x (a) For fx= x + 2 and gx= ------, determine: x + 1 (a-i) --5- (a-ii) -----17- (i) g f 3 (ii) f g 3 (iii) g f x (iv) f g x 2 4     (a-iii) 3------x + 6- (a-iv) 5------x + 2- x + 3 x + 1 (b) For f = 1 a c 5 xy and (b) gf = 1 a ct x 4 gaa=  c 3 5 t y 4 , determine: gf . Is fg defined? Justify your answer. The time has come to break out of our comfort zone and to consider functions of the form f: XY , where X or Y are no longer restricted to sets of real numbers. Here are some “exotic sets:”

DEFINITION 2.10 Let n  1 . The set of n-, denoted by n-tuples n , is given by: n  = a1a2  an ai   1 in Let n  1 and m  1 . The set of nm - Matrices matrices (pronounced n-by-m matrices). denoted by Mnm , is given by:

a11 a12  a1m

a21 a22  a2m

M =   nm 

an1 an2  anm

1 To illustrate:2 ---  2 and 3 250  4 . 2

4 25 01– 5  M22 , 0  M31 , and  M23 –91 31 4  66 Chapter 2 A Touch of Set Theory

As might be expected: Two n tuples a1a2  an , b1b2  bn are said to be equal if ai = bi for 1 in ; in other words: if the are one and the same. The same goes for elements of Mnm .

EXAMPLE 2.4 4 2 Let f:   M22 and g: M22  R be given by:

–2b a ab fabcd = and g= ad+  bc 3dc cd (a) Determine: 35 (i) f52–4 9  (ii) g (iii) gf 1253 24

(b) Find gf abcd

SOLUTION: (a-i) Just follow the “f-pattern:” a b c d f52–4 9  ==–252  –102 3 4 –9 12– 9

35 (a-ii) Follow the g-pattern: g==34+52  710 . 24 (a-iii) Both patterns come into play:

–22 gf 1253 ===gf1253 g318 – 95

–2b a (b) gf abcd ==g– b +6c ad 3dc

ONE-TO-ONE AND ONTO FUNCTIONS Our next goal is to single out some functions that count more than others; beginning with one-to-one functions: Equivalently, f is one-to-one if x  x  fx fx DEFINITION 2.11 A function f: XY is said to be one-to- 1 2 1 2 one (or an injection) if In words: ONE-TO-ONE different x’s are mapped fx==fx x x to different y’s. 1 2 1 2

In set notation:

x1 y  x2 y  fx 1 = x2 2.2 Functions 67

Figure 2.5 represents the action of two functions, f and g, from X = 012 to Y=  abcd . The function f on the left is one- to-one while the function g is not.

0. .a 0. .a

1. .b 1. .b .d .d 2 c 2 . f . . g .c one-to-one not one-to-one Figure 2.5

x EXAMPLE 2.5 (a) Show that the function fx= ------is one-to-one. 5x + 2 (b) Show that the function gx= x3 ++3x2 –75x is not one-to-one.

SOLUTION: (a) Appealing to Definition 2.10, we begin with fa= fb, and then go on to show that this can only hold if ab= : fa= fb fa= fb ------a = ------b 5a + 2 5b + 2 a5b + 2 = b5a + 2 5ab +52a = ab + 2b ab= 2a = 2b ab= (b) Rather than gx= x3 ++3x2 –75x , consider the function hx= x3 + 3x2 – x and corresponding equation x3 +03x2 – x = : 3 2 Recall that the solutions x +03x – x = of the : xx2 + 3x – 1 = 0 ax2 ++bx c = 0 –133  are given by the quadratic x ==0 x ------formula: 2 2 using the quadratic formula x = –------b  b – 4ac- 2a We see that the function hx= x3 + 3x2 – x is not one-to-one, for it maps three (need only two) different x’s to zero: 0, ------–133 + - , and 2 ------–133 – . It follows that the function gx= x3 ++3x2 –75x will 2 those same numbers to 75, and is therefore not one-to-one.

CHECK YOUR UNDERSTANDING 2.8 x (a) Show that the function fx= ------is one-to-one. x + 1 Answer: See page A-11. (b) Show that the function fx= x5 –x + 777 is not one-to-one. 68 Chapter 2 A Touch of Set Theory

In words: f: XY is onto if DEFINITION 2.12 A function f: XY is said to be onto (or a every element in Y is ONTO surjection) if for every yY there exists “hit” by some fx . xX such that fx= y .

In set notation: yYxXxy      f

Figure 2.6 represents the action of two functions, f and g, from X = 0123 to Y=  abc . The function f on the left is onto, while g is not. X Y X Y 0. .a 0..a 1 .b 1 .b 3 .. 3 .. 2. .c 2. .c f g onto not onto Figure 2.6

EXAMPLE 2.6 4 (a) Let f:   M22 be given by:

fxyzw = –2y x 3wz Show that f is both one-to-one and onto.

(a) Let g: M22  M22 be given by: ab ba g = cd cd+2b Show that f is neither one-to-one nor onto.

SOLUTION: (a) To show that f is one to one, we start with fxyzw = fxyzw and go on to show that xyzw = xyzw : We need to consider two ele- ments in 4 . They have to look fxyzw = fxyzw  –2y x = –2y x different; and so we called one 3wz 3wz of the elements xyzw and the other xyzw . (We could –y = –y  yy=  have labeled the other   ABCD , or whatever. The 2x = 2x  xx=  two 4-tuples just have to look   different, that’s all. 3w = 3w  ww=    zz=  zz= 

 xyzw = xyzw 2.2 Functions 69

ab To show that f is onto, we take an arbitrary element  M22 cd and set our sights on finding xyzw  R4 such that

fxyzw = ab : cd –y = a  ya= –    ab –2y x ab 2xb=  xb=  2  fxyzw =  =   cd 3wz cd 3wc=  wc=  3    zd=  zd=  The above argument shows that f will map the element

b c 4 ab ---–a d ---  R to  M22 . Let’s check it out: 2 3 cd

b –2–a --- b c 2 f ---–a d --- ==ab  2 3 c cd 3--- d 3

ab ba (b) g = is not one-to one, since: cd cd+2b

00 00 00 00 00 g==g  , and  . 10 01 10 10 01

The functiong is not onto, since no element ab is mapped to cd

10 ab b a 1 0 : f =  since 3 is not 2 times 1. 03 cd cd+ 2b 0 3

CHECK YOUR UNDERSTANDING 2.9

4 Show that the function f: M22  R given by:

ab f = dc –  3ab cd Answer: See page A-11. is one-to-one and onto. 70 Chapter 2 A Touch of Set Theory

BIJECTIONS AND THEIR INVERSES So: A f: XY pairs DEFINITION 2.13 A function f: XY that is both one-to-one off each element of X with an element of Y. BIJECTION and onto is said to be a bijection.

Consider the bijection f: 0123  abcd depicted in Fig- ure 2.7(a) and the function f –1: abcd  0123 in Figure 2.7(b). Figuratively speaking, f –1 , read “f inverse,” was obtained from f by “reversing” the direction of the arrows in Figure 2.7(a). On a more formal level: f = 0 a 1 b 2 c 3 d . and: f –1 = a 0 b 1 c. 2 d 3 3 .d 3 .d 2. .c 2. .c 1. .b 1. .b 0. a 0. a f . f –1 . (a) (b) Figure 2.7 In general: DEFINITION 2.14 The inverse of a bijection f: XY , is the –1 function f : YX given by: f –1y ==x where fx y More formally: f –1 = yx xy  f Figure 2.7 certainly suggests that if f is a bijection then so is f –1 , and –1 f that if you apply f and then f you are back to where you started from 3 d (see margin). This is indeed the case: 2 . .c 1 f –1 b 0 a THEOREM 2.4 Let f: XY be a bijection. Then: (a)f –1: YX is also a bijection. (b)f –1fx = x xX and ff–1y = y yY

–1 –1 –1 PROOF: (a) f is one-to-one: If f y1 ==f y2 x , then: Recall that to say that –1 –1 y1 x  f and y2 x  f fx= y is to say that xy  f (see Definition  xy 1  f and xy 2  f 2.8).  y1 = y2 (since f is a function) f –1 is onto: Let xX . Since f is onto, there exists yY such that xy  f . Then: yx  f –1 f –1y = x . 2.2 Functions 71

(b) Let xX . Since xfx  f , fx x  f –1 , which is to say: xf= –1fx . As for the other direction:

CHECK YOUR UNDERSTANDING 2.10 Verify that for any bijection f: XY : Answer: See page A-12. ff–1y = y for every yY

EXAMPLE 2.7 (a) Find the inverse of the binary function of 4 f:   M22 given by:

fxyzw = –2y x 3wz (see Example 2.6) (b) Show, directly, that –1ab ab ff = cd cd

ab SOLUTION: (a) For given we determine xyzw such that cd

fxyzw = ab: cd

–y = a  ya= –    ab –2y x ab 2xb=  xb=  2  fxyzw =  =   cd 3wz cd 3wc=  wc=  3    zd=  zd= 

–1ab b c Conclusion: f = ---–a d --- cd 2 3

Note: In Example 2.6 we showed that for fxyzw = –2y x : 3wz b c f ---–a d --- = ab 2 3 cd

–1ab b c That being the case: f = ---–a d --- . cd 2 3 72 Chapter 2 A Touch of Set Theory

b –2–a --- –1 b c 2 (b) ff ab ==f---–a d --- =ab   cd 2 3 c cd 3--- d 3 since fxyzw = –2y x 3wz

CHECK YOUR UNDERSTANDING 2.11

4 Find the inverse of the bijection f: M22  R given by ab f = dc –  3ab and verify, directly, that: Answer: cd –1 f xyzw = z  3 w –1 –1 ab ab –y x ffxyzw = xyzw and that f f = . For the rest: See page A-11. cd cd As it turns out, one-to-one and onto properties are preserved under composition:

THEOREM 2.5 Let f: XY and g: YW be functions with the range of f contained in the domain of g. Then: (a) If f and g are one-to-one, so is gf . (b) If f and g are onto, so is gf . (c) If f and g are , so is gf .

PROOF: (a) Assume that both f and g are one-to-one, and that:

gf x1 = gf x2

Which is to say: gfx1 = gfx2

Since g is one-to-one: fx1 = fx2

Since f is one-to-one: x1 = x2

(b) Assume that both f and g are onto, and let wW . We are to find xX such that gf x = w . Let’s do it: Since g is onto, there exists yY such that gy= w . Since f is onto, there exists xX such that fx= y . It follows that gf x ===gfx gy w .

(c) If f and g are both bijections then, by (a) and (b), so is gf . 2.2 Functions 73

Theorem 2.5(c) asserts that the composition gf of two bijections is again a bijection. As such, it has an inverse, and here is how it is related to the inverses of its components:

THEOREM 2.6 If f: XY and g: YW are bijections, This is an example of a so- called “shoe-sock theorem.” then: –1

Why the funny ? gf –1 = f g–1

The shoes come off and then the socks the then and off come shoes The

In the reverse process: reverse the In One puts on socks then shoes then socks on puts One PROOF: For given wW , let xX be such that gf x = w ; which is to say, that gf –1w = x . We complete the proof by showing that f –1g–1 w is also equal to x: gf –1w = x wg= f x wgfx=  g–1w = fx f –1g–1w = x f –1g–1 w = x 74 Chapter 2 A Touch of Set Theory

EXERCISES

Exercises 1-6. For X=  abcd and YABCD=  determine if the subset f of XY is a function f: XY . If so, find its range. 1.faA=  2. faA=  bA cA dA 3.faA=  aB bB cD dA 4. faA=  bB cC dD 5.faB=  bC cD dA 6. faB=  bA cD dC

Exercises 7-16. For X=  abcd , YABC=  , and Z = 123 , let f: XY , g: YZ , h: ZX , and k: ZZ be given by: faA=  bC cA dB gA=  1 B 3 C 2 h = 1 b 2 a 3 c k = 12 23 31 Indicate whether or not the function is one-to-one, and whether or not it is onto. 7.gf 8.gk 9.kk 10.hk 11. fh 12.kg 13.gf 14.fhk 15.kk k 16. fhkk

Exercises 17-20. Is f:  (a) One-to-one? (b) Onto? 3x – 7 x 17. fx= ------18.fx= x2 – 3 19. fx= ------20. fx= x3 –2x + x + 2 x2 + 1

Exercises 21-23. Is f:  2 (a) One-to-one? (b) Onto? 21. fx= xx 22. fx= x 1 23. fx= 2x –7x

Exercises 24-26. Is f: 2  2 (a) One-to-one? (b) Onto? 24.fxy = yx – 25.fxy = xx + y 26. fxy = 2xx + y

4 Exercises 27-28. Is f: M22   (a) One-to-one? (b) Onto?

ab ab 27.f = a–2b cc– d 28. f = ab– cdb– a cd cd

4 Exercises 29-30. Is f:   M22 (a) One-to-one? (b) Onto?

a ba+ ab ba+ 29. f abcd = 30. f abcd = cb+ da+ cb+ a2b2

Exercises 31-32. Is f: M31  M22 (a) One-to-one? (b) Onto?

a a  ab  ab 31.f b = 32. f b =  c 5  c abc++ c c 2.2 Functions 75

Exercises 34-42. Show that the given function f: XY is a bijection. Determine f –1: YX –1 –1 and show, directly, that f f x = x xX and that ff y = y yY . 33.X ==Y , and fx= 3x – 2 . x + 1 34.X ==–0  0  Y –1  1  , and fx= ------. x 2x 35.X ==–1 –  –1  Y –1  1  , and fx= ------. x + 1 36.XY==2 , and fab = –b a . 37.XY==2 , and fab = 5ab + 3 .

ab bc 38.XYM==22 , and f = . cd da

ab c 2d 39.XYM==22 , and f = . cd ab–

4 2bc+ 1 40.X ==  Y M22 , and fabcd = . da–

 a 3 41.XM= 31 Y = , and f b = 2aa – b bc+ .  c

42. Prove that a function f:  is one-to-one if and only if the function g:  given by gx= –fx is one-to-one. # 43. Prove that for any given f: XYg  : YS  and h: ST : hgf = hg f . 44. Let f: XYg  : XY  and h: YW be given, with h a bijection. (a) Prove that if hfh= g , then fg= . (b) Show, by means of an example, that (a) need not hold when h is not a bijection. 45. Let SX , Y   , and f: SY be given. Prove that there exists a function g: XY such that fx= gx for every xS . (That is, a function g which “extends” f to all of X.) 46. Let SX , Y   , and f: XY be given. Prove that there exists a function g: SY such that fx= gx for every xS . (That is, a function g which is the “restriction” of f to the subset S.) 76 Chapter 2 A Touch of Set Theory

Exercise. 48-53. (Algebra of Functions) For any set X, and functions f: X   and g: X   , f we define fg+ : X   fg– : X   fg : X   and ---: X   as follows: g fg+ x = fx+ gx fg– x = fx– gx fg x = fx fx f fx --- x = ------if gx 0 g gx

47. Prove that for any f: X   and g: X   : fg+ = gf+ and fg = gf . 48. Exhibit f:  , g:  , such that fg–  gf– . 49. Exhibit one-to-one functions f:  , g:  , such that fg+ is not one-to-one. 50. Exhibit onto functions f:  , g:  , such that fg+ is not onto. 51. Exhibit one-to-one functions f:  , g:  , such that fg is not one-to-one. 52. Exhibit onto functions f:  , g:  , such that fg+ is not onto.

PROVE OR GIVE A COUNTEREXAMPLE

53. For f:  and g:  , if fg+ is one-to-one, then both f and g must be one-to-one. 54. For f:  and g:  , if fg+ is one-to-one, then f or g must be one-to-one. 55. For f:  and g:  , if fg is one-to-one, then both f and g must be one-to-one. 56. For f:  and g:  , if gf is one-to-one, then f or g must be one-to-one.

57. (a) If f: XY is an onto function, then so is the function gf: XW onto for any func- tion g: YW . (b) If g: YW is an onto function, then so is the function gf: XW onto for any func- tion f: XY .

58. (a) Let f: XY and g: YW . If gf: XW is onto, then f must also be onto. (b) Let f: XY and g: YW . If gf: XW is onto then g must also be onto.

59. (a) If f: XY is one-to-one, then so is the function gf: XW for any g: YW . (b) If g: YW is one-to-one, then so is the function gf: XW for any f: XY .

60. For any X and Y   , there exists at least one function f: XY .

61. For any X   , there exists at least two functions f: XX . 2.3 Infinite Conunting 77

2

§3 INFINITE COUNTING Assume that you do not know how to count, and that you have two bags of marbles, one containing red marbles and the other blue. With- out counting, you can still determine whether or not the two bags con- tain the same number of marbles: Take a red marble from the one bag and a blue marble from the other, then put them aside. Continue this pairing process till one of the bags becomes empty. If, as the one bag becomes empty so does the other, then you can con- clude that the two bags had the same number of marbles (without knowing the number). If you let the bag of red marbles be a set R, and the bag of blue mar- bles be a set B, and if you think of the paring off process as representing a function f which assigns to each red marble a unique blue marble, then you can say that the two bags will have the same number of mar- bles if the function f is a bijection. Marbles aside: Note that if there exists a bijection f: AB then DEFINITION 2.15 Two sets A and B are of the same cardinal- there also exists a bijection ity, written CardA = CardB , if there going the other way; exists a bijection f: AB . namely: f –1: BA [see Theorem 2.4(a), page 70] In a sense, the term “same cardinality” can be interpreted to mean “same number of elements.” The classier terminology is used since the expression “same number of elements” suggests that we have associ- ated a number to each set, even those that are infinite. The following theorem should not surprise you. Intuitively, it says that if a set A has the same number of elements as a set B, and if B has the same number of elements as a set C, then A and C have the same number of elements. Intuition aside:

THEOREM 2.7 If CardA = CardB and CardB = CardC , then CardA = CardC .

PROOF: Applying Definition 2.15, we choose bijections f: AB and g: BC . By Theorem 2.5(c), page 72, gf: AC is also a bijection. Returning to Definition 2.15, we conclude that A and B are of the same cardinality. In our physical universe, if you take something away from a collec- tion of objects you will certainly end up with a smaller number of objects in the collection. When it comes to infinite sets, however, this need no longer be the case: 78 Chapter 2 A Touch of Set Theory

EXAMPLE 2.8 Let 2Z+ denote the set of even positive integers. Show that Card 2Z+ = CardZ+ .

SOLUTION: The diagram below illustrates how the elements of Z+ can be paired with those of 2Z+ : 1 2 3 4 5 6 7 8 9 . . .

2 4 6 8 10 12 14 16 18 . . . Let’s formalize the above observation by showing that the function f: Z+  2Z+ given by fn= 2n is a bijection. f is one-to-one:

fn1 ===fn2  2n1 2n2  n1 n2 f is onto: Let m  2Z+ . Since m is even, m = 2n for some nZ + . Then: fn==2nm.

CHECK YOUR UNDERSTANDING 2.12 Answer: See page A-12. Prove that: CardZ+ = Card0123 Roughly speaking, by a we simply mean a set whose elements can actually be “counted:” one element, two elements, three, and so on until you have counted them all, even though it may require all of Z+ to accomplish that feat. On a more formal footing: DEFINITION 2.16 A set is countable if it is finite or has the COUNTABLE SET same cardinality as Z+ . In particular, the set 123 is countable, as is the set 2Z+ of Exam- ple 2.9. Below, we show that the set Z of all integers is also countable. This may not be too surprising, since the feeling is that there should cer- tainly be enough positive integers to count anything. (That feeling is however wrong, as you will soon see.) EXAMPLE 2.9 Prove that the set Z of all integers is countable.

SOLUTION: The diagram below illustrates how the elements of Z+ can be paired off with those of Z: . . .

1 2 3 4 5 6 7 8 9 . . .

. . . -5 -4 -3 -2 -1 0 1 2 3 4 5 6 . . . 2.3 Infinite Conunting 79

Can we explicitly pin down the function represented in the above dia- To evaluate f, follow the  n + 1 stated instructions:  ------if n is odd If n is odd, say 7, then  2 use the upper “rule:” gram? Yes: fn=   n 71+ – ---1– if n is even f7 ==------4  2 2  If n is even, say 8, then f is one-to-one: Suppose that fn= fn . We first observe that go with the bottom 1 2 rule: since f takes even integers to non-positive integers, and odd inte- 8 gers to positive integers, n1 and n2 must both be even or both be f8 ==–---1– –3 2 odd. We consider both cases:

Case 1. n1 and n2 even. Taking the “low road” in the descrip- tion of f, we have: n n fn==fn –----1-1– = –----2-1–  n n 1 2 2 2 1 2 details omitted

Case 2. n1 and n2 odd. Taking the “high road” in the descrip- tion of f, we now have: n + 1 n + 1 fn==fn ------1 = ------2  n n 1 2 2 2 1 2 f is onto: For given z we are to find a positive integer n such that fn= z , and again break the argument into a couple of cases: Case 1. For zZ and positive, consider the odd positive inte- n + 1 z = ------ger n = 2z – 1 (see margin). Applying f we have: 2 2zn= + 1 2z – 1 + 1 2z fn==f2z – 1 ------==----- z n = 2z – 1 2 2

n z = –---1– Case 2. For zZ and non-positive, consider the even posi- 2 tive integer n = –22z + (see margin). Applying f we have: 2zn= –2+ –22z + fn==f–22z + –------– 1 =––11z + – =z n = –22z + 2

The set Q+ of positive rational numbers (fractions) appears to have a lot more elements that does Z+ . Not really:

EXAMPLE 2.10 Show that Q+ is countable.

SOLUTION: We call your attention to Figure 2.8. The positive integers occupy the first row of that figure. The positive rational numbers with denominator 2, which did not already appear in the first row, are listed in the second row. The positive rational numbers with a 3 in the denominator, which did not appear in either the first or second row, comprise the third row; and so on and so forth. Following the arrows in the figure, one is able to count each and every element of Q+ : The first number counted is the number 1. Moving along the sec- ond arrow we arrive at the second number counted: 2, and then the 80 Chapter 2 A Touch of Set Theory

1 third number: --- . Following the third arrow we count the fourth, 2 3 1 fifth, and sixth numbers: 3, --- , and --- , respectively. Can you follow 2 3 our procedure and determine the next four numbers counted? Sure, 5 2 1 they are pierced by the fourth arrow; namely: 4, --- , --- , and --- . 2 3 4

12345 6 7 8 

1 3 5 7 9 11 13 15 ------ 2 2 2 2 2 2 2 2

1 2 4 5 7 8 10 11 ------ 3 3 3 3 3 3 3 3

1 3 5 7 9 11 13 15 ------ 4 4 4 4 4 4 4 4

1 2 3 4 6 7 8 9 ------ 5 5 5 5 5 5 5 5 ...... Figure 2.8 Though not explicitly associated with a nice compact rule, the above counting method does establish a function f: NQ + : f1 = 1 , 1 3 f2 = 2 , f3 = --- , f4 = 3 , f5 = --- ; and, given enough time 2 2 we could, by brute force, grind out the value of f336 and beyond. Moreover, since each element of Q+ is ultimately pierced by one and only one arrow, f is a bijection, and Q+ is seen to be countable. In a more general setting: THEOREM 2.8 The countable union of countable sets is again countable.

PROOF: Your turn:

CHECK YOUR UNDERSTANDING 2.13

Let An = an1an2 an3  for n = 123 . Show that If: A = A – A  n n  i in AA=  n is countable.   n = 1 Then: A = A  n  n n = 1 n = 1 Note: You can assume, without loss of generality that Ai Aj = Answer: See page A-12. if ij (see margin). 2.3 Infinite Conunting 81

All infinite sets thus far considered are countable. Are there uncountable sets? In other words, sets so large that there are not enough integers with which to count them? Yes:

THEOREM 2.9 The interval J ==01 x 0 x 1 is uncountable.

PROOF: Assume that J is countable (we will arrive at a contradic- tion). Being countable, we can list its elements in sequential form (a first, a second, a third, etc.):

Jr= 1r2 r3 r4  Below, we again list the elements of J, but now in decimal form (the

first digit following the decimal point of the number rn is an1 , the

second digit is an2 , the third is an3 , and so on).

r1 = 0.a11a12a13a14

r2 = 0.a21a22a23a24

r3 = 0.a31a32a33a34 

rn = 0.an1an2an3an4

We now set to work on constructing a perfectly good number rJ which is not any of the numbers listed above — contradicting the assertion that the above list contains each and every element of J. To be a bit more specific, we construct a number rJ which differs th from the number ri of the above list in its i decimal place. To be entirely specific:  0ifa  0  ii r ==0.a1a2a3a4 where ai   1if aii=0

Note that r cannot be r1 , for if the first digit of r1 , namely a11 is any- thing but 0, then it cannot be r, as r’s first digit is 0; and if the first digit of r1 is zero, then it again cannot be r, as r’s first digit would now be 1. By the same token, r cannot be r2 , since the second digit of r will surely differ from a22 , the second digit of r2 . The same argu- ment can be used to establish the fact that r cannot be any ri , since th th r’s i digit differs from the i digit of ri . CHECK YOUR UNDERSTANDING 2.14 For ab and cd , let Lab=  and Mcd=  . Show that: CardL = CardM Answer: See page A-12. Suggestion: Consider line passing through L and M. 82 Chapter 2 A Touch of Set Theory

It is reasonable to say that a set A has cardinality less than or equal to that of a set B if the elements of A can be paired off with those of a subset of B: We also say that the car- dinality of B is greater The cardinality of a set A is less than or than or equal to that of DEFINITION 2.17 A, and write: equal to the cardinality of a set B, written CardB  CardA CardA  CardB , if there exists a one- to-one function f: AB . (The function need no longer be onto.) All of the following results might be anticipated, but only the first two are easily established. Alas, a proof of (c) lies outside the scope of this text. THEOREM 2.10 Let A, B, and C denote sets. (a) If AB , then CardA  CardB . (b) If CardA  CardB and CardB  CardC then CardA  CardC .

Georg Cantor (1845-1916) CANTOR-BERNSTEIN- (c) If CardA  CardB and CardB  CardA Felix Bernstein (1878-1956) SCHROEDER THEOREM Ernst Schroeder (1841-1902) then CardA = CardB

PROOF: (a) The IA: AB given by IAa = a is clearly one-to-one. (b) From the given conditions, we know that there exist one-to-one functions f: AB and g: BC . All we need do, to complete the proof, is to note that the composite function gf: AC is also one-to-one (Theorem 2.5(a), page 72).

CHECK YOUR UNDERSTANDING 2.15 Show that all finite intervals are of the same cardinality. Answer: See page A-13. Suggestions: Use the result of CYU 2.15 and Theorem 2.10. The above CYU assures us that all finite intervals are of the same car- dinality. What about the set of all real numbers  ? It is certainly bigger than every finite interval. Does its cardinality exceed that of the finite intervals? No:

THEOREM 2.11 The cardinality of  is the same as that of a finite interval.

PROOF: An analytical proof: The tangent function maps the open   interval –--- --- in a one-to-one fashion onto  (see margin). 2 2   –------A geometrical argument: Figure 2.9(a) displays a bijection from the 2 2 open interval I = –11 to the semicircle S of radius 1 [excluding the end points of its diameter: –01 and 10 ]. It follows that yx= tan CardI = CardS . 2.3 Infinite Conunting 83

1. 1. .. ..S –.1...... 1 ...... S (a) (b) Figure 2.9 Figure 2.9(b) depicts a bijection from S (moved up one unit) to  (the x-axis). It follows that CardS = Card . Applying Theo- rem 2.7 we conclude that CardI = Card .

We point out that  is said to have cardinality c (for the continuum). The assertion that there does not exists a set X such that CardZ  CardX  Card is called the .

CHECK YOUR UNDERSTANDING 2.16 (a) Show that the set –3  56   9  has cardinality c. (b) Prove that there cannot exist a bijection from the set of real num- Answer: See page A-13. bers to the set of rational numbers. At this point we only have two types of infinity: the countable type associated with Z+ , and the uncountable type associated with  . Is that all there is? Where should we look if we hope to find bigger infini- ties? Towards bigger sets, of course, and one that might come to mind is the familiar Cartesian plane (2 = xy xy   . No luck, for Card2 = Card [Exercise 25]. And if you are tempted to try the three dimensional space: 3 = xyz xyz   don’t bother, for it too has cardinality c. Indeed, for any n, the set n  = x1x2  xn xi   1 in also has cardinality c [Exer- cise 26.] But there is another avenue we might consider, one that involves the following concept: DEFINITION 2.18 The of a set A, denoted by PA, is the set of all subsets of A. For example, if A=  abc , then: PA=  a b c ab ac bc abc Observe that A has 3 element, and that PA has 23 = 8 elements. This is no fluke: Note that A has 3 element, and that PA has 23 = 8 elements. This is Due to this theorem, the no fluke: power set PA of a given set A is often denoted by THEOREM 2.12 If the set A has n elements, then its power set the symbol 2A . PA has 2n elements. 84 Chapter 2 A Touch of Set Theory

PROOF: (By induction). Since we are only interested in counting, we will assume, without loss of generality, that our set A of n elements is the specific set A = 123 n . I. The claim does hold at n = 1 , for the set P1 =   1 , contains 22= 1 elements.

k II. Assume that PAk contains 2 elements for Ak = 123k .

III. We complete the proof by showing that the set PAk + 1 contains k + 1 2 elements, where Ak + 1 = 123kk + 1 .

Clearly, every subset of Ak + 1 either contains the number k + 1 or it does not. Let X be the set of those subsets that do not contain k + 1 , and Y the set of those subsets that do contain k + 1 :

XSPA=  k + 1 k + 1  S and YSPA=  k + 1 k + 1  S

Noting that X is the set PAk , we conclude, by the induction hypothesis, that it contains 2k elements. But every element in Y can be obtained by simply adjoining k + 1 to some element in X: YSk=  + 1 SX As such, Y has the same number of elements as X, namely 2k .

Noting that PAk + 1 = XY and that XY= , we con- clude that the number of elements in PAk + 1 is 2k +22k = k + 1 , and III is established. The following wonderful theorem tells us that if you take the power set of any set A, even an infinite set, you will end up with a set whose cardinality is strictly greater than that of A. Wonderful, in that it tells us that there is no end to the different levels of infinity. THEOREM 2.13 For any set A: CANTOR CardA  CardPA

PROOF: Clearly CardA  CardPA (see margin). We complete The function f: APA  the proof by showing that the assumption CardA = CardPA given by fa = a is leads to a contradiction: certainly one-to-one. Suppose there exists a bijection g: APA  (we will arrive at a contradiction). For any aA , ga is a subset of A. As such, either aga  or aga  . Let us now focus on the following subset of A:

S0 = aAaga  

Since g is onto, there must exist some element of A, say a0 , such that: ga0 = S0 2.3 Infinite Conunting 85

Now, either a0  ga0 = S0 or a0  ga0 = S0 . Let’s see which:

If a0  S0 , then since S0 = aAaga   : a0  S0 .

If a0  S0 , then a0  ga0 ==S0 aAaga   . Since neither option can occur, CardA = CardPA must be false. The assertion that for any infinite set X there does not exist a set Y for which CardX  CardY  CardPX is called the generalized continuum hypothesis.

CHECK YOUR UNDERSTANDING 2.17 Prove that there cannot exist a set of all sets. Suggestion: Assume such a set S exists and consider the set TA=  along with Answer: See page A-13. AS Theorem 2.10(a) and Theorem 2.13 to arrive at a contradiction. 86 Chapter 2 A Touch of Set Theory

EXERCISES

Exercises 1-12. Establish a specific bijection from the set A to the set B.

1.AZ= + , B = 5nn Z+ 2.A = 5nn Z+ , BZ= +

3.AZ= + , B = 5nn Z 4.A = 5nn Z , BZ= +

5.AZ= + , Bn= + 2 nZ 6.Bn= + 2 nZ , AZ= +

7.A = 5nn Z+ , B = 10nn Z+ 8.AZ= + , B = 2n nZ

9. For given ab  Z+ : AannZ=  + , BbnnZ=  + .

10. For given ab  Z+ : AannZ=  , BbnnZ=  + .

11. For given st   , with s  0 and t  0 : AnsnZ=  + , BntnZ=  + .

12. For given st   , with s  0 and t  0 : Asrr=   , BtrrZ=  + .

Exercises 13-20. For abcd   , with ab and cd , exhibit a specific bijection, from: 13.ab to cd 14.ab to cd 15.ab  to cd  16.a b  to c d 

17.a  to b  18.ab to a  19.a  to b  20.a  to – b

Exercises 21-24. Prove that the given set is countable (see Definition 2.7, page 64).

21. Z+  Z+ 22. Z+  Z+  Z+ 23.QQ 24.QQ  Q

25. Prove that: Card2 = Card .

26. Use the Principle of Mathematical Induction to show that Cardn = Card for any positive integer n.

# 27. Prove that the set of intervals ab a and b are rational is countable.

28. Prove that there are only countably many with rational coefficients. 29. Prove that there are only countably many solutions to the set of all polynomials with rational coefficients. 30. Prove that the set of irrational numbers is not countable. 31. Prove that there are uncountably many lines in the plane. 32. Prove that there are only countably many lines of the form ymxb= + , where mb  Q . 2.3 Infinite Counting 87

33. Prove that there are uncountably many in the plane. 34. Let Ff= : 123  01 (functions that assigns to each integer in the set 123 the value of 0 or the value of 1. Prove that F contains 23 = 8 elements, and that therefore CardF = CardP123 35. Let Ff= : 1 2  n  01 (functions that assigns to each integer in the set 12 n the value of 0 or the value of 1). Use the Principle of Mathematical Induction to show that F contains 2n elements.

36. Let Ff= : Z+  01 (functions that assign to each positive integer either 0 or 1). Prove that CardF = Card . 37. Prove that for any given set X, Cardf: X  01 = CardPX.

PROVE OR GIVE A COUNTEREXAMPLE

38. If CardA  CardB and CardB  CardC , then CardA  CardC . 39. If CardA  CardB and CardB = CardC , then CardA  CardC . 40. If CardA = CardB and CardC = CardD , then CardAC = CardBD . 41. If CardA  CardB and CardC  CardD , then CardAC  CardBD . 42. If CardA  CardB and CardC = CardD , then CardAC  CardBD . 43. If CardA = CardB and CardC = CardD , then CardAC = CardBD . 44. If CardA  CardB and CardC  CardD , then CardAC  CardBD . 45. If CardA  CardB and CardC = CardD , then CardAC  CardBD . 46. The set of intervals ab a and b are rational is countable.

47. The set of intervals ab a is rational is countable. 48. There can be at most countably many mutually disjoint circles (with positive radius) in the plane. 49. There can be at most countably many mutually disjoint lines in the plane. 88 Chapter 2 A Touch of Set Theory

2

§4. EQUIVALENCE RELATIONS

Recall that XY , called In Section 2 we defined a function from a set X to a set Y to be a sub- the set fXY  such that: of X with Y, is the set of all ordered pairs xy , For every xX there exists a unique yY with xy  f . with xX and yY . Removing all restrictions, we arrive at a far more general concept than that of a function: DEFINITION 2.19 A relation E from a set X to a set Y is any RELATION subset EXY  . A relation from a set X to X is said to be a relation on X. Each and every subset of  , including the chaotic one in the margin, is a relation on  , suggesting that Definition 2.19 is a tad too general. Some restrictions are in order:

DEFINITION 2.20 A relation E on a set X is a subset EXX  and REFLEXIVE is said to be: Reflexive: if xx  E for every xX . SYMMETRIC (Every element of X is related to itself) Symmetric: if xy  E then yx  E . TRANSITIVE (If x is related to y, then y is related to x) Transitive: if xy  E and yz  E then xz  E . EQUIVALENCE (If x is related to y, and y is related to z, then x is related to z) RELATION An on a set X is a relation that is reflexive, symmetric and transitive. The notation x~y is often used to indicate that x is related to y with respect to some understood relation E. Utilizing that option, we can rephrase Definition 2.20 as follows: An equivalence relation ~ on a set X is a relation which is Reflexive: if x~x for every xX . Symmetric: if x~y , then y~x . and Transitive: if x~y and y~z , then x~z . 2.4 Equivalence Relations 89

Consider the set of points X = 1234 featured in the diagrams of Figures 2.10(a) and (b). For xy  X , let us say that x is related to y, x~y , if there is an arrow pointing from x to y. That being the case, the relation depicted in Figure 2.10(a) is an equivalence relation on X, as it is: Reflexive x~x . For every xX , there is an arrow (pointed loop) pointing from x to x. Symmetric (If x~y , then y~x ). If there is an arrow from x to y, then there is also an arrow from y to x. Transitive (If x~y and y~z , then x~z ). If there is an arrow from x to y and one from y to z, then there is also an arrow from x to z.

These illustrations are 1. 1 examples of directed . graphs with vertices 1 .4 .4 through 4. The “arrows” .2 2 between vertices are . called directed edges. 3. 3.

(a) (b) Figure 2.10 The relation depicted in Figure 2.10(b) is not reflexive: 2~2 , not symmetric: 3~1 but 1~3 , and not transitive: 1~2 2~3 but 1~3 .

CHECK YOUR UNDERSTANDING 2.18 (a) Is the given relation E on the set 123 an equivalence rela- tion? If not, indicate which of the three properties of Definition 2.20 fail to hold. (i) E = 11 22 33 12 23 13 (ii) E = 11 22 33 12 21 (iii) E = 11 22 33 12 21 13 31 (b) Let P denote the current world population. Determine if the given relation E is an equivalence relation on P. If not, indicate which of the three properties of Definition 2.20 fail to hold. (i) Exy=  xy  P Yes: (a-ii), (b-i), (b-ii). (ii) Exy=  xy  P such that x and y are of the same age No: (a-i). (a-iii), (b-iii). (iii) Exy=  xy  P such that x is strictly older than y 90 Chapter 2 A Touch of Set Theory

a c EXAMPLE 2.11 Show that the relation ---~--- if ad= bc is an b d equivalence relation on the set of rational numbers. SOLUTION: a a Reflexive: ---~--- since ab= ba . b b a c c a Symmetric: ---~---  ad ==bc cb da  ---~--- . b d d b a c c e Transitive: ---~--- and ---~--  ad ==bc and cf de b d d f (*) (**) a e We establish the fact that ---~-- by showing that af= be : b f bc bc de af ==------ f ------ ------=be d d c see (*) see (**)

In general, equivalence relations enable one to establish a somewhat “fuzzy” sense of equality — a “fuzzyness” which is all but ignored in the above example; for, as you know, when it comes to the set of rational numbers, one 3 15 3 15 simply writes --- = ------, even though the scribbles --- and ------7 35 7 35 certainly look different.

Recall that ab means EXAMPLE 2.12 (a) Let n be a fixed positive integer. Show that that a divides b (see Defi- the relation a~b if na– b is an equiva- nition 1.12, page 27). lence relation on Z. (b) Show that the relation a~b if 23a – 7b is an equivalence relation on Z. SOLUTION: (a) This turns out to be a particu- Reflexive: a~a since na– a . larly important equivalence relation. It is written: Symmetric: a~bnab –  nb– a b~a . ab mod n a~b and b~c  na– b and nb– c and is read: Transitive: a is congruent to b modulo n. Theorem 1.5(b), page 28:  nab– + bc–  na– c ac 2.4 Equivalence Relations 91

(b) The relation a~b if 23a – 7b is: Reflexive. a~a , since: 3a –47a ==– a 22– a here, a is playing the role of b

Symmetric. Assume that a~b , which is to say, that: 3a –27b = k for some kZ (*)

An expression of the We are to show that b~a , which is to say, that: 2k + 7b form a = ------is 3 3b –27a = h for some hZ unacceptable in the To do so, we start with 3b – 7a , and, by necessity, work (*) into solution process, since we are involved with the the picture: set Z of integers. 3b –37a = a – 7b + something To be more precise: 3b –37a = a –107b + –10a + b

Bringing us to: 3b –37a = a –107b + –10a + b The rest is straight forward: 3b –37a = a –107b + –10a + b = 2k + 25–5a + b ===2k –55a + b 2h, with hk–55a + b

CHECK YOUR UNDERSTANDING 2.19 Show that the relation A~B if CardA = CardB is an equiva- Answer: See page A-13. lence relation on PX for any set X. Suggestion: Consider Theorem 2.5, page 72.

DEFINITION 2.21 Let ~ be an equivalence relation on X. For each x  X , the equivalence of x , EQUIVALENCE 0 0 CLASS denoted by x0 , is the set: x0 = xXx ~x0

In words: The equivalence class of x0 consists of all elements of X that are related to x0 . We now show that any element in x0 will generate the same equivalence class: THEOREM 2.14 Let ~ be an equivalence relation on X. For any x1 x2  X : x1  x2  x1 = x2 92 Chapter 2 A Touch of Set Theory

PROOF: Assume that x1  x2 . We show that x1  x2 (a similar argument can be used to show that x2  x1 and that therefore x1 = x2 ): xx 1  xx 1

By transitivity, since x1~x2: xx 2  xx 2 Conversely: x1  x1 = x2  x1  x2  x1  x2

EXAMPLE 2.13 Example 2.12(a) assures us that the relation a~b if 3 ab– is an equivalence relation

on Z. Determine the set n nZ of its equivalence classes.

SOLUTION: Let’s start off with a = 0 . By definition: 0 ===b 30– b 3nn Z  –6,–3036 Since 1 is not a multiple of 3, 1 will differ from 0 (Theo- rem 2.14). Specifically: 1 ==b 31– b b 3 b – 1 = bb= 3n + 1, for nZ ==3n + 1 nZ  –5,–2147 Turning to 2 : 2 ==b 32– b 3n + 2 nZ = –4,–1258

In the above example we observed that the give equivalence relation chops up Z into disjoint equivalence classes; namely: Z = 0 1 2 with:

0 1 ==0  2  and  1  2 = This, as you will soon see, is no fluke, but first:

DEFINITION 2.22 A set of nonempty subsets S   A of a ARTITION See the indexing remarks P set X is said to be a partition of X if: that precede Theorem (i) XS= 2.3, page 58.     A

(ii) If S  S   then S = S

Figure 2.11(a) displays a 5-subset partition S1S2 S3 S4 S5 of the indicated set. A countably infinite partition of 0  is represented in  Figure 2.11(b): nn + 1 n = 0 2.4 Equivalence Relations 93

S1

S5  S2 [ )[ )[ )[ )[ )[ )[ S 0 1 2 3 4 5 6 4 S3 (a) (b) Figure 2.11

CHECK YOUR UNDERSTANDING 2.20 Determine if the given collection of subsets of  is a partition of  ?

(a) nn + 1 nZ

(a): No (b): Yes   (b) n nZ ii + 1 i = 0  –1i –  –i i = 0

There is an important connection between equivalence relations on a set X and partitions of X, and here it is: THEOREM 2.15 (a) If ~ is an equivalence relation on X, then the set of its equivalence classes, x xX , is a partition of X.

(b) If S   A is a partition of X, then the rela- tion x1~x2 if x1 x2  S for some   A is an equivalence relation on X.

PROOF: (a) We Show that: (i) Xx=   xX

and (ii) If x1  x2  , then x1 = x2 . (i): Since  is an equivalence relation, x~x for every xX . It follows that xx  for every xX , and that therefore Xx=  . xX

(ii): If x1  x2  , then there exists x0  x1  x2 .

Since x0  x1 and x0  x2 : x0~x1 and x0~x2 .

By symmetry and transitivity: x1~x2

By Theorem 2.14: x1 = x2

(b) Let S   A be a partition of X. We show that the relation:

x1~x2 if there exists   A such that x1 x2  S is an equiv- alence relation on X: 94 Chapter 2 A Touch of Set Theory

Reflexive: To say that x~x , is to say that x belongs to the same S as itself, and it certainly does.

