16 CHAPTER 1 PRELIMINARIES
n 13. Prove thatn<2 for alln N. 2 n 14. Prove that 2
Section 1.3 Finite and Infinite Sets
When we count the elements in a set, we say ‘‘one, two, three, . . . ,’’ stopping when we have exhausted the set. From a mathematical perspective, what we are doing is defining a bijective mapping between the set and a portion of the set of natural numbers. If the set is such that the counting does not terminate, such as the set of natural numbers itself, then we describe the set as being infinite. The notions of ‘‘finite’’ and ‘‘infinite’’ are extremely primitive, and it is very likely that the reader has never examined these notions very carefully. In this section we will define these terms precisely and establish a few basic results and state some other important results that seem obvious but whose proofs are a bit tricky. These proofs can be found in Appendix B and can be read later.
1.3.1 Definition (a) The empty set is said to have 0 elements. ; (b) Ifn N, a setS is said to haven elements if there exists a bijection from the set 2 Nn : 1;2;...;n ontoS. ¼ f g (c) A setS is said to befinite if it is either empty or it hasn elements for somen N. 2 (d) A setS is said to be infinite if it is notfinite.
Since the inverse of a bijection is a bijection, it is easy to see that a setS hasn elements if and only if there is a bijection fromS onto the set{1,2, .. .,n}. Also, since the composition of two bijections is a bijection, we see that a setS 1 hasn elements if and only if there is a bijection fromS 1 onto another setS 2 that hasn elements. Further, a setT 1 isfinite if and only if there is a bijection fromT 1 onto another setT 2 that isfinite. It is now necessary to establish some basic properties offinite sets to be sure that the definitions do not lead to conclusions that conflict with our experience of counting. From the definitions, it is not entirely clear that afinite set might not haven elements for more than one value ofn. Also it is conceivably possible that the setN: 1;2;3;... might be afinite set according to this definition. The reader will be relieved¼ that f these possibilitiesg do not occur, as the next two theorems state. The proofs of these assertions, which use the fundamental properties ofN described in Section 1.2, are given in Appendix B. 1.3.2 Uniqueness Theorem If S is afinite set, then the number of elements in S is a unique number inN. 1.3 FINITE AND INFINITE SETS 17
1.3.3 Theorem The setN of natural numbers is an infinite set.
The next result gives some elementary properties offinite and infinite sets.
1.3.4 Theorem (a) If A is a set with m elements and B is a set with n elements and if A B , then A B has m n elements. \ ¼; [ þ (b) If A is a set with m N elements and C A is a set with1 element, then A C is a set with m 1 elements. 2 n (c) If C is an infinite set and B is afinite set, then C B is an infinite set. n
Proof. (a) Letf be a bijection ofN m ontoA, and letg be a bijection ofN n ontoB. We defineh onN m n byh(i): f(i) fori 1;...;m andh(i) : g(i m) for i m 1;...;m n.þ We leave it¼ as an exercise¼ to show that¼h is a bijection from ¼ þ þ Nm n ontoA B. þ [ The proofs of parts (b) and (c) are left to the reader, see Exercise 2. Q.E.D.
It may seem ‘‘obvious’’ that a subset of afinite set is alsofinite, but the assertion must be deduced from the definitions. This and the corresponding statement for infinite sets are established next.
1.3.5 Theorem Suppose that S and T are sets and that T S. (a) If S is afinite set, then T is afinite set. (b) If T is an infinite set, then S is an infinite set.
Proof. (a) IfT , we already know thatT is afinite set. Thus we may suppose that T . The proof¼; is by induction on the number of elements inS. 6¼ ;IfS has 1 element, then the only nonempty subsetT ofS must coincide withS, soT is a finite set. Suppose that every nonempty subset of a set withk elements isfinite. Now letS be a set havingk 1 elements (so there exists a bijectionf ofN k 1 ontoS), and letT S. If þ þ f k 1 = T, we can considerT to be a subset of1 S: S fk 1 , which hask elementsþ ðÞ by Theorem2 1.3.4(b). Hence, by the induction hypothesis,¼ n fþ TðÞ is agfinite set. On the other hand, iff k 1 T, then1 :T T fk 1 is a subset ofS . Since þ ðÞ 2 ¼ n fþ ðÞ g 1 S1 hask elements, the induction hypothesis implies thatT 1 is afinite set. But this implies thatT T 1 fk 1 is also afinite set. ¼ [ þ ðÞfg (b) This assertion is the contrapositive of the assertion in (a). (See Appendix A for a discussion of the contrapositive.) Q.E.D.
Countable Sets We now introduce an important type of infinite set.
1.3.6 Definition (a) A setS is said to be denumerable (or countably infinite) if there exists a bijection ofN ontoS. (b) A setS is said to be countable if it is eitherfinite or denumerable. (c) A setS is said to be uncountable if it is not countable.