Symmetric: x~y    Ax  yS   yx  S  y~x Transitive: Assume x~y and y~z . We show that x~z :

Since x~y , xy  S for some   A . Since z~y , yz  S for some   A .  Since S S  (y is contained in both sets): S = S .     It follows that both x and z are in S (or in S if you prefer),   and that, consequently: x~z .

CHECK YOUR UNDERSTANDING 2.21 We showed, in Example 2.12(b), that the relation a~b if 23a – 7b is an equivalence relation on Z. Partition Z by the corresponding Answer: See page A-14. equivalence classes. 2.4 Equivalence Relations 95

EXERCISES

Exercises 1-6. Determine if the given relation on the set abcd is an equivalence relation. If not, specify the properties of an equivalence relation that are not satisfied. If so, list the equivalence classes. 1. aa bb cc 2. aa bb cc dd 3. aa bb cc dd ac ca 4. aa bb cc dd ac ca cb 5. aa bb cc dd ac ca cb ab 6. aa bb cc dd ac ca cb ab ba

Exercises 7-12. Give an example of a relation on abc that is: 7. Reflexive, but neither symmetric nor transitive. 8.. Symmetric, but neither reflexive nor transitive. 9. Transitive, but neither reflexive nor symmetric. 10. Reflexive and symmetric, but not transitive. 11. Symmetric and transitive, but not reflexive. 12. Transitive and reflexive, but not symmetric.

Exercises 13-15. Determine if the given directed graph represents an equivalence relation (con- sider Figure 2.10). If not, specify the properties of an equivalence relation that are not satisfied.

13. a. . b 14.a. . b 15. a. . b

c. .d c. .d c. .d

Exercises 16-21. Determine if the given relation on the set of all people is an equivalence relation. If not, specify the properties of an equivalence relation that are not satisfied. 16.a~b if a and b are of the same sex. 17.a~b if a is at least as old as b. 18.a~b if a and b have the same biological parents. 19.a~b if a and b have a common biological parent. 20.a~b if a and b are of the same blood-type. 21.a~b if a and b were born within three days of each other. 96 Chapter 2 A Touch of Set Theory

Exercises 22-27. Show that the given relation is an equivalence relation on Z. 22.a~b if ab= 23.a~b if a + 1 = b + 1 24.a~b if 5 a2 – b2 25.a~b if 2 a – 3b 26.a~b if 32ab+ 27. a~b if 4 a2 – b2 Exercises 28-30. Show that the given relation is not an equivalence relation on Z. 28. a~b if 2 ab 29. a~b if a2  b2 . 30. a~b if 6 a + 5b Exercises 31-36. Determine whether or not the given relation is and equivalence relation on Z. 31.a~b if 24a – 6b 32.a~b if 4 a – 5b 33.a~b if ca2 – b2 for a fixed c  0 34.a~b if 3 a – 5b 35.a~b if 2 a 36.a~b if anb= for some nZ Exercises 37-40. Show that the given relation is an equivalence relation on the set of rational num- bers reduced to lowest terms. a c a c 37.---~--- if ad–0 bc = 38.---~--- if 2 bd+ b d b d a c a c 39.---~--- if ad– bc b2 + d2 = 0 40.---~--- if ac= b d b d Exercises 41-42. Show that the given relation is not an equivalence relation on the set of rational numbers reduced to lowest terms.. a c a c 41.---~--- if ad+0 bc = 42.---~--- if ad= b d b d Exercises 43-46. Determine whether or not the given relation is and equivalence relation on the set of rational numbers reduced to lowest terms. a c a c 43.---~--- if bd= 44.---~--- if ac= bd b d b d a c a c 45.---~--- if ad 2 = bc 2 46.---~--- if a = 2d and c = 2b b d b d Exercises 47-52. Show that the given relation is an equivalence relation on  . 47.x~y if x2 = y2 48.x~y if xy= 49.x~y if x + 1 = y + 1 50.x~y if xy–  Z 51.x~y if x2 –0y2 = 52.x~y if sinx = siny + 2 Exercises 53-55. Show that the given relation is not an equivalence relation on  . 53. x~y if xy Z 54.x~y if x + 2yx 55.x~y if x – 3yxy + Exercises 56-61. Determine whether or not the given relation is and equivalence relation on  . 56.x~y if xy  0 57.x~y if xy 2  0 58.x~y if sinx = siny +  xy xy+ 1 x~y xny= nZ 59.x~y if ----- = x 60.x~y if ------------= x 61. if for some  y y + 1 Exercises 62-65. Show that the given relation is an equivalence relation on 2 .

62.x0 y0 ~x1 y1 if x0 + y0 = x1 + y1 . 63.x0 y0 ~x1 y1 if x0y0 = x1y1 . 2 2 2 2 64.x0 y0 ~x1 y1 if x0 + y0 = x1 + y1 . 65.x0 y0 ~x1 y1 if x0 = x1 . 2.4 Equivalence Relations 97

Exercises 66-69. Show that the given relation is not an equivalence relation on 2 .

66.x0 y0 ~x1 y1 if x0 = y1 . 67.x0 y0 ~x1 y1 if x0 – y1 = y0 – x1 .

68.x0 y0 ~x1 y1 if x0x1 = y0y1 . 69.x0 y0 ~x1 y1 if x0x1  0 and y0y1  0 .

Exercises 70-75. Determine whether or not the given relation is and equivalence relation on 2 .

70.x0 y0 ~x1 y1 if x0 + x1 = y0 + y1 . 71.x0 y0 ~x1 y1 if x1 – x0 = y1 – y0 .

72.x0 y0 ~x1 y1 if 2x0 = 3y0 73.x0 y0 ~x1 y1 if 3x1 – 3x0 = y1 – y0 2 2 2 2 74.x0 y0 ~x1 y1 if x0 – 1 +1y0 + 3 = and x1 – 1 +1y1 + 3 = . 2 2 2 2 75.x0 y0 ~x1 y1 if 2x0 – 4x0 +2y0 – 2y0 = x1 – 4x1 + y1 – 2y1 .

Exercises 76-81. Determine whether or not the given relation is an equivalence relation on 3 .

76.x0y0 z0 ~x1y1 z1 if y0 = y1 .

77.x0y0 z0 ~x1y1 z1 if x0 ++y0 z0 = x1 ++y1 z1 .

78.x0y0 z0 ~x1y1 z1 if x0 = y1 + z1 .

79. x0y0 z0 ~x1y1 z1 if x0z0 + 2y0 = x1z1 + 2y1

80.x0y0 z0 ~x1y1 z1 if x0 + 2y0 – 3z0 = x1 + 2y1 – 3z1 2 2 81. x0y0 z0 ~x1y1 z1 if x0 ++y0 z0  x1 ++y1 z1 Exercises 82-88. Show that the given relation is an equivalence relation on FZ= f: ZZ (the set of functions from Z to Z). 82.f ~ g if f1 = g1 . 83.f ~ g if fn= gn for every nZ . 84.f ~ g if fn= gn for every nZ . 85.f ~ g if 2 fn+ gn for every nZ . 86.f ~ g if fn+ m = gn+ m for every nm  Z . 87.f ~ g if 32fn+ gn for every nZ . 88.f ~ g if 32gf n + fn for every nZ . Exercises 89-92. Show that the given relation is not an equivalence relation on FZ= f: ZZ . 89.f ~ g if f2n = 2gn for every nZ . 90.f ~ g if fn= gn for some nZ . 91. 3 fn+ gn for some nZ . 92.f ~ g if fn gn for some nZ . Exercises 93-99. Determine whether or not the given relation is an equivalence relation on FZ= f: ZZ . 93.f ~ g if fngn 0 for every nZ . 94.f ~ g if 32fn+ gm for every nm  Z . 95.f ~ g if 32fn+ gn for some nZ . 96.f ~ g if 32gf n + fn for some nZ . 97.f ~ g if 32gf n + fg m for every nm  Z . 98 Chapter 2 A Touch of Set Theory

98.f ~ g if 32gf n + fg n for every nZ . 99.f ~ g if 32gf n + fg m for some nm  Z . Exercises 100-107. Describe the set of equivalence classes for the equivalence relation of: 100. Exercise 22 101.Exercise 23 102.Exercise 35 103.Exercise 40 104. Exercise 47 105.Exercise 48 106.Exercise 63 107.Exercise 65 Exercises 108-112. Show that the given collection S of subsets of the set X is a partition of X. 108.X =  , S = –0 0 0  . 109.XZ= , S = 3nn Z 3n + 1 nZ 3n + 2 nZ . 110.XZ= +  Z+ , Sab=  gcd a b = n . nZ +

111.X =  , Sxy=  yxb= + bR .

2 2 2 112.X =  , Sxy=  x + y = r r   . Exercises 113-116. Show that the given relation is an equivalence relation. Describe the set of its equivalence classes. 113. For ab  12 100 , a~b if a and b end in the same digit. 114. For ab  12 101 , a~b if a and b end in the same digit. 115. For ST  P123 , S~T if the number of elements in S equals the number of ele- ments in T. (See Definition 2.18, page 83.) 116. For ST  P123 , S~T if the sum of the elements in S equals the sum of the ele- ments in T. (See Definition 2.18, page 83.)

PROVE OR GIVE A COUNTEREXAMPLE

117. The union of any two equivalence relations on any given nonempty set X is again an equiv- alence relation on X. 118. The intersection of any two equivalence relations on any given nonempty set X is again an equivalence relation on X. 119. The union of any two reflexive relations on any given nonempty set X is again a reflexive relation on X. 120. The union of any two symmetric relations on any given set X is again a symmetric relation on X. 121. The union of any two transitive relations on any given set X is again a transitive relation on X. + 122. For abnm   Z , let Sn and Sm denote the set of equivalence classes associated with the equivalence relations a~b if na– b and a~b if ma– b , respectively. If nm , then

Sn  Sm . 123. If CX , A~B if AC = BC is an equivalence relation on PX . 124. There exists an equivalence relation on the set 12345 for which each equivalence class contains an even number of elements. 2.5 What is a Set? 99

2

§5. WHAT IS A SET? (OPTIONAL) The word “set” has been brandished about in previous sections. Here is an attempt to define that important concept: DEFINITION A(?) A set is a specified collection of objects. Attempt #1 Really? Who is to specify, and how? And anyway, what is a collec- tion? Okay, the word “set” is not precise. Yet, there is an important idea lurking within the above would-be definition, a mathematically indis- pensable idea. Can we zero in on it? We can try. Consider the variable proposition: x is a person born of my mother It appears that we can use the above to generate a well-defined collec- tion of objects; namely, those objects which when substituted for x ren- der the proposition True. To put it another way: those objects which satisfy the given proposition. This suggests another possible approach towards the definition of a set:

DEFINITION B(?) A set is a collection of objects satisfying a Attempt #2 variable proposition. Is this definition okay? Well, let’s begin by comparing it with our previous attempt: A set is a specified collection of objects. First of all, note that the word collection is used in a different sense in the two would-be definitions. In our first attempt, we tried to get by with a circular definition: a set is a collection, or, if you prefer, a collec- tion is a set. Now, however, the collection appears to be specifically determined by means of a variable proposition. Another objection we had with “Definition A” is that it appeared a bit personal. Who is to specify the collection of objects? We no longer appear to have this problem. For example, the statement: I can vote in a national election. is True or False, depending on my age, nationality, and other specified factors. But all of these factors are simply used to determine whether or not I am an object satisfying the variable proportion: x can vote in a national election. There is, of course, another problem; namely, how does one phrase a precise variable proposition to begin with? Let us reconsider “x can vote in a national election”. What does it mean to vote? What is a nation? The problem is that we are attempting to build something rigorous, a totally unambiguous statement, and we are trying to build it with non-rigorous material: fuzzy words from a common language. Still, the hope is that with extreme care one may be able to formulate a precise variable prop- 1872- osition, and by-pass that language barrier. But there is another fault with 1970). British Mathe- “Definition B” — a fatal fault, called the Russell : matician and Logician. 100 Chapter 2 A Touch of Set Theory

Assume that our second attempt does properly define the concept of a set. We could then consider the set S of objects satisfying the following variable proposition: X is a set which does not contain itself as an element: (*): SXX=  is a set  XX Now, one thing is clear, our set S either contains itself as an element SS or it does not SS . Let’s see which: If SS then, by (*): SS . Okay, our assumption that SS quickly leads us to the con- tradiction that SS , and therefore cannot hold. The other This argument is reminis- cent of that found in Can- option must hold. Let’s check, just to make sure: tor’s Theorem, page 84. If SS then, by (*): SS . What, again! This too cannot be. Clearly, our would-be “Definition B” will not do, for it leads us to a set which neither contains itself as an element nor does it not contain itself as an element. It’s back to the drawing board; but first, a paren- thetical remark, for what its worth:

We have spent considerable time and effort, and all for a couple of bad attempts in defining a concept. Not necessarily a waste of time. A consci- entious attack on a problem may turn the investigator in wrong directions, some of which may lead to places more interesting than the intended desti- nation. In the abstract universe of mathematics there is but one mode of transportation: thought. A lot of times it’s fun to let it take us where it may. Fine, but what is a set? Well, in the final analysis the concept of a set remains an undefined term. We can, however, tell you how certain sets have been created, and actually create a bunch of them right before your very eyes. We want to call something a set, but then again do not want to offend anything that does not think itself as such by calling it a set. So, what we do is to consider a collection of nothing, absolutely nothing, and call it the empty set, and denote it by the symbol  . It is unquestion- ably the most important set of all mathematics. We pass a formal law, lest there be any doubt as to the of  :

AXIOM 1  is a set. We now have a legitimate set, by law, and there are going to be a lot more. Anticipating their existence, let us come to some agreement as to when two sets are to be considered the same, or equal. 2.5 What is a Set? 101

Since sets are going to be collections of things, it seems reasonable to say that two sets are the same if they contain exactly the same things, in other words: DEFINITION 2.23 Two sets A and B are equal, written AB= This definition previously appeared on page 53. SET EQUALITY if: xA  xB and xB  xA or: xA  xB The above definition is certainly reasonable, and we would not want to change it for the world. It is, however, going to impose a restriction on all sets, wherever they may be. To make this point, let us assume momentarily the existence of a set containing an element, say the set a . Suppose, in addition, that we allow aa also to be a set. Defi- nition 2.23 would tell us that these two sets are equal, for every element in a is also an element in aa , and vice versa. This we do not want, right? And so we stipulate that no set can contain an element more than once. In particular aa is not an acceptable representation of a set. At this point, there is a conspicuous lack of sets. Let us remedy the situation by passing another law: 2 For any set A, A is also a set. That’s great! Prior to the above axiom we had but one set, and empty at that. We now have a lot of different sets. Witness a few:     The above sets are all different. For example,  and  are not the same since there is an element in the latter set, namely the element  , while there is absolutely nothing in the set  . Also     , since the only element in the set  is  , and the only element in the set  is  , and we already noted that   ; and so on down the line. We’ve come a long way from our initial set  , but there is room for improvement. After all, each of our current sets is either empty or is a set (contains but one element). Our set  , for exam- ple, contains only the element  . We are going to remedy this situation by passing other laws. But first, here is another parenthetical remark: What is our jurisdiction in passing these laws? In our abstract universe any one is entitled to create anything he or she wants. Some of those creations perish under an attack of reason, as was the case with our attempted cre- ation labeled “Definition B”. The laws that we are busy passing, or axioms, have survived in that they do not lead to any contradiction; or, at least, none has yet been unearthed. 102 Chapter 2 A Touch of Set Theory

We now introduce another law from which many more sets will evolve. First, however, let us agree to use the term collection of sets when referring to a set whose elements are themselves sets. AXIOM 3 For any collection C of sets, there exists a set consist- ing precisely of those elements that belong to at least one of the sets in C. The above set is called the union of the sets in C, and is denoted by the symbol:  A AC

When dealing with a collection of two sets, say CA= 1 A2 , we may choose to represent  A in the form A1  A2 . Similarly, the AC notation A1 A2 A3 can be used to denote the union of the sets A1A2 and A3 , and so on. Back to our set safari, and we begin by recalling our collection stem- ming from the first two axioms:     Axiom 3 now assures us that the following is also a set:    =   We can now employ Axiom 2 to arrive at yet another new set:   Back to Axiom 3:    =   And then Axiom 2 assures us that the set containing the above set; namely:   is also a set, and we could union it with our set   to obtain yet another set. There is a rather interesting development above which may be lost within the maze of commas and brackets. Let us try to clarify the situa- tion with the introduction of some notation. We start off by letting the symbol 0 denote our empty set: 0 =  (the above is not a set equality but simply designates a new symbol, 0 for the empty set) We then denote the set  by the symbol 1: 1 == 0 (once more, a matter of notation) Our set   will be denoted by 2: 2 ====  01 0  1 11  2.5 What is a Set? 103

The symbol 3 will be used to denote the set consisting of our sets 0,1, and 2: 3012== 01  2 =22  And in a similar manner we arrive at: 4== 0123 33  5== 01234 44  6== 012345 55  

135== 0 1 2  134 134  134

Let us give to the above sets. We will call the set 0, zero. The set 1 will be called one; 2 will be called two; 3 is three, and so on. Wait a minute, what’s going on? Are we talking about sets or are we talking about numbers? Both. What we have done is to define the num- ber 3, among others. We can now pick up this number 3 and cuddle it, if we want. It is no longer a vague concept, but is something concrete, at least in our abstract mathematical universe. Three is the set   or, equivalently, the set 012 . Either way, there it is, something. We hastily remark that a counts potatoes the same way everybody else does. He or she does not go around continuously aware of the fact that 3 is actually a set which is also an element of the set 9. Still, 3 is now something well-defined, and that is nice.

We can feel rather proud of ourselves. Just a few pages ago we had but one set, and empty at that. We now have infinitely many sets, and sets containing many elements. We now have, for example, sets con- taining 1,036,752 elements, among them is the set 1,036,752. But there is room for improvement. We do not, for instance, as yet have an infinite set. A particularly nice set we would like to construct is the one consisting of all of the sets we have already constructed: 0, 1, 2, 3, 4, and so on (which we will call the whole numbers). It is pretty clear, however, that our current axioms will not yield that collection since none of those axioms can boost us into the infinite. We need another law, and here is the most obvious one to pass: There exists a set consisting of all the whole numbers. Well, that’s right to the point alright; but, unfortunately, it will not do. We know exactly what the number zero is:  . We also know that: 10=   0, and that 943= 942  942 We can build like mad, for as long as we like, with as many gaps as we wish, but there will always be larger whole numbers yet to be con- structed. This concept of all the whole numbers is quite firm at one end, at the number 0, but kind of hangs loose at the other end. Some new ammunition is called for, and we start off with a definition: 104 Chapter 2 A Touch of Set Theory

In particular, if A = 01 , DEFINITION 2.24 For any set A, the set AA  is called the then the successor of A is the successor of A. set 01 01 . In particular, 1 is the successor of 0, 2 is the successor of 1, and 943 is the successor of 942 (recall that 943= 942  942 ). We are now in a position to pass our next law: AXIOM 4 There exists a set which contains 0 and contains the successor of each of its elements. Let us agree to call any set which satisfies the condition of the above axiom, a successor set. Clearly: ANY SUCCESSOR SET MUST CONTAIN EVERY WHOLE NUMBER Now we’ve gone ahead and done it, we created an infinite set, and maybe a lot of them. We also know that each of these successor sets contains all of the whole numbers. What we do not know, yet, is whether or not there exists a successor set which contains nothing other that the whole numbers. All successor sets might just be too big. Glancing back at our previous axioms, we see that none of them pro- vides us with the means of plucking smaller sets from a given set. And so we set to work on constructing an appropriate plucking mechanism, beginning with a definition: A set A is said to be a subset of a set B, writ- This definition previously DEFINITION 2.25 appeared on page 53. SUBSET ten AB , if each element in A is also con- AND tained in B. PROPER SUBSET If A is a subset of B, but is not equal to B, then A is said to be a proper subset of B; written: AB .

And so we have: 2638  196328 2638  196328 2638 = 2638 2638  2638 2638  2638

Let us denote this thing 0123 which we are trying to cre- ate by the letter N and ask ourselves the following question: If N were a set, how would it be distinguishable from all other sets? Answer: it should be a smallest successor set. Can we make the phrase smallest successor set precise? Yes: DEFINITION 2.26 A minimal (or smallest) successor set is a successor set which is a subset of every suc- cessor set. 2.5 What is a Set? 105

Now, if we can agree that N should satisfy the property that it is a minimal successor set, and if we can show that there is exactly one minimal successor set, then we will have our set N. Alright, let us first show that there is at most one minimal successor set, and then proceed to find it.

There is a distinction THEOREM 2.16 There exists at most one minimal successor set. between a Theorem and an Axiom. Axioms are PROOF: Let A and B be two minimal successor sets. dictated. They are the initial building blocks Since B is a successor set, and since A is a minimal successor set: from which mathemat- AB ics is constructed. Theo- rems, on the other hand, Since A is a successor set, and since B is a minimal successor set: are mathematical con- BA structions, built from axiomatic bricks and Employing Definition 2.23, we conclude that AB= . logical cement. Once established, they too We now know that there can be at most one minimal successor set. can be used as building blocks in the construc- Still to be demonstrated is that one exists. All in good time. First, tion of other . another law is passed: And so it goes, until there are so many blocks hanging around AXIOM 5 For any given set A and any variable proposition px , that one loses track of there exists a set (possibly empty) consisting of those which axiom is needed elements of A which satisfy px . for the proof of a given result. But it really does not matter, providing, of What is the difference between Axiom 5 and our defamed “Definition course, that we don’t B:” A set is a collection of objects satisfying a variable proposition permit a faulty block to creep into our collec- The main distinction is that while “Definition B” led to an absurdity, tion. If one does, then via Russell’s paradox, Axiom 5 does not (or, at least, none has as yet we will be in serious trouble, for any argu- been uncovered). The specific distinction between Axiom 5 and “Defi- ment built in part with a nition B” is the phrase “those elements of A” appearing in Axiom 5. We faulty block may itself are no longer permitting everything to be a candidate for satisfying a be faulty. variable proposition px , but are now restricting candidacy to those elements which are themselves elements of some existing set. Unlike the union concept which stemmed from Axiom 3, the inter- section concept rests on Axiom 5: DEFINITION 2.27 Let C be a non-empty collection of sets. The intersection of the sets in C, denoted by  A , is the set consisting of those ele- AC ments contained in every set in C:  A = xx A for every AC AC

Back to our search for a minimal successor set. Axiom 4 assures us of the existence of at least one successor set S : 0  S and xS  x  x S 106 Chapter 2 A Touch of Set Theory

Now, there are some subsets of S which are themselves successor sets (S itself, for example). The collection: CSS=  S is a successor set is therefore not empty, and we may consider the intersection of all the sets in C, which we optimistically call N: NS=  SC So, N is the intersection of all the subsets of the successor set S which are themselves successor sets. We now show that N is itself a successor set: Since each SC is a successor set, 0  S and NOTE: The adjacent argument shows that xx  S for every xS . Since NS=  , 0  N ; any intersection of suc- SC cessor sets is again a and for every xN = S , xx  N . successor set.  SC Alright, we have just established the fact that the intersection, N, of all subsets of the successor set S which are themselves successor sets is itself a successor set. We now define this particular set, N, to be the set of whole numbers. Objections! The above set N, being the intersection of all subsets of S and being itself a successor set, is clearly the smallest successor set in the particular set S . But how about successor sets other than S which may exist already, or others that may evolve if we continue creating additional sets? How do we know that no new successor set will evolve which is smaller than N? Objection overruled:

THEOREM 2.17 If A is a successor set, then NA .

PROOF: Consider once more the successor set S which led us to the definition of N above. Since, as we have noted, the intersection of successor sets is again a successor set, AS is also a successor set, and it is contained in S . But N sits inside every successor set con- tained in S . Thus: NASA   We now have the set N = 0123 and could proceed towards a rigorous set-theoretical construction of the real number sys- tem, and beyond. Instead, we chose to end the chapter with some gen- eral set theory remarks. First, however, a final result:

THEOREM 2.18 Mathgod = 

PROOF: Clear. 2.5 What is a Set? 107

SOME HISTORICAL REMARKS: The formal birth of set theory occurred in 1874 with the publication of the first purely set the- oretic work: Uber ene Eigenschaft des Inbegriffes aller Reelen Algebraishen Zahlen (On a Property of the Collection of all Real Algebraic Numbers). Within that paper, (1845-1916) distinguishes two infinite subsets of the reals that are not of the same cardinality, thereby pointing out the existence of different levels of infinity. He was not, however, the first to discover this staggering fact. Indeed, Galileo Galilei (1564-1642) as early as 1632 recorded the following interesting observation: There are as many squares as there are numbers because they are just as numerous as their roots. Later, Bernhard Bolzano (1781-1848), famil- iar with Galileo’s work, gave additional examples of bijections between infinite sets and some of their proper subsets. But these were isolated instances, and the creation of set theory as well as its fundamental development is justly accredited to Cantor. Cantor’s valuable contributions were far from universally accepted by the mathematical community of his time. Many disagreed vehemently with his work which, to them, appeared to rest on little more than intuition and empty fabrications based on nonconstructive reasoning. Some suggested that his work encroached on the domain of philosophers, while others even accused him of violating religious . Most noteworthy among his numerous critics of the day was one of Cantor’s former professors, Leopold Kronecher (1823-1891). Kronecher objected loudly and strenuously to Cantor’s uninhibited use of infinite sets. According to that noteworthy mathematician: Definitions must contain the means of reaching a decision in a finite number of steps, and existence proofs must be conducted so that the quantity in question can be calculated with any required degree of accuracy. Even the irrational numbers fail to satisfy his imposed criteria, and, along with the infinite, they too were disregarded by Kro- necher whose mathematical may best be read within his often quoted statement: God created the natural numbers, and all the rest is the work of man. And then came the , which naturally served to further fan the flames of discontent. Cantor’s reliance on precise statements to generate his sets simply would not do, and Bertrand Russell (1872-1970), in 1902 demonstrated that Cantor’s own definition of sets leads to a con- tradiction. Other similar paradoxes emerged, and they generated such turmoils that in 1908 Henri Poincare (1845-1912), a leading mathematician at the time, made the following state- ment at the International Congress of Mathematics: Later mathematicians will regard set the- ory as a disease from which one has recovered. The sensitive Cantor did not recover from the onslaught of criticisms from his peers. Particu- larly disturbed by what he felt to be Kronecher’s malicious and unjust persecutions, he suf- fered a complete nervous breakdown in 1884, and, to some degree, mental illness plagued him for the rest of his life. He died in the psychiatric clinic at Halle on January 6, 1918, but not before witnessing the beginning of the tremendous role his theory would play in mathematics, and realizing a belated recognition which he so justly deserved. 108 Chapter 2 A Touch of Set Theory

In 1908, (1871-1953) published his Untersuchungen ueber die Grundlagen der Mengenlehre (Investigations into the Foundations of Set Theory). In that work, an axiom- atic system for set theory is presented. After postulating the existence of certain sets, and with an additional undefined notion, that of membership ( ), he required but seven axioms to set the formal foundation for set theory and, indeed, for most of mathematics. These original axi- oms were later amended by (1891-1790), (1903-1957), and Kurt Gӧdel (1906-1978), among others, and has come to be called the Zermelo-Fraenkel . It remains the most widely used axiomatic system of the day. Yes, there is no universal axiomatic system. Though it is certainly true that the overwhelming majority of mathematics is globally accepted, there remain important results which depend on axioms accepted by some and rejected by others. In an attempt to clarify this statement, we turn to a few remarks concerning the axiomatic foundations of mathematics. Two axiomatic systems are equivalent if each axiom in either system is a consequence of those in the other. It follows that though the axioms in one system might be quite distinct from those in the other, both in form and number, any proposition which can be established to be True in one system can also be established to be True in the other. Thus, equivalent systems lead to the same theory. One system, however, might possess certain properties which makes it more appealing than another. One appealing property is that the system be efficient in the sense that each of its axioms is really needed — that none of its axioms is a consequence of the others. Such a system is said to be independent. Another nice property, admittedly more vague than the previous one, is that the system should, as much as possible, consist of “intuitively valid” axioms. After all, axioms are the building blocks from which the theory is developed, and, as such, they should be “believable” and “fundamental in nature.” The five axioms of this chapter are independent, and fundamental in nature. We began by stipulating the existence of a set, the empty set. Then, with the introduction of other axioms, arrived at an axiomatic system sufficiently rich to allow for the construction of the set of natu- ral numbers. We also established a few of the surface results founded on our meager collection of axioms. A richer theory might, however, evolve from an axiomatic system which properly contains ours. But how do we go about expanding our system? We expanded all along. Very little theory could be based on our first axiom:  is a set. And so we expanded by introducing the axiom on sets of sets, thus assuring the existence of a lot of singleton sets. There was then a need for a greater variety of sets, a need which was partially fulfilled by the introduction of the union axiom. We continued the process of satisfying needs, arriving finally at the five-axiom system of this section, which we now denote by the Greek letter :  . 2.5 What is a Set? 109

If we continue to play with  , we may very well come across some statement S which, for some reason or other, we would like to be True. If it can be determined that neither S nor its negation is a consequence of the axioms in  , then we say that S is independent of  , and may choose to add S to  . In particular, the statement: For any given set A there exists a set consisting of all the subsets of A can be shown to be independent of the five axioms in  . We could therefore add it to the five axioms in  and in so doing arrive at a richer system  . We now turn our attention to a couple of interesting statements that are independent of  . The first of these appeared as one of the axioms posed in Zermelo’s 1908 paper on the Founda- tions of Sets: THE : Given any non-empty collection C of non-empty sets, there exists a set consisting of exactly one element from each set in C. (This amounts to being able to “choose” and element from each of the sets in C) The motivation for the above axiom may, in part, be attributed to Cantor. In 1883 he asserted that every set can be well-ordered; which is to say, that an order relation can be imposed on any set, under which each of its non-empty subsets contains a smallest or first element. This, for example, is already the situation with any countable set. Cantor indicated that he would sub- stantiate this fundamental and truly remarkable fact at a future date. He did not, and, as Zer- melo subsequently showed, for good . The above WELL-ORDERING PRINCIPLE has such a wide range of applications that it was one of the famous twenty-three unsolved problems formally offered for consideration to the mathematical community by (1862-1943), at the 1900 International Congress of Mathematics. It was for Zermelo, in 1904, to offer a proof in the affirmative, a proof that depended on the principle set forth within his Axiom of Choice. In other words, given the Axiom of Choice, the Well-Ordering Principle follows. Moreover, it is easy to see that if the Well-Ordering Principle holds, then so does the Axiom of Choice (well order each of the sets in question, and select the first element of each). Thus, in the Zermelo-Fraenkel Axiomatic System, the Axiom of Choice is equivalent to the Well-Ordering Principle, in that the validity of either implies that of the other; or, if you prefer, if you don’t have the one then you don’t have the other (nor any of its other numerous equivalent formulations). In 1931, a twenty five year old student at the University of Vienna, Kurt Gӧdel (1906-1978) showed that if the Zermelo-Fraenkel axiom system excluding the Axiom of Choice is consis- tent, then adding the Axiom of Choice will not lead to a contradiction. It was not until 1966, however, that (1934-2007) succeeded in proving that the addition of its negation would also not lead to a contradiction. Thus, the Axiom of Choice is independent of the other axioms within the Zermelo-Fraenkel system, and whether to accept it or not boils down to a matter of personal inclinations. 110 Chapter 2 A Touch of Set Theory

This conviction of the solvability of every mathematical problem is a powerful incentive to the worker. We hear within us the perpetual call: There is the problem. Seek its solution. You can find it by pure reason, for in mathematics there is no: we will not know. Oh yes there is! For in 1931 Gӧdel published his famous Incompleteness Theorem: In any mathematical system rich enough to encompass the natural numbers, there is an assertion expressible within the system that is true, yet is not provable within that system. And just in case that was not bad enough, he then went on to prove that the of such a system is itself an undecidable proposition. What a double whammy! First, there will always be undecidable propositions. And, worse than that, we can never gain assurance that our mathematics is based on a firm foundation. It should be underlined, however, that just because we are not able to prove that our axiomatic system is consistent, that does not mean that it is not. Indeed, not many mathematicians lose much sleep over this issue, as most have the utmost faith that our Zermelo-Fraenkel axiomatic system, along with its variations, are indeed consistent systems. Yes, there are potential flaws in modern mathematics. Perhaps some drastic fundamental changes, possibly in the of logic, will remedy the situation. Or, perhaps, imperfection is within the very nature of things; not only in our physical universe, but in our expanding mathe- matical universe as well. At any rate, as things stand now, mathematics is not what we would call perfect, but it may very well be the closest thing to perfection around. 3.1 The Real Number System 111

3 CHAPTER 3 A Touch of Analysis The axiomatic structure of the real number system is introduced in Sec- tion 1 wherein the completion axiom taking center stage. Sequences, including Cauchy sequences, are featured in Section 2. The metric struc- ture of the real number system is discussed in Section 3, and the important concept of continuity is featured in Section 4. §1. THE REAL NUMBER SYSTEM If it looks like a duck, walks like a duck, and quacks like a duck, then it probably is a duck. Unlike ducks, which appear to require but three defining characteristics to distinguish them from all other worldly creatures, eleven characteristics (axioms) are needed to distinguish the set of real numbers from all other mathematical creatures. DEFINITION 3.1 A complete is a set F, along with two operations, called addition (+) and COMPLETE multiplication ( ), along with an order rela- ORDERED FIELD tion ( ), such that:

Commutative Axiom: (1)xy+ = yx+ xy  F.

Associative Axiom: (2)xyz+ + = xy+ + z and xyz  = xy  z xyz   F . Distributive Axiom: (3)xyz + = xy + xz xyz   F .

Additive Identity Axiom: (4) There exists an element which we will label 0  F such that x + 0 = x for all xF . Additive Inverse Axiom: (5) For each xF there exists an element which we will label –xF such that xx+0– = . Multiplicative Identity Axiom: (6) There exists an element which we will label 1  F with 10 such that 1  x = x for all xF . –1 Multiplicative Inverse Axiom: (7) For each x  0 there exists an element x–1  F such that xx = 1 . Trichotomy Axiom: (8) For any xy  F , either xy or yx . Additive Inequality Axiom: (9) If xy , then xzyz+  + for every zF . Multiplicative Inequality Axiom: (10) If xy and z  0 , then xz  yz Axiom: (11) Every nonempty subset of F that is bounded from above has a least upper (See Definition 3.2 below) bound. It can be shown that the set of real numbers  , with standard addition and standard multiplication, is a complete ordered field. It can also be shown that any other complete ordered field can only differ from  super- ficially (the number 5, for example, might be written as V). 112 Chapter 3 A Touch of Analysis

FOCUSING ON THE COMPLETION AXIOM You may be able to anticipate the meaning of the terminology in the completeness axiom of Definition 3.1 on your own; but just in case: DEFINITION 3.2 Let S be a subset of  .

UPPER BOUND a is an upper bound of S if as for every sS . S is bounded from above if S has an upper bound.

LOWER BOUND b is a lower bound of S if bs for every sS . S is bounded from below if S has a lower bound. S is bounded if S has both a lower and an upper bound. LEAST UPPER  is the least upper bound (or supremum) of BOUND In Exercise 12 you are S   if it is an upper bound of S and if it is invited to verify that if a less than or equal to every upper bound of S. least upper bound (or greatest lower bound) GREATEST LOWER  is the greatest lower bound (or infimum) of exists, then it is unique. BOUND S   if it is a lower bound of S and if it is greater than or equal to every lower bound of S.

The notation lub S and glb S is used to denote the least upper bound and greatest lower bound of S, respectively. For example: glb125 = 1 and lub125 = 5 glb –2 4 = –2 and lub –2 4 = 4 lub–7 ==lub–7  7 and glb 3  ==glb 3  3 As is illustrated above, neither the least upper bound nor the greatest lower bound of a set S need be an element of that set. If it is, then it is said to be the maximum or greatest member of S, and the minimum or smallest member of S, respectively.

CHECK YOUR UNDERSTANDING 3.1 Determine the least upper bound, the greatest lower bound, the maxi- (a) lub: 7, glb: 3, Max: 7 mum, and the minimum element of the given set, if they exist. (b) lub: 9, Max: 9 (c) lub: 2 , glb: – 2 , (a) 35  47 (b) –0 13 9 Max: 2 (c) x  0 x2  2  x  0 x2  2 3.1 The Real Number System 113

The following result provides a useful characterization for the least upper bound and greatest lower bound of a given set.

THEOREM 3.1 (a) A number  is the least upper bound of S   if and only if it satisfies the follow- ing two properties: (i) is an upper bound of S. The Greek letter epsilon  is  generally used to denote a S . . (ii) For any given   0 there exists some “small” unspecified number. s  sS (which depends on  ) such that Note that: s  – .  is FIRST given. (That is: – is not itself an upper bound) THEN: s is to be found to accommodate that (b) A number  is the greatest lower bound of particular  . S   if and only if it satisfies the follow- ing two properties: . . S (i) is a lower bound of S.  s (ii) For every   0 there exists some sS (which depends on  ) such that s  + . (That is:  +  is not itself a lower bound)

PROOF: (a) If  = lub S then  is, in particular, an upper bound. To s  show that (ii) also holds, we consider a given   0 . S . . Since –  , and since nothing smaller than  can be an upper bound of S, there must exist some s  S to the right of – – (see margin). Conversely, suppose  satisfies (i) and (ii). To show that  is the least upper bound of S we consider an arbitrary upper bound a of S, and show that   a : Assume, to the contrary, that   a . Let = – a  0 . By (ii):  sSs   – ==– – a a Contradicting the assumption that a is an upper bound of S.

CHECK YOUR UNDERSTANDING 3.2 Answer: See page A-14. Prove Theorem 3.1(b).

EXAMPLE 3.1 We remind you that: 1 2 3 4 n + + Let S ==--------------- ------nZ Z = 123 8 9 10 11 n + 7 Show that lub S = 1 . 114 Chapter 3 A Touch of Analysis

n SOLUTION: (a) Since the denominator of ------is always greater than n + 7 the numerator, 1 is seen to be an upper bound of S. To show that it also satisfies (ii) of Theorem 3.1(a), we observe that for any given   0 : n ------ 1 –   nn + 7 1 –  n + 7  nnn –77 + –   n  77–  77–  Since   0:  n  ------ Cutting out the middle steps in the above development we conclude 77–  n that for any integer n  ------, the element ------of S will lie to the  n + 7 right of 1 –  .

Note that the smaller  is, the larger ------77– - becomes, and that therefore  no one element of S will lie to the right of 1 –  for all   0 . But for any given   0 an element of S does exist to “accommodate” that  .

CHECK YOUR UNDERSTANDING 3.3 Find the smallest element of the set S of Example 3.1 which lies to the right of: 99 99.9 (a) ------(b) ------(a) ------694 - (b) ------6994 - 694+ 7 6949+ 7 100 100 Suggestion: Write the given number in the form 1 –  .

THEOREM 3.2 For any given ab   with a  0 , there exists ARCHIMEDEAN a positive integer n such that na b . PRINCIPLE

PROOF: Let SnanZ=  + . We first show that S is not bounded above: Assume, to the contrary, that S is bounded above. By the com- pletion axiom, S has a least upper bound  . Letting a play the role of  in Theorem 3.1(a-ii), we conclude that there exists an

element of S, say n0a , which lies to the right of  – a . But if

 – a  n0a , then   n0aa+ = n0 + 1 aS — contra- dicting the fact that  is an upper bound of S. To complete the proof we need but note that since S is not bounded above, the given b in the statement of the theorem is not an upper bound of S. It follows that some element na of S must lie to the right of b. 3.1 The Real Number System 115

CHECK YOUR UNDERSTANDING 3.4 (a) Prove that for any given number b   there exists a positive integer n such that nb . (b) Prove that for any given   0 there exists a positive integer n 1 Answer: See page A-14. such that ---   . n Here is particularly useful consequence of the completion axiom:

THEOREM 3.3  If In = an bn n = 1 is a collection of non- NESTED empty closed intervals such that CLOSED INTERVAL  PROPERTY an + 1 bn + 1  an bn , then  In   . n = 1

+ PROOF: Consider the set L = an nZ of left endpoints of the given closed intervals an bn :

  a1 a2 a3 b3 b2 b1

Since L is bounded above (byb1 , for example), it has a least upper bound  . We complete the proof by showing that  is contained in each In :

Since each bn is an upper bound of L, and since  = lub L ,   bn for every n. We also know that   an for ever n ( is an upper bound of L). It follows that an  bn for every n,

and that therefore    In . n = 1

CHECK YOUR UNDERSTANDING 3.5 Note that these nested  intervals are not closed. 1 Let J = 0 --- for nZ + . Show that J =  . n  n  n Answer: See page A-15. n = 1 Suggestion: Use CYU 3.4(b).

So, in the real number system: THEOREM 3.4 For any given x  0 there exists   0 such x exists for every x  0 . that 2 = x .

PROOF: The set Ss=  0 s2  x is nonempty as it contains 0. It is also bounded above (by 1 if 0 x 1 , and by x if x  1 ). As such 116 Chapter 3 A Touch of Analysis

 = lub S exists. We show that 2 = x by eliminating the two other possibilities, 2  x and 2  x : Assume that 2  x . We will exhibit an element of S that is greater than  , thereby contradicting the fact that  is an upper bound of S. Our first step is to consider the square of numbers a bit larger than  . For any n  1 : 1 2 2 1  + --- = 2 ++------ 2 n n n 2 1 1 2 ++------= 2 + ---2 + 1 (*) n n n Since   0 and 2  x : 1 x – 2 x – 2 Choosing N sufficiently large so that ----  ------(see ------ 0 N 2 + 1 2 + 1 It follows, from CYU 3.4(b), margin), and appealing to (*), we have: 1 x – 2 that ----  ------for some N. 1 2 1 x – 2 N 2 + 1  + ----  2 + ----2 + 1  2 + ------2 + 1 = x N N 2 + 1 1 So,  + ----  S , and is larger than  — a contradiction. N Having ruled out the possibility 2  x , we now do the same for 2  x , and do so by looking at the square of numbers a bit smaller than x: 1 2 2 1 2  – --- = 2 – ------- + -----  2 – ------- (**)  2 n n n n 1 2 – x Choosing N , sufficiently large, so that ----  ------and N 2 appealing to (**), we have: 1 2 1 2 – x  – ----  2 – ----2  2 – ------2 = x N N 2 1 2 Since  – ----  x , and since (by assumption) s2  x for every N 1 element in S,  – ---- is an upper bound of S — contradicting the N fact that  is the least upper bound of S.

a We remind you that a rational number is a number of the form --- , b where a and b are integers, with b  0 . A number that is not rational is said to be irrational. You are not insulting a number by calling it irrational — you are just saying that it is not the ratio of two integers. 3.1 The Real Number System 117

Not every number is rational: a THEOREM 3.5 There is no rational number --- such that b a 2 --- = 2 . (In other words: 2 is irrational). b a PROOF: (By contradiction) Assume there exists a rational number --- b a 2 in lowest terms (margin) such that --- = 2 . Then: b Any rational number --a- b a 2 a2 can be expressed in low- --- = 2  ----- = 2 est terms (a and b share b b2 no common factor).  a2 = 2b2 (*) 2  a is even Theorem 1.10,page 47:  a is even  a = 2k for some integer k Substituting 2k for a in (*) we have: 2k 2 =42b2  k2 =22b2  b2 = 4k2  b2 = 2k2  b2 is even Theorem 1.10:  b is even  b = 2h for some integer h a 2k a Bringing us to: --- = ------— contradicting our stated condition that --- b 2h b is in lowest terms. Since the assumption that 2 is rational led to a contradiction, we con- clude that 2 must be irrational.

EXAMPLE 3.2 Show that the sum of any irrational number and any rational number is irrational.

SOLUTION: (By contradiction) Let x be irrational. Assume there a a exists a rational number --- such that x + --- is rational, say: b b x + --a- = --c- b d Then: x = --c- – --a- d b cb– ad x = ------(*) db Since cb– ad is an integer and db is an integer distinct from 0 (why?), (*) tells us that x is rational — contradicting our stated condi- tion that x is irrational. 118 Chapter 3 A Touch of Analysis

CHECK YOUR UNDERSTANDING 3.6 (a) Prove that the product of any irrational number with any nonzero rational number is irrational Answer: See page A-15. (b) Can the sum of two irrational number be rational? Justify your answer.

DENSE SUBSETS OF 

In other words: between DEFINITION 3.3 A set D of real numbers is dense in  if for any two distinct real num- bers, one can always find DENSE SUBSET any given real numbers a and b with ab an element of D. there exists dD such that adb . Here are two important dense subsets of  : THEOREM 3.6 The set of rational numbers and the set or irra- tional numbers are dense in  .

PROOF: To show that the rationals are dense in  we exhibit, for any given ab   with ab , a rational number lying between a and b: 0 Case 1. a 0 b : Not much to be done here, since 0 = --- is a 1 rational number. m Case 2. 0  ab : Our goal is to find a rational number ---- such that: n m a ---- b n We begin by choosing a positive integer n such that: 1 1 ---  ba– or: ab – --- (*) n n See CYU 3.4(b) We then consider the non-empty set: t StZ=  + ---  a n The Well-Ordering Principle (page 39), assures us that S has a smallest (first) element, which we will call m. m m Since mS : ----  a . If we can show that ----  b then we will be n n done. We can: since m – 1 is less than m: m – 1  S (*) m – 1 1 ------ ab – --- n n m 1 1 ---- – ---  b – --- n n n m ----  b n 3.1 The Real Number System 119

Case 3. ab  0 : Noting that 0  –ba – , we appeal to the previ- m m ous case and choose a rational number ---- such that –b ---- –a . n n m Multiplying through by –1 yields the desired result: a  –----  b . n Using the established fact that the rationals are dense in  we now show that the irrationals are also dense in  : m Let ab be given. Choose a rational number ---- such that: n m a – 2 ---- b – 2 n m or: a  ----2+  b n This completes the proof, since the sum of a rational number and an irrational number is itself irrational (Example 3.2).

CHECK YOUR UNDERSTANDING 3.7 (a) Prove that no finite subset of  is dense in  . (b) Let S be dense in  . Show that any open interval ab must Answer: See page A-15. contain infinitely many elements of S. 120 Chapter 3 A Touch of Analysis

EXERCISES

Exercises 1-6. Determine the least upper bound and greatest lower bound of the given set, and its maximum and minimum element, if they exist. 1.05  7 2.05  7 3. –3  410  4. –3  4  5.Z+  x   x  7 6. Z+  x   x  7 Exercises 7-11. Determine the least upper bound of the set S. Justify your claim.

1 1 + 2n 7. S = 1 – --- nZ + 8. S = 1 – ----- nZ 9. S = ------nZ + n n2 n + 1

n2 – 1 20sinn 1 10.S = ------nZ + 11. S = ------n  10  5 – ----- 1 n 5 2n2 5n n2

12. (a) Prove that if a subset S of  has a least upper bound, then it is unique. (b) Prove that if a subset S of  has a greatest lower bound, then it is unique.

13. (a) Prove that if a subset S of  has a maximum element, then it equals lub S . (b) Prove that if a subset S of  has a minimum element, then it equals glb S .

14. Prove that any finite subset of  contains a maximum and minimum element.

15. Prove that every nonempty subset of  that is bounded from below has a greatest lower bound.

16. Prove that if A is a nonempty bounded subset of  , then glb A  lub A .

17. (a) Let A and B be nonempty subsets of  , bounded above, with AB . Show that lub A  lub B . (b) Give an example of nonempty sets A and B, bounded above, with AB and lub A = lub B .

18. For A   and B   , let AB+ = abaAbB+    . Prove that: (a) If A and B are bounded above, then so is AB+ . (b) If A and B are bounded below, then so is AB+ .

19. (a) Prove that if a subset S of  has a least upper bound  , then the set of all upper bounds of S is  . (b) Prove that if a subset S of  has a greatest lower bound  , then the set of all lower bounds of S is –  .

20. Prove that every number is both an upper bound and a lower bound of  .

21. Prove that A is bounded if and only Aa= aA is bounded. 3.1 The Real Number System 121

22. (a) Prove that if AB , with A   , then lub A  lub B . (b) Give an example where AB and lub A = lub B .

23. (a) Prove that if AB , with A   , then glb A  glb B . (b) Give an example where AB and glb A = glb B .

24. (a) Let A and B be nonempty bounded sets of real numbers such that for every aA there exists bB such that ab , and for every bB there exists aA such that ba . Prove that lub A = lub B and glb A = glb B . (b) Give an example of sets A and B satisfying the conditions of part (a), with AB .

25. (a) Let A and B be nonempty bounded sets of real numbers such that ab for every aA and every bB . Prove that lub A  glb A . (b) Give an example of sets A and B satisfying the conditions of part (a), with lub A = glb A .

26. Let A be bounded above, and let x   . Prove that: (a) lub xaaA+  = x + lub A (b)lub xa a A = xlub A if x  0 , and that glbxa a A = xglbA if x  0 .

27. Let A be a nonempty subset of  which is bounded above but does not have a maximum ele- ment. Prove that A cannot be finite.

28. Let A be a nonempty subset of  which is bounded below but does not have a minimum ele- ment. Prove that A cannot be finite.

29. Let A be a nonempty subset of  which is bounded above but does not have a maximum ele- ment. Prove that for any aA , lub A = lub Aa–  .

30. Let A be a nonempty subset of  which is bounded below but does not have a minimum ele- ment. Prove that for any aA , lglb A = glb Aa–  .

31. Let A be a nonempty subset of  which is bounded above but does not have a maximum ele- ment, and let a1a2  an  A . Show that lub A = lub Aa– 1a2  an .

32. Let A be a nonempty subset of  which is bounded below but does not have a minimum ele- ment, and let a1a2  an  A . Show that glb A = glb Aa– 1a2  an .

33. Show that any subset of  that contains a dense subset of  is itself dense.

34. Give an example of an infinite subset of  that is not dense in  .

35. Prove that if x and y are rational and z is irrational then xy+ z is irrational.

36. Prove that for any positive integer n, n is rational if and only if n is an integer.

37. Show that there exists irrational numbers x and y such that xy is rational. Suggestion: Consider the number 2 2 . If it is rational, then the claim is seen to hold. If it is irrational, then consider raising it to the power 2 . 122 Chapter 3 A Touch of Analysis

PROVE OR GIVE A COUNTEREXAMPLE

38. If AB , and if B is bounded, then A is bounded. 39. If AB , and if B is bounded, then lub A  lub B or glb A  glb B . 40. If lub A  lub B , then there exists an element bB that is an upper bound of A. 41. If glb A  glb B , then there exists an element aB that is a lower bound of B.

42. For A   and B   , let AB+ = abaAbB+    . If A = lub A and

B = lub B , then A +B = lub AB+ .

43. For A   and B   , let AB+ = abaAbB+    . If A = glb A and

B = glb B , then A +glbB = AB+ . 44. For A   and B   , let AB = ab a A b B . If A and B are bounded, then so is AB .

45. For A   and B   , let AB = ab a A b B . If A = lub A and B = lub B ,

then A  B = lub AB .

46. For A   and B   , let AB = ab a A b B . If A = glb A and B = glb B ,

then A  B = glb AB . 47. Every infinite subset of  is dense in  . 48. Let A and B be subsets of  . If AB is dense in  then A or B must be dense in  . 49. Let A and B be subsets of  . If AB is dense in  then A and B must be dense in  . 50. Let A and B be subsets of  . If A and B are dense in  then AB is also dense in  . 51. Let A and B be subsets of  . If A and B are dense in  then AB is also dense in  . 52. The product of any irrational numbers and any rational number is again irrational. 53. If x and y are irrational and z is rational then xyz++ is irrational. 54. If x and y are irrational and z is rational then xy+ z is irrational. 55. If x and y are irrational and z is rational then xy+ z is rational. ax+ b 56. If ------1= then x must be rational. cx+ d

57. (a) If one solution of the quadratic equation ax2 ++bx c = 0 is rational then the other solu- tion is also rational. (b) If one solution of the quadratic equation ax2 ++bx c = 0 is rational, and if the coeffi- cients a, and c are integers, then the other solution is also rational. 3.2 Sequences 123

3

§2. SEQUENCES Formally: DEFINITION 3.4 A of real numbers is a real-valued SEQUENCE function with domain the set of positive inte- gers. Formality aside, one seldom represents a sequence in the function- +  form f: Z   but, rather, as an infinite string of numbers, or terms Unlike the set a , ele- n n = 1   a a1a2 a3  or an n = 1 ments in a sequence n n = 1 th can appear more than once, as with n -term an . is the case with the sequence 010101 . Consider the sequences: 1 1 n + 1  (a) 1------ (b) ------and (c) 121212 2 3 n n = 1 While the sequence in (a) appears to be heading to 0, and that of (b) to 1, the sequence in (c) does not look to be going anywhere in particular, as its terms keep jumping back and forth between 1 and 2. Appearances are well and good, but mathematics demands precision;

DEFINITION 3.5  A sequence an n = 1 converges to the num- CONVERGENT ber  if for any given   0 there exists a SEQUENCE positive integer N (which depends on  ) such that: nN  an –    A sequence that does not converge is said to diverge. In words: By going far enough in the sequence, nN , you can get the

terms of the sequence to be as close as you want to  , an –    . The next theorem asserts that a sequence cannot converge to two dif-  We remind you that the abso- ferent numbers. That being the case, if an n = 1 converges to  , we lute value function   a if x  0 are justified in saying that  is the limit of an n = 1 , and write: a =   –ifa x  0 lim an =  , or liman =  , or simply an   . denotes the distance between n   the number a and the origin on the number line, and that ab– represents the dis- Additional Notation: For given a   and   0 , we will tance between the numbers a use the symbol Sa to denote the set of numbers that lie and b. For example: within  units of a: 27–5= is the distance   } between 2 and 7, while S a = x  x – a  : }  ( . ) 34+34==–7– is the a –  a a +  distance between 3 and –4 . In anticipation of higher dimensional spaces, we call Sa the (open) of radius  about a. 124 Chapter 3 A Touch of Analysis

THEOREM 3.7 A sequence can have at most one limit.

 ROOF P : Assume that a sequence an n = 1 converges to two different numbers  and  (we will arrive at a contradiction). Letting  – denote the larger of the two numbers, we consider  = ------: 2   ( . )( . ) { { S S  Since an n = 1 converges to  and to  , there exist integers N and N such that nN   an –    and nN   an –    . Let N be the larger of N and N . As such, the term aN + 1 of the sequence must lie in both S and S — a contradiction, since S S = .

EXAMPLE 3.3 (a) Prove that lim------n + 1- = 1 . n   n (b) Prove that for any constant c the sequence cccc converges to c. (c) Show that the sequence 121212 diverges.

SOLUTION: (a) Let   0 be given. We are to find N such that n + 1 nN  ------1–   . Let’s do it: n n + 1 We want: nN  ------– 1   n n + 1 – n Let’s rewrite the goal: nN  ------  n again: nN  --1-   n 1 and again: nN  ---   n 1 and finally: nN  n  ---  n + 1 So, to find an N such that nN  ------– 1   is to find an N n Note how N is dependent 1 on  — the smaller the such that nN  n  --- . Piece of cake: let N be the first integer given  , the larger the N.  1 greater than --- . 

3.2 Sequences 125

(b) For cn = cccc and any given   0 let N = 1 . Then: nN  cn – c ==cc–0  . (c) We show that 121212 diverges by demonstrating that no fixed but arbitrary r   can be the limit of the sequence: Let N be any positive integer. Since any two numbers in  1 1 --1- --- S r = xx– r --- are less than one unit apart (see mar- 2 2 --1- 2 ( | ) 2  r gin), both a and a cannot be contained in S r , as N + 1 N + 2 --1- 2 one of the numbers is 1 while the other is 2. This shows that no 1 N “works” for  = --- , and that, consequently, the arbitrarily 2 chosen number r cannot be a limit of 121212 .

CHECK YOUR UNDERSTANDING 3.8

 101  (a) Let a = 7 – ------. n n = 1  n n = 1

(i) Prove that liman = 7 . n   (ii) Find the smallest positive integer N such that 1 nN  a– 7  ------. n 100

  ------n – 5- Answer: See page A-15. (b) Show that the sequence an n = 1 = diverges. 333 n = 1

DEFINITION 3.6  A sequence an n = 1 is: INCREASING Increasing if an  an + 1 ECREASING D Decreasing if an  an + 1 MONOTONE Monotone if it is either increasing or decreasing. BOUNDED Bounded if there exists a real number M such

that an  M for every n. Example 3.3(c) shows that not every bounded sequence converges. However: THEOREM 3.8 Every increasing (decreasing) sequence that is bounded from above (below) converges. 126 Chapter 3 A Touch of Analysis

 ROOF P : Let an n = 1 be increasing and bounded from above. The completion axiom assures us that the set an has a least upper  bound:  . We show that an n = 1 converges to  : Let   0 be given. Theorem 3.1 (a-ii), page 113, tells us that there exists a term a such that a  – . Since a  is an increas-  N N n n = 1 ing sequence, an  – for every nN . Since  = lub an , .– .a . n an   + for every n. It follows that an –    for every an –  nN (see margin). A similar argument can be used to show that every decreasing sequence bounded from below converges (Exercise 35).

CHECK YOUR UNDERSTANDING 3.9

Answer: See page A-16. Prove that if a sequence converges, then it is bounded.

THE ALGEBRA OF SEQUENCES When it comes to sums, differences, products, and quotients, sequences behave nicely:

THEOREM 3.9 If lim an =  and lim bn =  , then:

(a)lim can = c , for any c   .

(b) lim an + bn = + (The limit of a sum equals the sum of the limits)

(c) lim anbn = (The limit of a product equals the product of the limits)

an  (d)lim ----- = --- , providing no bn = 0 and   0 . bn  (The limit of a quotient equals the quotient of the limits )

PROOF: (a) Case 1. c = 0 .

lim can ====lim 0  an 00  c Case 2.c  0 . For given   0 we are to exhibit an N such that

nN  can – c  

i.e: nN  can –     i.e: nN  a–   ----- n c

3.2 Sequences 127

Since lim an =  , we know that for any   0 there exists an N such that nN  a–    . In particular, for  = ----- we can choose N n c such that nN  a–   ----- , and we are done. n c (b) Let   0 . We are to find N such that

nN   an + bn – +   (*) Note that:

an + bn – + = an –  + bn –   an –  + bn –  inequality

So, if we can arrange things so that both an –  and bn –  are less  than --- , then (*) will hold. Let’s arrange things: 2  Since a   , there exists N such that: nN  a –   --- . n   n 2  Since b   , there exists N such that: nN  b –   --- . n   n 2

Letting N = maxN N (the larger of N and N ), we find that for nN : a + b – + a –  + b –   --- + --- = n n n n 2 2

(c) Let   0 . We are to find N such that:

nN  anbn –   

In order to get an –  and bn –  into the picture (for we have con- trol over those two expressions), we insert the clever zero – an + an in the expression anbn –  :

anbn –  = anbn – an + an – 

= anbn – an + an–

 anbn – an + an– = an bn –  +  an –  } } (i) (ii)  The next step is to find an N such that both (i) and (ii) are less than --- . 2

Focusing on (i): an bn –  :  The temptation is to let N be such that nN   bn –   ------2 an   (yielding an bn –   an ------= --- ). No can do. For one thing, if 2 an 2  an = 0 , then the expression ------is undefined. More importantly: 2 an 128 Chapter 3 A Touch of Analysis

 ------is NOT A CONSTANT! We can, however, take advantage of the 2 an Let the positive number ------ - 2M fact that there exists an M  0 such that an  M for every n (see  play the role of  in Defini- CYU 3.9), and choose N such that nN  b –   ------(see tion 3.5.   n 2M  margin). Then: nN  a b –   M------= --- .  n n 2M 2

Focusing on (ii):  an –  .  Wanting  a –  to be less than --- , one might be tempted to n 2  choose N such that nN  a –   ------. But what if  = 0 ?   n 2  To get around this potential problem we choose N such that  nN  a –   ------. No problem now:  n 2  + 1   nN a –    ------ ---  n 2  + 1 2   since ------ ------2  + 1 2 

Letting N = maxN N , we see that, for nN : a b –  a b –  +  a –   --- + --- = n n n n n 2 2

a (d) Appealing to (c), we establish the fact that lim ----n- = --- , by show- bn  1 1 ing that lim ----- = --- : bn  Let   0 be given. We are to find N such that: 1 1  – b nN  ----- – --- = ------n   (*) bn  bn  Since bn    0 , we can choose N1 such that:  nN  b –   ------. 1 n 2  – b   – b  n n For nN 1 we also have: bn  ------(see margin). Consequently,  2 = b –   ------n 2 For nN :  1  b  ------n 2  – b 1  – bn 1  – bn 2 ------n ==------ ------ ------ ------b –  (**) b    2 n bn  n ------ 2

Since bn   , we can choose N2 such that:  2 nN  b –   ------ (***) 2 n 2

3.2 Sequences 129

Letting N = maxN1 N2 , we find that, for nN :  – b 2 2  2 ------n ------b –  ------ ------ =  2 n 2 bn    2 By (**) By (***)

thereby establishing (*).

CHECK YOUR UNDERSTANDING 3.10

  (a) Let an n = 1 and bn n = 1 be such that 0 an bn for every n. Show that if an   and bn   , then  .   (b) Give an example of two sequences an n = 1 and bn n = 1 with Answer: See page A-16. 0  an  bn and such that lim an = lim bn .

SUBSEQUENCES  Roughly speaking, to get a subsequence of an n = 1 , simply discard some of its terms in an orderly fashion. Formally:

DEFINITION 3.7 a  = a a a  is a subse- nk k = 1 n1 n2 n3 SUBSEQUENCE quence of a  if each a is a term of n n = 1 nk  an , and n1 n2 n3 .

For example, 9111315 is a subsequences of 123 , while 4286 is not.

CHECK YOUR UNDERSTANDING 3.11 Coin a function-form definition of a subsequence of a sequence (see Definition 3.4). We’ll get you started: The sequence g: Z+   is a subsequence of the sequence Answer: See page A-16. f: Z+   if ...... There are sequences, like 1234 , that do not contain any convergent subsequences. However: Bernhard Bolzano (1781- THEOREM 3.10 Every bounded sequence contains a conver- 1841), Karl Weierstrass (1815-1897). BOLZANO-WEIERSTRASS gent subsequence.  ROOF P : Let an n = 1 be a bounded sequence. Being bounded, there exists M  0 such that an  –M M for every n. 130 Chapter 3 A Touch of Analysis

Cut –M M into two equal pieces: –0M and 0 M . Select one M of those two intervals (of length ----- ) which contains an infinite num- 2  ber of elements of the sequence an n = 1 , and call it I1 . Next cut I1 1 M M in half and let I be one of those halves (now of length ---  ----- = ----- ) 2 2 2 22  which still contains infinitely many elements of an n = 1 . Continue in this fashion to generate a nested sequence of closed intervals In of M  length ----- , each containing infinitely many terms of a . 2n n n = 1

Theorem 3.3, page 115, assures us that  In   . Let x0   In . n= 1 n = 1 We now construct a subsequence of an n = 1 converging to x0 : Let:a be any term of a  in I n1 n n = 1 1 a be any term in I with n  n n2 2 2 1

(this we can do since I2 contains infinitely many entries of an ) a be any term in I with n  n n3 3 3 2  Proceeding in the above fashion we arrive at a subsequence an n = 1 with a  I . As for convergence: nk k M Let   0 be given. Choose N such that ------  . Since x and 2N 0 a are contained in I , for n  N : a – x   . nk k k nk 0

Augustin Louis Cauchy CAUCHY SEQUENCES (1789-1857).

To say that lim an =  is to say that the ans eventually get arbi- n   trarily close to  . We now consider sequences whose terms eventually get arbitrarily close to each other (with no mention made of any limit whatsoever):

DEFINITION 3.8  A sequence an n = 1 is a Cauchy sequence CAUCHY SEQUENCE if, for any given   0 there exists a positive integer N (which depends on  ) such that if nm  N, then an – am   .

3.2 Sequences 131

CHECK YOUR UNDERSTANDING 3.12 Answer: See page A-16. Prove that every Cauchy sequence is bounded. It may not be surprising to find that every convergent sequence is Cauchy. After all, if the terms of a sequence are “bunching up” around a number  , then they must certainly be getting close to each other. A bit more surprising is that the converse also holds: THEOREM 3.11 A sequence converges if and only if it is Cauchy.

  ROOF P : Assume that an n = 1 converges to  . We show an n = 1 is Cauchy: Let   0 be given. Choose N such that nN implies  a –   ---. Then, for nm  N : n 2   a – a ==a –  +  – a a –  + a –   --- + --- . n m n m n m 2 2

 Conversely, assume that an n = 1 is Cauchy.  By CYU 3.12, an n = 1 is bounded and, as such it has a conver-  gent subsequence an (Theorem 3.10). Let lim an =  . k k = 1 k   k

We complete the proof by showing that lim an =  : n    Let   0 be given. Since an n = 1 is Cauchy, there exists N  such that a – a  --- for all nm  N (in particular for all n m 2 nn  N ). Since a   , we can choose a term a of k nk nk  a , with n  N , such that: a –   --- . For nN we nk K nK 2 then have: a –  = a – a + a –  n n nK nK  a – a + a –   --- + --- =  n nK nK 2 2

CHECK YOUR UNDERSTANDING 3.13

 Let X be a nonempty subset of  . A sequence an n = 1 is said to be

a sequence in X if each an  X . A sequence in X is said to converge

in X if liman =   X . Construct a Cauchy sequence in X = 0  which does not Answer: See page A-16. converge in X. 3.2 Sequences 132

EXERCISES

Exercises 1-6. Find a formula for the nth term of the given sequence.

1 2 3 4 2 3 4 5 1. --------- 2. ---–------–------ 2 3 4 5 3 9 27 81

3.121212 4. 1232345678910  5. 112439416  6. aba+ ba 2 b2 a2 + b2 a3 b3 a3 + b3  Exercises 7-15. Find the limit of the given sequence. Use Definition 3.5 to justify your claim. 1   1  7. --- 8. 5 + -----1- 9. – 1 – ------   n n = 1 2n n = 1 n n = 1 n  n2  3n2 + n  10. ------11. ------12. ------2 2 n + 1 n = 1 n + 1 n = 1 n n = 1

2n2 + 10  n  n2 + 10n  13. ------14. r n = 1 , for r  1 15. ------2 2 n + 1 n = 1 n + 1 n = 1

Exercises 16-21. Show that the given sequence diverges. 1 1 1 1 n  n  16. ------1 ------1 ------1 ------1  17. r n = 1 , for r  1 18. –1 n = 1 10 20 30 40

n  2  n!  ------–1 n ------n - 21. ----- 19. 20. n n + 1 n = 1 n + 100 n = 1 2 n = 1

22. Prove that liman = 0 if and only if liman = 0 . n   n  

Exercises 23-31. Find the limit of the given sequence if it exists. You may use Theorem 3.9 and the result of Exercise 22 to justify your claim. n  sinn  n  –1 24. ------–1 2n 23. ------ 25. ------n + 5 n = 1 n n = 1 n + 1 n = 1

n2 + 5  n3 ++10n2 –1n  n2 cosn – nsin2n  ------27. ------28. ------26. 3 3 n2 + 5 n = 1 n n = 1 n + 5 n = 1 sinnncos  cosn –1 n  n + 1 3 n3 + n – 1  29. ------30. ------– ------31. ------------3 n + 1 n = 1 n + 1 n n = 1 n – 3 2n n = 1

32. Let 0  an  bn for nN . Prove that if bn  0 , then an  0 . 133 Chapter 3 A Touch of Analysis

Exercise 33-34. Find the limit of the given sequence. You may use Exercises 22 and 32 to justify your claim. sinn  –100 n  33. ------34. ------n! n = 1 n! n = 1

35. Prove that every decreasing sequence, bounded below, converges.

  36. (a) Give an example of two converging sequences an n = 1 and bn n = 1 such that liman + bn = 5 . n     (b) Give an example of two divergent sequences an n = 1 and bn n = 1 such that liman + bn = 5 . n  

n  37. Prove that the sequence r n = 1 converges if and only for –1  r  1 . # 38. Prove that lim an =  if and only if, for any given   0 , S contains all but finitely n    many terms of an n = 1 .

39. Prove that if lim an =  and if an  0 for all nN , then   0 . n  

  40. Prove that a sequence an n = 1 converges if and only if an nN= converges for any positive integer N.

41. Prove that if liman = 0 and if rb– n  an then lim bn = r . n   n  

 1 42. Write down the first four terms of the sequence an n = 1 , if a1 = 2 and an + 1 = ------. 5 – an Show that the sequence converges. Suggestion: Consider Theorem 3.8.

 43. (a) Prove that if liman = 0 and if bn is bounded, then lim anbn = 0 . n   n = 1 n     (b) Give an example of sequences an and bn , such that liman = 0 and n = 1 n = 1 n    anbn n = 1 diverges.   (c) Give an example of a convergent sequence an n = 1 and a bounded sequence bn n = 1 ,  such that anbn n = 1 diverges. 44. Establish the following “Squeeze Theorem:”

If an bn cn for nN , and if lim an ==lim cn  , then lim bn =  . n   n   n  

 45. Prove that if lim an =  , then every subsequence of an also converges to  . n   n = 1 46. Prove that every subsequence of a Cauchy sequence is itself a Cauchy sequence. 3.2 Sequences 134

47. Prove that if a sequence contains two subsequences with different limits, then the sequence diverges. 48. Give an example of a divergent sequence that contains: (a) Two subsequences with different limits. (b) Three subsequences with different limit. (c) Infinitely subsequences with different limits.

PROVE OR GIVE A COUNTEREXAMPLE

49. Every bounded sequence converges.

  50. If an n = 1 converges, then so does an n = 1 .

  51. If an n = 1 converges, then so does an n = 1 .

   52. If an n = 1 and an + bn n = 1 converge, then bn n = 1 must converge.

   53. If an + bn n = 1 converges, then an n = 1 and bn n = 1 must both converge.

   54. If an + bn n = 1 diverges, then an n = 1 and bn n = 1 must both diverge.

55. If lim an =  , then lim an =  . n   n  

56. If lim an =  , then lim an =  . n   n  

57. If lim an =  and if an  0 for all nN , then   0 . n  

58. If lim an =  and lim bn =  , and if an  bn for all nN , then  . n   n  

   59. If an n = 1 and bn n = 1 are Cauchy sequences, then an + bn n = 1 is a Cauchy sequence.

   60. If an n = 1 and bn n = 1 are Cauchy sequences, then anbn n = 1 is a Cauchy sequence. 3.3 Metric Space Structure of  135

3

§3. METRIC SPACE STRUCTURE OF  Roughly speaking, a metric space is a set with an imposed notion of distance. Our concern in this section is with the standard or Euclidean metric d on  ; that function which assigns to any two numbers in  the distance between them: dxy = xy– The following distance-properties of the real number system will evolve into the defining axioms of an abstract metric space in the next chapter:

THEOREM 3.12 (i)xy   : xy–0 , with xy–0= if and only if xy= . The distance between two numbers is never negative, and is 0 only if the two numbers are one and the same. (ii)xy   : xy– = yx– The distance between x and y is the same as that from y to x. (iii)xyz  : xy–  xz– + zy– The triangle inequality. If a  0 and b  0 then: a2  b2  ab PROOF: The first two properties are direct consequences of the defi- nition of the absolute value function. We establish (iii) by showing Here: axy= – and that xy– 2  xz– + zy– 2 (see margin): bxz= – + zy– . xz– + zy– 2 = xz– 2 ++2 xz– zy– zy– 2

a 2 = a2: = xz– 2 ++2 xz– zy– zy– 2

aa :  xz– 2 ++2xz– zy– zy– 2 expanding: = x2 – 2xz ++z2 2xz –22xy –2z2 +zy + z2 – 2zy +y2 ===x2 – 2xy + y2 xy– 2 xy– 2

CHECK YOUR UNDERSTANDING 3.14 Answer: See page A-16. Prove that for any xy   , xy–  xy–

DEFINITION 3.9 A set O   is open if for every aO there We remind you that exists an   0 such that: Sa = xx– a  Sa  O

EXAMPLE 3.4 (a) Show that the interval 15 is open. (b) Show that the interval 1 5  is not open. 136 Chapter 3 A Touch of Analysis

SOLUTION: (a) For any given a  15 we are to find   0 such   that S a  15 . This is easy to do: take  to be the smaller of the

}  ( } ( .a ) two numbers a – 1 and 5 – a (see margin). 1 5 Any smaller   0 will do just as well; but no larger  will work.    (b) For any given   0 , 5 + ---  5 . It follows that there does not } } 2 ( ].  5 + --- 1 5 2 exist an   0 such that S5  1 5  .

CHECK YOUR UNDERSTANDING 3.15 (a) Show that ab = xa x b is open for any ab . (b) Show that – a  and a are open for any a   . Answer: See page A-16. (c) Show that no finite subset of  is open. If you peek ahead to page 171 you will see that the following three properties of open sets are transformed into the three defining axioms of a .

THEOREM 3.13 (i) and  are open sets. (ii) Arbitrary unions of open sets are open. (iii) Finite intersections of open sets are open.

PROOF:

(i) is open: For any x   , S1x   .  is open (by default): For any a   there exists an   0 such

that Sa   — for the simple reason that no such a exists.

See the indexing remarks (ii) Let O be a collection of open sets, and let xO . that follow CYU 2.3,    A   page 58.   A Since x is in the union of the Os , there must exist some   A such that xO . Since O is open, we can choose 0 0 0   0 such that S x  O . Then:  0 xS x O O  0     A n (iii) Let Oi i = 1 be a collection of open sets.

n For xO and 1 in , choose  such that S x  O .  i i i i i = 1 and let  = min12  n . Since Sx  Oi for each i, n xS x   Oi . i = 1 3.3 Metric Space Structure of  137

CHECK YOUR UNDERSTANDING 3.16 Give an example illustrating the fact that an infinite intersection of Answer: See page A-17. open sets need not be open.

We remind you that for a DEFINITION 3.10 A set H   is closed if its complement given set A, Ac denotes c the complement of A. H = x   xH is open.

EXAMPLE 3.5 (a) Show that 15 = x 1 x 5 is closed. (b) Show that 1  is closed. (c) Show that – 1   5  is not closed.

SOLUTION: (a) Since 15 c = – 1  5  is open [see CYU 3.15(b) and Theorem 3.13(ii)], 15 is closed. (b) Since 1 c = –1 is open, 1  is closed.

(c) Since – 1   5  c = 15  is not open [Example 3.4(b)], – 1   5  is not closed.

CHECK YOUR UNDERSTANDING 3.17 (a) Show that ab = xa x b is closed for any ab . Answer: See page A-17. (b) Is 1 3  closed? Justify your answer. Here is the “closed-version” of Theorem 3.13:

THEOREM 3.14 (i) and  are closed sets. (ii) Any intersection of closed sets is again a closed set. (iii) Any finite union of closed sets is again a closed set.

PROOF: (i) Since both c =  and c =  are open [Theorem 3.13(i)], both  and  are closed.

(ii) Let H   A be a collection of closed sets. Since: c c c H = H , and since each H is     A   A Theorem 2.3(b), page 58

open (Definition 3.10),  H is closed [Theorem 3.13(ii) and   A Definition 3.10]. 138 Chapter 3 A Touch of Analysis

n (iii) Let Hi i = 1 be a collection of closed sets. Since c n n n H = H c is open, H is closed.  i  i  i i = 1 i = 1 i = 1 Theorem 2.3(a), page 58

CHECK YOUR UNDERSTANDING 3.18

Answer: See page A-17. Give an example illustrating the fact that the infinite union of closed sets need not be closed.

A sequence S1S2 S3  of sets is said to be nested if Si  Sj for S S  1 ij (see margin). ... n S2 In the exercises you are asked to show that the intersection of a nested sequence of nonempty bounded sets may turn out to be empty, and that the intersection of a nested sequence of nonempty closed sets may also be empty. However: This is a generalization of THEOREM 3.15  Theorem 3.3, page 115. If Hi i = 1 is a nested sequence of nonempty

closed bounded sets, then  Hi   . i = 1

PROOF: For each i, select an element ai  Hi . The Bolzano-Weiers- tass theorem of page 129 assures us that the bounded sequence a  contains a subsequence a  that converges to some i i = 1 ik k = 1 number  . We show that   Hi for every i: Suppose, to the contrary, that   H for some i . Since i0 0   H c , and since H c is open, we can find   0 such i0 i0 c  that S   H . It follows that S  H = (see .  i0  i0 H i0 margin). Since the H s are nested, a  H for every i  i , i ik i0 k 0 which implies that S contains only finitely many terms of the sequence a  , contradicting the assumption that ik k = 1 a converges to  (see Exercise 38, page 133). ik k = 1 3.3 Metric Space Structure of  139

DEFINITION 3.11 O   A is an open cover of S   if each OPEN COVER O is open and SO   .   A

An open cover O   A of S is said to con- FINITE SUBCOVER tain a finite subcover if there exists n     A such that SO . 1 2 n  i i = 1 COMPACT A subset K of  is said to be compact if every open cover of K has a finite subcover.

EXAMPLE 3.6 (a) Show that no unbounded subset of  is compact. (b) Show that –11 is not compact. SOLUTION: (a) Let B be an unbounded subset of  . Consider the open cover  In n = 1 of B, where In = –n n . Since ni  nj implies that N I  I , for any N: I = I . Being unbounded, BI for ni nj  ni nN nN i = 1 N any N, and is therefore not contained in I for any N.  ni i = 1 1 1 (b) For each integer n  2 , let I = –1 + --- 1 – --- . In the exercises n n n  you are asked to show that In n = 2 is an open cover of –11 . That cover has no finite subcover since, for any given N, N 1 1 1 – ----  –11 and 1 – ----  I . N N  n n = 2 The following result tells us that to challenge the compactness of S   one need only consider countable open covers of S. This theorem does not hold in a general topological THEOREM 3.16 S   is compact if and only if every count- space. able open cover of S has a finite subcover.

PROOF: If S is compact, then every open cover of S has a finite sub- cover. In particular, every countable open cover of S has a finite sub- cover. 140 Chapter 3 A Touch of Analysis

For the converse, assume that every countable open cover of S has a

finite subcover, and let O   A be an arbitrary open cover of S.

We show that O   A contains a finite subcover: For each xS , choose O  O such that xO . x    A x Since O is open, and since the rationals are dense in  , we x For S x  O , choose  x can find an open interval Ix with rational endpoints such that rational numbers r1 and xI x  O (see margin). Since there are but a countable r2 such that: x number of intervals with rational endpoints (Exercise 27, x –  r1 xr2 x +  Then: page 86), Ix xS is a countable open cover of S. By our xr  r  O 1 2 x assumption, a finite number of those intervals, I I  I x1 x2 xn cover S. It follows, since Ix  O , that O O  O i xi x1 x2 xn also covers S.

Heinrich Heine (1821- THEOREM 3.17 Every closed bounded subset of  is compact. 1881). Emile Borel (1871-1958) HEINE-BOREL

PROOF: Let S be a closed bounded subset of  . Employing The- orem 3.16 we establish compactness by showing that every  countable open cover On n = 1 of S has a finite subcover.  Assume there exists a countable open cover On n = 1 of S H = S  O O O c containing no finite subcover. To arrive at a contradiction, con- 3 1 2 3  sider the nested sequence Hi i = 1 of nonempty closed bounded sets, where (see margin): i c S O2  O 1 (*) H = SO(*) i   n the complement of O3 the intersection of two n = 1 closed sets an open set

Choose xH  i (see Theorem 3.15). Since each Hi  S , i 1 xS . Moreover, since x is contained in each Hi , x cannot be contained in any Oi [note the complement operator in (*)], con-  tradicting the assumption that On n = 1 is an open cover of S. The converse of Theorem 3.17 also holds:

THEOREM 3.18 Every compact subset of  is closed and bounded.

PROOF: Let K   be compact. We observed, in Example 3.6(a), that K must be bounded, and now establish that K is closed by show- ing that its complement is open: 3.3 Metric Space Structure of  141

c xk– | x – k Let xK . For each kK let  = ------(see margin). Since - k  = ------k |x .x k 4 4 K is compact, the open cover S k has a finite subcover, k kK k. K k say: S k1 S k2 S kn . Noting that every ele- k1 k2 kn

ment of K is contained in some S ki , and that ki S k S x = , we see that x  S x  Kc . k i k   i i ki i = 1 an open set Combining Theorems 3.17 and 3.18 we have:

THEOREM 3.19 A subset of  is compact if and only if it is both closed and bounded.

CHECK YOUR UNDERSTANDING 3.19 Determine if the given subset of  is compact. Justify your answer.   1 1 (a) --- (b) 0  --- (a) No (b) Yes n n n = 1 n = 1 In the real number system, one can also take a sequential approach to compactness, namely:

THEOREM 3.20 K   is compact if and only if every sequence in K contains a subsequence that converges to a point in K.

 ROOF P : Assume that K is compact and let kn n = 1 be a sequence

with each kn  K . Since K is bounded (Theorem 3.18) the sequence contains a convergent subsequence k  (Theorem 3.10, page ni i = 1 129). Can the limit of that subsequence lie outside of K? No: Let aK c. Since K is closed (Theorem 3.18), K c is an open set containing a which contains no element of the k  . It fol- ni i = 1 lows that a is not the limit of k  . ni i = 1 Conversely, assume that every sequence in K contains a subsequence that converges to a point in K. We show that K must be bounded and closed, and therefore compact (Theorem 3.18). 142 Chapter 3 A Touch of Analysis

Assume that K is not bounded (we will arrive at a contradic- tion).

Choose k1  K . Since K is not bounded, we can choose

k2  K such that k2 –1k1  , and k3  K such that

k3 –1k1  and k3 –1k2  , etc. Having chosen

k1k2 k3  kn with no two of the numbers within one unit

of each other, we can still find a number kn + 1 that is more than one unit from any of its predecessors (the

k1k2 k3  kn is bounded, and K is not). The constructed  sequence kn n = 1 has no convergent subsequence since, for

any a   , the open sphere S1a can contain at most one element of the sequence. Assume that K is not closed (we will arrive at a contradiction). Since K c is not open, we can choose aK c such that + Sa K  for all   0 . In particular, for any nZ we 1 can choose k  K such that k – a  --- . The sequence n n n  kn n = 1 cannot contain a subsequence converging to a point in K, as by construction it converges to a which lies outside of K.

CHECK YOUR UNDERSTANDING 3.20 Prove that K   is compact if and only if every Cauchy sequence Answer: See page A-17. in K converges to a point in K. 3.3 Metric Space Structure of  143

EXERCISES

Exercises 1-9. Determine if the given subset of  is: (a) Open, closed, or neither. (b) Bounded above, below, or neither. (c) Compact. 1.01  1 2  2.02   23 3. 02   23

+ 4. Z 1 1 5.--- nZ + 6. 0  --- nZ + n n 8. Q (the set of rational numbers) 9. Qc (the set of irrational numbers) 7. 02  3

10. Prove that for any two distinct real numbers, x and y, there exist open sets Ox and Oy such that xO x , yO y , and Ox Oy = . 11. Show, by means of an example, that the intersection of a nested sequence of nonempty bounded sets may turn out to be empty. 12. Show, by means of an example, that the intersection of a nested sequence of nonempty closed sets may be empty.

13. (a) Prove that every finite subset of  is compact. (b) Exhibit a countable subset of  that is not compact.

 14. Give an example of a nested sequence of nonempty open sets O1 O2 O3 for which:  

(a)  On =  . (b)  On is closed and nonempty. n = 1 n = 1

 1 1 15. Show that –1 + --- 1 – --- is an open cover of –11 n n n = 2

16. Construct a cover (necessarily not open) of the compact set 01 which does not contain a countable subcover.

17. Prove that if K1 K2 3  is a nested sequence of nonempty compact sets, then   Kn   . n = 1

18. Prove that if K is compact, then lub K and glb K exist and are contained in K. 144 Chapter 3 A Touch of Analysis

19. (a) Prove that if K is compact, and if C is closed, then KC is compact. (b) Show, by means of an example, that (a) need not hold if K is not compact. (b) Show, by means of an example, that (a) need not hold if C is not closed.

20. (a) Let B be a closed subset of  . Let x be such that Sx B  for every   0 . Prove that xB . (b) Show, by means of an example, that (a) need not hold if B is not closed.

21. (a) Let O be an open subset of  . Prove that if lim an =   O , then all but a finite num- n    ber of the terms in an n = 1 are contained in O. (b) Show, by means of an example, that (a) need not hold if O is not open.

22. Let S be a subset of  , bounded below. Let glb SO , with O open. Show that OS .

23. Let S be a subset of  , bounded above. Let lub SO , with O open. Show that OS .

 24. Prove that if lim an =  , then an   is compact. n   n = 1

25. Let K1 and K2 be compact subsets of  with K1 K2 = . Prove that there exist open sets O1 and O2 with K1  O1 , K2  O2 , and O1 O2 = .

26. Prove that  and  are the only subsets of  that are both open and closed.

PROVE OR GIVE A COUNTEREXAMPLE

27. Every open cover of any subset S of  contains a countable subcover.

28. If O is an open subset of  which contains every rational number, then O =  .

29. If O is an open subset of  which contains every integer, then O =  .

30. If B is a closed subset of  which contains every rational number, then B =  .

  31. If the sequence an n = 1 converges, then the set an n = 1 is compact.

  32. If the sequence an n = 1 converges to  , then the set   an n = 1 is compact.

  33. If the sequence an n = 1 diverges, then the set an n = 1 cannot be compact.

34. Let B be a closed subset of  . Prove that if the sequence an has a subsequence converging to  , then   B . 3.3 Metric Space Structure of  145

35. If AB is compact, than either A or B is compact. 36. A finite intersection of compact sets is again compact. 37. An arbitrary intersection of compact sets is again compact.

38. If AB is compact, than either A or B is compact.

39. If AB is compact, than both A are B are compact. 40. A finite union of compact sets is again compact. 41. An arbitrary union of compact sets is again compact.

42. Let B be a closed subset of  . Prove that if the sequence an has a subsequence converging to  , then   B .

43. If S is bounded below and if glb SK , with K compact, then OS .

44. If S1 and S2 are subsets of  for which S1  S2 is open, then S1 and S2 must both be open.

45. If S1 and S2 are subsets of  for which S1  S2 is open, then S1 or S2 must be open.

46. If S1 and S2 are subsets of  for which S1  S2 is open, then S1 and S2 must both be open.

47. If S1 and S2 are subsets of  for which S1  S2 is open, then S1 or S2 must be open.

48. If S1 and S2 are subsets of  for which S1  S2 is closed, then S1 and S2 must both be closed.

49. If S1 and S2 are subsets of  for which S1  S2 is closed, then S1 or S2 must be closed.

50. If S1 and S2 are subsets of  for which S1  S2 is closed, then S1 and S2 must both be closed.

51. If S1 and S2 are subsets of  for which S1  S2 is closed, then S1 or S2 must be closed.

52. If O1 and O2 are open subsets of  , then O1  O2 is also open.

53. If H1 and H2 are closed subsets of  , then H1  H2 is also closed.

n 54. For any nZ + , if O n is a collection of open subsets of  , then O is also open. i i = 1  i i = 1 146 Chapter 3 A Touch of Analysis

n 55. For any nZ + , if H n is a collection of closed subsets of  , then H is also closed. i i = 1  i i = 1

 56. If O  is a collection of open subsets of  , then O is also open. i i = 1  i i = 1

 57. If H  is a collection of closed subsets of  , then H is also closed. i i = 1  i i = 1

58. If O1 and O2 are open subsets of  , then O1  O2 is also open.

59. If H1 and H2 are closed subsets of  , then H1  H2 is also closed.

n 60. For any nZ + , if O n is a collection of open subsets of  , then O is also open. i i = 1  i i = 1

n 61. For any nZ + , if H n is a collection of closed subsets of  , then H is also closed. i i = 1  i i = 1

 62. If O  is a collection of open subsets of  , then O is also open. i i = 1  i i = 1

 63. If H  is a collection of closed subsets of  , then H is also closed. i i = 1  i i = 1 3.4 Continuity 147

3

§4. CONTINUITY Let’s begin with a formal definition:

The Greek letter  is pro- DEFINITION 3.12 A function f: D   is continuous at nounced “delta.” Note the CONTINUITY AT A cD if for any given   0 there exists a similarities between this POINT definition and Definition   0 such that: 3.5, page 123. xc–    fx– fc  with xD In a calculus course conti- nuity is typically defined What a beautiful definition! Here is what it is saying: in terms of the limit con- cept. Specifically: The distance between fx and fccan be made as small as you wish f is continuous at c if lim fx= gc xc xc–    fx– fc 

providing x is sufficiently close to c

From a geometrical point of view: y

 ) fc If a particular  “works” . for a given  , then any ( smaller  will also work ends up in here  for that  . However, a x in here  fx smaller  , may call for a (  ) smaller  . . x c Given   0   0  xS c  fx Sfc

EXAMPLE 3.7 Show that the function fx= 2x + 5 is con- tinuous at x = 3 .

SOLUTION: For a given   0 we are to find   0 such that: Compare with Example 3.3(a), page 124. x – 3    fx– f3   i.e: x – 3    2x + 5 – 11   same x – 3    2x – 6   x – 3    2 x – 3    x – 3    x – 3  --- 2 While a choice of  for which x – 3    2x + 5 – f3   may not be so apparent, it is trivial to find a  that works if the task’s   rewritten form x – 3    x – 3  --- , namely:  = --- . 2 2 148 Chapter 3 A Touch of Analysis

CHECK YOUR UNDERSTANDING 3.21 Answer: See page A-17. Prove that the function fx= 5x + 1 is continuous at x = 2 .

EXAMPLE 3.8 Show that the function fx= x2 is continuous at x = 3 .

SOLUTION: For a given   0 we are to find   0 such that: x – 3    x2 – 9   x – 3    x + 3 x – 3  x – 3    x + 3 x – 3  

The proof will be complete once we find a   0 for which: x – 3    x + 3 x – 3   (*)

While it is tempting to choose  = ------ - , that temptation must be x + 3 suppressed, for  has to be a positive number and not a function of x. Since we are interested in what happens near x = 3 , we decide to focus on the interval: 24 = xx–13  Within that interval x +73  (see margin). Consequently, within yx= + 3 that interval:

7 x + 3 x –73  x – 3 5  Taking  to be the smaller of the two numbers 1 and --- [written: 7    = min 1 --- ], we are assured that both x +73  and that   --- ,  2 3 4 7 7 and this enables us to meet our goal:  x – 3    x2 – 9 ==x + 3 x –73  --- 7

CHECK YOUR UNDERSTANDING 3.22 Answer: See page A-18. Show that the function fx= x2 is continuous at x = 2

DEFINITION 3.13 A is a function that is CONTINUOUS FUNCTION continuous at every point in its domain. 3.4 Continuity 149

EXAMPLE 3.9 Show that the function fx= x is continuous.

SOLUTION: Let x0  Df (which is to say: x0  0 ), and let   0 be given. We are to find   0 such that:

xx– 0    xx– 0  

i.e: xx+ 0 xx– 0    xx– 0   (*)

Assume that x0  0 . Since we are interested in what happens near x0 , x x x 3x we decide to focus on the interval x – ----0- x + ----0- = ----0- ------0 . 0 2 0 2 2 2 Within that interval (see margin):

x0 x0 3 x xx+  ----- + x  ------+ x = --- x ----0- + x 0 0 0 2 0 2 0 2 2

x 3 Letting  = min ----0- --- x  , we see that (*) is satisfied: 2 2 0 x0 3x0 ----- x0 ------2 2 3 xx+ 0 xx– 0    xx+ 0 xx– 0  --- x0 yxx= + 0 2 3 3 --- x  --- x  2 0 2 0  xx–  ------------= 0 3 xx+ 0 --- x 2 0

As for the x0 = 0 case:

CHECK YOUR UNDERSTANDING 3.23

(a) Show that fx= x is continuous at x = 0 .

Answer: See page A-18. (b) Show that fx= x2 is a continuous function. Roughly speaking, a function is continuous if you can sketch its graph without lifting your writing utensil. In particular, we now show that the yx= 2 + 3  2x if x  1 function fx=  (margin) is not continuous at x = 1 : 4  x2 +if3 x  1 3 2 . Let  = 1 and let  be ANY positive number whatsoever. Take 1 ( )1 +  1 any x  11 +  . Clearly x – 1   , but: y = 2x fx– f1 ===x2 + 3 – 21 x2 +11  The above argument shows that there does not exist a   0 such that x – 1    fx–1f1  . Since the continuity challenge fails for  = 1 (as it would for any 0    2 ), f is not continuous at 1, and is therefore not a continuous function. 150 Chapter 3 A Touch of Analysis

CHECK YOUR UNDERSTANDING 3.24

  x if x  1 Prove that the function fx=  999 is not continuous.  ------if x = 1 Answer: See page A-18.  1000

THE ALGEBRA OF FUNCTIONS Just as you can perform algebraic operations on numbers to arrive at other numbers, so then can one perform algebraic operations on func- tions to arrive at other functions, and in a most natural fashion: DEFINITION 3.14 For real-valued functions f and g with domains Df and Dg respectively, we define the func- f tions fg+ , fg– , fg , and --- as follows: g SUM OF fg+ : Df  Dg   with: FUNCTIONS fg+ x = fx+ gx DIFFERENCE OF fg– : D  D   with: FUNCTIONS f g fg– x = fx– gx PRODUCT OF fg: D  D   with: FUNCTIONS f g fg x = fxgx

QUOTIENT OF f ---: D  D – xgx 0 with: FUNCTIONS g f g f fx --- x = ------ g gx

Continuity at a point is preserved under the above operations:

Consequently: If f and g are THEOREM 3.21 If f and g are continuous at c then so are the continuous, then so are the functions: functions f (a) fg+ (b) fg– fg+ fg– fg, and --- g f (c) fg (d) --- (if gc 0 ) for g  0 g

PROOF: We establish (a) and (c). You are invited to establish (b) in CYU 3.25 below, and (d) in the exercises. 3.4 Continuity 151

(a) For a given   0 we are to find   0 such that: Throughout this devel- xc–    fg+ x – fg+ c   opment we are assuming that the variable x is con- We begin by manipulating fg+ x – fg+ c   into a form tained in the domain of both f and g. which displays the expressions fx– fc and gx– gc , as the continuity of f and g at c enables us to make those two expressions as small as we wish: xc–    fg+ x – fg+ c   xc–    fx+ gx– fc– gc  xc–    fx– fc+ gx– gc   (*) Since fx– fc+ gx– gc  fx– fc+ gx– gc, We now set our sights on finding a   0 for which xc–   implies   that both fx– fc --- and gx– gc --- : 2 2

Since f and g are continuous at c, there exist 1  0 such that  xc–    fx– fc --- and   0 such that 1 2 2  xc–    gx– gc --- . It follows that (*) holds for 2 2  = min1 2 : xc–    fx– fc+ gx– gc fx– fc+ gx– gc --- + --- = 2 2

(c) For a given   0 we are to find   0 such that xc–    fg x – fg c  

Note the similarity Again, we somehow need to work fx– fc and gx– gc into between this proof and the picture, and do so by introducing the clever zero, that of the sequence ver- sion of Theorem 3.9(c), – fxgc+ fxgc within the expression fg x – fg c : page 126. xc–    fg x – fg c   xc–    fxgx– fcgc  xc–    fxgx– fxgc+ fxgc– fcgc  xc–    fxgx– gc+ gcfx– fc   (*)

Noting that: fxgx– gc+ gcfx– fc  fxgx– gc+ gcfx– fc we set our sights on finding a   0 such that xc–   implies that:   (i) fxgx– gc --- and (ii) gcfx– fc --- . 2 2 152 Chapter 3 A Touch of Analysis

For (i): Since f is continuous at c, we can choose f such that:

xc–  f  fx–1fc

could choose any positive number Since fx– fc fx– fc (CYU 3.14, page 135): fx–1fc  fx– fc 1  fx 1 + fc (**)

Since g is continuous at c, we can choose g  0 such that:  xc–    gx– gc ------g 2fc+ 1 (***) can’t be zero

Letting  = minf  g , we see that (i) is satisfied:

(**)  xc–    fxgx–1gc + fc  ------= --- 2fc+ 1 2 (***)  For (ii): gcfx– fc --- . 2 We use gc+ 1 instead Choose   0 such that: of simply gc in the gc  denominator since  xc–    fx– fc ------might be zero. 2gc+ 1 (margin) Then:   xc–    gcfx– fc gc ------ --- 2gc+ 1 2 gc gc+ 1

End result: For  = min , (*) is satisfied.

CHECK YOUR UNDERSTANDING 3.25 Answer: See page A-19. Prove Theorem 3.19(b)

CONTINUITY, FROM A TOPOLOGICAL POINT OF VIEW The abstract setting for the concept of continuity resides in the field of Topology, which is introduced in the next chapter. To provide a bridge to that field we now focus our attention on real-valued functions with domain all of  . yfx=  fS DEFINITION 3.15 For f:  , and S  : MAGE OF S I S The image of S under f, denoted by fS , is the set fS= fxxS . yfx=  S PRE-IMAGE OF S The pre-image of S under f, denoted by –1 –1 –1 f S f S , is the set f S = xXfx  S . 3.4 Continuity 153

Using the above notation we can restate Definition 3.12 as follows:

A function f:  is continuous at c if for any given   0 there exists a   0 such that fSc  Sfc The following result will enable us, in the next chapter, to extend the concept of continuity to topological spaces:

THEOREM 3.22 A function f:  is continuous if and only if: f –1O is open  open set O

PROOF: Assume that f:  is continuous, and let O be open. We establish the fact that f–1O is open by showing that:

For any given cf –1O there exists –1   0 such that Sc  f O . (*) Since fc O and since O is open, there exists   0 such that S fc O . Since f is continuous, we can choose   0 such –1 O  f O that: fSc  S fc O (see margin). It follows that f fc    c  . . –1 Sc  f O . –1 For the sake of clarity, we take the We found a   0 for which Sc  f O . Replacing  liberty of representing subsets of  as two-dimensional objects. with the symbol “ ” (just for the sake of appearance), we see that (*) is established. Conversely, assume that for every open set O, f –1O is open. We show that f is continuous at c   by showing that for given   0 there exists a   0 such that fSc  Sfc : –1 Since Sfc is open, f Sfc is also open, and it clearly contains c . That being the case, we can choose   0 –1 such that Sc  f Sfc . It follows that:

fSc  Sfc (see Exercise 24). We take advantage of Theorem 3.22 to establish the following important result:.

THEOREM 3.23 If f:  and g:  are continuous, See Definition 2.9, page 64. then the composite function gf:  is also continuous. 154 Chapter 3 A Touch of Analysis

PROOF: Let O be open in  . Since g is continuous, g–1O is open. Since f is continuous, f –1g–1O is also open.The desired result now follows from Theorem 2.6, page 73 which asserts that –1 –1 –1 gf O = f g O .

CHECK YOUR UNDERSTANDING 3.26 Answer: See page A-19. Prove Theorem 3.23 using Definition 3.12 directly. 3.4 Continuity 155

EXERCISES

Exercises 1-8. Prove that the given function is continuous at the given point. 1. fx= –35x + at x = 4 1 2.fx= ---x + 3 at x = –1 2 3. fx= 3x2 + 1 at x = 2 4.fx= 2x2 – x at x = 1 1 3 5. fx= ------at x = 5 6.fx= ------at x = 0 2x 2x + 1 x 7. fx= ------at x = 0 8.fx= 32x – 2 at x = 8 2x + 1 Exercises 9-12. Prove that the given function is not continuous at the given point.

 2x +if3 x  0  x2 if x  1 9. fx=  at x = 0 10.fx=  at x = 1  –4x +ifx  0  x +if1 x  1

 2x +if3 x  0  1 11. fx=  at x = –1  ------if x  0  –3x +ifx  0 12.fx=  x + 1 at x = 1   x if x  0 Exercises 13-18. Prove that the given function is continuous. (Recall that a function is continuous if it is continuous at each point in its domain.) 13. fx= 3x + 5 1 2 14. fx= –1---x + 15. fx= x + 2x 4 2 1 16. fx= –4x – 17. fx= ---  2x +if3 x  0 x 18. fx=   –3x +ifx  0 Exercises 19-22. Display a function with domain  which fails to be continuous only at the num- bers in S. 19.S = –1 20. S = –101 21. SZ= + 22.SZ=

23. Prove Theorem 3.21(d). –1 24. Let f:  . Prove that if Af B , then fA B (see Definition 3.15). 25. Use the Principle of Mathematical Induction to prove that if f is continuous at c, then so is the function f n given by f nx = fxn .

+ 26. Use the Principle of Mathematical Induction to prove that, for all nZ , if the functions fn are continuous at c, then so is the function:   (a) f1 ++f2 +fn (b) f1 f2 fn n n – 1 27. Prove that every function, px= anx ++++an – 1x  a1xx0 , is continu- ous. 156 Chapter 3 A Touch of Analysis

28. Use the Principle of Mathematical Induction to prove that if fi:  is continuous for  1 in, then so is the function fnfn – 1  f2f1 . px 29. Prove that every , ------where px and qx are polynomials with qx qx 0 , is continuous. (Recall that a function is continuous if it is continuous at each point in its domain.) 30. Prove that every function f: Z   is continuous.

 1 31. Prove that every function f: ---  is continuous. n n = 1

 1 32. Give an example of a function f: 0  ---  that is not continuous. n n = 1

33. Let S   . Prove that a function f: S   is continuous at c if and only if lim xn = c , n   with each xn  S implies lim fxn = fc . n  

34. (a) Let H   be closed and let f: H   be continuous. Prove that if xn is a convergent sequence with each xn  H , then the sequence fxn must converge. (b) Show, by means of an example, that (a) need not hold if the set H is not closed.

35. Prove that H   is closed if and only if every convergent sequence xn , with each xn  H , converges to a point in H. 36. Let f:  and g:  be continuous. Prove that the set xfx= gx is closed. 37. Let f:  and g:  be continuous. Prove that if fx= gx for every rational number x, then fg= . 38. Display a function with domain  which fails to be continuous everywhere. Suggestion: Consider the dense sets of the rational numbers and of the irrational numbers.

PROVE OR GIVE A COUNTEREXAMPLE

39. For given functions f and g, if fg+ is continuous at c, then both f and g must be continuous at c. 40. For given functions f and g, if fg+ is continuous at c, then f or g must be continuous at c. 41. For given functions f and g, if fg+ and f are continuous at c, then g is continuous at c. 42. For given functions f and g, if fg is continuous at c, then both f and g must be continuous at c. 43. For given functions f and g, if fg is continuous at c, then f or g must be continuous at c. 44. For given functions f and g, if fg and f are continuous at c, and g must be continuous at c. 3.4 Continuity 157

f 45. For given functions f and g, if --- is continuous at c, then both f and g must be continuous at c. g f 46. For given functions f and g, if --- is continuous at c, then f or g must be continuous at c. g f 47. For given functions f and g, if --- and f are continuous at c, and g is continuous at c. g

48. For given functions f and g, if fg is continuous at c, then g must be continuous at c, and f at gc.

49. For given functions f and g, if fg and g are continuous at c, then f must be continuous at gc.

50. For given functions f and g, if fg is continuous at c, and f is continuous at gc, then g must be continuous at c. 51.f:  is continuous if and only if f –1H is closed for every closed set H. 158 Chapter 3 A Touch of Analysis 4.1 Metric Spaces 159

4 CHAPTER 4 A Touch of Topology While a number of refer- The distance concept of the previous chapter will lead us to the ences to Chapter 3 appear in this chapter, they are abstract setting of a metric space in Section 1. A further abstraction, only included to under- from metric spaces to topological spaces takes place in Section 2. The line the fact that an abstract metric space all-important notion of continuity blossoms to that of a continuous func- stems from the standard tion between topological spaces in Section 3. Cartesian products and  . quotient spaces of topological spaces are introduced in Section 4.

§1. METRIC SPACES As you know, the absolute value of a   , denoted by a , is given by:

 a if a  0 a =   –ifa a  0 You can, and should, interpret a as representing the distance (num- ber of units) between a and 0 on the number line. For example, both 5 and –5 are 5 units from the origin, and we have: 55= and –55 = When you subtract one number from another the result is either plus or minus the distance (number of units) between those numbers on the number line. For example, 72–5= while 27–5= – . In either case, the absolute value of the difference is 5, the distance between the two numbers: 72–55== and 27–5==–5 In general: ab– represents the distance (number of units) between a and b. Roughly speaking, a metric space is a set with an imposed notion of distance — one inspired by the following result:

This theorem appears in THEOREM 4.1 (i)xy   : xy–0 , with xy–0= if Chapter 3, page 135. It is reproduced here for the sake and only if xy= . of “chapter-independence.” The distance between two numbers is never negative, and is 0 only if the two numbers are one and the same. (ii)xy   : xy– = yx– The distance between x and y is the same as that from y to x. (iii)xyz  : xy–  xz– + zy– The triangle inequality.

PROOF: The first two properties are direct consequences of the defi- nition of the absolute value function. We establish (iii) by showing that xy– 2  xz– + zy– 2 (this will do the trick, since neither side of the inequality can be negative): 160 Chapter 4 A Touch of Topology

xz– + zy– 2 = xz– 2 ++2 xz– zy– zy– 2 2 2 a 2 = a2: = xz– ++2 xz– zy– zy–

aa :  xz– 2 ++2xz– zy– zy– 2 expanding: = x2 – 2xz ++z2 2xz –22xy –2z2 +zy + z2 – 2zy +y2 ===x2 – 2xy + y2 xy– 2 xy– 2

CHECK YOUR UNDERSTANDING 4.1 Answer: See page A-19. Prove that for any xy   , xy–  xy– As you can see, the three properties of Theorem 4.1 morph into the defining axioms of a metric space: DEFINITION 4.1 A metric on a set X is a function METRIC d: XX   which satisfies the following three properties: (i)xy  X : dxy  0 , and dxy = 0 if and only if xy= . (ii)xy  X : dxy = dyx (iii)xyz   X : dxy  dxz + dzy A pair Xd consisting of a set X and a metric d is said to be a metric space, and may simply be denoted by X. Theorem 4.1 tells us that  d , with dxy = xy– is a metric space (called the one-dimensional Euclidean space). In the exercises you 2 = xy xy are asked to verify that 2 d and 3 d , with and 2 2 dx1 y1  x2 y2 = x1 – x2 + y1 – y2 and 3 = xyz xyz  2 2 2 dx1y1 z1  x2y2 z2 = x1 – x2 ++y1 – y2 z1 – z2 are also metric spaces (called the two- and three-dimensional Euclidean spaces, respectively).

EXAMPLE 4.1 Verify that the function d , given by:

dx1 y1  x2 y2 = x1 – x2 + y1 – y2 is a metric on the set 2 = xy xy   .

SOLUTION: We show that d satisfies each of the three defining axi- oms of Definition 4.1. 2 (i) For every x1 y1  x2 y2 

dx1 y1  x2 y2 = x1 – x2 + y1 –0y2  since a  0 for every a   4.1 Metric Spaces 161

Moreover, dx1 y1  x2 y2 = 0 only if both x1 –0x2 = and y1 –0y2 = , and this can only happen if x1 = x2 and y1 = y2 ; which is to say, if x1 y1 = x2 y2 .

2 (ii) For every x1 y1  x2 y2  :

dx1 y1  x2 y2 = x1 – x2 + y1 – y2

==x2 – x1 + y2 – y1 dx2 y2  x1 y1

(iii) Let Ax= 1 y1 , Bx= 2 y2 , and Cx= 3 y3 . We show that dAC  dAB + dBC : Note how the definition of dAC = dx1 y1  x3 y3 d ; in used in both directions in this development. = x – x + y – y One direction: 1 3 1 3 dx1 y1  x3 y3 Theorem 4.1(iii): = x – x + y – y 1 3 1 3  x1 – x2 +++x2 – x3 y1 – y2 y2 – y3 Other direction: x – x + y – y 1 2 1 2 regroup: = x1 – x2 +++y1 – y2 x2 – x3 y2 – y3

= dx1 y1  x2 y2 = dx1 y1  x2 y2 + dx2 y2  x3 y3 Yes: DEFINITIONS RULE = dAB + dBC

CHECK YOUR UNDERSTANDING 4.2 (a) Let X be any set. Show that the function d: XX   given by:  1 if xy dxy =   0 if xy= is a metric on X. The metric d is said to be the discrete metric on X, Answer: See page A-19. and Xd is said to be a discrete space.

DEFINITION 4.2 Let Xd be a metric space. For x0  X and OPEN SPHERE r  0 , the open sphere of radius r about x0 Compare with the notation introduced at the bottom of is the set: page 123. Srx0 = xXdxx  0  r Open in metric spaces need not even be y round. Consider, for example the adjacent sphere 1 of radius 1, centered at the origin, in the metric space of Example 4.1, wherein: x 1 dx1 y1  x2 y2 = x1 – x2 + y1 – y2

so that: S100 = xy xy+  1 162 Chapter 4 A Touch of Topology

CHECK YOUR UNDERSTANDING 4.3

Determine the open spheres S15 and S51 in the discrete metric space Z+ , wherein:  1 if nm Answer: See page A-20. dnm =  (see CYU 4.2),  0 if nm=

DEFINITION 4.3 Let Xd be a metric space.

Open and Closed subsets of OPEN SUBSET OF A OX is open in X if for every x0  O the Euclidean space  were METRIC SPACE there exists r  0 such that S x  O . introduced in the previous r 0 chapter (Definition 3.9, and CLOSED SUBSET OF A Definition3.10). HX is closed in X if its complement METRIC SPACE Hc = xXxH  is open in X.

EXAMPLE 4.2 (a) Show that the interval 15 is open in the Euclidean space  . (b) Show that the square Dxy=  0 x 1  0 y 1 is closed in the Euclidean space 2 . (c) Show that every subset of a discrete space X is both open and closed. (See CYU 4.2.)   SOLUTION: (a) For any given a  15 we are to find   0 such } ( } that S a  15 . This is easy to do: take  to be the smaller of the ( .a )  1 5 two numbers a – 1 and 5 – a (see margin). Any smaller   0 will do just as well; but no larger  will work.

(b) We show Dc is open:

r x0 y0 If x0 y0  D , then either x0  01 or y0  01 . For definite- 1 . D ness, assume that x0  01 . Then, for r = minx0  x0 – 1 : S x  y D = (indeed S x  y 01  = ). 1 r 0 0 r 0 0 (c) Let XXd=  be discrete, which is to say:  1 if xy dxy =   0 if xy=

For any DX and any xD : S1x  D = x  D . It follows that every subset D of X is open. That being the case, every subset D must also be closed, and, for the simple reason that Dc is open. 4.1 Metric Spaces 163

CHECK YOUR UNDERSTANDING 4.4 (a) Show that the interval 1 5  is neither open nor closed in the Euclidean space  . (a) See page A-20. n (b) Let X be a discrete metric space consisting of n elements. Deter- (b) 2 mine the number of open subsets of X. The following theorem will lead us to the definition of a topological space in the next section. THEOREM 4.2 For any metric space X: (i)X and  are open subsets of X. (ii) Arbitrary unions of open sets are open. (iii) Finite intersections of open sets are open.

This is the identical proof PROOF: offered for Theorem 3.13, (i)X is open: For any xX , S1x  X . page 136.  is open (by default): For any a   there exists an   0 such

that Sa   , for the simple reason that no such a exists. (ii) Let O be a collection of open sets, and let xO .    A     A Since x is in the union of the Os , there must exist some   A such that xO . Since O is open, we can choose 0 0 0   0 such that S x  O . Then:  0 xS x O O  0     A n (iii) Let O n . be a collection of open sets, and let xO . i i = 1  i i = 1 For each i, choose  such that S x  O , and let i i i  = min12  n . Since Sx  Oi for each i, n

xS x   Oi . i = 1

CHECK YOUR UNDERSTANDING 4.5 Let X be a metric space. Prove that: (i) X and  are both closed in X. (ii) The arbitrary intersection of closed sets is closed. Answer: See page A-20. (iii)The finite union of closed sets is closed.

An order relation (ab ) need not reside in an abstract metric space. There is, however, a sense of finite containment: 164 Chapter 4 A Touch of Topology

DEFINITION 4.4 A subset S of a metric space Xd is BOUNDED SUBSETS OF bounded if there exists MZ + such that A METRIC SPACE dx1 x2  M for every x1 x2  S . Any such M is said to be a bound for S. In particular, if there exists MZ + such that dx1 x2  M for every x1 x2  X , then the space X is said to be bounded.

CHECK YOUR UNDERSTANDING 4.6

Prove or give a counterexample. In a metric space X (a) The union of two bounded subsets is bounded. (b) The arbitrary union of bounded subsets is bounded. Answer: See page A-20. (c) The arbitrary intersection of bounded subsets is bounded.

COMPACT SPACES

See the indexing remarks A collection O of open subsets of a metric space X is said to that follow CYU 2.3,   A page 58. be an open cover of a subset SX if SO   . The open cover is   A said to have a finite subcover if there exists 12 n  A n such that SO .  i i = 1

DEFINITION 4.5 A subset K of a metric space X is compact if every open cover of K has a finite subcover. Compare with Definition 3.11, COMPACT page 139 In particular, a metric space X is said to be compact if every open cover of X has a finite subcover.

Compare with Theorem 3.18, THEOREM 4.3 Compact subsets of metric spaces are closed page 140. and bounded.

PROOF: Let K be compact. Assume that K is not closed (we will arrive at a contradiction).

Choose x0  X such that for every   0 , Sx0  K   (if this were not the case, then the complement of K would be open, and K dxx would therefore be closed). For each xK , let  = ------0- , and x 2 consider the open cover S x of K. Being compact that x xK n open cover has a finite subcover S xi . xi i = 1 4.1 Metric Spaces 165

Let  = minx x x . Since Sx0 S xi = 1 2 n xni radius: radius: x for 1 in (see margin), and since KS x : 0  x i . i

 i = 1 x

i Sx0 K = — a contradiction. K x.i = ------d  x

i Assume that K is not bounded (we will arrive at a contradiction). 2  x 0 + Choose x0  K . Since K is not bounded, for any nZ we can choose xn  K such that dxn x0  n . Since every ele- ment of K is a finite distance from x , S x is an 0 n 0 nZ + open cover of K, which has no finite subcover — contradict- ing the assumption that K is compact.

CHECK YOUR UNDERSTANDING 4.7

Answer: See page A-21. Give an example of a closed bounded subset of a metric space X that is not compact. (Note Theorem 3.17, page 140.)

CONTINUITY We begin by introducing some useful set-theory concepts:

This is a generalization of DEFINITION 4.6 For f: XY , and SX we define the image Definition 3.15, page 152. IMAGE OF S of S under f, denoted by fS , to be the set fS= fxxS (see margin).

PRE-IMAGE OF S For f: XY , and SY we define the pre- image of S under f, denoted by f –1S , to be the set f –1S = xXfx  S (see margin).

Your turn: CHECK YOUR UNDERSTANDING 4.8 (a) Let f: XY . Prove that (i) fA B = fA fB for any subsets A, B of X.

(ii) f –1AB = f –1A  f –1B for any subsets A, B of Y.

(iii) f –1AB = f –1A  f –1B for any subsets A, B of Y. (iv) f –1Ac = f –1A c for any AY . (b) Show that if f: XY is a bijection, then fA= fAc c for Answer: See page A-21. every AX . Since the distance notion resides in metric spaces, the dreaded “ - method” could be used to define the concept of continuity in the current metric space environment. We take a different approach: 166 Chapter 4 A Touch of Topology

Motivated by Theorem DEFINITION 4.7 Let X and Y be metric spaces. A function –1 3.22, page 153. CONTINUOUS f: XY is continuous if f O is open in FUNCTION X for every set O open in Y.

EXAMPLE 4.3 Let d be the standard Euclidean metric on 2 = xy xy   : 2 2 dx1 y1  x2 y2 = x1 – x2 + y1 – y2 and let d be the metric of Example 4.1:

dx1 y1  x2 y2 = x1 – x2 + y1 – y2 Show that the : I: 2 d  2 d given by Ixy = xy is continuous. –1 SOLUTION: Let O be open in 2 d . Since I O = O , the conti- nuity of I will be established once we show that the set O, open in 2 d , is also open in the Euclidean space 2 d . Let’s do it:

For given x0 y0  O choose   0 be such that

xy xx– 0 + yy– 0    O 2 xy open sphere of radius  centered at x0 y0 in   d xy – x0 y0 . yy Since xy – x0 y0  xx– 0 + yy– 0 (see margin), the

0 open sphere x  y xy xy – x  y   0 0 . 0 0 xx– 0 2 2 dxy  x0 y0 = xx– 0 + yy– 0 in the Euclidean space 2 d is contained in O. It follows that O is open in 2 d .

CHECK YOUR UNDERSTANDING 4.9 Referring to the notation of the above example, show that the identity function that “goes the other way:” I: 2 d  2 d is also Answer: See page A-21. continuous. If you worked your way through the previous chapter you discovered that if f:  and g:  are continuous, then the composite function gf:  is also continuous. That result extends to arbi- trary metric spaces:

THEOREM 4.4 Let X, Y, and Z be metric spaces. If f: XY and g: YZ are continuous, then the composite function gf: XZ is also continuous. 4.1 Metric Spaces 167

PROOF: Let O be open in Z. Since g is continuous, g–1O is open in This is the proof of Theorem Y. Since f is continuous, f –1g–1O is open in X. The desired result 3.23, page 153. now follows from Theorem 2.6, page 73 which asserts that –1 –1 –1 gf O = f g O . CHECK YOUR UNDERSTANDING 4.10 Let X, Y, and Z be metric spaces. For given functions f: XY and g: YZ , if gf: XZ is continuous must both f and g be contin- Answer: See page A-21. uous? Justify your answer.

ISOMETRIES In any abstract setting, one attempts to define a notion of “sameness” for its denizens. Bringing us to: DEFINITION 4.8 A metric space Xd is isometric to a space ISOMETRIC SPACES If two spaces are isometric, Yd if there exists a bijection f: XY then they are the “same, up to appearances” (the naming of such that dx1 x2 = dfx1  fx2 for elements, for example). every x1 x2  X . Such a function f is said to be an isometry from X to Y.

EXAMPLE 4.4 Show that the metric space Zd with dab = ab– is isometric to the space 2Zd , where 2Z = 2nn Z and 1 d2a 2b = --- 2a – 2b . 2 SOLUTION: The function f: Z 2Z given by fn= 2n is easily seen to be a bijection: f is one-to-one: fa===fb 2a 2ba b . f is onto: For any given 2a  2Z , we have fa= 2a . f also preserves the distance between any two points ab  Z : 1 dab ==ab– ---2a – 2b 2 1 ===--- 2a – 2b d2a 2b dfa– fb 2

CHECK YOUR UNDERSTANDING 4.11 Show that isometry is an equivalence relation on any set of metric Answer: See page A-22. spaces. (See Definition 2.20, page 88.) 168 Chapter 4 A Touch of Topology

EXERCISES

Exercise 1-5. Verify that each of the following is a metric space.

1.2 d , where 2 = xy xy and:

2 2 dx1 y1  x2 y2 = x1 – x2 + y1 – y2

2.3 d , where 3 = xyz xyz  and:

2 2 2 dx1y1 z1  x2y2 z2 = x1 – x2 ++y1 – y2 z1 – z2

n n 3.  d , where  = x1x2  xn xi   1 in and:

2 2  2 dx1x2  xn  y1y2  yn = x1 – y1 +++x2 – y2 xn – yn

4.2 d , where 2 = xy xy and:

dx1 y1  x2 y2 = maxx1 – x2  y1 – y2

2 Sketch the unit sphere S100 ==xy dxy  00 1 in   d .

5. (Calculus Dependent). Cd , where Cf= : 0 1  f is continuous and: 1 dfg =  fx– gxdx 0

Exercise 6-11. Explain why the given function is NOT a metric on  .

6. dxy = xy+ 7.dxy = x2 – y2 8. dxy = x2 + y2

9.dxy = xy 10.dxy = xy– 11. dxy = xy–

12. Let xy be any two distinct elements of a metric space X. Show that there are disjoint open

sets Ox Oy with xO x and yO y . # 13. Prove that if a metric space contains at least two points, then it must contain at least four dis- tinct open sets.

14. Let S be an unbounded subset of a metric space Xd . Prove that for any x0  S and any + given nZ there exists xn  S such that dxx 0  n .

15. Let X be a discrete space. Prove that every function f: XY is continuous for any metric space Y. 4.1 Metric Spaces 169

16. Let X, and Y be metric spaces. Prove that for any y  Y the f : XY 0 y0 given by f x = y for every xX is continuous. y0 0 17. Prove that the continuous image of a compact metric space is compact.

18. Let Xd X and Yd Y be two metric spaces. Prove that XY  d , where

XY = xy xX and yY and dx1 y1  x2 y2 = dXx1 x2 + dYy1 y2 is a metric space.

19. Prove that a space X is compact if and only for any collection H   A of closed subsets of n X with H =  one can choose a finite subcollection H n such that H =    i i = 1  i   A i = 1 20. Let H be a compact subset of a metric space X. Show that for any given xH there exist dis-

joint open sets Ox OH with xO x and HO H .

21. Let H1 and H2 be two disjoint compact subsets of a space X. Show that there exist two dis-

joint open sets O1 O2 with H1  O1 and H2  O2 . 22. Let X and Y be isometric spaces.Prove that X is bounded if and only if Y is bounded. 23. Let X and Y be isometric spaces.Prove that X is compact if and only if Y is compact.

# 24. (Set theory). Letf: XY . Prove: n n (a) For S  X , 1 in : fS= fS . (b) For SX : ff–1S  S . i  i  i i = 1 i = 1

# 25. (Set theory). Let f: XY . Prove that forS  X ,   A :

–1 –1 –1 –1 (a) f S = f S (b) f S = f S .       A   A   A   A

Exercise 26-32. () For A a subset of a metric space X, let CA = HX AH and H is closed in X . The set AH=  is called the closure of A. HC A 26. Determine A , for the given subset A of the Euclidean space  . (a) A = 13 (b) AZ= + (c) AQ= (the set of rational numbers) 1 (d) A = 13  5 (e) A = --- nZ + n 170 Chapter 4 A Touch of Topology

27. Prove that for AX , A is the smallest closed subset of X that contains A, in that A is a subset of every closed set containing A.

28. Prove that A is closed for every AX .

29. Prove that if A is closed if and only if AA= .

30. Prove that xA if and only if OA   for every open set O containing x.

31. Let x1 x2 be any two distinct elements of a metric space X. Show that there exists r  

such that Srx1 Srx2 = .

32. Let x be an element of an open subset O of a metric space X. Show that there exists r  

such that Srx  O .

PROVE OR GIVE A COUNTEREXAMPLE

33. d , where dxy = rx– y is a metric space for any r   .

2 2 34.  d , where  = xy xy and dx1 y1  x2 y2 = x1 – x2 – y1 – y2 is a metric space.

35. A function f from a metric space X to a metric space Y is continuous if and only if f –1H is closed in X for every closed subsets H of Y.

36. Let f: XY be a bijection from the metric space X to the metric space Y. If f is continuous, then f is an isometry.

37. Let f: XY be a bijection from the metric space X to the metric space Y. If f and f –1 are continuous, then f is an isometry.

+ 38. (Closure: See Exercises 26-32) If Si is a subset of a metric space X, for iZ , then:

c c c (a) S1  S2 = S1  S2 (b) S1  S2 = S1  S2 (c)S1  S2 = S1  S2

  c c (d) S1  S2 = S1  S2 (e)  Si =  Si (f)  Si =  Si i = 1 i = 1 i = 1 i = 1 c c   (g) S = S  i  i i = 1 i = 1 4.2 Topological Spaces 171

4

§2 TOPOLOGICAL SPACES Theorem 4.2, page 163, inspires the definition of a topological space: DEFINITION 4.9 A topology on a set X is a collection  of sub- TOPOLOGY sets of X, called open subsets of X, satisfying The Greek letter  , spelled “tau” — rhymes the following properties: with “cow.” (i) X and  are in  . (ii) Any union of elements of  is again in  . ( is closed under arbitrary unions) (iii) Any finite intersection of elements of  is again in  . ( is closed under finite intersections) TOPOLOGICAL SPACE A pair X  consisting of a set X and a topol- ogy  is called a topological space, and may at times simply be denoted by X.

CHECK YOUR UNDERSTANDING 4.12

For any set X show that:

(a)0 = X  is a topology on X [it is called the indiscrete topology on X].

(b)1 = SS X is a topology on X [it is called the discrete Answer: See page A-22. topology on X].

EXAMPLE 4.5 For any set X, let c FINITE-COMPLEMENT  =   SXS is finite SPACE Show that X  is a topological space.

SOLUTION: We verify that the three axioms of Definition 4.9 hold for the family  . (i) We are given that    . In addition, since X c =  is a finite subset of X, X   .

(ii) Consider the set O   A with each O   . Since c c O = O (Theorem 2.3, page 58), the number of ele-     A   A 172 Chapter 4 A Touch of Topology

c ments in O cannot exceed the number of elements in any of    A c the O , all of which are finite. It follows that  O   .   A

n (iii) Consider the finite set Oi i = 1 with each Oi   . Since c n n O = Oc (Theorem 2.3, page 58), the number of ele-  i  i i = 1 i = 1 c n ments in O cannot exceed the sum of the finite number of  i i = 1 elements in the sets Oi 1 in ; and is therefore finite. It follows n that  Oi   . i = 1

CHECK YOUR UNDERSTANDING 4.13 Let Xab=  and  = Xa . Show that X  is a topo- Answer: See page A-22. logical space (called the Sierpinski space).

Every metric space gives rise to a topological space. Specifically: In the Euclidean space  , THEOREM 4.5 For any metric space X:  = O   O is open in  is called the standard or  = OXO is open in the metric space X Euclidean topology on  ). is a topology on X.

PROOF: Theorem 4.2, page 163.

DEFINITION 4.10 A topological space X  is said to be metriz- able if there exists a metric d on X such that = OXO is open in the metric space X  Not every topological space is metrizable:

CHECK YOUR UNDERSTANDING 4.14 Show that the Sierpinski space [CYU 4.13] is not metrizable. Answer: See page A-22. Suggestion: consider Exercise 13, page 168. 4.2 Topological Spaces 173

The complement of open sets in a metric space were called closed sets. Following suit: DEFINITION 4.11 Let X  be a topological space. HX CLOSED SUBSET OF A is closed if its complement Hc is open TOPOLOGICAL SPACE (that is: Hc   ). As is the case in any metric space (see CYU 4.5, page 163): THEOREM 4.6 In any topological space X: (i) X and  are closed in X (they are also open). (ii) Any intersection of closed sets is again a closed set. (iii) Any finite union of closed sets is again a closed set.

PROOF: (i) Since Xc =  and c = X are open (Definition 4.9), both X and  are closed.

(ii) Let H be a collection of closed sets. We show H is    A     A c c closed by showing that its complement H = H is     A   A open: c Since each H is closed, H   A is a collection of c open sets. Since unions of open sets are open,  H is   A open. n (iii) Let Hi i = 1 be a collection of closed sets. Since c n n H = H c , and since finite intersections of open sets are  i  i i = 1 i = 1 c n n open, H is open; which is to say: H is closed.  i  i  i = 1 i = 1

CHECK YOUR UNDERSTANDING 4.15 Characterize the closed subsets of the given topological space.

(a) A discrete space X 1 . (See CYU 4.12.)

(b) An indiscrete space X 0 . (See CYU 4.12) Answer: See page A-22. (c) The Sierpinski space of CYU 4.13. 174 Chapter 4 A Touch of Topology

SUBSPACES OF A TOPOLOGICAL SPACE

DEFINITION 4.12 For X  a topological space and SX let SUBSPACE S = OSO   The S S is said to be a subspace of the space X  and the ele-

ments of S are said to be open in S. As might be expected: THEOREM 4.7 Let X  be a topological space, SX . The subspace S S is also a topological space.

PROOF: We verify that the three axioms of Definition 4.9 hold for the family S . (i) Since  and X are open in X  , both  S = and XS = S are open in S S .

(ii) Let U   A be a family of open sets in S S . We show that  U is again in S :   A

For each   A choose O   such that O  S = U . Since  is a topology,  O   . The desired result now follows   A from:  U ==O  S O  S      A   A   A Exercise 82(a), page 61 As for (iii): CHECK YOUR UNDERSTANDING 4.16 Let X  be a topological space and SX . Show that Answer: See page A-26. S = OSO   is closed under finite intersections. 4.2 Topological Spaces 175

BASES AND SUBBASES When analyzing a topological space it is often sufficient to consider the following subsets of its topology: DEFINITION 4.13 A base for a space X  is a subset  of BASE  which satisfies the following property: x O   B    xBO  (In other words, every open set is a union of elements from  ) SUBBASE A subbase for X is a subset  of  (gamma) is the Greek   letter for C. such that the set of finite intersections of elements of  is a base for the topology.

EXAMPLE 4.6 (a) Show that  = Srx xXr   0 is a basis for any metric space X. (b)Show that = –  b b    a  a   is a subbase for the Euclidean space  OX is open in X if OLUTION for every x  O there S : (a) A direct consequence of Definition 4.3, page 162 (see 0 margin). exists r  0 such that Srx0  O . (b) Since every open interval in  is the intersection of two elements in = –  b b    a  a   , and since Srx is the open interval xr–  xr+ ,  is a subbase for the Euclidean topol- ogy on  .

CHECK YOUR UNDERSTANDING 4.17

+ Let X be a metric space. Show that  = Srx xXrQ   , where + Answer: See page A-23. Q denotes the set of positive rational numbers, is a base for the topology of X. In the exercises you are invited to show that for any collection

SS=    A of subsets of X, the set:  n  = S S  Sn  Z+  i i  i = 1 is a subbase for a topology S on X, called the topology generated by S. In other words:

176 Chapter 4 A Touch of Topology

If you start with any collection S of subsets of X, and then take all unions of the finite intersections of the sets in S, you will end up with a topology S on X.

In particular, consider the set Snm=  nm Znm in  . Since r = r – 1 r  rr + 1 , every subset of  , being the union

of the open sets r r   , is open in S . In other words:

S is the discrete topology on  [see CYU 4.12(b)]

CHECK YOUR UNDERSTANDING 4.18

Let  denote the standard Euclidean topology on  , and let S denote the topology on  generated by Sxy=    xy . Show

that  S . Answer: See page A-23. (H =  S is called the half-open-interval space)

COMPACT SPACES Reminiscence of Definition 4.5, page 164: DEFINITION 4.14 A subset K of a topological space X is com- COMPACT pact if every open cover of K has a finite subcover. In particular, the space X is said to be com- pact if every open cover of X has a finite sub- cover. Theorem 4.3, page 164, asserts that compact subsets of metric spaces are closed and bounded. This result cannot carry over to general topo- logical spaces, for the notion of “bounded” involves the concept of dis- tance which does not reside in the general setting of topological spaces. Moreover, as it turns out, compact subsets of a general topological space need not even be closed:

EXAMPLE 4.7 Let  denote the finite complement topology on the set Z of integers of Example 4.5:  =   SZS c is finite Exhibit a compact subset of Z  that is not closed. SOLUTION: As it turns out, every single subset HZ of Z  is, in fact, compact: 4.2 Topological Spaces 177

Let O   A be an open cover of H. Since the empty set is certainly compact, we can assume that H   and choose an arbitrary element h0  H . Some element of the open cover, say O contains h . By the very definition of  , only 0 0 finitely many elements of Z lie outside of O . A fortiori, 0 there are only a finite number of elements of H that are not contained in O , say h n . For each 1 in choose an 0 i i = 1 element O of the given cover which contains h . Clearly i i n Oi i = 0 covers H. We next show that the compact subset Z+ of Z is not closed by show- ing that its complement is not open: Let O be any open set containing 0  Z+ c . Since Oc is finite, OZ +   . Thus Z+ c is not open.

CHECK YOUR UNDERSTANDING 4.19 Answer: See page A-23. Prove that every closed subset of a is compact. In the exercises, you are asked to show that a space X is compact if every open cover of X by sets taken from a base  of X has a finite sub- cover. We now state, without proof, a stronger result:

James W. Alexander THEOREM 4.8 Let X be a topological space and let  be a (1888-1971). ALEXANDER’S subbase for the topology of X. If every open SUBBASE cover of X by sets in  has a finite subcover, THEOREM then X is compact.

The above result can be shown to be equivalent to the Axiom of Choice (see page 109).

Felix Hausdorff HAUSDORFF SPACES (1868-1942). In the previous section, analytical properties of the real number system directed us to the definition of a metric space. The concept of a metric space was then generalized further to that of a topological space. This generalization may, in fact, be a bit too general in that there are some rather uninteresting topological spaces hanging around. For example, a set with indiscrete topology has very little “topological personality.” What distinguishes a topological space from a plain old set is its collec- tion of open sets, and while one does not like to degrade any structure, it is nonetheless difficult to think of an indiscrete space X  , with  =  X , as having evolved far from its underlying set X. 178 Chapter 4 A Touch of Topology

The following important class of spaces contain sufficiently many open sets to “separate points:” Distinct points in a Haus- DEFINITION 4.15 X  is a if for any two dorff space can be sepa- HAUSDORFF SPACE rated by disjoint open distinct points xy in X there exists disjoint sets. Additional separa- O O x .x . y y tion properties are intro- open sets Ox Oy with xO x and yO y . duced in the exercises. An open set in a space X that contains a point xX is said to be an (open) neighborhood of x. With this terminology at hand, we can say that a space is Hausdorff if any two points in the space reside in disjoint neighborhoods.

CHECK YOUR UNDERSTANDING 4.20 Answer: See page A-23. Prove that every is Hausdorff. We previously observed that compact subsets of a general topological space need not be closed. However: This result along with Any compact subset of a Hausdorff space is CYU 4.19 tell us that: THEOREM 4.9 A subset of a compact closed. Hausdorff space is compact if and only if PROOF: Suppose K is a compact subset of a Hausdorff space X it is closed.  that is not closed (we will arrive at a contradiction): Since K is not closed, its complement is not open, and we can choose x0  X such that Ox0  K   for every open neigh- .x0 borhood Ox of x (if this were not the case, then the com- Ox 0 0 x. Ox0 plement of K would be open). For each xK , choose disjoint open sets Ox and Oxx0 containing x and x0 , respectively K (margin), and consider the open cover Ox xK of K. Being compact that open cover has a finite subcover O n . It fol- xi i = 1 n lows that the neighborhood O x of x contains no ele-  xi 0 0 i = 1 ment of K — a contradiction.

CHECK YOUR UNDERSTANDING 4.21 Let K be a compact set in a Hausdorff space X. Show that for any given x  K there exists disjoint open sets O and O containing 0 x0 K Answer: See page A-24. x0 and K, respectively. 4.2 Topological Spaces 179

. EXERCISES

1. Let X=  abcd . Determine whether or not the given collection of subsets of X is a topol- ogy on X. (a) X (b) Xab (c) Xa ab ac (d)a b c d (e) Xa (f) Xa b (g)X abc abd abc (h) Xa ab acd 2. Show that the topology of Example 4.6 is a proper subset of the standard Euclidean topology on  .

3. Prove that if  is a base for a discrete space X, then x   for every xX .

4. Prove that if  is a base for a topological space X  , and if    , then  is also a base for  .

5. Let S S be a subspace of a topological space X  . Prove that DS is closed in S S if and only if there exists a closed subset H of X such that HS = D .

6. Exhibit three 12 3 on the set X=  abc with 1 2 3 .

7. (a) Let 1 2 be topologies on a set X. Prove that 1  2 is also a topology on X. n n (b) Let i i = 1 be a collection of topologies on a set X. Prove that  i is also a topology on X. i = 1 (b) Let    A be a collection of topologies on a set X. Prove that   is also a topol- ogy on X.   A

# 8. Let S be open in the topological space X. Prove that OS is open in the subspace S if and only if O is open in X.

+ + 9. Show that  =   O + where O = iZ in is a topology on Z . n nZ n

10. Let X be an . Prove that the collection  = OXO c is countable is a topology on X. # 11. Prove, without appealing to Theorem 4.8, that a space X is compact if every open cover of X by sets taken from a base  of X has a finite subcover.

12. Prove that = –  b bQ  a  aQ is a subbase for  .

13. Prove that a subbase of a topological space is a base for the space if an only if it is closed under finite intersections. 180 Chapter 4 A Touch of Topology

# 14. Show that for any collection SS=    A of subsets of a set X, the set:  n  = S S  Sn  Z+  i i i = 1 is a subbase for a topology on X. 

15. (Rational-Real Topology) Let  denote the standard Euclidean topology on  and Q the set of rational numbers. Let  denote the topology on  generated by   Q . Prove that: (a)  . (b) = ab ab  ab  Q ab is a base for  . (b) = a  a   –  a a   Q is a subbase for  . 16. (Half-Open Interval Topology) (a) Show that Hab=    ab is closed under finite intersections, and is therefore a base for a topology  on  .

(b) Show that    , where  denotes the Euclidean topology on  . 17. (Tangent Disc Topology) Consider the upper closed plane 2 Txy=   R y  0 . For any pxy=  with y  0 .p

let Sp= Srp ry , and let Sq for any qx=  0 consist of those sets of the form q  D where D is an . open disk in T tangent to q (see adjacent figure). q

(a) Show that SS= xy xy  T is closed under finite intersections, and is therefore a base for a topology  on T. [T  is called the Moor plane.]

(b) Is the point q open in  ? Justify your answer.

(c) Is the topology  the same as the T , where  denotes the standard Euclidean topology on 2 ? Justify your answer.

18. Let f: XY be a function from a non-empty set X to a topological space Y  . Show that f –1O O   is a topology on X. (See Definition 3.15, page 152.) 4.2 Topological Spaces 181

19. (a) Let X1 1 , X2 2 be two topological spaces. Prove that X1  X2  , where

 = O1  O2 O1  1 and O2  2 is also a topological space.

(b) If 1 and 2 are base for X1 1 and X2 2 , is 1  2 a base for X1  X2  ? Jus- tify your answer. (c) State and establish a generalization of part (a) to accommodate a collection of n topological n spaces Xi i i = 1 . 20. Let S be a subset of a topological space X. Prove that the subspace S is compact if and only if every cover of S by sets open in X contains a finite subcover. 21. (a) Show that every infinite subset S of a compact space X contains a point whose every neighborhood contains infinitely many elements of S. (b) Show, by means of an example, that (a) need not hold if X is not compact. (c) Show that the converse of (a) does not hold.

22. Prove that a topological space X is compact if and only if for any given collection H   A of closed sets such that H =  , there exists a finite subcollection H n with   i i = 1 n   A H =  .  i i = 1

 Exercise 23-24. (Sequences) A sequence xi i = 1 in a topological space X is said to be a conver- gent sequence which converges to x0  X if for any neighborhood O of x0 there exists N such that nN  xn  O . In the event that the sequence converges to x0 we write lim xn = x0 , n   and say that x0 is a limit point of the sequence. 23. (a) Give an example of a convergent sequence in a topological space which has more than one limit point. (b) Prove that in a Hausdorff space a convergent sequence has a unique limit point.

 24. (a) Let H be a closed subset of a topological space. Prove that if a sequence xi i = 1 with each xi  H converges to x0 , then x0  H . (b) Prove that (a) need not hold if H is not closed in X. Suggestion: Think indiscreetly. (c) Show, by means of an example, that the condition in (a) can hold without H being closed. 182 Chapter 4 A Touch of Topology

Exercise 25-33. (Closure) For S a subset of a topological space X, let

CS = HX SH and H is closed in X . The set SH=  is called the closure of A. HC S 25. Prove that xS if and only if OA   for every open set O containing x.

+ 26. For the topology =  Z  nn+ 1 n + 2  + on Z :  nZ (a) List the closed subsets of Z+ . (b) Determine the closure of the sets 51113 and 3579 .

27. For the topology  = Xa ab acd abe abcd on X=  abcde : (a) List the closed subsets of X. (b) Determine the closure of the sets a , b , c , d , e , ce , and be .

28. Prove that S is closed for every SX .

29. Prove that SS= for every SX .

30. Prove that if SH with H closed, then SH .

31. Prove that if S is closed if and only if SS= .

n 32. (a) Prove that for any finite collection Si i = 1 of subsets of a topological space X: n n

 Si =  Si i = 1 i = 1 (b) Prove that for any collection S , S  S .    A      A   A (c) Show, by means of an example, that (a) need not hold if the collection is not finite.

(d) Prove that for any collection S , S  S .    A       A   A n (e) Show, by means of an example, that even for a finite collection Si i = 1 of subsets, equality in (d) need not hold.

33. Determine A , for the given subset A of the topological space  . (a) A = 13 (b) AZ= + (c) AQ= (the set of rational numbers)

1 (d) A = 13  5 (e) A = --- nZ + n 4.2 Topological Spaces 183

Exercise 34-37. (Dense) A subset A of a topological space X is said to be dense in X if AX= (see Exercise 24-32). 34. Prove that AX is dense if and given any xX and any neighborhood O of x, OA .

35. Prove that the set Q of rational numbers is dense in  .

36. For the topology  = Xa ab acd abe abcd on X=  abcde , determine the dense subsets of X.

+ 37. For the topology =  Z  nn+ 1 n + 2  + on Z , determine the dense  nZ subset of Z+ . Exercise 38-43. (Separation Properties)

38. A T0-space is a topological space in which for any two distinct points there is an open set containing one of the points and not the other.

(a) Show that  is a T0-space .

(b) Give an example of a topological space that is not a T0-space .

39. A T1-space is a topological space in which for any two distinct points there is a neighbor- hood of each point not containing the other point.

(a) Show that  is a T1-space .

(b) Give an example of a T0-space that is not a T1-space (see Exercise 38).

40.T2-space or Hausdorff space (see Definition 4.14).

(a) Show that  is a T2-space .

(b) Give an example of a T1-space that is not a T2-space (see Exercise 39).

41. A T3-space or regular space is a T1-space (see Exercise 40) in which for any closed set H and any point xH there exist disjoint open sets UV with xU and HV

(a) Show that  is a T3-space .

(b) Give an example of a T2-space that is not a T3-space (see Exercise 40).

42. A T4-space or is a T1-space (see Exercise 40) in which for any two disjoint

closed sets H1 H2 there exist disjoint open sets O1 O2 with H1  O1 and H2  O2 .

(a) Show that  is a T4-space .

(b) Give an example of a T4-space that is not a T3-space (see Exercise 41). 43. Prove that any compact Hausdorff space is normal (see Exercise 42). 184 Chapter 4 A Touch of Topology

Exercise 44-49. (Connected Spaces) A topological space is connected if it is not the union of two nonempty disjoint open sets. 44. Prove that a topological space X is connected if and only if the only subsets of X that are both open and closed are  and X. 45. Prove that any discrete space consisting of 2 or more elements is not connected.

46. Show that the space X  , where X=  abcd and  = Xab is con- nected.

47. Show that  with the standard Euclidean topology is connected. (You will need to invoke the completion axiom of Definition 3.1, page 111.)

48. Show that the subspace Q (rational numbers) of the Euclidean space  is not connected. 49. Prove that if A is a connected subspace of a space X, then the closure of A in X is also con- nected. (See Exercises 25-33).

PROVE OR GIVE A COUNTEREXAMPLE

50. If 1 2 are topologies on a set X, then so is 1  2 a topology on X.

51. If  is a subbase for a topological space X  , and if  , then  is also a subbase for  .

52. If  is a base for a topological space X  , and if    , then  is also a base for  . 53. If X is an indiscrete space (CYU 4.11), then any base for X must contain two elements. 54. If X is an indiscrete space (CYU 4.11), then any base for X cannot contain more than two ele- ments.

55. (See Exercises 26-34) For subsets S and T of a topological space X:

(a) ST c = Sc  Tc (b) ST c = S c  T c (c) Sc c = S (d) If ST , then ST (e) If ST= , then ST= 56. (See Exercises 44-49) If A and B are connected subspaces of a space X, then: (a) AB is also connected. (b) AB is also connected. (c) If AB   , then AB is also connected. 57. The set of open subsets of a topological space and that of the closed subsets of the space are of the same cardinality (see Definition 2.15, page 77). 4.3 Continuous Functions and Homeomorphisms 185

4

§3 CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS We extend Definition 4.7 of page 166 to accommodate general topo- logical spaces: DEFINITION 4.16 Let X and Y be topological spaces. A func- –1 CONTINUITY tion f: XY is continuous if f O is open in X for every O open in Y.

EXAMPLE 4.8 Show that: (a) If X is a discrete topological space (CYU 4.12, page 171), then every function f: XY is continuous for every topological space Y. (b) Let X and Y be topological spaces. For any y  Y the constant function f : XY 0 y0 given by f x = y for every xX is con- y0 0 tinuous. (c) For any subspace S of a topological space X, the inclusion function IS: SX given by ISs = s for every sS is continuous.

SOLUTION: (a) Since the discrete topology consists of all subsets of X, f –1O is open in X for every open set O in Y. (b) Let O be open in Y. We consider two cases.

Case 1: y0  O . Since, by the definition of f, every xX maps to y , f –1O = X — an open subset of X. 0 y0 Case 2: y  O . Since no element in X maps to y , f –1O =  0 0 y0 — an open subset of X. (c) Let O be open in X. By Definition 4.12, page 174, OS is open –1 in S. Noting that IS O = OS , we conclude that IS: SX is continuous.

CHECK YOUR UNDERSTANDING 4.22 Exhibit a continuous function f from the Sierpinski space of CYU 4.13 (page 172) to  , and a non-continuous function g from the Answer: See page A-24. Sierpinski space to  . 186 Chapter 4 A Touch of Topology

Here are several equivalent characterization of the continuity con- cept: THEOREM 4.10 Let f be a function from a topological space X to a topological space Y. The following are equivalent: (i) f is continuous. (ii)f –1H is closed in X for every H closed in Y. (iii)f –1S is open in X for every S in a sub- base for the topology of Y. (iv)f –1B is open in X for every B in a base for the topology of Y.

PROOF: We show i ii iii iv i .

(i) (ii) .Let H be closed in Y. Since Hc is open, f –1H c is open in X (by continuity). But f –1H c = f –1H c [CYU 4.8(a-iv), page 165]. It follows that f –1H is closed in X. (ii) (iii) . Let  be a subbase for Y, with S   . Since S is open in Y, Sc is closed. By (ii), f –1Sc is closed in X. But f –1S c = f –1Sc [CYU 4.8(a-iv), page 165]. Taking the comple- ment of both sides we have f –1S = f –1Sc c . Being the comple- ment of a closed set, f –1S is open in X. (iii) (iv). Follows directly from the observation that every base for a topology is also a subbase for the topology.

(iv) (i) .Let  be a base for Y, and let O be open in Y. If f –1O = , then it is open in X. –1 –1 For f O   and x0  f O choose y0  O such that

fx0 = y0 . Let B   be such that y0  BO . We then have –1 –1 –1 –1 x0  f B  f O . Since, by assumption, f B is open: f O is open.

CHECK YOUR UNDERSTANDING 4.23 Let X be an arbitrary topological space and Y an indiscrete topologi- cal space (see CYU 4.12, page 171). Prove that every function Answer: See page A-24. f: XY is continuous, 4.3 Continuous Functions and Homeomorphisms 187

Theorem 4.4, page 166, extends to general topological spaces: THEOREM 4.11 Let X, Y, and Z be any topological spaces. If f: XY and g: YZ are continuous, then the composite function gf: XZ is also continuous.

PROOF: Let O be open in Z. Since g is continuous, g–1O is open in We just copied the proof of Theorem 4.4. Y. Since f is continuous, f –1g–1O is open in X. The desired result now follows from Theorem 2.6 of page 73 which asserts that –1 –1 –1 gf O = f g O .

CHECK YOUR UNDERSTANDING 4.24 Construct functions f: XY and g: YZ (where X, Y, and Z are topological spaces) such that: (a) f is continuous and gf: XZ is not continuous. (b) g is continuous and gf: XZ is not continuous. Answer: See page A-24. (c) gf: XZ is continuous with neither f nor g continuous. The following result is of particular importance in analysis: THEOREM 4.12 Let f: XY be continuous. If K is a com- pact subset of X, then fK is a compact sub- set of Y.

ROOF P : Let K be a compact subset of X and let V   A be an open –1 cover of fK . Since f is continuous, f V   A is an open cover of K. Since K is compact, that cover contains a finite subcover n f –1V . We then have: i i = 1 n n n n –1 –1 –1 Kf  V  fK ff V =  ff V   V i i i i i = 1 i = 1 i = 1 i = 1 Exercise 24, page 169

CHECK YOUR UNDERSTANDING 4.25

PROVE OR GIVE A COUNTEREXAMPLE: Let f: XY be continuous and let SX . If fS is a compact sub- Answer: See page A-24. set of Y, then S is a compact subset of X. 188 Chapter 4 A Touch of Topology

OPEN AND CLOSED FUNCTIONS The definition of a continuous function f: XY kind of “works in reverse” in that it hinges on the behavior of f –1: YX : f –1O is open in X for every O open in Y. Here are a couple of “forward looking” concepts for your consideration: DEFINITION 4.17 Let X and Y be topological spaces. A func- OPEN AND CLOSED tion f: XY is said to be: FUNCTIONS Open if fO is open in Y for every O open in X. Closed if fC is closed in Y for every C closed in X. The following example shows the independent nature of continuous, open, and closed functions. EXAMPLE 4.9 (a) Give an example of a continuous function that is neither open nor closed. (b) Give an example of an open function that is neither continuous nor closed. (c) Give an example of a closed function that is neither continuous nor open.

SOLUTION: (a) Let Xab=  . Let 1 denote the discrete topology on X and 0 the indiscrete topology (CYU 4.12, page 171). It is easy

to see that the identity function IX: X 1  X 0 , given by

IXx = x for every xX , is a continuous function which is nei- ther open nor closed.

(b) Let X be the subspace 01 of  , and let Y be the subspace 01 of the half-open interval space H of CYU 4.18, page 176. Consider the inclusion function J: XY , given by Jx= x for every xX . J is open: Let O be open in X. Since X = 01 is open in  , O is open in  (Exercise 8, page 179). Since the topology of H contains the Euclidean topology (CYU 4.18), O is open in H. It follows that JO= OY is open in the subspace Y = 01 of H. J is not continuous: The set --1- 1  is open in Y but 2 J–1--1- 1  = --1- 1  is not open in X. 2 2 4.3 Continuous Functions and Homeomorphisms 189

J is not closed: X = 01 is certainly closed in X. However, J01 = 01 is not closed in Y, since every open neighbor- hood of 0 in the space Y meets 01 . (c)Let X be the following subspace of  1 1 X = 0 --- nZ + 2 – --- nZ + 2 not open in X n n X . Let Y be the same as the topological space X, but with 2 added .0.... . 1 .1 . 3 .. ...2 ------2 2 to its topology [2 is open in Y (see margin)]. Consider the func- f Y ...... tion f: XY given by: .0 1 1 3 .2 ------ 2 2 x if x  1 open in Y fx=   0ifx  1 f is closed: For C closed in X, we show that fCc is open in Y. Let yfC  . Case 1: y  0 . Since Y – 0 is discrete (recall that 2 , is open in Y) we have that y is an open set in Y containing no element of fC . Case 2: y = 0 . Since f0 = 0 and yfC  , 0  C . Since C is closed in X, we can choose an integer N such that

1 0  --- nN C = n The desired result now follows from the observation that 1 0  --- nN is also open in Y. n f is not continuous: The set 2 , is open in Y but not in X, and f –12 = 2 . 1 1 f is not open: The set --- is open in X, but f --- = 0 is not 2 2 open in Y.  

CHECK YOUR UNDERSTANDING 4.26

Answer: See page A-24. Give an example of a function f: XY that is open and closed, but not continuous. 190 Chapter 4 A Touch of Topology

HOMEOMORPHIC SPACES Two topological spaces are considered to be “the same” if there exists a bijection from one to the other which preserves the topological struc- ture of those spaces. To be more precise:

DEFINITION 4.18 A topological space X X is homeomor- HOMEOMORPHIC phic to a topological space Y Y , written Recall that f: XY is a SPACES XY , if there exists a bijection f: XY bijection if it is both one- such that: to-one and onto (Defini- tion 2.13, page 70). fX ==fOO  X Y Such a function f is said to be a homeomor- phism from X to Y.

CHECK YOUR UNDERSTANDING 4.27 Prove that a bijection f: XY is a if and only if Answer: See page A-24. both f and f –1: YX are continuous.

THEOREM 4.13 Any continuous closed bijection is a homeo- .

PROOF: Let f: XY be a continuous closed bijection. Appealing to CYU 4.26, we show that f –1: YX is also continuous, and do so by –1 showing that for any O, open in X, f –1 O is open in Y: Noting that the inverse of the inverse function f –1 is the func- tion f, we have f –1 –1O ==fO fOc c . CYU 4.8(b), page 165 And why is fOc c open? Because: O open  Oc is closed  fOc is closed  fOc c is open f is closed

CHECK YOUR UNDERSTANDING 4.28 (a) Prove that any continuous open bijection is a homeomorphism.

Answer: See page A-25. (b) Give an example of a continuous bijection f: XY which is not a homeomorphism.

THEOREM 4.14 Any continuous bijection from a compact space to a Hausdorff space is a homeomor- phism. 4.3 Continuous Functions and Homeomorphisms 191

PROOF: Let f: XY be a continuous bijection from a compact space X to a Hausdorff space Y. We complete the proof by showing that f is closed (see Theorem 4.13): Let C be closed in X. Since X is compact, C is compact (see CYU 4.19, page 177). By Theorem 4.12, fC is compact, and therefore closed [Theorem 4.9 (page 178)].

CHECK YOUR UNDERSTANDING 4.29 Give an example of a bijection that is both open and closed from a Answer: See page A-25. compact space to a Hausdorff space that is not a homeomorphism.

TOPOLOGICAL PROPERTIES A major goal in topology is to determine when two spaces are homeo- morphic. Finding an explicit homeomorphism may not be easy. There is, however, a useful technique that can at times be used to show that two spaces are not homeomorphic: Properties of a topological space which are preserved under homeomorphisms are said to be topological invari- ant properties. Exhibiting a topologically invariant prop- erty possessed by one space and not by another can serve to show that the two spaces are not homeomorphic. In particular, if two topological spaces X and Y are not of the same car- dinality then they cannot be homeomorphic. Why not? Because any f: XY must be a bijection. Here is a less obvious topological invariant property: DEFINITION 4.19 A path in a topological space X is the image f X PATH of a continuous function f:01  X . A space X is path connected if for any || . a b 01 . PATH CONNECTED two given points ab  X there exists a path: f: 0 1  X with f0 ==a and f1 b.

THEOREM 4.15 Path connectedness is a topological invariant property.

PROOF: Let X be a path , and let Y be homeomorphic to X. We show that Y must also be path connected: f .x2 Since XY , there exists a homeomorphism g: XY . For any . . two given points y  y  Y , let x  x  X be such that 0 1 . 1 2 1 2 x 1 g fx1 = y1 and fx2 = y2 . Since X is path connected, we know y that there exists a continuous function f: 0 1  X such that . 2 f0 = x1 and f1 = x2 . The continuous function y1. gf: 0 1  Y (Theorem 4.11), produces a path joining y1 to y2 : fg 0 ===fg0 fx1 y1 and fg 1 = y2 . 192 Chapter 4 A Touch of Topology

In the exercises you are asked to verify that the subspace 13 of  is path connected, and that 01  23 is not. It follows, from the previous theorem, that the two spaces are not homeomorphic.

CHECK YOUR UNDERSTANDING 4.30 (a) Prove that compactness is a topological invariant property. Answer: See page A-25. (b) Are the subspaces 01 and 01 of  homeomorphic? Justify your answer.

SOME CASUAL REMARKS Roughly speaking two spaces are homeomorphic if by stretching one of them you can arrive at the other. Indeed, topology is often called the “rubber-sheet geometry.” In particular, one can stretch the “rubber”   interval 13 to get to the interval 15 , and the interval –--- --- to 2 2 the interval –  (just keep on stretching):

3- . x 1 yfx== --- + --- 1- 2 2 .   –------||2 2 1 5

f: 1 5  13   A homeomorphism [Exercise 17(a)] fx= tanx: –--- ---   2 2 A homeomorphism [Exercise 17(b)] One cannot, however, stretch the non-compact interval 13 to the compact interval 15 [see CYU 4.28(a)]. In addition, the path con- nected interval 13 is not homeomorphic to the non path connected 12  23 (see Theorem 4.15).

CHECK YOUR UNDERSTANDING 4.31 Prove that homeomorphic is an equivalence relation ( ) on any set Answer: See page A-25. of topological spaces (see Definition 2.20, page 88). 4.3 Continuous Functions and Homeomorphisms 193

EXERCISES

1. Let f be a function from a topological space X to  . Prove that f is continuous if and only if for any a   both the sets xfx a and xfx a are open.

2. Let Y be any space that is not indiscrete. Show there exists a space X and a function f: XY that is not continuous.

3. Let X be any space that is not discrete. Show there exists a space Y and a function f: XY that is not continuous. 4. Let X  be a topological space and let Yd be a metric space. Prove that a function f: XY is continuous at xX if and only if for any   0 there exists a neighborhood O of x such that dfx fy for every yO .

5. Let f and g be continuous functions from a topological space X to  . Prove that: (a)af+ bg: X   is continuous for any ab   . (a)fg: X   is continuous. f (a)---: X   is continuous if gx 0 for any xX . g

6. Give an example of a function f: XY that is continuous and open, but not closed.

7. Give an example of a function f: XY that is continuous and closed, but not open.

8. Let f: XY be a bijection. Prove that f is open if and only if f –1: YX is continuous.

9. (a) Show that a function f: XY is open in Y if fB is open in Y for every B in a base for the topology of X. (b) Show, by means of an example, that (a) need not hold when the word “base” is replaced with “subbase.” (c) Show that in the event that f is a bijection, then (a) will hold when the word “base” is replaced with “subbase.”

10. (a) Prove that a bijection f: XY is a closed function if and only if it is an open function. (b) Give and example of an open onto function that is neither closed nor continuous. (c) Give and example of an open one-to-one function that is neither closed nor continuous. (d) Give and example of a closed onto function that is neither open nor continuous. (e) Give and example of a closed one-to-one function that is neither open nor continuous.

n 11. Let fi: Xi  Xi + 1 i = 1 be a collection of continuous functions. Use the Principle of Mathe-  matical Induction to show that fnfn – 1 f2f1:X1  Xn + 1 is also continuous. 194 Chapter 4 A Touch of Topology

12. Closure (See Exercises 25-33, page 182)]. Establish the equivalence of the following three properties: (i)f: XY is continuous.

(ii)fA fA for every AX .

(iii)fA fA for every AX .

13.[Closure (See Exercises 25-33, page 182)]. Prove that f: XY is closed if and only if fA fA for every AX .

14. Let f: XY be closed. Show that for any AY and any open set U containing f –1A there exists an open set V containing A such that f –1V  U .

15. Let f: XY be open. Show that for any AY and any closed set H containing f –1A there exists a closed set V containing A such that f –1V  H .

16. Give an example of a set X, and two topologies  and  such that the identity function

IX: X   Y  is: (a) Continuous but not open. (b) Continuous but not closed. (c) Continuous but not a homeomorphism. (d) Open but not closed. (e) Open but not continuous. (f) Open but not a homeomorphism. (g) Closed but not open. (h) Closed but not continuous. (i) Closed but not a homeomorphism.

x 1 17. (a) Show that f: 1 5  13 given by fx= --- + --- is a homeomorphism. 2 2   (b) Show that f: –--- ---   given by fx= tanx is a homeomorphism. 2 2

18. Is the closed 01 homeomorphic to the open interval –11 ? Justify your answer. 19. Give an example of an open bijection from a compact Hausdorff space to a Hausdorff space that is not a homeomorphism. 20. Give an example of a closed bijection from a compact Hausdorff space to a Hausdorff space that is not a homeomorphism.

21.[Let f: XY be a bijection. Show that the following properties are equivalent: (i) f is a homeomorphism. (ii) f is continuous and open. (ii) f is continuous and closed. (iii)fA= fA for every AX . 4.3 Continuous Functions and Homeomorphisms 195

22. Show that the Sierpinski space (CYU 4.13, page 172) is not homeomorphic to the discrete space of two points (CYU 4.12, page 171).

23. Show that the subspace 13 of  is path connected, and that 01  23 is not. 24. (Fixed Point Property). A nonempty space X satisfies the fixed point property if for any continuous function f: XX there exists xX such that fx= x . Prove that the fixed point property is a topological invariant. 25. (a) Prove that the existence of a proper subset of a topological space X that is both open and closed is a topological invariant property. (b) Use (a) to show that the subspaces 01 and 01  2 of  are not homeomorphic. 26. Prove that connectedness is a topological invariant property (see Exercise 44-49, page 184). 27. Prove that the cardinality of the set of subsets of a topological space that are both open and closed is a topological invariant property. 28. Prove that metrizable is a topological invariant property.

29. Prove that each of the separation properties: T0T1 T2 T3 T4 of Exercises 38-42, page 183, is a topological invariant property.

PROVE OR GIVE A COUNTEREXAMPLE

30. If Y is an indiscrete space, then every function f: XY is continuous for every space X. 31. Any continuous open bijection is a homeomorphism. 32. Two topological spaces, X and Y, are homeomorphic if and only if CardX = CardY . 33. If the space X is homeomorphic to a space Y, then CardX = CardY . 34. (See Exercises 34-37, page 183.) If f: XY is a homeomorphism, and if A is dense in Y, then fA is dense in Y. 35. (See Exercises 34-37, page 183.) If f: XY is continuous, and if A is dense in X, then fA is dense in Y. 36. (See Exercises 34-37, page 183.) If f: XY is onto and continuous, and if A is dense in X, then fA is dense in Y. 37. (See Exercises 34-37, page 183.) If f: XY is continuous, and if A is dense in X, then fA is dense in Y. 196 Chapter 4 A Touch of Topology

4

§4 PRODUCT AND QUOTIENT SPACES

See Definition 2.7, page 63. Let’s impose a topology on the Cartesian product of two topological spaces:

DEFINITION 4.20 Let X X and Y Y be topological spaces. The on PRODUCT TOPOLOGY ON TWO SPACES XY = xy xXyY   is that topology with basis:

 = UV U  X V  Y Note that while the above collection  is closed under finite intersec- tions (Exercise 2), it need not be closed under unions. For example, while the intersection of the two rectangles 13  12 and 24  03 in Figure 4.1 is again of the form UV with U and V open in  , their union is not. It follows that  is not a topology on  ; but then again, it does not profess to be a topology, but rather a basis for the product topology on  : the collection of all arbitrary unions of elements from  .

3 13  12 24  03 2 V UV 1

U 0 1 2 3 4 Figure 4.1

CHECK YOUR UNDERSTANDING 4.32

(a) Show that if X and Y are bases for the topological spaces X and Y, respectively, then  = DE D  X and E  Y is a basis for the product topology on XY . (b) Show that the product topology on  coincides with the Answer: See page A-26. Euclidean topology on 2 . 4.4 Product and Quotient Spaces 197

As you will see, the following result continues to hold for any collec- tion of compact spaces. THEOREM 4.16 If X and Y are compact spaces, then the prod- uct space XY is also compact.

PROOF: If either X or Y is empty, then so is XY , and, as such, is compact. That being the case, we need only consider the case where neither X nor Y is empty. Appealing to Exercise 11, page 179, we show that XY is compact by showing that every open cover of XY , taken from the basis UV , where U is open in X and V is open in Y, has a finite subcover:

Let U  V   A be a cover of XY , where each U is open in X, and each V is open in Y. For any xX and yY , the element xy is contained in some U  V . It follows

that U   A and V   A are open covers of X and Y, respectively. By compactness, we can choose finite subcovers U n and V m of X and Y, respectively. Noting that i i = 1 i i = 1 for any xy  XY there exist elements U and V of i j those subcovers containing x and y respectively, we conclude that the finite collection U  V 1 in 1 jm covers i j XY .

CHECK YOUR UNDERSTANDING 4.33 Establish the converse of Theorem 4.16: If XY is compact, then X Answer: See page A-26. and Y are compact.

Y DEFINITION 4.21 Let X and Y be nonempty sets. The func- tions  : XY  X and  : XY  Y  xy PROJECTION 1 2 y 2 . . FUNCTIONS given by  xy = x and  xy = y  1 2 1 are called the projection functions onto X X .x and Y, respectively. THEOREM 4.17 If X and Y are nonempty topological spaces, the projection functions 1 : XY  X and 2 : XY  Y are continuous.

PROOF: To see that 1 is continuous you need but note that if U is –1 open in X, then 1 U = UY is open in XY (see Definition 4.20). The same argument can be applied to 2 : XY  Y . 198 Chapter 4 A Touch of Topology

CHECK YOUR UNDERSTANDING 4.34 Are the projection functions of Theorem 4.17 necessarily open? Answer: See page A-26. Closed? Justify your answer. In Exercise 14, page 180, you were asked to show that any collection  of subsets of a given set generates a topology on that set; namely: the set of arbitrary unions of finite intersections of elements taken from  . As it turns out:

THEOREM 4.18 Let X and Y be topological spaces. The prod- uct topology on XY is generated by:

–1 –1 1 U U is open in X  2 V V is open in Y  (In other words: the above collection consti- tutes a subbase for the product topology.)

PROOF: The claim is a consequence of Definition 4.20 and the obser- V UV vation that for any open set U in X, and any open set V in Y: –1 –1 UV = 1 U  2 V (see margin) U GENERAL PRODUCT SPACES The Cartesian product of two sets can easily be extended to accom-

modate a finite number of sets X1X2  Xn : n  Xi = x1x2  xn xi  Xi i = 1 A countable number of sets:   Xi = x1x2  xn  xi  Xi i = 1 Any collection of sets:

 X = x x  X   A By the same token, Definition 4.29 can be generalized to accommo- date any collection of topological spaces. Let’s first turn to a finite col- lection:

DEFINITION 4.22 n Let Xi Xi i = 1 be a collection of PRODUCT TOPOLOGY topological spaces. The product topology n ON n SPACES on  Xi is that topology with basis: i = 1

 = U1 U2   Un Ui  Xi Which is to say: The set consisting of arbi- trary unions of elements from  . 4.4 Product and Quotient Spaces 199

The following generalization of Definition 4.21 will enable us to extend the concept of a product topology to any collection of topologi- cal spaces:

DEFINITION 4.23 Let X   A be a collection of nonempty sets. PROJECTION For   A , the function FUNCTION  : X  X given by  x = x         A    A is called the projection function onto the th component of  X .   A

DEFINITION 4.24 Let X X   A be a collection of topo- PRODUCT TOPOLOGY logical spaces. The product topology on

 X is that topology with subbase   A –1  =  Ua Ua  X   A Y While one is able to geometrically depict the Cartesian product of two y .xy or three sets (see margin), the same cannot be said for four or more sets. We can, however, provide a visual sense of the product topology X x  X of an arbitrary collection of spaces by first representing each   A Z X as a vertical line, as is done in Figure 4.2. An open set O is also z xyz . depicted (in blue) in the figure. Note that being a union of finite inter- x X –1 y sections of the subbase elements  Ua , only a finite number of the Y X lines can fail to be “entirely blue.” An open set O Y ( ) ) ( ) ( ) ...... ( ...... ( ) ( ) ( ( ) ( ... ( ) ...... The Xs Figure 4.2 200 Chapter 4 A Touch of Topology

Theorem 4.16 assured us that the product of two compact spaces is again compact. Much more can be said: Andrey Tychonoff THEOREM 4.19 (1906-1993) Any product of compact spaces is compact. TYCHONOFF’S PRODUCT THEOREM PROOF: Appealing to Alexander’s Subbase Theorem of page 177, we assume that F  –1U U  is a cover of    X   A

XX=   that has no finite subcover, and arrive at a contradiction:   A –1 For any   A , U  U  F cannot cover the compact

space X . For if it did, then there would exist a finite subfamily n U –1U  F which covers X , resulting in a finite i  i i = 1  subfamily of F which covers X; namely: –1U n . Conse-  i i = 1

quently, for every   A , we can choose an element x  X –1 which is not contained in any U with  U  F . Since the

particular point x   A  X is not contained in any element of F, F does not cover X — a contradiction.

CHECK YOUR UNDERSTANDING 4.35

Show that if XX=  is compact then each X must be compact. Answer: See page A-27.    A

QUOTIENT SPACES We remind you that: •An equivalence relation on a set X is a relation, ~, that is reflexive, symmetric, and transitive (Definition 2.20, page 88).

• The equivalence class of x0  X , denoted by x0 , is given by

x0 = xXx ~x0 (Definition 2.21, page 91). We will denote the set of equivalence classes associated with an equivalence relation ~ on a topological space X by the symbol X ~ :

X  ~ = x xX In addition, the function : XX ~ , given by x = x , will be called the projection function of X onto X ~ . 4.4 Product and Quotient Spaces 201

As it turns out, the topology on X imposes a topology on the set X ~ : THEOREM 4.20 If ~ be an equivalence relation on a topologi- cal space X  , then

QUOTIENT TOPOLOGY ˜  = –1O  X O   is a topology on X ~ (called a quotient topology). ˜ The resulting topological space X  ~, is QUOTIENT SPACE  said to be a quotient space of X.

PROOF: We verify that the three defining axioms of a topological space (Definition 4.9, page 171) are satisfied: (i) Since  and X are open in X, –1 =  and –1X = X  ~ are open in X ~ . (ii) Arbitrary unions of open sets in X ~ are again open.  –1 –1  O =  O     A   A Exercise 25, page 169 open in X , (iii) Finite intersections of open sets in X ~ are again open. n n –1O = –1O  i  i i = 1 i = 1 Exercise 25, page 169 open in X

Here are a couple of specific quotient spaces for your consideration:

While every point in the (a) Let ~ be the equivalence relation on the subspace 02  open interval 02  is of  represented by the partition x = x for every identified with itself only, the two end points x  02  , and 0 ==2 02  . As is sug- of the closed interval are gested in Figure 4.3, the quotient space 02  ~ is identified with each other. The visual effect homeomorphic to the unit is that of gluing one end S1 = xy  2 x2 + y2 = 1 — a fact that can be rig- point of the interval to the other. orously established by showing that the function f: 0 2 ~  S1 given by fx = cosxx sin is a homeomorphism. 202 Chapter 4 A Touch of Topology 

.0 =  2  ) ( . ) (  0 2 . Figure 4.3 (b) Consider the space X ~ where X = 02   01 , and ~ is the relation represented by the partition xy = xy for 0 x 2 , 0 y 1 , and 0 y ==1 y 0 y  1 y for 0 y 1 . As is Visually, we are gluing the left and right edges suggested in Figure 4.4, the quotient space X ~ is of the square together. homeomorphic to the cylinder CS= 1  01 — a fact that can be rigorously established by showing that the function f: X  ~  C given by fxy = cosxx sin  y is a homeomorphism.

01 2 1

0 y . .2 y . .xy .

00 2 0

Figure 4.4

CHECK YOUR UNDERSTANDING 4.36 Let X denote the rectangle of Figure. 4.4. Define an equivalence relation ~ on X in

Answer: See page A-27. such a way so that X ~ is homeomorphic to a torus. 4.4 Product and Quotient Spaces 203

EXERCISES

1. Let 1 = Xab and 2 = Xa ab ac be two topologies on the set X=  abc . Determine the topology on the product space:

(a) X 1  X 1 (b) X 1  X 2 (c) X 2  X 2

2. (a) Let X X and Y Y be topological spaces. Prove that SUV=  U  X V  Y is closed under finite intersections. n (b) Let Xi Xi i = 1 be a collection of topological spaces. Prove that

 = U1 U2   Un Ui  Xi is closed under finite intersections.

3. Prove that a space X is Hausdorff if and only if xx xX is closed in XX .

4. (a) Prove that the product of two Hausdorff spaces is Hausdorff. (b) Prove that the product of Hausdorff spaces is Hausdorff. 5. (a) Prove that the product of two regular spaces is regular (see Exercise 41, page 183). (b) Prove that the product of regular spaces is regular (see Exercise 41, page 183).

6. Let f and g be functions from a topological space X to a topological space Y. Let h:XYY  be given by hx= fx gx . Prove that h is continuous if and only if both f and g are con- tinuous.

7. For given functions f: X1  Y1 and g: X2  Y2 , let fg : X1  X2  Y1  Y2 be given by fg x1 x2 = fx1  gx2 . Prove that fg is continuous if and only if f and g are continuous.

8. (a) Let U and V be a subspaces of X and Y, respectively. Prove that UV is a subspace of XY . n n

(b) Let Si be a subspace of a space Xi , 1 in. Prove that  Si is a subspace of  Xi . i = 1 i = 1 (c) Let S be a subspace of a space X ,   A . Prove that  S is a subspace of  X .   A   A 204 Chapter 4 A Touch of Topology

9. (a) Let U and V be subsets of the spaces X and Y, respectively. Prove that UV is dense in XY if and only if U is dense in X, and V is dense in Y. (See Exercises 34-37, page 183.) n n

(b) Let Si be a subset of a space Xi , 1 in . Prove that  Si is dense in  Xi if and i = 1 i = 1 only if each Si is dense in Xi . (c) Let S be a subset of a space X ,   A . Prove that  S is dense in  X if and   A   A only if each S is dense in X .

10. (a) Let X and Y be topological spaces. Prove that the projection functions 1 : XY  X

and 2 : XY  Y are open.

(b) Prove that each projection function :  X  X is open.   A

11. (a) Prove that for any functions f: X1  Y1 and g: X2  Y2 , the function

h: X1  X2  Y1  Y2 given by hx1 x2 = fx1  gx2 is continuous if and only if f and g are both continuous.

(b) Let f: X  Y be given, for   A . Prove that the function h:  X   Y   A   A given by hx   X = fx   A is continuous if and only if each f is continuous.

12. (a) Let X, Y1 , and Y2 be topological spaces. Prove that a function f: XY 1  Y2 is continu-

ous if and only if 1f: XY 1 and 2f: XY 2 are continuous.

(b) Prove that a function f: XY   is continuous if and only if the functions   A  f: XY  are continuous for every   A .

13. (a) Let S1 and S2 be closed subsets of the topological spaces X1 and X2 , respectively. Prove

that S1  S2 is a closed subset of X1  X2 .

(b) Let S be a closed subset of X , for   A . Prove that  S =  X .   A   A

14. (a) Let S1 and S2 be subsets of the topological spaces X1 and X2 , respectively. Prove that

S1  S2 = S1  S2 . (See Exercises 25-33, page 182.)

(b) Let S be a subset of X , for   A . Prove that  S =  S .   A   A 15. Prove that for any two topological spaces X and Y, XYYX   . 4.4 Product and Quotient Spaces 205

16. (a) Prove that if X1 is homeomorphic to Y1 , and if X2 is homeomorphic to Y2 , then

X1  X2 is homeomorphic to Y1  Y2 .

(b) Prove that if X is homeomorphic to Y , for   A , then  X is homeomorphic to   A  Y .   A

17. (a) Let ~ be an equivalence relation on a space X. Prove that X ~ is T1 if and only if each equivalence class x is closed in X. (See Exercise 39, page 183.)

(b) Give an example of a T1 -space X and an equivalence relation ~ on X such that X ~ is not T1 .

18. Let ~ be the equivalence relation on the space X = 01  01 given by

x1 y1 ~x2 y2 if and only if y1 = y2  0 . Describe the quotient space X ~ and show that it is not a Hausdorff space. 19. (a) Let X and Y be topological spaces. Let ~ be the equivalence relation given by

x1 y1 ~x2 y2 if and only if x1 = x2 . Prove that XY  ~ is homeomorphic to X.

n (b) Generalize (a) to accommodate a collection Xi i = 1 of spaces.

(c) Generalize (a) to accommodate a collection X   A of spaces.

20. Let ~ be an equivalence relation on a compact space X. Prove that X ~ is compact.

21. Let ~ be the equivalence relation on the subspace 02  of  induced by the partition x = x for every x  02  , and 0 ==2 02  (see Figure 4.4). Show that

the function f: 0 2  ~  S1 given by fx = cosxx sin is a homeomorphism.

22. Let ~ be the equivalence relation on the space X = 02   01 induced by the partition xy = xy for 0 x 2 , 0 y 1 , and 0 y ==1 y 0 y  1 y for 0 y 1 (see Figure 4.4). Let S1 denote the unit circle: S1 = xy  2 x2 + y2 = 1 . Show that the function f: X  ~  S1  01 given by fxy = cosxx sin  y is a homeomorphism. 206 Chapter 4 A Touch of Topology

PROVE OR GIVE A COUNTEREXAMPLE

23. The Cartesian Product of two metrizable spaces is again metrizable. 24. The Cartesian Product of any collection of metrizable spaces is again metrizable.

25. For any two topological spaces X and Y, the projection functions 1 : XY  X and

2 : XY  Y are closed.

26. For any three topological spaces X, Y1 , and Y2 , a function f: XY 1  Y2 is open if and only

if 1f: XY 1 and 2f: XY 2 are open.

27. For any four nonempty topological spaces X1 , X2 , Y1 , and Y2 , if X1  Y1 is homeomorphic

to X2  Y2 , then X1 is homeomorphic to X2 and Y1 is homeomorphic to Y2 .

28. For any three nonempty topological spaces X, Y1 , and Y2 , if Y1  X is homeomorphic to

Y2  X , then Y1 is homeomorphic to Y2 . 5.1 Definitions and Examples 207

5 CHAPTER 5 A Touch of

§1. DEFINITIONS AND EXAMPLES The following familiar properties reside in the set Z of integers: Property Example:

Closure 1.ab+  Z a bZ 57+  Z

Associative 2.abc+ + = ab+ + c a bZ 541+54+ = + + 1

Identity 3.a + 0 = a a  Z 40+4=

Inverse 4.aa+0– = a  Z 55+0– =

A generalization of the above properties bring us to the definition of a group — an abstract structure upon which rests a rich theory, with numerous applications throughout mathematics, the sciences, architecture, music, the visual arts, and elsewhere: A binary operator on a set X is a function that assigns to DEFINITION 5.1 A group G * , or simply G, is a nonempty any two elements in X an ele- ment in X. Since the function GROUP set G together with a binary operator, *, (see value resides back in X, one margin) such that: says that the operator is closed. Associative Axiom: 1.a*b*c = a*b *c for every abc   G .

Evariste Galois defined Identity Axiom: 2. There exists an element in G, which we will the concept of a group in label e, such that a*ea= for every aG . 1831 at the age of 20. He was killed in a duel one Inverse Axiom: 3. For every aG there exists an element, year later, while attempt- a  G such that a a = e . ing to defend the honor of * a prostitute. In particular, Z + is a group; with “+, 0, and –a ” playing the role of We show, in the next section, “*, e, and a ” in the above definition. that both the identity element e and the inverse element a Is the set of integers under multiplication a group? No: of Axioms 2 and 3 are, in fact, both unique and “ambidex- While “regular” multiplications is an associative trous:” binary operator on Z, with 1 as identity, no integer a*ee==*aa other than 1 has a multiplicative inverse in Z.  a*a ==a*ae Yes, there is a number whose prod- uct with 2 is 1: 1 1 2  --- = 1 , but ---  Z . 2 2 Bottom line: The set of integers under multiplication is not a group. 208 Chapter 5 A Touch of Group Theory

CHECK YOUR UNDERSTANDING 5.1 Determine if the given set is a group under the given operation. If not, specify which of the axioms of Definition 2.1 do not hold. (a) The set Q of rational numbers under addition. (b)The set  of real numbers under addition. (c)The set  of real numbers under multiplication. (a), (b), and (d) are groups. (d)The set + = r   r  0 of positive real numbers under (c) is not a group. multiplication.

ADDITION MODULO n

For any integer n  1 , we define a binary operator +n on the set Z = 012 n – 1 , called addition modulo n, as follows: See the Division Algo- n rithm on page 43. a +n br== where: ab+ qn+ r 0  rn

In particular, if n = 6 , then Z6 = 012345 and we have:

2+6 3=== 5 since 2+506 3  + 5

2+6 5 ===1 since 2+716 5  + 1

5+6 5 ===4 since 5+1016 5  + 4

3+6 3 ===0 since 3+616 3  + 0

THEOREM 5.1 For any integer n  1 , Zn +n is a group.

PROOF: +n is an operator on Zn : Let ab  Zn = 012 n – 1 . Think remainder! Since a +n br= , with 0  rn : a +n bZ n .

Zero Axiom: 0  Zn , and for every aZ n :

a+n 0 = a , as 0  an ).

Inverse Axiom: Let aZ n = 012 n – 1 .

If a = 0 , then a is its own inverse: 0+n 00= .

If 0 an , then na–  Zn is its inverse: a+n na– = 0 .

Associative Axiom: For abc  Zn , we are to show that

a+n b+n c = a+nb +nc Getting geared up for that challenge we start with:

(1) a+n br==1 so that: ab+ q1nr+ 1 with 0  r1  n

(2) b+n cr==2 so that: bc+ q2nr+ 2 with 0  r2  n

(3) a+n r2 ==r3 so that: ar+ 2 q3nr+ 3 with 0  r3  n

(4) r1+n cr==4 so that: r1 + c q4nr+ 4 with 0  r4  n 5.1 Definitions and Examples 209

So:

a+n b+n c ==a+n r2 r3 and a+nb +ncr==1+n cr4

We complete the proof by showing that in Zn , r3 – r4 = qn :

r3 ==ar+ 2 – q3n q1nr+ 1 – b + r2 – q3n (3) (1) and: r ==cr+ – q n q nr+ – b + r – q n . Thus: 4 (4) 1 4 (2) 2 2 1 4

r3 – r4 = q1nr+ 1 – b + r2 – q3n – q2nr+ 2 – b + r1 – q4n

==q1 – q3 – q2 + q4 nqn.

CHECK YOUR UNDERSTANDING 5.2

Complete the following (self-explanatory) group table for Z4 +4 .

+4 0 1 2 3

0 3 since 0+433= 1 2

21since 2+431=

3 02since 3+432=

Answer: See page A-27. since 1+412= since 3+410=

Groups containing infinitely many elements, like Z + and  + , are said to be infinite groups. Those containing finite may elements,

like Zn +n which contains n elements, are said to be finite groups. DEFINITION 5.2 Let G be a finite group. The number of ele- ORDER OF A GROUP ments in G is called the order of G, and is denoted by G .

GROUP TABLES AND BEYOND

The group Z4 , with table depicted in Figure 5.1(a), has order 4. Another group of order 4, the so-called Klein 4-group, appears in Figure 5.1(b).

+4 0123 * eabc Z4: K: 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae (a) (b) Figure 5.1 210 Chapter 5 A Touch of Group Theory

Is K really a group? Well, the above table leaves no doubt that the closure and identity axioms are satisfied (e is the identity element). Moreover, each element has an inverse, namely itself: ee=== e aa ebb e and cc =e . Finally, though a bit tedious, you can check directly that the holds [for example: ab acab== and aba==ac b ]. You can also easily verify that K is an abelian group; where:

Abelian groups are also said DEFINITION 5.3 A group G * is abelian if to be commutative groups. ABELIAN GROUP a*bb= *a for every ab  G

We will soon show that Z4 and K are the only groups of order 4, but first: THEOREM 5.2 Every element of a finite group G must appear once and only once in each row and each column of its group table.

th PROOF: Let Gea= 1 a2  an – 1 . By construction, the i row

of G’s group table is precisely aieaia1 aia2  aian – 1 . The fact that every element of G appears exactly one time in that row is a con- sequence of Exercise 37, which asserts that the function f : GG ai given by f g = a g is a bijection. As for the columns: ai i

CHECK YOUR UNDERSTANDING 5.3 Answer: See page A-27. Complete the proof of Theorem 5.2. We now show that the two groups in Figure 5.1 represent all groups of order four. To begin with, we note that any group table featuring the four elements eabc must “start off” as in T in Figure 5.2, for e represents the identity element.

* eabc eeabc T: aa bb cc

* eabc * eabc * eabc eeabc eeabc eeabc E: aae B: aab C: aac bb bb bb cc cc cc

Figure 5.2 5.1 Definitions and Examples 211

Since no element of a group can occur more than once in any row or column of the table, the -box in T can only be inhabited by e, b or c, with each of those possibilities displayed as E, B, and C in Figure 5.2. Repeatedly reemploying Theorem 5.2, we observe that while E leads to two possible group tables, both B and C can only be completed in one way (see Figure 5.3).

* eabc eeabc aaecbE bbcae 1 * eabc * eabc * eabc two options eeabc eeabc eeabc ccbea E: aae aaecb aaecb bb bb bbc * eabc cc cc ccb eeabc aaecb only option only option E bbcea 2 ccbae

* eabc * eabc e a b c * eabc eeabc eeabc e e a b c eeabc B: aab aabce a a b c e aabceB bb bb b b c bbcea cc cc c c e cceab only option only option only option

* eabc * eabc * eabc * eabc eeabc eeabc eeabc eeabc C: aac aaceb aaceb aacebC bb bb bbe bbeca cc cc ccb ccbae only option only option only option Figure 5.3 At this point we know that there can be at most four groups of order

4. Their group tables, labeled E1E2 B , and C appear in Figure 5.4.

The group tables for Z4 and K of Figure 5.1 are also displayed at the top of that figure. 212 Chapter 5 A Touch of Group Theory

+4 0123 * eabc Z4: K: 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae

eabc * E2: * eabc E1: eeabc eeabc aaecb aaecb bbcae bbcea ccbea ccbae

* eabc eeabc B: aabce bbcea cceab

* eabc eeabc aaceb C: bbeca ccbae Figure 5.4

While table E2 and the Klein group K tables in Figure 5.4 are identi- cal, the same cannot be said for the four tables on the left side of the figure. But looks can be deceiving. To show, for example, that Z4 and E1 only differ superficially, we begin by reordering the elements in the first row and first column of Z4 in Figure 5.5(a) from 0, 1, 2, 3 to 0, 2, 1, 3, [see Figure 5.5 (b)]. We then transform Figure 5.5(b) to E1 in (c) by replacing the symbols “0, 2, 1, 3” with the symbols “e, a, b, c,”

respectively, and the operator symbol “+4 ” with “*”. + 0123 Z4: 4 +4 0213 E1: * eabc 00123 00213 eeabc 11230 22031 aaecb 22301 11320 bbcae 33012 33102 ccbea (a) (b) (c) Figure 5.5 This “appearances aside” concept is formalized in So, appearances aside, the group structure of E1 coincides with that of Section 4. Z4 . In a similar fashion you can verify that tables B and C of Figure 5.4 only differ from table Z4 syntactically.

5.1 Definitions and Examples 213

PERMUTATIONS AND SYMMETRIC GROUPS

For any non-empty set X, let SX = f: XXf is a bijection . We then have: The composition operator “ ” is defined on page 3. THEOREM 5.3 For any non-empty set X, SX  is a group.

PROOF: Turning to Definition 5.1:

Operator. fg  SX : gfS X [Theorem 2.5(c), page 72].

Associative. fgh   SX : hgf = hg f [Exercise 44, page 75].

Identity. fS X : fIX ==IX ff , where IX: XX is the iden- tity function: IX x = x for every xX . –1 Inverse. fS X : ff = IX [Theorem 2.4(b), page 70].

The elements (functions) in SX are said to be permutations (on X), and SX  is said to be the symmetric group on X.

IN PARTICULAR:

For X = 12 n , SX  is called the symmetric group of degree n, and will be denoted by Sn .

Let’s get our feet wet by considering the symmetric croup S3 , the set of all permutations on X = 123 . Given that there are n! ways of ordering n objects (Exercise 35, page 42), we conclude that the group S3 consists of 3!== 1 2 3 6 elements: e  1 2 3 4 5 11 12 13 11 13 12 22 23 21 23 22 21 33 31 32 32 31 33 In a more compact (and more standard) form (see margin), we write: Directly below each ele- 1 2 3 1 2 3 1 2 3 ments of the first row e = , 1 = , 2 = appears its image under the 1 2 3 2 3 1 3 1 2 permutations. The fact that 3 lies below 1 in  , for 1 2 3 1 2 3 1 2 3 4  = ,  = ,  =  example, simply indicates 3 1 3 2 4 3 2 1 5 2 1 3 that the permutation 4 maps 1 to 3: 13 . Note that 0 is the identity function e: 01 = 1 02 =2 and 03 = 3

The symmetric group S3 Figure 5.6 214 Chapter 5 A Touch of Group Theory

Generalizing the above observation we have:

THEOREM 5.4 The symmetric group Sn of degree n contains n! elements.

EXAMPLE 5.1 Referring to the group S3 featured in Figure 5.6, Determine:  –1 (a) 24 (b) 42 (c) 2

SOLUTION: (a) To find 24 we first perform 4 and then apply 2 to the resulting function values: From Figure 5.6: 4 2 2  4 132 1  2 13 13 21 22 221  24 ==2  1 5 32 31 313 3  3

(b) Using the standard form we show that 42 = 3 :

1 2 3 1 2 3 first :  2  1 2 3 3 1 2 3 1 2    4 2 ==3 1 2 3  1 3 2 then 4: 1 3 2 3 2 1

1 2 3 (c) Tor arrive at the inverse of the permutation  =  , 2 3 1 2 simply reverse its action: 1 2 3 –1 3 1 2 1 2 3 –1 ==== 2 3 1 2 1 2 3 2 3 1 1

CHECK YOUR UNDERSTANDING 5.4 Answer: 1 2 3 4 5 With to the symmetric group S5 , determine  and  ,  :   5 4 3 2 1 where: 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 :  = and  = 4 3 2 5 1 1 5 2 3 4 5 3 2 1 4 Adhering to , we will start using ab (rather than a*b ) to denote the in a generic group. Under this notation, the symbol a–1 (rather than a ) is used to denote the inverse of a, while e continues to represent the identity element. In a generic abelian group, however, the symbol “+” is typically used to represent the binary oper- ator, with 0 denoting the identity element, and –a denoting the inverse of a. To summarize: 5.1 Definitions and Examples 215

Original Form Product Form Sum Form (Reserved only for abelian groups) 1.a*bG ab  G 1.ab G 1.ab+  G 2.a*b*c = a*b * c 2.abc= ab c 2.abc+ + = ab+ + c 3. a*ea= 3.ae= a 3.a + 0 = a

4.a*a = e 4. aa–1 = e 4. aa+0– = SOME ADDITIONAL NOTATION: Referring to the product form, For any positive integer n: For any positive integer n: –n do not express a in the form an represents aaaa na represents aaa+++ +a 1 ----- (there is no “division” in n a’s n a’s an –n –1 n the group). and a = a and –n ana= – From its very definition we We also define a0 to be e. We also define 0a to be 0. find that the following expo- nent rules hold in any group G: Utilizing the above notation: For anynm  Z : anam = anm+ DEFINITION 5.4 (Product form) A group G is cyclic if there an m = anm n CYCLIC GROUP exists aG such that Ga= nZ . In the sum form, it is accept- able utilize the notation ab– . (Sum form) An abelian group G is cyclic if By definition: ab– = ab+ – . there exists aG such that GnanZ=  .

GENERATOR In either case we say that the element a is a gen- erator of G, and write Ga=  .

EXAMPLE 5.2 Show that: (a) Z6 is cyclic (b) S3 is not cyclic.

SOLUTION: (a) Clearly Z6 = 1 . In fact, as we now show, 5 is also a generator of Z6 (don’t forget that we are summing modulo 6): Note that: 15 = 5 505=  + 5

25 ==5+65 4 10= 1 6 + 4

35 ==5+65+65 3 15= 2 6 + 3

45 ==5+65+65+65 2 20= 3 6 + 2

55 ==5+65+65+65+65 1 25= 4 6 + 1

65 ==5+65+65+65+655+ 6 0 30= 5 6 + 0

Since every element of Z6 = 012345 is a multiple of 5, we conclude that Z6 = 5 . (b) We could use a brute-force method to verify, directly, that no element of S3 generates all of S3 . Instead, we appeal to the following theorem [and Example 5.2(b)] to draw the desired conclusion. 216 Chapter 5 A Touch of Group Theory

THEOREM 5.5 Every cyclic group is abelian.

PROOF: Let Ga== an nZ . For any two elements as and at in G (not necessarily distinct) we have: asat ===ast+ ats+ atas

CHECK YOUR UNDERSTANDING 5.5

(a) Show that 1 and 5 are the only generators of Z6 .

(b) Show that S2 is cyclic.

Answer: See page A-28. (c) Show that Sn is not cyclic for any n  2 . At this point we have two groups of order 6 at our disposal:

Z6 +n and S3 Do these groups differ only superficially, or are they really different in some algebraic sense? They do differ algebraically in that one is cyclic while the other is not, and also in that one is abelian while the other is not. 5.1 Definitions and Examples 217

EXERCISES

Exercise 1-11. Determine if the given set is a group under the given operator. If not, specify which of the axioms of Definition 5.1 do not hold. If it is, indicate whether or not the group is abe- lian, and whether or not it is cyclic. If it is cyclic, find a generator for the group. 1. The set 2nn Z of even integers under addition. 2. The set 2n + 1 nZ of odd integers under addition.

3. The set of integers Z, with a*bc= , where c is the smaller of the two numbers a and b (the common value if ab= ) ab 4. The set Q+ of positive rational numbers, with a b = ------. * 2 a2 5. The set x   x  0 , with a b = ----- . * b 6. The set 02468 under the operation of addition modulo 10. 7. The set 0123 under multiplication modulo 4 (For example: 2*3 = 2 , since 23 ==614 + 2 , and 3*3 = 1 since 33 ==924 + 1 .) 8. The set 01234 under multiplication modulo 5. (See Exercise 7.) 9. The set ab+ 2 ab  Z under addition.

10. The set ab+ 2 ab  Q with not both a and b equal to 0 under the usual multiplica- tion of real numbers. 11. The set ZZ = ab ab  Z , with ab *cd = ac+  bd+ .

Exercise 12-23. Referring to the group S3 : 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3  = ,  = ,  =  = ,  = , =  0 1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 4 3 2 1 5 2 1 3 determine: 12.  and   2 3 n + 2 4 4 2 13.3 and 3 14.3 for nZ .

n + –2 –3 –n + 15. 1 for nZ . 16.3 and 3 17.3 for nZ .

–n + 2 3 n + 18.3 for nZ . 19.2 and 2 20.2 for nZ .

n + –2 –3 –n + 21. 2 for nZ . 22.2 and 2 23.2 for nZ . 218 Chapter 5 A Touch of Group Theory

Exercise 24-33. For

123456 123456 123456  =   =   =  234561 214365 654321 Determine: 24.  25.  26.  27.  28.  29. 5 30. 100 31. 101 32. 100 33. 101

34. Let S = 1 . Show that S * with 1*11= is a group. Is the group abelian?

35. Let Sxy=  xy   . Show that S * with xy *xy = xx+ – 1 yy++1 is a group. Is the group abelian?

ab 36. Let M22 = abcd   (the set of two-by-two matrices). Show that M22  * cd

ab a b aa+ bb+  with * = is a group. Is the group abelian? cd c d cc+  dd+ 

37. For n  0 , let Pn denote the set of polynomials of degree less than or equal to n. Show that n n n i i i Pn * with a x * b x = a + b x is a group. Is the group abelian?  i  i  i i i = 0 i = 0 i = 0

38. Let Sxy=  xy   . Show that S * with xy *xy = xx++2 yy+ is a group. Is the group abelian?

39. Let Q denote the set of rational numbers. Show that Q * with a*b= abab++ is not a group.

40. Let QaQa=   –1 . Show that Q * with a*b= abab++ is a group. Is the group abelian?

41. (a) Show, by means of an example, that the exponent law ab n = anbn need not hold in a group G. (b) Prove that the exponential law ab n = anbn does hold if the group G is abelian. (c) Express the property ab n = anbn in sum-notation form.

42. Let G be a group and abc  G . Show that if ab= ac , then bc= , and if ba= ca , then bc= .

+ 43. (a) Show that the group Zn of Theorem 5.1 is cyclic for any nZ .

(b) Prove that mZ n is a generator of Zn if and only if m and n are relatively prime. Suggestion: Consider Theorem 1.7, page 45. 5.1 Definitions and Examples 219

44. Let F denote the set of all real-valued functions. For f and g in F , let fg+ be given by fg+ x = fx+ gx . Show that F  + is a group. Is the group abelian?

45. Let B denote the set of all bijections f:  . Show that B   is a group. Is the group abelian?

46. Let G and H be groups. Let GH = gh gGhH   with:

gh *gh = gg hh (a) Show that GH  * is a group. (b) Prove that GH  * is abelian if and only both G and H are abelian.

PROVE OR GIVE A COUNTEREXAMPLE

47. Let G be a group and abc  G . If bc , then ab ac .

2 48. The group S3 contains four elements  such that  = e and three elements  such that 3 = e .

49. Let G be a group and ab  G . If ab= b , then ac= c for every cG .

50. Let G be a group and ab  G . If ab= ba , then ac= ca for every cG .

51. The cyclic group Z + has exactly two distinct generators. 220 Chapter 5 A Touch of Group Theory

5

§2 ELEMENTARY PROPERTIES OF GROUPS We begin by recalling the group axioms, featuring both the product and the sum notation: Product Form Sum Form (Typically reserved for abelian groups) Closure ab G ab+  G Associative Axiom 1.abc= ab c 1.abc+ + = ab+ + c Identity Axiom 2.ae= a 2.a + 0 = a Inverse Axiom 3. aa–1 = e 3. aa+0– =

For aesthetic reasons, a set of Actually, as we show below, both the identity element of Axiom 2 and axioms should be indepen- dent, in that no axiom or part of the inverse elements of Axiom 3 work on both sides; but first: an axiom is a consequence of 2 the rest. One should not, for LEMMA 5.1 Let G be a group. If aG is such that a = a , example, replace Axiom 2 in Definition 5.1, page 207: then ae= . eGaeaaG  =   PROOF: with: –1 –1 –1 e  Gaeea = = aa  G aa= a  aa a = aa  aaa= eaeea =  = e Axiom 1 Axiom 3 Axiom 2

THEOREM 5.6 Let G be a group. ForaG : (a) aa–1 = ea –1ae= (b) ae= a ea = a

PROOF:

(a) a–1a a–1a ===a–1aa–1 aa–1e aa–1a

Axiom 1 Axiom 3 Axiom 2 Since a–1a a–1a = a–1a : a–1ae= (see Lemma 2.1).

(b) ea==== aa–1 aaa–1a ae a

Axiom 3 Axiom 1 part (a) Axiom 3

Axioms 2 and 3 stipulates the existence of an identity in a group, and of an inverse for each element of the group. Are they necessarily unique? Yes: THEOREM 5.7 (a) There is but one identity in a group G. (b) Every element in G has a unique inverse. 5.2 Elementary Properties of Groups 221

PROOF: (a) We assume that e and e are identities, and go on to show that ee= : Since e is an identity eeee== Since e is an identity

(b) We assume that a–1 and a–1 are inverses of aG , and show that a–1 = a–1 . Since aa–1 and aa–1 are both equal to the identity they must be equal to each other: aa–1 = aa–1 Multiply both sides by a–1: a–1aa–1 = a–1aa–1 Associativity: a–1a a–1 = a–1a a–1 ea–1 = ea–1 a–1 = a–1

CHECK YOUR UNDERSTANDING 5.6

Show that if abc are elements of a group such that abc= e , then Answer: See page A-28. bca= e.

The left and right cancellation laws hold in groups. In any group G: Sum form: THEOREM 5.8 ab+ = cb+  a = c (a) If ab = cb , then ac= . ba+ ==bc+  a c (b) If ba = bc , then ac=

PROOF: Just in case you are ask- (a) ab= cb (b) ba= bc ing yourself: What if b is 0 and –1 –1 –1 –1 has no inverse? ab b = cb b b ba = b bc Tisk, every element in a –1 –1 group has an inverse. abb–1 = cbb–1 b b ab= b c ae= ce ea= ec ac= ac=

CHECK YOUR UNDERSTANDING 5.7

PROVE OR GIVE A COUNTEREXAMPLE: (a) In any group G, if ab= bc then ac= . Answer: See page A-28. (b) In any abelian group G, if ab+ = bc+ then ac= . 222 Chapter 5 A Touch of Group Theory

In the real number system, do linear equations ax= b have unique solutions for every ab   ? No — the equation 0x = 5 has no solu- tion, while the equation 0x = 0 has infinitely many solutions. This observation assures us that the reals is not a group under multiplication, for:

THEOREM 5.9 Let G be a group. For any ab  G , the linear equations ax= b and ya= b have unique solutions in G.

PROOF: Existence: (a) ax= b (b) ya= b a–1ax = a–1b ya a–1 = ba–1 a–1a xa= –1b yaa–1 = ba–1 ex= a–1b ye= ba–1 xa= –1b yba= –1

Uniqueness: We assume (as usual) that there are two solutions, and then proceed to show that they are equal: Theorem 5.8(b)

(a) ax1 ==b and ax2 b  ax1 ==ax2  x1 x2 Theorem 5.8(a)

(b) y1ab== and y2ab y1a ==y2ay 1 y2

CHECK YOUR UNDERSTANDING 5.8 Since the set of real numbers under addition is a group, Theorem 5.9 applies. Show, directly, that any linear equation in  + has a Answer: See page A-29. unique solution. The inverse of a product is the product of the inverses, but in reverse order:

This is another shoe-sock THEOREM 5.10 For every ab in a group G: theorem (see page 73). ab –1 = b–1a–1

PROOF: To show that b–1a–1 is the inverse of ab is to show that b–1a–1 ab = e . No problem:

b–1a–1 ab ====b–1a–1a bb–1eb b–1be 5.2 Elementary Properties of Groups 223

CHECK YOUR UNDERSTANDING 5.9 Give an example of a group G for which ab –1 = a–1b–1 does not Answer: See page A-29. hold for every ab  G . The associative axiom of a group G assures us that an expression such as abc , sans parentheses, is unambiguous [since ab c and abc yield the same result]. It is plausible to expect that this nicety

extends to any product a1a2an of elements of G. Plausible, to be sure; but more importantly, True:

THEOREM 5.11 Let a1a2an  G . The product expression

a1a2an is unambiguous in that its value is independent of the order in which adjacent factors are multiplied.

PROOF: [By induction (page 33)]: I. The claim holds for n = 3 (the associative axiom). II. Assume the claim holds for nk= , with k  3 . III. (Now for the fun part) We show the claim holds for nk= + 1 :

Let x denote the product a1a2ak + 1 under a certain pair- ing of its elements, and y the product under another pairing of its elements. We are to show that xy= . Assume that one starts the multiplication process with the following pairing for x and y: A B C D xa==1a2ai ai + 1ak + 1 and ya1a2aj aj + 1ak + 1 Case 1. ij= : By the induction hypothesis (II), no matter how the products in A and C are performed, A will equal C. The same can be said concerning B and D. Consequently xABCDy== =. Case Assume, without loss of generality, thatij . Break- ing the “longer” product B into two pieces M and D we have: A M D xa= 1a2ai ai + 1aj aj + 1ak + 1 By the induction hypothesis, A, M, and D are well defined (independent of the pairing of its elements in their products). Bringing us to: I: Claim holds for n = 3 xABAMD== =AM DCDy = = 224 Chapter 5 A Touch of Group Theory

CHECK YOUR UNDERSTANDING 5.10 Use the Principle of Mathematical Induction, to show that for any a1a2an  G : Answer: See page A-29. –1 –1 –1 –1 ana2a1 = a1 a2 an

THEOREM 5.12 For any given element a of a finite group G: am = e for some mZ + .

PROOF: Let G be of order n. Surely not all of the n + 1 elements aa2 a3  an + 1 can be distinct. Choose 1  stn  + 1 such that at = as . Since ata–s ==ats– e : am = e , for mts= – .

DEFINITION 5.5 Let G be a finite group, and let aG In the additive notation, am = e translates to ORDER OF AN The smallest positive integer m such that na = 0 ; which is to say: ELEMENT OF G am = e is called the order of a and is aa++ +a = 0 sum of n as denoted by oa . If no such element exists, then a is said to have infinite order.

EXAMPLE 5.3 (a) Determine the order of the element 4 in the group Z6 +6 .

(b) Determine the order of the element 1 2 3 4 5  =  3 2 4 1 5

in the symmetric group S5 . SOLUTION: (a) Since: 14= 4

24==4+642

34===4+64+642+640

The element 4 has order 3 in Z6 . (b) Since: 1 2 3 4 5   3 2 4 1 5 2 e 4 2 1 3 5 3   1 2 3 4 5 5.2 Elementary Properties of Groups 225

1 2 3 4 5 The element  =  has order 3 in S . 3 2 4 1 5 5

CHECK YOUR UNDERSTANDING 5.11 1 2 3 4 (a) Determine the order of the element  =  in S . 2 3 4 1 4

Answer: (a) 4 (b) 6 (b) Determine the order of the element 4 in Z24 . 226 Chapter 5 A Touch of Group Theory

EXERCISES

1. Let G be a group and abc  G . Solve for x, if: (a) axa–1 = e (b) axa–1 = a (c) axb= c (d) ba–1xab–1 = ba

2. Let G be a group. Prove that a–1 –1 = a for every aG .

3. Prove that for any element a in a group G the functions fa: GG given by fab = ab and

the function ga: GG given by gab = ba are bijections.

4. Let G be a finite group. Show that for any aG there exists mZ + such that am = e .

5. Prove that a group G is abelian if and only if ab –1 = a–1b–1 for every ab  G .

6. Let G be group for which a–1 = a for every aG . Prove that G is abelian.

7. Let G be group for which ab 2 = a2b2 for every ab  G . Prove that G is abelian.

8. Let G be a finite group consisting of an even number of elements. Show that there exists aG , ae , such that a2 = e .

9. (a) Let G be a group. Show that if, for any ab  G , there exist three consecutive integers i such that ab i = aibi then G is abelian. (b) Let G be an abelian group. Show that for any aG and nZ : –n ana= – .

–1 10. Let G be a group and aG . Define a new operation * on G by b*cba= c for all bc  G . show that G * is a group.

11. Let G be a group and ab  G . Use the Principle of Mathematical Induction to show that for any positive integer n: a–1ba n = a–1bna .

12. Let aG be of order n. Find a–1 .

13. List the order of each element in the Symmetric group S3 of Figure 2.6, page 7.

14. Let aG be of order n. Prove that as = at if and only if n divides st– .

15. Prove that if a2 = e for every element a in a group G, then G is abelian. 5.2 Elementary Properties of Groups 227

PROVE OR GIVE A COUNTEREXAMPLE

16. If abc are elements of a group such that abc= e , then cba= e .

17. In any group G there exists exactly one element a such that a2 = a .

18. In any group G, ab –2 = b–2a–2 .

19. If ab 2 = a2b2 in a group G, then ab= ba .

20. Let G be a group. If abc= bac then ab= ba .

21. Let G be a group. If abcd= bacd then ab= ba .

22. Let G be a group. If abc –1 = a–1b–1c–1 then ac= . 228 Chapter 5 A Touch of Group Theory

5

§3. SUBGROUPS

DEFINITION 5.6 A subgroup of a group G is a nonempty SUBGROUP subset H of G which is itself a group under the imposed binary operation of G. As it turns out, apart from closure, to challenge whether or not a non- GROUP AXIOMS empty subset of a group is a subgroup you need but challenge Axiom3: Closure: ab G Axiom 1. abc= ab c THEOREM 5.13 A nonempty subset S of a group G is a subgroup Axiom 2. ae= a of G if and only if: Axiom 3. aa–1 = e (i) S is closed with respect to the operation in G. (ii) aS implies that a–1  S .

PROOF: If S is a subgroup, then (i) and (ii) must certainly be satisfied. Conversely, if (i) and (ii) hold in S, then Axioms 1 and 2 hold: Axiom 1: Since abc= ab c holds for every abc  G , that associative property must surely hold for every abc  S. Axiom 2: Since ae= a for every aG , then surely ae= a for every aS . It remains to be shown that eS . Lets do it: Choose any aS . By (ii): a–1  S . By (i): aa–1 = eS .

When challenging if SG is a subgroup, we suggest that you first determine if it contains the identity element. For if not, then S is not a subgroup, period. If it does, then S   and you can then proceed to challenge (i) and (ii) of Theorem 2.13.

EXAMPLE 5.4 Show that for any fixed nZ the subset For example: 5Z = –510 –0510 nZ=  nm m Z is a subgroup of Z + .

SOLUTION: Since 0 = n  0  nZ , nZ   . (i) nZ is closed under addition:

nm1 + nm2 = nm1 + m2  nZ

We remind you that, under (ii) For any nm nZ : addition, –a rather than –nm = nm–  nZ a–1 is used to denote the inverse of a. Conclusion: nZ is a subgroup of Z (Theorem 5.13). 5.3 Subgroups 229

CHECK YOUR UNDERSTANDING 5.12 The previous example assures us that 3Z is a subgroup of Z + . As Answer: See page A-29. such, it is itself a group. Show that 6Z is a subgroup 3Z . You are invited to show in the exercises that the following result holds for any collection of subgroups of a given group: THEOREM 5.14 If H and K are subgroups of a group G, then HK is also a subgroup of G.

PROOF: Since H and K are subgroups, each contains the identity ele- ment. It follows that eHK  and that therefore HK . We now verify that conditions (i) and (ii) of Theorem 5.13 are satisfied: (i) (Closure) If ab  H K , then ab  H and ab  K . Since H and K are subgroups, ab H and ab K . It follows that ab H K. (ii) (Inverses) If aHK  , then aH and aK . Since H and K are subgroups, a–1  H and a–1  K . consequently, a–1  HK .

CHECK YOUR UNDERSTANDING 5.13 PROVE OR GIVE A COUNTEREXAMPLE: If H and K are subgroups of a group G, then HK is also a sub- Answer: See page A-30. group of G. We recall the definition of a cyclic group appearing on page 215: A group G is cyclic if there exists aG such that Ga= n nZ .

DEFINITION 5.7 Let G be a group, and aG . The cyclic group a = an nZ is called the cyclic subgroup of G generated by a. (In sum form: a = na n Z )

CHECK YOUR UNDERSTANDING 5.14 Answer: 3 = Z8 4 = 04 For GZ= 8 , determine 3 and  4 . (Use sum notation.) 230 Chapter 5 A Touch of Group Theory

THEOREM 5.15 Any subgroup of a cyclic group is cyclic.

PROOF: Let G be a cyclic group generated by a: Ga= . Let H be a subgroup of G. Case 1. He=  , then He=  is cyclic with generator e. Case 2. He  . Let m be the smallest positive integer such that am  H . We show Ha= m by showing that every hH is a power of am : Let ha= n  H . Employing the Division Algorithm of page 43, we chose integers q and r, with 0  rm , such that: nmqr= + . And so we have: h ==an amq+ r =am qar (*) or: ar = am –qan (**) Since an and am are both in H, and since H is a group: am –qan  H . Consequently, from (**): ar  H . Since 0  rm and since m is the smallest positive integer such that am  H : r = 0 . Consequently, from (*): ha==n am qa0 =am q — a power of am .

Here is a particularly important result: THEOREM 5.16 If G is a finite group and H is a subgroup of Joseph-Louis Lagrange G, then the order of H divides the order of G: (1736-1813) (Lagrange) H G (see Definition 5.2, page 210) While the converse of The- orem 5.17 holds for abelian groups, it does not hold in To illustrate: If a group G contains 35 elements, it cannot general. In particular, there contain a subgroup of 8 elements, as 8 does not divide 35. exists a group of order 12 that does not contain a sub- group of order 6 (The so A proof of Lagrange’s Theorem is offered at the end of the section.We called alternating group of now turn to a few of its consequences, beginning with: degree 4). THEOREM 5.17 Any group G of prime order is cyclic.

PROOF: Let Gp= , where p is prime. Since p  2 , we can choose an element aG distinct from e . By Lagrange’s theorem, the order of the cyclic group a = an nZ must divide p. But only 1 and p divide p, and since a contains more than one element, it must con- tain p elements, and is therefore all of G. 5.3 Subgroups 231

The symmetric group S 3 THEOREM 5.18 Every group of order less than 6 is abelian. is an example of a non- abelian group of order 6. PROOF: We know that Z4 and the Klein group K are the only groups of order 4, and each is abelian. Clearly the trivial group e of order 1 is abelian. Any group or order 2 or 3, being of prime order, must be cyclic (Theorem 5.17), and therefore abelian (Theorem 5.5, page 216).

We remind you that oa THEOREM 5.19 For any element a in a finite group G: denotes the order of a (Definition 5.5, page 224). oa G

PROOF: If oa= m , then a = aa2  am – 1 am = e is a subgroup of G consisting of m elements. Consequently: oa G .

If n = 1 , then Ge=  THEOREM 5.20 If G is a finite group of order n, then an = e for every aG .

PROOF: Let aG , with oa= m . Since m divides n (Lagrange’s Theorem), ntm= for some tZ + . Thus: an ==atm am t ==et e

CHECK YOUR UNDERSTANDING 5.15 PROVE OR GIVE A COUNTEREXAMPLE Let G be a group with ab  G . If oa= n and ob= m , then Answer: See page A-30. ab nm = e .

PROOF OF LAGRANGE’S THEOREM We begin by recalling some material from Chapter 1: An equivalence relation ~ on a set X is a relation which is See Definition 2.20 page 88. Reflexive: x~x for every xX , Symmetric: If x~y , then y~x , Transitive: If x~y and y~z , then x~z .

See Definition 2.21 page 91. For x0  X the equivalence class of x0 is the set:

x0 = xXx ~x0 .

LEMMA 5.2 Let H be a subgroup of a group G. The relation a~b if ab–1  H is an equivalence relation on G. Moreover: a = ha h H 232 Chapter 5 A Touch of Group Theory

PROOF: ~ is reflexive: x~x since xx–1 = eH . ~ is symmetric:. a~b  ab–1 = h for some hH H is a group  ab–1 –1 = h–1 Theorem 5.11, page 223:  b–1 –1a–1 = h–1 Exercise 1, page 226:  ba–1 = h–1  b~a since h–1  H ~ is transitive: If a~b and b~c , then: ab–1  H and bc–1  H  ab–1 bc–1  H  ab –1b c–1  H

 aec–1  H  ac–1  H  a~c Having established the equivalence part of the theorem, we now ver- ify that a = ha h H : ba  b~a  ba–1 = h for some hH  b = ha for some hH  bhahH 

NOTE: The above set ha h H will be denoted by Ha : Ha=  ha h H

We are now in a position to offer a proof of Lagrange’s Theorem: If H is a subgroup of a finite group G, then H G .

PROOF: Theorem 2.15(a), page 93, and Lemma 5.2, tell us that sets Ha a G partition G. Since G is finite, we can choose k a1a2  ak such that GHa=  i with Hai Haj = if ij . i = 1 We now show that each Hai has the same number of elements as H

by verifying that the function fi: HHa i given by fih = hai is a bijection:

fi is one-to-one: fih1 = fih2  h1a = h2a –1 –1  h1a a = h2a a  h aa–1 ==h aa–1  h h 1 2 1 2 5.3 Subgroups 233

fi is onto: For any given hai  Hai , fih = hai .

Since G is the disjoint union of the k sets Ha1Ha2  Hak , and since each of those sets contains H elements: GkH= , and there- fore: H G . 234 Chapter 5 A Touch of Group Theory

EXERCISES

Exercise 1-5. Determine if the given subset S is a subgroup of Z + . 1.Snn=  is even 2.Snn=  is odd 3. Snn=  is odd

4.Snn=  is divisible by 2 and 3 5. Snn=  is divisible by 2 or 3

Exercise 6-8. Determine if the given subset S is a subgroup of Z8 +n (see Theorem 5.1, page 208). 6.S = 0246 7.S = 036 8. S = 0234

Exercise 9-12. Determine if the given subset S is a subgroup of  + . 9.Sxx= = 7y for y   10. Sxx= = 7y for y  0

11.Sxx= = 7 + y for y   12. Sxx= = 7 + y for y  0

Exercise 13-18. Determine if the given subset S is a subgroup of S3  where: 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3  = ,  = ,  =  = ,  = , =  0 1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 4 3 2 1 5 2 1 3

13.S = 0 1 14.S = 0 2 15. S = 0 2

16.S = 01 2 17.S = 03 5 18. S = 12 3 4 5

19. Prove that if e and G are the only subgroups of a group G, then G is cyclic.

20. Prove that if e and G are the only subgroups of a group G, then G is cyclic of order p, for p prime.

21. Show that if S is a nonempty subset of a group G for which ab–1  S for every ab  S , then S is a subgroup of G.

22. Let G be an abelian group. Show that aGa 2 = e is a subgroup of G.

23. Let G be an abelian group. Show that for any integer n, aGa n = e is a subgroup of G. 24. Prove that the subset of elements of finite order in an abelian group G is a subgroup of G (called the torsion subgroup of G). 25. Prove that if H is a subgroup of a subgroup K of a group G, then H is a subgroup of G. 26. Let a be an element of a group G. The set of all elements of G which commute with a: Ca= bGab = ba is called the centralizer of a in G. Prove that Ca is a subgroup of G.

27. Let G be a group and ab  G . Prove that Cb–1ab = b–1Cab , where b–1Cabb= –1cb c C a (see Exercise 26.) 5.3 Subgroups 235

28. The center ZG of a group G is the set of all elements in G that commute with ever ele- ment of G: ZG= aGab = ba for all bG . (a) Prove that ZG is a subgroup of G. (c) Prove that aZG  if and only if Ca= G . (c) Prove that aZG  if and only if Ca= G (see Exercise 26.) (b) Prove that ZG=  Ca . aG 29. Let H be a subgroup of a group G. The centralizer CH of H is the set of all elements of G that commute with every element of H: CH= aGah = ha for all hH . Prove that CH is a subgroup of G. 30. Show that Table C in Figure 5.4, page 212, can be derived from Table B by appropriately relabeling the letters e, a, b, c in B.

31. Let H and K be subgroups of an abelian group G. Verify that HK=  hk h H and kK is a subgroup of G.

32. Let H and K be subgroups of a group G such that k–1Hk H for every kK . Show that HK=  hk h H and kK is a subgroup of G.

33. Let G be a finite group, and ab  G . (a) Prove that the elements aa –1 and bab–1 have the same order. (b) Prove that ab and ba have the same order. Suggestion: Apply (a) to ab= a ba a–1

34. Prove that H is a subgroup of a group G if and only if HH–1 = ab–1 ab  H  H .

35. Let H and K be subgroups of an abelian group G of orders n and m respectively. Show that if HK = e , then HK=  hk h H and kK is a subgroup of G of order nm.

36. (a) Prove that the group Z + contains an infinite number of subgroups. (b) Prove that any infinite group contains an infinite number of subgroups.

37. Let S be a finite subset of a group G. Prove that S is a subgroup of G if and only if ab S for every ab  S .

n 38. (a) H n be subgroups of a group G. Show that H is also a subgroup of G. i i = 1  i i = 1  (b) Let H  be a collection of subgroups of a group G. Show that H is also a i i = 1  i subgroup of G. i = 1 (c) Let H be a collection of subgroups of a group G. Show that H is also a    A     A subgroup of G. 236 Chapter 5 A Touch of Group Theory

PROVE OR GIVE A COUNTEREXAMPLE

39. In any group G, aGa n = e for some nZ is a subgroup of G.

40. In any abelian group G, aGa n = e for some nZ is a subgroup of G.

41. If H and K are subgroups of a group G, then HK=  hk h H and kK is also a subgroup of G.

42. In any group G, aGa 3 = e is a subgroup of G.

43. No nontrivial group can be expressed as the union of two disjoint subgroups. 5.4 and Isomorphisms 237

5

§4. HOMOMORPHISMS AND ISOMORPHISMS Up until now we have focused our attention exclusively on the inter- nal nature of a given group. The time has come to consider links between groups: The word homomorphism comes from the Greek DEFINITION 5.8 A function : GG  from a group G to a homo meaning “same” and HOMOMORPHISM group G is said to be a homomorphism if morph meaning “shape.” ab = a b for every ab  G . Let’s focus a bit on the equation: ab = a b The operation, ab, on the left side of the equation is taking place in the group G while the one on the right, a b , takes place in the group G . What the statement is saying is that you can perform the product in G and then carry the result over to the group G (via  ), or you can first carry a and b over to G and then perform the product in that group. Those groups and products, however, need not resemble each other. Consider the following examples:

You can easily verify that EXAMPLE 5.5 Let GZ=  + , and let G = –11 under G = –11 , under stan- dard multiplication standard integer multiplication (see margin). Show that f: GG  given by: *1– 1 11– 1  1if n is even –11 –1 n =  is a group.  –if1 n is odd is a homomorphism.

SOLUTION: We consider three cases: Case 1. (Both integers are even). If a = 2n and b = 2m , then: ab+ ==2n + 2m 1 (since 2n + 2m is even) And also: a b === 2n 2m 11 1 . Case 2. (Both are odd). If a = 2n + 1 and b = 2m + 1 , then: ab+ ===2n + 1 + 2m + 1 2n ++2m 2 1 And also: a b === 2n + 1 2m + 1 –1 –1 1 .

Since Z + is abelian, we Case 3. (Even and odd). If a = 2n and b = 2m + 1 , then: need not consider a = 2n + 1 ab+ ===2n + 2m + 1 2nm+ + 1 –1 and b = 2m And also: a b === 2n 2m + 1 1 –1 –1 . 238 Chapter 5 A Touch of Group Theory

See page 208 for a discus- EXAMPLE 5.6 Show that the function : Z +  Z  +n sion of the group Zn +n . n given by m = r where mnqr= + with 0  rn is a homomorphism.

SOLUTION: Let anq= 1 + r1 with 0  r1  n , bnq= 2 + r2 with 0  r2  n , and r1 + r2 = nq3 + r3 with 0  r3  n . Then:

ab+ = nq1 + r1 + nq2 + r2

= nq1 + q2 + r1 + r2

= nq1 + q2 + nq3 + r3 with 0  r3  n

==nq1 ++q2 q3 + r3 r3 (since 0  r3  n) same

And: a +n b = nq1 + r1 +n nq2 + r2

==r1+n r2 r3 (since r1 + r2 = nq3 + r3 with 0  r3  n 

EXAMPLE 5.7 For any fixed element a in a group G, let f : G  G be given by f g = ag . Show See page 213 for a discussion on a a the symmetric group S  . G  that the function : GS G  given by a = fa is a one-to-one homomorphism.

SOLUTION:  is one-to-one:

a = b  fa = fb  fae = fbe  ae ==be  a b in particular To show that  is a homomorphism we need to show that

ab = a b , which is to say, that the function fab: G  G is equal to the function fafb: G  G . Let’s do it: For any xG : fabx = ab x

and fafb x ===fafbx fabx abx By associativity, ab xabx=  , and we are done.

CHECK YOUR UNDERSTANDING 5.16 (a) Show that for any two groups G and G the function : G  G given by a = e for every aG is a homomorphism (called the trivial homomorphism from G to G ). (b) Let + denote the group of positive real numbers under multipli- cation. Prove that the function : Z +  + , given by Answer: See page A-30. n = n is a homomorphism. 5.4 Homomorphisms and Isomorphisms 239

Homomorphisms preserve identities, inverses, and subgroups: THEOREM 5.21 Let : GG  be a homomorphism. Then: (a) e = e (b) a–1 =  a –1 (c) If H is a subgroup of G, then H is a subgroup of G . (d) If H is a subgroup of G , then –1H is a subgroup of G .

PROOF: (a) Since  is a homomorphism: e == ee e e . Multiplying both sides by e –1 yields the desired result: e –1e =  e –1e e e= e –1e e e ==ee e

(b) Since a–1 a === a–1a e e : (a) a–1 =  a –1 . (c) We use Theorem 5.13, page 228, to show that the nonempty set H is a subgroup of G : Since a b = ab :H is closed with respect to the operation in G . Since, for any aG , a–1 =  a –1 : a –1  H for every a  H . (d) We use Theorem 5.13, page 228, to show that the nonempty set –1H is a subspace of G : Let ab  –1H . To say that ab  –1H is to say that ab  H , and it is: Since ab = a b , and since H , being a subgroup of G , is closed with respect to the operations in G : ab  H . Let a  –1H . To say that a–1  –1H is to say that a–1  H , and it is: Since a–1 =  a –1 , and since H contains the inverse of each of its elements: a–1  H . CHECK YOUR UNDERSTANDING 5.17 Let : GG  and : G  G be homomorphisms. Prove that the Answer: See page A-30. composite function : GG  is also a homeomorphism. 240 Chapter 5 A Touch of Group Theory

IMAGE AND KERNEL For any given homomorphism : GG  , we define the kernel of  to be the set of elements in G which map to the identity e  G [see Figure 5.7(a)]. We define the set of all elements in G which are “hit” by some a to be the image of  [see Figure 5.7(b)].

G G G G  .  e Kernel of  Image of  (a) (b) Figure 5.7 More formally:

DEFINITION 5.9 Let : GG  be a homomorphism. KERNEL The kernel of  , denoted by Ker , is Utilizing the notation of Definition 2.8, page 63: given by: Ker = –1e Ker = aG a = e Im = G IMAGE The image of  , denoted by Im , is given by: Im =  a aG

Both the kernel and image of a homomorphism turn out to be sub- groups of their respective groups:

THEOREM 5.22 Let : GG  be a homomorphism. Then: (a)Ker is a subgroup of G. (b)Im is a subgroup of G .

PROOF: We appeal to Theorem 5.14, page 228 to establish the desired results. (a) Nonempty: Since e = e [Theorem 5.22(a)], e  Ker .

Closure: ab Ker ab ===a b ee e  is a homomorphism

Inverses: For a  Ker , we are to show that a–1  Ker . Let’s do it: a–1 === a –1 e –1 e Theorem 5.22(b) 5.4 Homomorphisms and Isomorphisms 241

(b) Nonempty: Since e = e , e  Im Closure: For a b  Im choose ab  G such that a = a and b = b . Then: ab == a b ab  ab  Im .  is a homomorphism Inverses: For a  Im choose aG such that a = a . Then: a–1 == a –1 a –1  a –1  Im Theorem 5.22(b)

CHECK YOUR UNDERSTANDING 5.18 Show that the function : 2Z  4Z given by 2n = 8n is a Answer: See page A-31. homomorphism. Determine the kernel and image of  . Definition 5.9 tells us that a homomorphism : GG  is onto if and

A homomorphism : GG  only if Im = G . The following result is a bit more interesting, in must map e to e . What this that it asserts that in order for a homomorphism to be one-to-one, it theorem is saying is that if e is need only behave “one-to-one-ish” at e (see margin): the only element that goes to e , then no element of G is THEOREM 5.23 A homomorphism : GG  is one-to-one going to be hit by more that one element of G. This is certainly if and only if Ker = e . not true for arbitrary functions: y PROOF: Suppose  is one-to-one. If a  Ker , then both a = e and e = e [Theorem 5.22(a)]. Consequently ae= (since  is assumed to be one-to-one). Hence: Ker = e . x Conversely, assume that Ker = e . We need to show that if fx= x a = b , then ab= . Let’s do it: a = b a b –1 = e Theorem 5.22(b): a b–1 = e –1  is a homomorphism: ab = e Ker = e : ab–1 = e ab–1 beb= ab=

CHECK YOUR UNDERSTANDING 5.19

Prove that a homomorphism : GG  is one-to-one if and only if there exists an element aG (not necessarily the identity e) such that if a = b then ab= . In other words: for a homomorphism : GG  to be one-to-one, it Answer: See page A-31. need only behave “one-to-one-ish” at any one-point in G.” 242 Chapter 5 A Touch of Group Theory

The word ISOMORPHISMS comes from the Greek iso meaning “equal” and As previously noted, a homomorphism : GG  preserves the morph meaning “shape.” in that ab = a b . An isomorphism also preserves the set structures in that it pairs up the elements of G with those of G . More formally:

DEFINITION 5.10 A homomorphism : GG  which is ISOMORPHISM also a bijection is said to be an isomor- phism from the group G to the group G . ISOMORPHIC Two groups G and G are isomorphic, written GG  , if there exists an isomor- phism from one of the groups to the other.

EXAMPLE 5.8 Show that the group  + of real numbers under addition is isomorphic to the group + . of positive real numbers under multi- plication. SOLUTION: We show that the function :  +  + . given by In this discussion we are not a = ea is a homomorphism that is also a bijection: using e to denote the identity element in + . (which is  is a homomorphism: 1). Here, e is the transcen- ab+ ===eab+ eaeb a b dental number e  2.718 .  is one-to-one:

The identity in + . The identity in  + a ===1  ea 1  a 0

 is a onto: For a  + . , we have: lna ==elna a .

CHECK YOUR UNDERSTANDING 5.20 (a) Prove that  is an equivalence relation on any set of groups (see Answer: See page A-31. Definition 2.19, page 88). (b) Let gG . Prove that the map ig: GG given by –1 igx = gxg xG is an automorphism (called an inner automorphism.) 5.4 Homomorphisms and Isomorphisms 243

A ROSE BY ANY OTHER NAME

Let : GG  be an isomorphism. Being a bijection it links every element in G with a unique element in G (every element in G has its own G counterpart, and vice versa). Moreover, if you know how to –1 a  a function algebraically in G, then you can also figure out how to func- –1 b b tion algebraically in G (and vice versa). Suppose, for example, that ab ab you forgot how to multiply in the group G , but remember how to mul-  tiply in G. To figure out ab in G you can take the “–1 bridge” back to G to find the elements a and b such that a = a and b = b , perform the product ab in G, and then take the “ bridge” back toG to find the product ab : ab . Basically, if a group G is isomorphic to G , then the two groups can only differ in appearance, but not “algebraically.” Consider, for exam- ple, the following two groups which appeared previously in Figure 5.1, page 210:

+4 0123 * eabc Z4: K: 00123 eeabc 11230 aaecb 22301 bbcea 33012 ccbae (a) (b) Both contain four elements ({0,1,2,3} and {e,a,b,c}); so, as far as sets go, they “are one and the same” (different element-names, that’s all). But as far as groups go, they are not the same (not isomorphic). Here are two algebraic differences (either one of which would serve to prove that the two groups are not isomorphic):

1.Z4 is cyclic while the Klein 4-group, K, is not. 2. There exist three elements in K of order 2 (see Definition

5.5, page 224), while Z4 contains but one (the element 2). To better substantiate the above claims:

THEOREM 5.24 If GG  , then: (a) G is cyclic if and only if G is cyclic. (b)For any given integer n, there exists an element aG such that an = e if and only if there exists an element a  G such that a n = e . 244 Chapter 5 A Touch of Group Theory

PROOF: Let : GG  be an isomorphism. (a) Suppose G is cyclic, with Ga=  . We show G= a by showing that for any b  G , there exists nZ such that b= a n : Let bG be such that b = b . Since Ga=  , there exists nZ such that ba= n . Then: b==b an = a n Exercise 13 The “only-if” part follows from the fact that if G is isomorphic to G , then G is isomorphic to G (see CYU 5.20). (b)Let aG be such that an = e . Then: a n ===an e e The “only-if” part follows from CYU 5.20.

CHECK YOUR UNDERSTANDING 5.21 Answer: See page A-31. Prove that if GG  , then G is abelian if and only if G is abelian. A property of a group G that is shared by any other group isomorphic to G is said to be an invariant property of G. For example, G , abelian, and cyclic, are group-invariant properties. Other group invariant proper- ties are cited in the exercises. In general, one can show that two groups are not isomorphic by exhib- iting a group invariant property that holds in one of the groups but not in the other.

The following results underlines the importance of symmetric groups (see discussion on page 213).

Arthur Cayley (1821-1885) THEOREM 5.25 Every group is isomorphic to a subgroup of (Cayley) a symmetric group.

PROOF: Let G be a group. We show that G is isomorphic to a subgroup

of the symmetric group SG  .

Step 1. The function : GS G given by

g ==fg: GG where fgx gx (xG  is a homomorphism:

gg = fgg and g g = fgfg Since, xG :

fggx ===gg xggx fggx =fgfg x , gg = g g 5.4 Homomorphisms and Isomorphisms 245

Step 2. The function  is one-to-one: g == g  gx gx xG  ge ==ge g g Note: Step 3: G is isomorphic to the subgroup G of the symmetric G = ImG = g gG group SG . Indeed, in general: CHECK YOUR UNDERSTANDING 5.22 If G and G are groups, and if : GG  be a one-to-one homo- Answer: See page A-31. morphism, then : G  G is an isomorphism. 246 Chapter 5 A Touch of Group Theory

EXERCISES

Exercise 1-11. Determine if the given function : GG  is a homomorphism.

1.GG== Z + and n = 2n .

2.GG== Z + and n = n + 1 .

3.G == +  G Z + and x = n where n is the smallest integer greater than or equal to x.

4.GZ== + G  + and n = n .

5.GZ== +  G S3 and n = r where n = 3mr+ with 0  r  3 .

6.GZ== +  G –11  . with n = 1 if n is even and fn= –1 if n is odd.

7.GG=  and a = a–1 for aG .

8.GG=  with G abelian, and a = a–1 for aG .

9.GG=  with G abelian, nZ + , and a = an for aG .

10.GZ= 6 , G = Z2 and n = r where n = 2dr+ with 0  r  2 (see Theorem 1.6, page 43).

11.GZ= 5 , G = Z2 and n = r where n = 2dr+ with 0  r  2 (see Theorem 1.6, page 43).

12. Let  + denote the group of all real numbers under addition, and + . the group of all positive real numbers under multiplication. Show that the map : +   given by x = lnx is an isomorphism.

13. Let : GG  be a homomorphism and let aG . Prove that an =  a n for every nZ .

14. Let : GG  be a homomorphism. Prove that for all ab  G : ab–1 ==a b –1 and a–1b a –1b

15. Prove that a group G is abelian if and only if the function f: GG given by fg= g–1 is a homomorphism.

16. Let Ga=  be cyclic and let G be any group. Let : GG  be a homomorphism. Prove that Im is cyclic.

17. Let : GG  be a homomorphism. Show that if k  Ker , then gkg–1  Ker for every gG . 5.4 Homomorphisms and Isomorphisms 247

18. Let Ga=  be cyclic and let H be any group. Prove that for any chosen hH there exists a unique homomorphism : GH such that a = h . So, a homomorphism on a cyclic group Ga=  is completely determined by its action on a.

19. Let : GG  be a homomorphism. Prove that, for any given xG : gGg = x = xk k  Ker

20. (a) Show that the set ZZ = ab ab  Z , with ab *cd = ac+  bd+ is a group.

(b)Verify that the functions 1: ZZ  Z and 2: ZZ  Z given by 1ab = a and

2ab = b , respectively, are homomorphisms.

(c) Show that the function : ZZ  Z given by ab = 21ab + 32ab is a homomorphism.

(d)Show that the function : ZZ  ZZ given by ab =  2ab  1ab is an isomorphism.

21. For mZ , let m: ZZ be given by mn = mn .

(a) Show that m is a one-to-one homomorphism.

(b) Show that m is an isomorphism if and only if m = –1 .

22. Let F denote the additive group of real valued function (see Exercise 44, page 218), and let  denote the additive group of real numbers. Prove that for any c   the function c: F   given by cf = fc for fF  is a homomorphism (called an evalu- ation homomorphism.)

23. Let D denote the set of differentiable functions from  to  . (a) Show that D  + is a group.

(b) Show that for any c   the function c: D   given by cf = fc is a homomorphism.

(c) Is c one-to-one for any c?

(d) Is c onto for any c?

24. Let I denote the set of continuous real valued functions. (a) Show that I  + is a group. (b) Show that for any closed interval ab in  the function : I   given by b f = fxdx is a homomorphism. a 1 3 (c) Show that the function : I   given by f = fxdx + 2 fxdx is a 0 2 homomorphism. 248 Chapter 5 A Touch of Group Theory

25. Show that for any   S3 , the function : S3  S3 given by =  is a homomor- phism. Is it necessarily an isomorphism?

i if i  4 26. Show that the function : S3  S4 given by i = is a homomor- 4ifi = 4 phism. Determine the image and kernel of  Exercise 27-28. (Automorphism) An isomorphism from a group to itself is said to be an auto- morphism.

27. Let gG . Prove that : GG given by x = gxg–1 for every xG is an automor- phism.

28. Let G be a group. Prove that AutG =  : GG is an automorphism   is a group. Exercise 29-34. Show that the give property on a finite group G is an invariant.

29.G — the order of G. 30. G contains a nontrivial cyclic subgroup.

31. G contains an element of order n for given n  1 .

32. G contains a subgroup of order of order n for given n  1 .

33. The number of elements in Tn = gGog = n (see Definition 2.5, page 56).

34. The number of elements in ZG — the center of G. (See Exercise 28, page 67.)

PROVE OR GIVE A COUNTEREXAMPLE

35. The additive group  is isomorphic to the additive group Q of rational numbers) 36. The additive group Z is isomorphic to the additive group Q of rational numbers)

37. If  is a homomorphism from a group G to a cyclic group G = a , then Ker is a cyclic subgroup of G.

38. If  is an isomorphism from a group G to a cyclic group G = a , then Ker is a cyclic subgroup of G.

39. For I the group of continuous real valued functions under addition the function 1 3 : I   given by f = fxdx fxdx is a homomorphism. 0 2

40. If nm , Sn and Sm are not isomorphic. CYU SOLUTIONS A-1

CHECK YOUR UNDERSTANDING SOLUTIONS CHAPTER 1 A LOGICAL BEGINNING 1.1 PROPOSITIONS

CYU 1.1 (a) Since q is True (35+8= ), pq is True, independently of p. (b) Since p is False (75= ), pq is False, independently of q.

CYU 1.2 (a) Since p is True (53 ), its negation ~p is False. (b) Since q is False (35= ), its negation p is True, so ~p is False.

CYU 1.3 (a) Since p and q are True, so is pq . It follows that ~pq is False. (b) Since p and q are True, so is pq . It follows that ~pq is False. (c) Since q is True, so is sq . It follows that ~sq is False. (d) Since s is False, so is sq . It follows that ~sq is True. (e) Since p is True, ~p is False. It follows that ~pq is False. (f) Since q is True, so is  pq . (g) Since s is False, ~s is True. It follows that  sq is True. (h) Since both ~s and q are True, so is  sq . (i) Since s is False, so is qs . It follows that ps  qs is False. (j) Since p is True, so is ps . It follows that ps  qs is True. (k) Since p is True, so is ps . It follows that ~ps is False, as is  ps  qs .

CYU 1.4 We show that pq  ~q  ~p is a tautology:

p q pq ~q pq  ~q ~p pq  ~q  ~p TT T F F F T TF F T F F T FT T F F T T FF T T T T T A- 2 APPENDIX A

CYU 1.5 (a) Negating Mary is going to a movie or she is going shopping: ~[(Mary is going to a movie) ˅ (Mary is going shopping)] Theorem 1.1(b): ~(Mary is going to a movie) ˄ ~(Mary is going shopping) Mary is neither going to a movie nor going shopping. (b) Negating Bill weighs more than 200 pounds and is less than 6 feet tall: ~[(Bill weighs more that 200 pounds) ˄ (Bill is less than 6 feet tall)] Theorem 1.1(a): ~(Bill weighs more that 200 pounds) ˅ ~(Bill is less than 6 feet tall) Bill weighs at most 200 pounds or is at least 6 feet tall.

(c) Negating x  0 or x  –5 : ~(x  0 or x  –5 ) Theorem 1.1(b): ~(x  0 ) and ~(x  –5 )

x  0 and x  –5 –5  x  0

CYU 1.6 (a) ~~p p p ~p ~~p TF T FT F

(b) pq  ~p  q p q pq ~p ~p  q TT TF T TF FF F FT TT T FF TT T

(c) ~pq  ~~p q p q pq ~pq  p ~p  q ~~p q TT T FFT F TF F TFF T FT T FTT F FF T FTT F

(d) pq   pq  qp p q pq pq qp pq  qp TT TTT T TF FFT F FT FTF F FF TTT T CYU SOLUTIONS A-3

CYU 1.7 (a) pq  ~s  s  ~p  ~q : p q s pq ~s pq  ~s ~p ~q ~p  ~q s  ~p  ~q TTT T F FFFF F TTF T T TFFF T TFT T F FFTF F TFF T T TFTF T FTT T F FTFF F FTF T T TTFF T FFT F F TTTT T FFF F T TTTT T

(b) pq  ~s ~(~s) ~pq s  ~p  ~q

Theorem 1.4 CYU 1.6(a) and Theorem 1.1(b)

CYU 1.8 Go with a Truth Table if you wish. For our part: qp  ~p ~q by Theorem 1.4.

1.2 QUANTIFIERS CYU 1.9 (a) All months have at least thirty days is False: February has 28 (or 29) days. (b) Every month contains (at least) three Sundays is True.

(c)nm  Z+ nm+  Z+ is True.

(d)nZ + and m  Zn + m Z+ is False: 34+1– = –  Z+ .

CYU 1.10 (a) There exists a month with more than thirty days is True: January has 31 days. (b) There exists a week with more than seven day is False. (c) nm  Z+  nm+ = 100 is True: 50+100 50 = . (d) nm  Z+  nm = nm+ is True: 22 = 22+ .

CYU 1.11 (a)nm  Z+ sZ +  snm is True: Let snm= + 1 .

(b)sZ + nm  Z+  snm is False: If s = 1 , then snm for every nm  Z+ .

(c)x    nZ + xn  x is False: x   x1  x .

(d)nZ +  mZ + nm = n is True: 1m = 1mZ + A- 4 APPENDIX A

CYU 1.12 (a) Negation of All college students study hard: Some college student does not study hard. (b) Negation of Everyone takes a bath at least once a week: Someone does not take a bath at least once a week. (c) Negation of x  Xpx  qx : xX  ~px ~qx . (d) Negation of x  Xpx  qx : xX  ~px ~qx .

CYU 1.13 (a) Negation of There are days when I don’t want to get up: Every day I want to get up. (b) Negation of x  Xpx  qx : xX  ~px ~qx . (b) Negation of x  Xpx  qx : xX  ~px ~qx .

CYU 1.14 (a) Negation of For every xX there exists a yY such that y blips at x: There exists some xX such that no yY blips at that x.

(b) Negation of x   nZ +  x = 2n : x    nZ + x  2n .

CYU 1.15 (a) Negation of There is a motorcycle that gets better mileage than any car: For every motorcycle there is some car that gets the same or better mileage than that motorcycle.

(b) Negation of nZ   mZ +nm : nZ mZ +  nm .

1.3 METHODS OF PROOF

CYU 1.16 (a) Let n ==2k +21 and m h + 1 . We show that nm+ is even: nm+2==k + 1 +22h + 1 kh++1 . (b) The sum of any even integer with any odd integer is odd. Proof: For n ==2k and m 2h + 1 , nm+2==k +22h + 1 kh+ + 1 .

CYU 1.17 2mn+ is even 2mn+2= k  n = 2k – 2m  n = 2km–  n is even CYU SOLUTIONS A-5

CYU 1.18 (a) 3n is odd if and only if n is odd n odd 3n odd: (a direct proof): n odd  n ==2k +3213 n k + 1  3n = 6k + 3  3n = 6k + 2 + 1  3n = 23k + 1 + 13 n odd 3n odd  n odd (a contrapositive proof): n not odd  n even  n = 2k  3n = 6k  3n = 23k  3n even  3n not odd

(b) n3 is odd if and only if n is odd n odd  n3 odd: (a direct proof): n odd  n = 2k + 1  n3 ==2k + 1 3 8k3 +++12k2 6k 1  n3 = 24k3 ++6k2 3k + 1  n3 odd n3 odd  n odd (a contrapositive proof): n not odd  n even  n = 2k

3 3 3 3  n ==8k 24k  n even  n not odd

CYU 1.19 A direct proof can be use to show that n odd 3n + 2 odd : n odd  n = 2k + 1

 3n + 2 ===32k + 1 +62 k +235 k + 2 + 13 n + 2 odd An attempt to show, directly, that 3n + 2 odd  n odd leads us to a dead end: 3n +22 ==k +31  n 2k – 1 — Now what? We try something else, that’s what: 3n + 2 odd  n odd Contrapositive Proof: Proof by Contradiction: n even 3n + 2 even Let 3n + 2 be odd (given condition) n even  n = 2k (for some k) Assume that n is even, say n = 2k . Then:  3n + 2 = 32k + 2 3n +322 ==k +232 k + 1 even = 23k + 1 — contradicting the stated condition that  3n + 2 even 3n + 2 is odd. A- 6 APPENDIX A

CYU 1.20 (a-i) If an and am , then nha= and mka= for some h and k. Consequently: nm+ ==ha+ ka hk+ aanm + (a-ii)na a= nh for some h. For cZ we then have: ca== c nh ch nnca (b-i) If ab or ac , then ab+ c is False: Counterexample: ab==2 and c = 1 . Since 22 : 22 or 21 is True. However 22+ 1 is False. (b-ii) If a and b are even and if ab+ c , then c must be even is True:

Proof: Let k1k2 h be such that a = 2k1 b = 2k2 , and bc+ = ha . ab+ c (*) (**)

Then: ha== b+2 c k2 + c  2k2 + c =ha (**) (*)  2k2 + c ==h2k1  c 2hk1 – 2k2 (*)  c = 2hk1 – k2 (even)

(b-iii) If ab+ cd+ , then there exist k such that ak+++ bk c d = 0 is True. Proof: Let h be such that cd+ = ha+ b . Then: cdha+0– – hb = , or ak+++ bk c d = 0 , where kh= – .

1.4 PRINCIPLE OF MATHEMATICAL INDUCTION

CYU 1.21 (a) The equation 24+= 1234+++–  13+ illustrate that the sum of the first two even integers can be expressed as the sum of the first four integers minus the sum of the first two odd integer. Generalizing, we anticipate that the sum of the first n even integers is the sum of the first 2n integers minus the sum of the first n odd integers; leading us to the conjecture that the sum of the first n even integers equals n2 + n : 2n2n + 1 ------–2n2 ==n2 + nn– 2 n2 + n 2 sum or first 2n integers sum of first n odd integers (Eample 1.16) (page 34)

(b) Let Pn be the proposition that the sum of the first n even integers equals n2 + n . I. Since the sum of the first 1 even integers is 2, P1 ==12 +21 is true. II. Assume Pk is true; that is: 246+++ +2k = k2 + k . III. We complete the proof by verifying that Pk+ 1 is true; which is to say, that 246+++ +2k +2k + 1 = k + 1 2 + k + 1 : by II 246+++ +2k +2k + 1 = k2 ++k 2k + 1 ==k2 ++2k 1 + k + 1 k + 1 2 + k + 1 CYU SOLUTIONS A-7

CYU 1.22 Let Pn be the proposition that 6 n3 + 5n for all integers n  1 . I. True at n = 1 : 613 + 51 . II. Assume Pk is true; that is: 6 k3 + 5k . III. To establish that 6 k + 1 3 + 5k + 1 , we begin by noting that k + 1 3 + 5k + 1 = k3 ++3k2 8k + 6 and then set our sights on showing that 6 k3 ++3k2 8k (for clearly 66 ). Wanting to get II into play we rewrite k3 ++3k2 8k in the form k3 + 5k + 3k2 + 3k . Our induction hypothesis allows us to assume that 6 k3 + 5k . If we can show that 6 3k2 + 3k , then we will be done, by virtue of Theorem 1.6(b), page 28. Let’s do it: Since 3k2 +33k = kk+ 1 , and since either k or k + 1 is even: 6 is a factor of 3x2 + 3k .

CYU 1.23 Let Pn be the proposition that 2n  n + 2 ! for all integers n  0 I. Since 20  02+ ! , P0 is true. II. Assume Pk is true; that is: 2k  k + 2 ! . III. We complete the proof by showing that 2k + 1  k + 1 + 2 ! = k + 3 ! : 2k + 1 ==22 k  k + 3 k + 2 ! k + 3 ! II

CYU 1.24 Let Pn be the proposition that the n lines, no two of which are parallel and no three n2 ++n 2 of which pass through a common point, will separate the plane into ------regions. 2 2 I. One line does separate the plane into ------1 ++------12-2= regions. 2 II. Assume Pk is true. III. Consider k + 1 lines. The k + 1 th line will pass line 2 r line 3 through region-line-region-line-region... (one line 1 egio n more region than the k lines) and will split each . regi . on of the encountered regions into two parts, adding reg 2 ion k + 1 regions to the k------++k 2- preceding regions 2 generated by the first k lines. (see above figure for k = 3 ). It follows that the number of regions stemming from the k + 1 lines is given by: 2 2 2 ------k ++k------2 ++k 1 ==k------+++k ------22------k +2------k + 1 ------++k +------1 2- 2 2 2 II A- 8 APPENDIX A

CYU 1.25 Let Pn be a proposition for which P1 is True and for which the validity at k implies the validity at k + 1 . We are to show, using the Well-Ordering Principle, that Pn is True for all n. Suppose not (we will arrive at a contradiction): Let SnZ=  + Pn is False . Since P1 is True, S   . The Well-Ordering

Principle tells us that S contains a least element, n0 . But since the validity at n0 – 1

implies the validity at n0 , n0 – 1 must be in S — contradicting the minimality of n0 .

1.5 THE DIVISION ALGORITHM AND BEYOND

CYU 1.26 The division algorithm tells us that n must be of the form 3m , or 3m + 1 , or 3m + 2 , for some integer m. We show that, in each case, n2 = 3q or n2 = 3q + 1 for some integer q: If n = 3m , then n2 ==9m2 3q with q = 3m2 . If n = 3m + 1 , then n2 ==9m2 ++6m 1 33m2 + 2m +31 =q + 1 . If n = 3m + 2 , then n2 ==9m2 ++312m 4 3m2 ++4m 1 +31 =q + 1 .

CYU 1.27 (a) As in Example 1.22: (1) 5605= 2 1870 + 1865 (2) 1870= 1865+ 5 (3) 1865== 373 5 +50  gcd1870 5605

above line 2 above line 1 (b) 5== 1870– 1865 1870– 5605– 2 1870 = 3 1870 + –1 5605 (c) We simply show that c  0 divides nZ if and only if cn : cknc==== kn c kn c hn where hk =

since c  0

CYU 1.28 Proof by contradiction: Assume that gcd a c = 1 . From Theorem 1.9: if abc , and if gcd a c = 1 , then ab — contradicting the given condition that ab .

 CYU 1.29 Let Pn be the proposition that if pa1a2 an , then pai for some 1 in . I.P1 is trivially True.  II. Assume Pk is True: If pa1a2 ak , then pai for some 1 ik .   III. Suppose pa1a2 akak + 1 ; or, to write it another way: pa1a2 ak ak + 1 .  If pak + 1 then we are done. If not, then by Theorem 1.9: pa1a2 ak .

Invoking II we conclude that pai for some 1 ik . CYU SOLUTIONS A-9

CYU 1.30 CYU 1.20(iii) enables us to restrict our attention to the case where a  1 and b  1 .

r1 r2 rs m1 m2 mt Let ap= 1 p2 ps  b = q1 q2 qt be the prime decompositions of a and b, with distinct primes p1p2  ps , and distinct primes q1q2  qt .

r1 r2 rs ri Since an : nakp==1 p2 ps  k . It follows that and each pi must appear in the prime decomposition of n, for 1 is (with possibly additional pi ’s appearing

mi in the prime decomposition of k). Similarly, since bn , each qi must appear in the prime decomposition of n, for 1 it .

Since a and b are relatively prime, none of the pis is equal to any of the qis . It fol-

r1 r2 rs m1 m2 mt lows that p1 p2 ps q1 q2 qt appears in the prime decomposition of n, and

r1 r2 rs m1 m2 mt that therefore ab= p1 p2 ps q1 q2 qt divides n.

CHAPTER 2 A TOUCH OF SET THEORY 2.1 BASIC DEFINITIONS

CYU 2.1 (a) (i) AB c ==123457 c 6 (ii) AB c  AB c ==135  156  6 156 (iii) AB– c  C ==15 c  345 23467  345 =34 (iv) ABC–  c ===135 – 34 c 15 c 23467 (v) xU xy= + 2 yB = 456 (b-i) True. (b-ii) True. (b-iii) True. (b-iv) True: Every element in  is contained in 12 , by default, since  contains no element. (b-v) False: The empty set is not an element of 12 — it is not contained in 12 . (b-vi) False:  is not an element of  . (b-vii) True.

CYU 2.2 (a) The proposition ABC–  = AB–  AC– is False. Here is a counterex- ample: For A = 123 , B = 34 , and C = 23 : ABC– =  123 – 34  23 ==123 –123  While: AB–  AC– = 123 – 34  123 – 23 ==12  1 1 A- 10 APPENDIX A

(b) The propositionAB c c  B = Ac  B is True:

A B Bc AB c AB c  B Ac Ac  B 110 0 000 101 1 101 011 0 111 000 0 011 same

CYU 2.3 (a) ABAc Bc AB AB c Ac  Bc 1100 1 0 0 1001 0 1 1 0110 0 1 1 0011 0 1 1

(b) xAB  c  xAB   xA and xB  xA c and xB c  xA c  Bc

c CYU 2.4 xS   xS   xS  for some 0  A   0   A   A  x  Sc for some   A  xS c 0 0     A

CYU 2.5 (a) –22  05 = 02  (b) –13 c  5  = 5 

(c) –20 –21 35 = –22   35

2.2 FUNCTIONS

CYU 2.6 (a) No: 23  f and 26  f (b) Yes. Range: 2  (c) No: 3  X but there is no 3--  f (d) No: 4  X

15 5 CYU 2.7 (a-i) g f 3 ====gf3 g5 ------ 6 2

9 9 17 (a-ii) f g 3 ====fg3 f ------2+ ------ 4 4 4

3 x + 2 3x + 6 (a-iii) g f x ==gfx gx+ 2 =------------=------ x + 2 + 1 x + 3

3x 3x 5x + 2 (a-iv) f g x ==fgx f ------=------2+ =------ x + 1 x + 1 x + 1 CYU SOLUTIONS A-11

(b) gf = 1 a ct x 4 . fg is not defined. [ga , for example, is not in the domain of f]. a b CYU 2.8 (a) fa= fb ------= ------ ab+ 1 = ba+ 1 ab+ a = ba+ b a = b a + 1 b + 1 (b) The function gx==x5 – x xx4 – 1 is easily seen not to be one-to-one, since g0 ==g1 0 . It follows that the function fx= x5 –x + 777 is also not one-to-one, since f0 ==f1 777 .

ab ab CYU 2.9 One-to-one: f ==f  dc –  3ab dc –  3ab cd cd dd= –c = –c ab ab   = 3a = 3a cd cd bb= ab ab Onto: For given xyzw , we find such that f = xyzw : cd cd

dx=  dx=    ab –c = y  cy= –  f ==xyzw  dc–3ab xyzw   cd 3az=  az=  3    bw=  bw= 

z  3 w Hence: f = xyzw –y x

CYU 2.10 Let yY . Since yf –1y  f –1 , f –1y  y  f , which is to say: ff–1y = y .

 4 ab CYU 2.11The function f: M22  R given by f = dc –  3ab is a bijection [see cd ab ab CYU 2.9]. To find its inverse we determine for which f = xyzw : cd cd dx=  az=  3    ab –c = y  bw=  f ==xyzw  dc –  3ab xyzw   cd 3az=  cy= –    bw=  dx= 

Conclusion: f –1xyzw = z  3 w –y x A- 12 APPENDIX A

· –1 z  3 w z Moreover: ffxyzw ==f xy–3– ---  w =xyzw –y x 3

–1 ab –1 3a  3 b ab and: f f ===f dc –  3ab cd ––c d cd

2.3 INFINITE COUNTING

CYU 2.12 The function f: Z+  0123 given by fn= n – 1 is a bijection: On-to-one: fa= fb a – 1 ==b – 1  a b Onto: For any a  0123 , a + 1  Z+ and fa+ 1 = a .

CYU 2.13 a11 a12 a13 a14 

a21 a22 a23 a24 

a31 a32 a33 a34 

a41 a42 a43 a44  

dc– CYU 2.14 We show that the function f: ab  cd given by y = ------xi+ ba– dc– fx= ------xi+ (inspired by the adjacent figure) is a bijection. ba– d. . y-intercept fx One-to-one: fx1 = fx2 c. . dc– dc– i . ------x1 + i = ------x2 + i . . ba– ba– a x b

dc– x1 = dc– x2

x1 = x2

ONTO: For y0  cd , we are to find xab  such that fx= y0 . Let’s do it: dc– fx==y  ------xi+ y 0 ba– 0  dc– xiba+ – = y0ba– y ba– – ib– a ba–  x ==------0 ------y – i ------dc– 0 cd–

ba– We complete the proof by showing that ay – i ------ b : 0 cd– CYU SOLUTIONS A-13

dc– dc– dc– dc– fay fb------ai+ y ------bi+ ------ay – i  ------b 0 ba– 0 ba– ba– 0 ba– c d ba–  ay – i ------ b 0 dc–

CYU 2.15 For any given finite interval F, choose closed intervals Lab=  , Mcd=  such that ab Fcd . CYU 2.15 tells us that CardL = CardM . It follows, from Theorem 2.9, that the finite interval F is of the same cardinality as that of (any) finite closed interval. That being the case, any two finite intervals must be of the same cardinality (Theorem 2.7).

CYU 2.16 (a) Since –3 –3 56   9  and since Card–3 ==Card c , Card–3  56   9  = c (b) Since CardQ  Card , there cannot exist a bijection from  to Q.

CYU 2.17 Assume the set of all sets exists. Let’s call it S . Consider the set TA=  , and its AS power set PT . Since S contains all sets, PT S . As such, PT T . But then: CardPT CardT [Theorem 2.10(a)] — a contradiction [Theorem 2.13].

2.4 EQUIVALENCE RELATIONS CYU 2.18 (a-i) Not symmetric: 1~2 but 2~1 . (a-ii) Yes. (a-iii) Not transitive: 2~1 and 1~3 but 2~3 . (b-i) Yes. (b-ii) Yes. (b-iii) Not reflexive nor symmetric.

CYU 2.19 Reflexive: Let SPX  . Since I: SS , given by Is= s sS is a bijection, SS . Symmetric: If S~T for ST  PX , then there exists a bijection f: ST . Theorem 2.4(a), page 70, tells us that f –1: TS is a bijection. Hence, TS Transitive: If S~T and T~W with STW   PX , then there exists bijections f: ST and g: TW . Theorem 2.5(c), page 72, tells us that gf: SW is a bijec- tion. Hence, S~W .

CYU 2.20 (a) No: 12 23  . (b) Yes: Every element of  is either an integer or is contained in some ii + 1 for some integer i  0 or in some –i –1i – for some i  1 . Moreover the sets in   n nZ ii + 1 i = 0  –i –1i – i = 1 are mutually disjoint. A- 14 APPENDIX A

CYU 2.21 Let’s find the equivalence classes, starting with 0 : 0 ===b 230– 7b b 7b is even even integers Moving on to 1 : 1 ===b 23– 7b b 7b is odd odd integers Conclusion: The equivalence relation partitions the integers into two disjoint pieces: the class of even integers and the class of odd integers. CHAPTER 3 A TOUCH OF ANALYSIS CYU 3.1 (a)lub 3 5  47 =glb357   47 =3 . Maximum: 7. No minimum. (b)lub–0 13 9 = 9 . Maximum: 9. Greatest lower bound and mini- mum do not exist. (c) lubx  0 x2  2  x  0 x2  2 =2

glbx  0 x2  2  x  0 x2  2 =– 2 . Maximum: 2 . No minimum.

CYU 3.2 If  is the greatest lower bound, then it is a lower bound. To establish (b-ii), we con- sider a given   0 . Since +  , and since nothing greater than  can be a lower bound of S, there must exist some sS to the left of + . Conversely, suppose  satisfies (b-i) and (b-ii). To show that  is the greatest lower bound of S we consider an arbitrary lower bound b of S, and show that   b : Assume, to the contrary, that   b . Let  = b –   0 . By (b-ii)  sSs   + ==+ b –  b Contradicting the assumption that b is a lower bound of S.

n 77–  CYU 3.3 As observed in Example 3.1: ------ 1 –   n  ------. In particular: n + 7  7 7 – ------n 99 1 100 694 (a)------ ------==1 –------ ------693 . It follows that ------is the smallest n + 7 100 100 ------1 - 694+ 7 100 99 element of S which lies to the right of ------. 100 7 n 99.9 7 – ------6994 (b) ------ ------ n  ------1000- = 6993 . It follows that ------is the smallest ele- 1 n + 7 100 ------6949+ 7 1000 99.9 ment of the set S which lies to the right of ------. 100

CYU 3.4 (a) Theorem 3.2, with a = 1 , assures us that such an integer n exists. 1 (b) There exists a positive integer n such that ---   if and only if there exists a posi- n tive integer n such that 1  n . Theorem 3.2, with 1 playing the role of b and  that of a, assures us that such an integer n exists. CYU SOLUTIONS A-15

CYU 3.5 Let x  01  be given. CYU 3.4(b) assures us of the existence of a positive integer 1 nx such that -----  x (let x play the role of  ). In particular, x is not contained in nx 1 1 J = 0 ----- . It follows that no x  01  is contained in every J = 0 --- ; or, to nx  n  nx n

put it another way:  Jn =  . n = 1 a a CYU 3.6 (a) Let x be irrational, and let --- be a rational number with a  0 . Assume that ---x is b b a c c b cb rational, say ---x = --- . Then: x ==---  ------— contradicting the given condition b d d a da that x is irrational. (b) The sum of two irrational numbers can be rational: 22–21==+0– 2

CYU 3.7 (a) If S is a finite subset of  , then SNN –  for some NZ + . Since no element of S is contained in NN + 1 , S is not dense in  . (b) Suppose there exists an interval ab containing only finitely many elements of S. Let c be the smallest of those finitely many elements of S. It follows that the inter- val ac contains no element of S — contradicting the given condition that S is dense in  . 101 CYU 3.8 (a-i) Let   0 be given. We want N such that: nN  7 –7------–   n 101  ------  n 101  n  ------ 101 From the above, we see that if N is any integer greater than ------, then  101 nN  7 –7------–   . n 1 101 101 (a-ii) If  = ------, then ------==------10,100 . It follows, from (i) that 100  ------1 - 100 101 N = 10,101 is the smallest integer for which nN  7 –7------–   . n n – 5 (b) For any given r   we observe that: ------ r + 1  n – 5 333r + 333 333  n  333r + 338 n – 5 It follows that the sequence ------will be larger than r + 1 for any n  333r + 338 , 333 and that, consequently, the sequence does not converge to r. Since r was arbitrary, the sequence does not converge, period. n – 5 n – 5 (Why not start the above argument with ------ r rather than ------ r + 1 ? 333 333 A- 16 APPENDIX A

CYU 3.9 Let an   . Taking  = 1 we choose N such that nN  an –1  . Noting that only the elements a1a2  aN can be more than 1 unit from  , we let N K = maxai –  i = 1 . Taking M to be the larger of the two numbers, 1 and K, we see that an –   M for all n. That is, the sequence is bounded below by  – M and above by  + M . – CYU 3.10 (a) Suppose, to the contrary, that  . For  = ------- , let N be such that 2 nN  an –    and nN  bn –    . It follows that aN + 1  bN + 1 :

  ( . )( . ) : a contradiction.  

1 1 (b) One possible answer: For a = 000 and b = 1------ : n n 2 3 0  an  bn and lim an ==lim bn 0 .

CYU 3.11 The sequence g: Z+   is a subsequence of the sequence f: Z+   if gf= h , where h: Z+  Z+ is a strictly increasing function.

 CYU 3.12 Let an n = 1 be Cauchy. Choose N such that nm  N an –1am  . In particular: an –1aN + 1  for every nN . It follows that an  1 + aN + 1 for every nN . N Letting K = maxai i = 1 and M = maxK 1 + aN + 1 , we find that –Man M for every an . 1  CYU 3.13 The sequence --- is Cauchy but does not converge in X = 0  , since the n n = 1 number 0 is not in X.

triangle inequality CYU 3.14 xxy= – + y  xy– + yxxy  – + yxy –  xy– .

CYU 3.15 (a) For xab  let  = minxabx–  – . Then Sx  ab .

(b) For x  – a let  = ax– . Then Sx  –  a .

For xa   let  = xa– . Then Sx  a  .

(c) Let Sx= 1x2  xn and let xi  S . For any given   0 , since Sx is infinite and S is finite Sx – S   . It follows that Sx is not contained in S. CYU SOLUTIONS A-17

 1 1 CYU 3.16 For O = –--- --- , O = 0 (not open) . i i i  i i = 1

CYU 3.17 (a)ab c = – a  b  . Since both – a and b  are open [CYU 3.15(b)] and since unions of open sets are open [Theorem 3.13(ii)], ab c is open. Consequently, ab is closed. c (b) Since 13  = –1   3  is not open [no S1 is contained in –1   3  ], 13  is not closed.

1 CYU 3.18 If H = --- 1 , then H = 01  — not closed since: i i  i i = 1 c 01  = – 0  1  is not open [no S0 is contained in – 0  1  ].

  1 1 CYU 3.19 (a) B = --- is not compact: The open cover S --- of B has no finite n -----1-n n = 1 2n n = 1 1 1 subcover (note that S ---  B = --- ). -----1-n n 2n 

 1 (b) B = 0  --- is compact: For UO=  an open cover of B, let O n    A 0 n = 1 be an element of U which contains 0, and let   0 be such that 0  S 0  O .  0 1 Choose N such that ----   [CYU 3.4(b), page 115]. For each nN let O be an ele- N n 1 N ment of U containing --- . It follows that O covers B. n i i = 0

CYU 3.20 A consequence of CYU 3.12 (page 131), Theorem 3.10 (page 129), and Theorem 3.19.

CYU 3.21 For a given   0 we are to find   0 such that: x – 2    5x + 1 – 11   i.e: x – 2    5x – 10   x – 2    5 x – 2    x – 2    x – 2  --- Let  = --- 5 5 A- 18 APPENDIX A

CYU 3.22 For given   0 we are to find   0 such that: x – 2    x2 + 1 – 5   x – 2    x + 2 x – 2  x – 2    x + 2 x – 2   (*) Since we are interested in what happens near x = 2 , we decide to focus on the interval: 13 = xx–12  . Within that interval 5 yx= + 2 x +52  . Consequently, within that interval: x + 2 x –52  x – 2 .  We observe that (*) is satisfied for  = min 1 --- : 5 1 3  x – 2   x + 2 x –52  5--- = 5

CYU 3.23 (a) Let   0 be given. We are to find   0 such that: x – 0  x – 0

x  0 and x  0: x    x   x    x  2 Let = 2

(b) Let x0 be a fixed but arbitrary number. We are to find   0 such that: 2 2 xx– 0    x – x0  

i.e: xx– 0    xx– 0 xx+ 0   We observe (see adjacent figure) that in the chosen y interval x0 – 1 x0 + 1 : . xx+ 0  M = max 2x0 +21  x0 – 1  It follows that for  = min 1 ----- : . x M –x0 x0 x0 + 1 x 0

 – xx– 0   xx– 0 xx+ 0  -----  M = 1 M yxx= + 0

  x if x  1 CYU 3.24 We show that fx=  999 is not continuous at x = 1 :  ------if x = 1  1000 999 1 Let  ==1 – ------(the “jump” at 1), and let  be ANY positive number whatsoever. 1000 1000 999 1 Since, for every x   1 , fx– f1 ==x – ------------, f is not continuous at 1. 1000 1000 CYU SOLUTIONS A-19

CYU 3.25 For given   0 we are to find   0 such that: xc–    fg– x – fg– c   xc–    fx– gx– fc+ gc  xc–    fx– fc+ gc– gx   (*)  Let   0 be such that xc–    fx– fc --- , and let   0 such that 1 1 2 2  xc–    gc– gx --- . Letting  = min   , we see that if xc–   , then (*) 2 2 1 2 holds: fx– fc+ gc– gx  fx– fc+ gx– gc --- + --- =  2 2 triangle inequality

CYU 3.26 Let f:  and g:  be continuous. We show that gf:  is also con- tinuous at an arbitrarily chosen x0   : For given   0 we are to find   0 such that:

xx– 0    gf xg– f x0  

i.e: xx– 0    gfx– gfx0  

Since g is continuous at fx0 , we can find   0 such that:

yfx– 0    gy– gfx0   (*)

Since f is continuous at x0 , we can find   0 such that:

xx– 0    fx– fx0   (**) (**) (*)

Consequently: xx– 0    fx– fx0    gfx– gfx0   y

CHAPTER 4 A TOUCH OF TOPOLOGY

triangle inequality CYU 4.1 xxy= – + y  xy– + yxxy  – + yxy –  xy–

 1 if xy CYU 4.2 For: dxy =  :  0 if xy= : (i)dxy = 0 if and only if xy= (see above definition of d). (ii)dxy = dyx (see above definition of d). (iii) Let xyz  X . If xyz== , then: dxz ==dxy +0dyz . If no two of the elements are equal, then: dxz ==111 + dxy + dyz If two of the elements are equal but differ from the third element, say xy= and yz , then: dxz ==101+ =dxy + dyz A- 20 APPENDIX A

+ + CYU 4.3 Since S15 = nZ dn 5  1 , and since 5 is the only element of Z which is within one unit of 5: S15 = 5 . + + Since every element of Z falls within five units of 1: S51 = Z .

CYU 4.4 (a) Since for all   0 : S5  1 5  , 1 5  is not open. c c c Since 15  = –  1   5  , and since for all   0 : S1  1 5  , 1 5  is not open. It follows that 1 5  is not closed. (b).Since every subset of X is open [Example 4.2(c)], and since a set of n element has 2n subsets (Theorem 2.12, page 84), X has 2n open subsets.

CYU 4.5 (i) Since both Xc =  and c = X are open, both X and  are closed. c c (ii) Let H be a collection of closed sets. Since: H = H   A     A   A c [Theorem 2.3(b), page 58], and since each H is open,  H is closed.   A c n n (iii) Let H n be a collection of closed sets. Since H = H c [Theo- i i = 1  i  i i = 1 i = 1 n n c rem 2.2(a), page 57] and since  Hi is open,  Hi is closed. i = 1 i = 1 CYU 4.6 (a) True. Let A and B be bounded subsets of Xd , bounded by MA and MB , respec- tively. If either A or B is empty, then AB is bounded by the bound of the other. If neither A nor B is empty, then choose elements aA and bB . We show that MA ++dab MB is a bound for AB : Let xy  A B . If both x and y are in either A or B, then clearly dxy  MA ++dab MB . For xA and yB we have:

dxy dxa ++dab dby MA ++dab MB triangle inequality  (b) False. Each of the sets ii + 1 i = 0 in the Euclidean space  is bounded by 1,  but the set  ii + 1 = 0  is not bounded. i = 0 (c) True. Let S be a collection of bounded sets in a space X. Since S is    A     A contained in each S , the bound M of any chosen S will be a bound for  S .   A CYU SOLUTIONS A-21

CYU 4.7 (One possible answer): Consider the discrete metric d on Z+ = 1234 ;  1 if nm + namely: dnm =  (see CYU 4.2). Z  d is a subset of itself, that is  0 if nm=  closed and bounded by 1. Since the open cover S1n n = 1 contains no finite sub- cover, it is not compact.

CYU 4.8 (a-i) yfAB   y = fx for xAB   y = fx for xA or xB  yfA  or yfB  yfA  fB –1 (a-ii) xf AB  fx AB  fx A or fx B

 xf –1A or xf –1B  xf –1A  f –1B

–1 (a-iii) xf AB  fx AB  fx A and fx B

 xf –1A and xf –1B  xf –1A  f –1B

(a-iv) xf –1Ac fx Ac fx Axf  –1A  xf –1A c

(b) fA fAc c : yfA  y = fx for xA  yfx  for xA c  yfA c  yfA c c since f is one-to-one fAc c  fA : yfA c c  yfA c  y = fx for xA  yfA  since f is onto

CYU 4.9 To show that I: 2 d  2 d is continuous, we take an arbitrary open set O in 2 d and go on to show that I–1O = O is open in 2 d :

For given x0 y0  O let   0 be such that xy xy dxy  x0 y0  is contained in the open subset O of  2   d .Since dx0 y0  xy ==xx– 0 + yy– 0  + 2

x0 y0 the open sphere of radius 2 centered at x0 y0 in the space xx– 0   2 d is contained in O.

CYU 4.10 No: Let X be the set of real numbers with the discrete metric and let X and Y be the Euclidean space  . Let f: XY be the identity map fx= x for all xX and let 1 if x = 0 g: YZ be given by gx=  . The composite function gf: XZ is  0 if x  0 –1 continuous, since every subset of the discrete space X [in particular, gf O is open 1 3 for every O open in Z]. The function g is not continuous since g–1--- --- = 0 is not 2 2 open. A- 22 APPENDIX A

CYU 4.11 Let S be a collection of metric spaces. We show that the relation X~Y , if X is isometric to Y, is an equivalence relation on S. (i) For every XS , X~X : The identity function I: XX is a bijection which satisfies the condition that dx1 x2 = dIx1  Ix2 . (ii) For Xd and Yd in S, if X~Y then Y~X : Let f: XY be a bijection such that –1 dx1 x2 = dfx1  fx2 (*). The function f : YX is a bijection (Theorem 2.4, page –1 –1 –1 –1 70). Moreover: dy1 y2 ==dffy1  ffy2 dfy1  f y2 [read (*) from right to left]. (iii) For Xd , Yd , Zd ˆ in S, if X~Y and Y~Z then X~Z : Let f: XY and g: YZ be ˆ bijections such that dx1 x2 = dfx1  fx2 (*) and dy1 y2 = dgy1  gy2 (**) . The function gf: XZ is a bijection (Theorem 2.5, page 72). Moreover: ˆ ˆ dx1 x2 ==dfx1  fx2 dgfx1  gfx1 =dgf x1  gf x1 (*) (**)

CYU 4.12 (a) For 0 = X  . (i) X and  are certainly in 0 .

(ii) XX = X  0 , X   = X  0 , =  0 .

(iii) XX = X  0 , X  =  0 , =  0 .

(b) For 1 = SS X :(i) Since 1 contains all subsets of X, it certainly contains X and  . (ii) and (iii) are a consequence that unions and intersections of subsets of X are again subsets of X.

CYU 4.13 We are given that  and X are contained in  . Since  contains but three elements, one can easily check directly that it is closed under unions and intersection.

CYU 4.14 Since any metric space containing at least two points must contain at least four distinct open sets (Exercise 13, page 168), the Sierpinski space is not metrizable.

CYU 4.15 (a) Since every subset of a discrete space is open, the complement of every subset must be open. Consequently, every subset of a discrete space is closed. (b) Since the only open subsets of an indiscrete space X are X and  , the only closed subsets of X are Xc =  and c = X . (c) The Surpinski space is Xab=  with topology  = Xa . Taking complements of the elements of  we obtain the set of closed subsets of X: cXc a c = X b . CYU SOLUTIONS A-23

n CYU 4.16 Let Ui i = 1 be a family of open sets in S S . For each i choose Oi   such that n

Oi  S = Ui . Since  is a topology,  Oi   . The desired result now follows i = 1 n n n from the fact that: U ==O  S O  S.  i  i  i i = 1 i = 1 i = 1 Exercise 82(b), page 61

CYU 4.17 Let O be open in X, and xO . We already know that Srx xXr   0 is a basis for X [Example 4.6(a)]. Since the rationals are dense in  (Theorem 3.6, page 118), + there exists rQ such that 0 rr . It follows that xS rx  O .

CYU 4.18 Let O   be open in the Euclidean topology. For given xO choose r  0 such r r that xS x  O ; which is to say: xr–  xr+  O . Since xx – --- x + ---  O , r 2 2

O  S . At this point we know that  S . Since 01   S and 01    ,

 S .

CYU 4.19 If O   A is an open cover of a closed subset H of a compact space X, then c O   A  H is an open cover of X. Since X is compact, that cover has a finite sub- cover O n  Hc . It follows that HO n . i i = 1 i i = 1 dxy CYU 4.20 Let d be a metric on X X  . For any two distinct point x and y in X, let r = ------ . 2 Then Srx Sry = .

CYU 4.21 For each xK choose disjoint open sets O , O containing x and x , respectively. x x0 x 0 Since K is compact the open cover O of K has a finite subcover O n . It x xK xi i = 1 n follows that the open neighborhood O of x is disjoint from the open set  x0 xi 0 i = 1 n O = O (note that KO ). K  xi K i = 1 A- 24 APPENDIX A

CYU 4.22 For Xab=  with topology Xa . the function f: X   given by fa==fb 0 is continuous [Example 4.8(b)]. The function g: X   given by ga==0 gb 2 is not continuous, since g–113 = b is not open in X.

CYU 4.23 Let f be any function from a topological space X to an indiscrete space Y. Since f –1 =  and f –1Y = X are open in X, and since  and Y are the only open sets in Y, f is continuous.

CYU 4.24 (a) Let XY== ab with indiscrete topology, and let Zab=  with discrete topology (see CYU 4.12, page 171). Let f: XY and g: YZ be identity maps. f: XY is continuous but gf: XZ is not, since a is open in Z but –1 gf a = a is not open in X. (b) Let Xab=  with indiscrete topology, and let YZ== ab with discrete topology. Let f: XY and g: YZ be identity maps. g: YZ is continuous –1 but gf: XZ is not, since a is open in Z but gf a = a is not open in X. (c) Let XYZ=== ab with topology Xa (three copies of the Sier- pinski space). Let both f: XY and g: YZ be the function which maps a to b and b to a. Neither of these function is continuous since the inverse of the open set a is the set b which is not open in the Sierpinski space. On the other hand gf: XZ is the identity map, which is continuous.

CYU 4.25 False. Let f:  be the constant function fx= 0 x   . f = 0 is compact, but  is not.

CYU 4.26 Let f:   Z+ be given by fnn  + 1  = n xnn   + 1  . Since the subspace Z+ of  is discrete, f is open and closed. It is, however, not continuous since f -10 = 01  is not open in  .

CYU 4.27 Let f: XY be a homeomorphism.

f is continuous: If U is open in Y, then U  Y . As such, there exists O  X with

fO= U [recall that fX ==fOO  X Y ]. We then have: –1 f U ==f –1fO O (recall that f is a bijection).

–1 f is continuous: If O is open in X, then O  X . As such fO= U  Y . We –1 then have: f –1U ==f fO O . CYU SOLUTIONS A-25

Conversely, assume that f: XY is a continuous bijection such that f –1: YX is

also continuous. We show fX ==fOO  X Y : –1 Let U  Y . Is there an O  X such that fO= U ? Yes, since Uff= U and –1 Of= U  X by virtue of the continuity of f. So fX  Y .

–1 –1 –1 Let O  X . Is fO Y ? Yes, since f O = fO, and f is given to be

continuous. So, fX  Y .

CYU 4.28 (a) Let f: XY be a continuous open bijection. We show that f –1: YX is also continuous: For any O, open in X. Since the inverse of the inverse function f –1 is the function f, –1 we have f –1 O = fO which is open in Y (since f is an open map).

(b) Let Xab=  with discrete topology, and Yab=  with indiscrete topology. The identity function I: XY is a continuous bijection which is not a homomor- phism since {a} is open in X but I –1a = a is not open in Y.

CYU 4.29 Let Xab=  with indiscrete topology, and let Yab=  with discrete topology. X is compact and Y is Hausdorff. The identity function I: XY is an open and closed bijection which is not a homomorphism.

CYU 4.30 (a) Let XY with X compact. We show that Y is also compact:

Let f: XY be a homeomorphism. If O   A is an open cover of Y then, –1 since f is continuous, f O   A is an open cover of X. Since X is compact, n that cover contains a finite subcover f –1O . It follows that i i = 1 n ff–1O = O n covers Y. i i = 1 i i = 1 (b) Since 01 is compact and 01 is not, the two spaces are not homeomorphic.

CYU 4.31 For topological spaces X, Y, and Z: (i) Since the identity map I: XX is a homeomorphism: XX . (ii) If XY , then there exist a homomorphism f: XY . CYU 4.25 assures us –1 that f : YX is also a homeomorphism, and that consequently: YX . (iii) If XY and YZ , then there exist homeomorphisms f: XY and –1 –1 –1 g: YZ . Since both gf: XZ and gf = f g : ZX are continuous bijections: XZ : A- 26 APPENDIX A

CYU 4.32 (a) We show that every open set in the basis  = UV U  X V  Y is the union of elements of  = DE A  X and B  Y :

For xy  UV , with U  X V  Y , choose D  X and E  Y with xD  U and yE  V . Then xy  DE  UV .

+ 2 (b) The open spheres  = Srxy r   xy  2 is a basis for the topology on  , and the open rectangles  = UV U and V are open in  is a basis for  . We establish the fact that the product topology on  coincides with the Euclidean 2 topology on  , by showing that: (i) every Srxy is a union of the elements in  and that: (ii) every UV is a union of elements in  (in other words, that you can build open spheres using open rectangles, and vice versa):

(i) For ab  Srxy , choose   0 such that Sab  Srxy .   ab . Then: ab  UV  Srxy for Ua= – ------ a + ------and xy . 2 2 r   Vb= – ------ b + ------. 2 2

(ii) For ab  UV let r denote the shortest distance between ab r ab and the boundary of the rectangle UV . V . U Thenab  Srab  UV . CYU 4.33 Suppose that XY is compact and that either X or Y is not compact (we will arrive at

a contradiction). For definiteness, assume that X is not compact, and let O   A be

an open cover of X containing no finite subcover. This implies that O  Y   A is an open cover of XY containing no finite subcover — contradicting the given con- dition that XY is compact.

CYU 4.34 We show that 1 : XY  X is both open and closed. A similar argument can be use dot show that 2 : XY  Y is also open and closed.

1 : XY  X is open: Let O be open in XY , and let x  1O . Choose yY

such that xy  O . Let U  X V  Y be such that xy  UV  O . Then,

xU  1O .

c 1 : XY  X is closed: Let H be closed in XY . We show 1H is open in X:

For any x  1H , x  Y H = . Choose yY such that xy  H . c Since H is closed there exist U  X V  Y such that xy  UV  H . It c follows that xU  1H . CYU SOLUTIONS A-27

CYU 4.35 Since X = X , and since projection maps are continuous, X is compact (Theo- rem 4.12, page 187).

CYU 4.36 Start with the space X = 02   01 and let ~ be the equivalence relation repre- sented by the partition xy = xy for x  02  and y  01 , and 0 y = 0 y  1 y along with x 1 = x 1  x 0 .

CHAPTER 5 A TOUCH OF GROUP THEORY n n CYU 5.1 (a)Q + is a group, with identity 0 and –---- the inverse of ----  Q . m m (b) + is a group, with identity 0 and –r the inverse of r   . (c) . is not a group. It does have a multiplicative identity; namely 1: r  1 = 1  r r   . However, 0   does not have a multiplicative inverse: there does not exist r   such that 0  r = 1 . (d) + . is a group. Unlike the situation in (c) every element in + = r   r  0 1 1 1 does have an inverse; namely --- : r  --- ==1 and ---  r 1 . r r r

CYU 5.2 The values in column a follow from the observation that 0+n nn= for 0 n 3 . As for column b, row 3: 3+ 1 = 0 , since 31+414== + 0 4 a b c d As for column c, rows 2 and 3: 2+42 = 0 and 3+42 = 1 , since: + 4 0123 22+414== + 0 and 32+514== + 1 . 0 0123 As for column d, rows 1, 2, and 3: 1+ 3 = 0 , 2+ 3 = 1 , 4 4 1 1230 and 3+ 3 = 2 , since: 13+414== + 0 , 4 22301 23+514== + 1 , 33+614== + 2 . 33012

th CYU 5.3 Let Gea= 1 a2  an – 1 . By construction, the i column of G’s group table is precisely eaia1ai a2ai  an – 1ai . The fact that every element of G appears exactly one time in that row is a consequence of Exercise 37, which asserts that the function k : GG given by ai k g = ga is a bijection. ai i A- 28 APPENDIX A

1 2 3 4 5 1 2 3 4 5 CYU 5.4 From  =  and  = we have: 1 5 2 3 4 5 3 2 1 4 1 2 3 4 5  1 2 3 4 5  1 5 2 3 4  :  5 4 3 2 1 5 4 3 2 1

1 2 3 4 5  1 2 3 4 5  5 4 3 2 1  :  4 3 2 5 1 4 3 2 5 1

CYU 5.5 (a) We know that 1 and 5 are generators of Z6 [Example 5.2(a)]. The remaining 4 ele-

ments in Z6 are not:

2+ 24= 4+ 42= 0+600= 6 3+630= 6 2+62+620= 4+64+640=

0 1 11 12 121 (b)  is cyclic, with generator  :   == . S2:  1 1 1 0 22 21 212

(c) For n  2 , consider the following bijections   Sn : 1 ==2 2 = 1 and i i for 3 in 2 ==3 3 = 2 and i i for i not equal to 2 or 3

Since  1 === 1 2 3 and  1 === 1 1 2 ,

Sn is not abelian, and therefore not cyclic.

CYU 5.6 Since bca bca ===bc abc abcea bca : bca= e (Theorem 5.6).

12 12 13

CYU 5. 7 (a) False: For    S3 given by : 23 , : 21 and : 22 we have: 31 33 31     121 131 : 233 and : 223 as well. 312 312

(b) True: ab+ = bc+  ba+ ==bc+  a c commutativity Theorem5.9 CYU SOLUTIONS A-29

CYU 5.8 We show that the equation ax+ = b has a unique solution in  + : Existence: ax+ = b – a + ax+ = –ab+ – a + a + x = – a + b  0 + x ==– a + bx – a + b Uniqueness: If x and x then: ax+ = b and ax+ = b  ax+ ==ax+  x x Theorem 5.8

CYU 5.9 We know that we have to consider a non-abelian group, and turn to our friend S3 . Spe- 13 13 13 –1 cifically for   S3 given by : 22 and  = 21 we have:  = 22 , 31 32 31   12 132 12 –1  = 23 , and  ==221 21 , so that: 31 313 33 –1 –1 12 131 11 –1 –1 –1  ===21 while   223 23 33 312 32

–1 –1 –1 –1 CYU 5.10 I. ana2a1 = a1 a2 an clearly holds for n = 1 .  –1 –1 –1 –1 II. Assume ak a2a1 = a1 a2 ak . Then: III.  –1  –1 ak + 1  ak a2a1 = ak + 1  ak a2a1 –1 –1 Theorem 2.10:  –1 –1 –1 –1 ==ak a2a1 ak + 1 a1 a2 ak  ak + 1 II

CYU 5.11 (a) 1 2 3 4 (b) 14= 4  24==4+ 48 2 3 4 1 24 2  34==8+ 412 3 4 1 3 e 24 3   44==12+24416 4 1 2 3 4 54==16+24420 1 2 3 4 64==20+244 0: 4 has order 6  has order 4

CYU5.12 We already know that 6Z is a subgroup of Z. To show that it is a subgroup of 3Z we need but observe that 6Z  3Z : n  6Z  n = 6m for mZ  n = 32m  n  3Z A- 30 APPENDIX A

CYU 5.13 False: 2Z and 3Z are groups, but 2Z  3Z is not: 23  2Z  3Z while 23+52=  Z  3Z .

CYU 5.14 We show that 3 ==Z8 01234567 by demonstrating that every ele-

ment of Z8 is a multiple of 3:

13 ======3 2 3 3 +836 33 3 +83 +831 43 3 +83 +83 +834 =

53 =====3 +83 +83 +83 +837 63 2 7 3 5 8 3 0

Claim: 4 = 04 : 14 ===4 2 4 4 +840 . Fine, but can we pick up other ele- ments of Z8 by taking additional multiples of 4? No: The division algorithm assures us that nq= 4 + r for any nZ , with 0  r  4 . From the above we know that 14 and 24 are in 04 , and surely 04   04 . The only possible loose end is 34 . Let’s tie it up:

34 ===4 +84 +840 +844

123 123 CYU 5.15 False. In S3 both the permutations  =  and  =  have order 2. 321 213

Does =  have order 4? No, it has order 3:

 123 123     123 231     = Then: 2 e  213  312   231 3  231 123

(By the way, as you can easily verify, if G is abelian, then the assertion in CYU 5.15 does hold)

CYU 5.16 (a) Let : G  G be given by a = e . Since for every ab  G , ab ==eee =a b ,  is a homomorphism. (b) Let : Z +  + be given by n = n . Since for every nm  Z , nm+ ===nm+ nm n m ,  is a homomorphism.

CYU 5.17 For ab  G we have:  ab+ ==ab+ a + b =a + b

Definition of composition =  a +  b CYU SOLUTIONS A-31

CYU 5.18 Homomorphism:

2n1 + 2n2 ====2n1 + n2 8n1 + n2 8n1 + 8n2 2n1 + 2n1 Ker ===2n 2n = 0 2n 8n = 0 0 . Im === 2n 8n 8Z

CYU 5.19 Let a  G be such that b == a  b a (*) . We establish that  is one -to- one by showing that Ker = e (Theorem 5.24): Assume that c = e (we want to show that ce= ). Consider the element ca . Since  is a homomorphism: ca === c a ea a . Letting ca play the role of b in (*) we have: caa=== ca a–1 ec e

CYU 5.20 (a) We show that the relation  given by GG  if G is isomorphic to G is an equivalence relation (see Definition 2.20, page 88): Reflexive GG since the identity map Ig= g is clearly an isomorphism. Symmetric GG   G  G : Let : GG  be an isomorphism. Theorem 1.1(a), page 5, assures us that the map –1: G  G is a bijection. We show that it is also a homomorphism: For a b  G let ab  G be such that a = a and b = b . Since ab = ab , We then have: –1ab ==ab –1a –1b .

Transitive G1  G2 and G2  G3  G1  G3 : Follows from Theorem 1.2(c), page 7, and CYU 2.19.

(b) Let gG . The map ig: GG is a bijection: –1 –1 –1 –1 –1 –1 One-to one: iga = igb  gag = gbg  g gag g ==g gbg ga b –1 –1 –1 Onto: For aG , igg ag ==ggag g a –1 –1 –1 ig: GG is a homomorphism: igab ==gabg gag gbg =iga igb

CYU 5.21 Let : GG  be an isomorphism. Assume that G is abelian. For a b  G let ab  G be the elements in G such that a = a and b = b . Then: ab=====a b ab ba b a ba The same argument can be used to show that if G is abelian, then so is G.

CYU 5.22 We first point out that G , being a subgroup of G (Theorem 5.23), is itself a group. While the one-to-one homomorphism : GG  need not be onto all of G , it is certainly onto G = Im . As such, : G  G is an isomorphism. A- 32 APPENDIX A ANSWERS B-1

Appendix B Answers to Selected Exercises Chapter 1 A logical Beginning 1.1 PROPOSITIONS 1. True 3. True 5. True 7. False 9. False 11. False 13. False 15. True 17. False 19. False 21. True 23. True 25. True 27. False 29. False 31. False 33. Yes 35. No 37. Yes 39. Yes 45. No 47. No 49. Yes 51. No 53. No 55. No 57. No 59. No 63. True 65. False 67. Joe or Mary is not a math major. 69. Joe is a math or biology major. 71. 3x + 55 or x and y are not both integers. 73. x is divisible by 2 and 3 or it is not divisible by 7.

Statement: contrapositive: converse inverse pq ~q  ~p qp ~p  ~q 75. If it rains, then I will If it dose not rain, If I stay home, then it If it does not rain then stay home. then I will not stay will rain. I will not stay home. home. 77. If Nina feels better, if Nina does not go to If Nina goes to the If Nina does not feel then she will either go to the library or go shop- library or shopping, better then she will not the library or go shop- ping, then she will tot then Nina feels better. go to the library nor go ping. feel better. shopping.

79. If XZ= then MN . If MN then XZ . If MN then XZ= . If XZ then MN .

81. If XZ then M or N If M and N are not M or N is a solution If XZ= then neither is a solution of the equa- solutions of the equa- of the equation, then nor N is a solution of tion. tion, then XZ= . XZ . the equation.

83. If XZ then the If the equation has a If the equation has no If XZ then the equation has no solution. solution, then XZ . solution, then XZ . equation has a solution

1.2 QUANTIFIERS 1. False 3. True 5. False 7. True 9. False 11. True 13. True 15. False 17. False 19. False 21. False 23. True 25. True 27. True 29. False 31. True 33. True 35. False 37. True 39. False 41. True 43. False 45. False 47. False 49. False 51. False 53. There is a road that does not lead to Rome. 59. Sometimes it rains but not pennies from heaven. 57. nZna   59. nZna   or nb 61. xX  ~px ~qx ~sx 63. x  X ~px ~qx ~sx 65. For every boy there is not some girl. 67. n  Zn  a 69. n  Zn  a or nb 71. x  X ~px ~qx ~sx ANSWERS B-2

73. x  X ~px ~qx ~sx 75. There is somebody that loves nobody. 77. There is someday with no special moment. 79. x  Xy   Yx + y 0 81. a bS   m n  Za + b mn 83. x  Xa   b  Ya xb+ and b ax+ 85. For every opera there exists a symphony not longer than that opera. 87. For every person there is someone that is greater than that person. 89. x  Xy   Yxy +  0 91. a bSm    n  Zab +  mn 93. x  Xa   b  Ya xb+ and b ax+

1.3 METHODS OF PROOF Each exercise calls for a verification or proof.

1.4 PRINCIPLE OF MATHEMATICAL INDUCTION Each exercise calls for a verification or proof.

1.5 THE DIVISION ALGORITHM AND BEYOND 1. qr==0 3. q ==–027 r 5. 1

Chapter 2 A Touch of Set Theory 2.1 BASIC DEFINITIONS 1. U 3. 15nn U 5. B 7. D 9. U 11. C 13. F 15. D 17. 12457810111314 19. 16 17 18  21. 1 2 12 23.  

2.2 FUNCTIONS 1. Not a function 3. Range: ABD 5. Range: ABCD 7. Not one-to-one, not onto. 9. Both one-to-one and onto. 11. One-to-one and not onto. 13. Onto and not one-to-one. 15. Both one-to-one and onto. 17. One-to-one and not onto. 19. Not one-to-one, not onto. 21. One-to-one and not onto. 23. One-to-one and not onto. 25. Both one-to-one and onto. 27. Not one-to-one, not onto. 29. Not one-to-one, not onto. –1 y + 2 –1 y 31. Not one-to-one, not onto. 33. f y = ------35. f y = ------3 2 – y

a --- cd– 2 –1 a –1 ab –1 1 37. f ab = --- b – 3 39. f = b 41. f abc = ---s – 2b 5 cd a --- 2 2 1 ---–2a + b – 2c 2 ANSWERS B-3

2.3 INFINITE COUNTING Each exercise calls for a verification or proof.

2.4 EQUIVALENCE RELATIONS 1. No 3. Yes 5. No 13. Yes 15. No 17. No 19. No 31. Yes 33. Yes 35. No 43. Yes 45. Yes 57. Yes 59. Yes 61. Yes 71. Yes 73. No 75. Yes 77. Yes 79. Yes 81. No 93. Yes 95. No 97. Yes 99. No a a 101. a = aa –2– 103. --- = --- gcd a d = 1 105. r = –rr b d 107. xy = xy y   113. n = n ++10kn 10k  100 115.  =  , 123 = 123 , for n = 12or 3: n = n , 12 ===13 23 12 13 23

Chapter 3 A Touch of Analysis 3.1 THE REAL NUMBER SYSTEM 1. Least upper bound: 7, no max or min. 3. Least upper bound: 10, no max or min. 124 5. No Least upper bound: 7, no max, min: 7. 7. 1 9. 2 11. ------25

3.2 SEQUENCES  n 2 --- if n is even n 1 if n is even 2 1. ------3.  5.  7. 0 9. –1 11. 1 n + 1  2 if n is odd  n + 1 ------if n is odd  2  1 13. 2 15. 1 23. 0 25. Does not exist 27. 1 29. 0 31. --- 2

3.3 METRIC SPACE STRUCTURE OF  1. Neither open nor closed, bounded below and above, not compact. 3. Open, bounded below and above, not compact. 5. Open, bounded below and above, not compact. 7. Closed, bounded below and above, compact.

3.4 CONTINUITY Each exercise calls for a verification or proof. ANSWERS B-4

Chapter 4 A Touch of Topology 4.1 METRIC SPACES 7. Fails property (i) of Definition 4.1. 9. Fails Properties (i) and (iii) of Definition 4.1. 9. Fails Properties (i), (ii), and (iii) of Definition 4.1.

4.2 TOPOLOGICAL SPACES 1(a). Yes 1(b). Yes 1(c). No 1d). No 1(e). Yes 1(f). No 1(g). No 1(h). Yes 17(b) No 17(c) No 27. a = abcd , b = be , c = cd , d = dc , e = e , ce = ced , be = be

1 33. (a) 13 (b) Z+ (c)  (d) 13  5 (e) 0  --- nZ + 37. Z+ n

4.3 CONTINUOUS FUNCTIONS AND HOMOMORPHISMS Each exercise calls for a verification or proof.

4.4 PRODUCT AND QUOTIENT SPACES

1(a). 1  1 1(b). 1  2 1(c). 2  2 Chapter 5 A Touch of Group Theory 5.1 DEFINITIONS AND EXAMPLES 1. A cyclic group with generator 2. 2. Not a group. It does not contain an identity. 3. Not a group. It does not contain an identity. 5. Not a group. It does not contain an identity. 7. Not a group. 1 is the identity, but 2 has no inverse. 9. Abelian group. Not cyclic. 11. Abelian group. Not cyclic.

 e if n  0 mod 3 2 3 n  –n  e if n is even 13. 3 = e , 3 = 3 15. 1 =  1 if n  1 mod 3 17. 3 =    3 if n is odd  2 if n  2 mod 3

 e if n  0 mod 3  e if n  0 mod 3 2 3 n  –n  19. 2 = 1 , 2 = e 21. 2 =  2 if n  1 mod 3 23. 2 =  1 if n  1 mod 3    1 if n  2 mod 3  2 if n  2 mod 3 ANSWERS B-5

123456 123456 123456 25.  =  27.  =  29. 5 ==–1  133652 563412 612345

123456 123456 31. 101 ==5 –1 = 33. 101 ==5 –1 = 612345 214365 35. Abelian 37. Abelian 45. Not Abelian

5.2 ELEMENTARY PROPERTIES OF GROUPS 1.(a) e (b) a (c) a–1cb–1 (d) aba–1

5.3 SUBGROUPS 1. Yes 3. No 5. Yes 7. Yes 9. Yes 11. Yes 13. No 15. Yes 17. No

5.4 HOMOMORPHISMS AND ISOMORPHISMS 1. Yes 3. No 5. Yes 7. Yes 9. Yes 11. Yes ANSWERS B-6 Index I-1

A Continuous Function, 149, 166, 185 Abelian Group, 210 Contrapositive, 8 Addition Modulo n, 209 Converse, 9 Alexander’s Subbase Theorem, 177 Countable Set, 78 Algebra of Functions, 150 Convergent Sequence, 123 Algebra of Sequences, 126 Cyclic Group, 215 Alternate Principle of Induction, 38 Generator, 215 Archimedian Principle, 114 Axiom of Choice, 110 D Decreasing Sequence, 125 B De Morgan’s Laws, 6, 57, 58 Base, 175 Dense Subset, 118 Biconditional Statement, 4 Discrete Metric Space, 161 Bijection, 70 Discrete Topological Space, 171 Binary Operator, 2 Disjunction, 2 Bolzano-Weierstrass Theorem, 129 Division Algorithm, 43 Bound, 112 Divisibility, 27 Greatest Lower, 112 Domain, 63 Least Upper, 112 Lower Bound, 112 E Upper Bound, 112 Equivalence Class, 91 Bounded Sequence, 125 Equivalence Relation, 88 Bounded Set, 164 Euclidean Metric, 135 Even Integer, 23 C Existential Proposition, 16 Cardinality, 77 Cartesian Product, 63 F Cayley’s Theorem, 244 Finite Complement Space, 171 Cauchy Sequence, 130 Finite Subcover, 139, 164 Closed Function, 188 Function, 63 Closed Set, 137, 162, 176 Bijection, 70 Compact, 139, 164, 181 Closed, 188 Complete Ordered Field, 111 Continuous, 149, 166, 170 Completion Axiom, 112 Domain, 63 Composition, 64 Composition, 64 Compound Proposition, 2 Homeomorphism, 199 Conditional Statement, 3 Homomorphism, 237 Converse, 9 Inverse, 70 Inverse, 9 Isomorphism, 241 Negation, 8 One-to-One, 65 Congruence Modulo n, 90 Onto, 66 Conjunction, 2 Open, 188 Cantor Theorem, 85 Range, 63 Cantor-Bernstein-Schroder Theorem, 82 Fundamental Counting Principle, 7 Continuity at a point, 147 Fundamental Theorem of Arithmetic, 48 I-2 Index

G K Generator, 230 Kernel, 239 Greatest Common Divisor, 44 Klein Group, 210 Greatest Lower Bound, 112 Group, 207 L Abelian, 210 Lagrange’s Theorem, 230 Cyclic, 215 Least Upper Bound, 112 Generator, 215, 230 , 123 Elementary Properties, 220 Logically Equivalent, 5 Klein, 210 Lower Bound, 112 Order, 209 Subgroup, 228 M Symmetric, 213 , 68 Table, 210 Maximum, 112 Metric, 160 Metric Space, 159 H Compact, 16a Hausdorff Space, 178 Discrete, 161 Heine-Borell Theorem, 140 Euclidean, 135 Homeomorphic Spaces, 190 Metrizable Space, 172 Homeomorphism, 190 Minimum, 112 Homomorphism, 237 Monotone Sequence, 125 Image, 239 Kernel, 239 N Hypothesis, 3 n-, 68 Negation of a Proposition, 2 I of a Conditional Proposition, 8 Image, 152, 165 of a Quantified Proposition, 18 Incompleteness Theorem, 110 Nested Closed Interval Property, 115 Increasing Sequence, 125 Indiscrete Topological Spaces, 171 O Induction 33 Odd Integer, 23 Integer, 23 One-to-One, 65 Even, 23 Onto, 66 Odd, 23 Open Cover, 139, 164 Prime, 47 Open Sphere, 161 Relatively Prime, 46 Open Function, 188 Inverse Function, 70 Open Set, 135, 162 Isometry, 167 Order of a Group, 209 Isometric Spaces, 167 Order of an Element, 224 Isomorphism, 241 Isomorphic Spaces, 241 P Interval Notation, 59 Partition, 92 Invariant Property, 191, 243 Path Connected Spaces, 191 Inverse Function, 70 Permutation, 213 Index I-3

Power Set, 83 Set, 10 Pre-Image, 152, 165 Bounded, 164 Proposition, 1 Closed, 137, 162, 173 Compound, 2 Compact, 139, 164, 176 Power Set, 83 Complement, 54 Prime, 47 Dense, 118 Principle of Mathematical Induction, 33 Disjoint, 54 Product Space, 2196, 198, 199 Equality, 53 Projection Function, 199, 200 Intersection, 53 Proof, 23 Notation, 10 by Contradiction, 26 Open, 139, 169, 178 Contrapositive, 24 Subset, 53 Direct, 23 Proper, 53 Proposition, 1 Union, 53 Compound, 2 Sierpinski Space, 172 Containing Multiple Quantifiers, 16 Sphere of Radius of r Srx , 123, 161 Existential, 15 Statement, 1 Universal, 14 Subbase, 175 Subgroup, 228 Q Subsequence, 129 Quantifiers, 14 Subspace, 174 Existential, 15 Subset, 55 Universal, 14 Successor Set, 103 Quotient Space, 201 Symmetric Group, 213 Quotient Topology, 201 T R Tautology, 4 Range, 63 Topology, 171 Relation, 88 Topological Space, 171 Equivalence, 88 Base, 175 Reflexive, 88 Compact, 139 Symmetric, 88 Discrete, 164, 171 Relatively Prime, 46 Finite-Complement, 171 Hausdorff, 178 S Half-Open, 176 Sequence, 122, 166 Indiscrete, 171 Algebra of, 126 Metrizable, 172 Bounded, 125 Sierpinski, 172 Cauchy, 130, 167 Subbase, 175 Convergent, 123, 166 Subspace 174 Decreasing, 125 Topologically Invariant Property, 191 Divergent, 123 Tychonoff’s Product Theorem, 200 Increasing, 125 Monotone, 125 Subsequence, 129 I-4 Index

U Uncountable Set, 81 Union, 53 Unary Operation, 2 Universal Proposition, 14 Upper Bound, 112 W Well Ordering Principle, 39