From the properties of bijections, it is clear thatS is denumerable if and only if there exists a bijection ofS ontoN. Also a setS 1 is denumerable if and only if there exists a 18 CHAPTER 1 PRELIMINARIES
bijection fromS 1 onto a setS 2 that is denumerable. Further, a setT 1 is countable if and only if there exists a bijection fromT 1 onto a setT 2 that is countable. Finally, an infinite countable set is denumerable.
1.3.7 Examples (a) The setE: 2n:n N of even natural numbers is denumerable, ¼ f2 g since the mappingf:N E defined byf(n): 2n forn N is a bijection ofN ontoE. ! ¼ 2 Similarly, the setO: 2n 1:n N of odd natural numbers is denumerable. ¼ f 2 g (b) The setZ of all integers is denumerable. To construct a bijection ofN ontoZ, we map 1 onto 0, we map the set of even natural numbers onto the setN of positive integers, and we map the set of odd natural numbers onto the negative integers. This mapping can be displayed by the enumeration:
Z 0;1; 1;2; 2;3; 3;... : ¼ f g (c) The union of two disjoint denumerable sets is denumerable. Indeed, ifA a 1;a 2;a 3;... andB b 1;b 2;b 3;... , we can enumerate the elements ofA B ¼as: f g ¼ f g [ & a1;b 1;a 2;b 2;a 3;b 3;...:
1.3.8 Theorem The setN N is denumerable. Informal Proof. Recall thatN N consists of all ordered pairs (m,n), wherem,n N. We can enumerate these pairs as: 2 1;1 ;1;2 ;2;1 ;1;3 ;2;2 ;3;1 ;1;4 ;...; ðÞ ðÞ ðÞ ðÞ ðÞ ðÞ ðÞ according to increasing summ n, and increasingm. (See Figure 1.3.1.) Q.E.D. þ The enumeration just described is an instance of a ‘‘diagonal procedure,’’ since we move along diagonals that each containfinitely many terms as illustrated in Figure 1.3.1. The bijection indicated by the diagram can be derived as follows. Wefirst notice that thefirst diagonal has one point, the second diagonal has two points, and so on, withk points in thekth diagonal. Applying the formula in Example 1.2.4(a), we see that the total number of points in diagonals 1 throughk is given by ck 1 2 k 1 k k 1 ðÞ ¼ þ þ þ ¼ 2 þ ðÞ
Figure 1.3.1 The setN N 1.3 FINITE AND INFINITE SETS 19
The point (m,n) lies in thekth diagonal whenk m n 1, and it is themth point in that diagonal as we move downward from left to right.¼ þ (For example, the point (3, 2) lies in the 4th diagonal since 3 2 1 4, and it is the 3rd point in that diagonal.) Therefore, in the counting scheme displayedþ ¼ by Figure 1.3.1, we count the point (m,n) byfirst counting the points in thefirstk 1 m n 2 diagonals and then addingm. Therefore, ¼ þ the counting functionh:N N N is given by ! h m;n : cm n 2 m ðÞ ¼ þ ðÞ þ 1 m n 2 m n 1 m: ¼ 2 þ ðÞ þ ðÞ þ 1 For example, the point (3, 2) is counted as numberh3;2 2 3 4 3 9, as shown by Figure 1.3.1. Similarly, the point (17, 25) is counted asðÞ number ¼h (17, þ 25)¼ c(40) 17 837. ¼ þ This¼ geometric argument leading to the counting formula has been suggestive and convincing, but it remains to be proved thath is, in fact, a bijection ofN N ontoN. A detailed proof is given in Appendix B. The construction of an explicit bijection between sets is often complicated. The next two results are useful in establishing the countability of sets, since they do not involve showing that certain mappings are bijections. Thefirst result may seem intuitively clear, but its proof is rather technical; it will be given in Appendix B.
1.3.9 Theorem Suppose that S and T are sets and that T S. (a) If S is a countable set, then T is a countable set. (b) If T is an uncountable set, then S is an uncountable set.
1.3.10 Theorem The following statements are equivalent: (a) S is a countable set. (b) There exists a surjection ofN onto S. (c) There exists an injection of S intoN.
Proof. (a) (b) IfS isfinite, there exists a bijectionh of some setN n ontoS and we defineH on) by N h k fork 1;...;n; H k : ð Þ ¼ ð Þ ¼ h n fork>n: ð Þ ThenH is a surjection ofN ontoS. IfS is denumerable, there exists a bijectionH ofN ontoS, which is also a surjection of N ontoS. (b) (c) IfH is a surjection ofN ontoS, we defineH 1 :S N by lettingH 1(s) be the ) 1 ! least element in the setH s : n N:H n s . To see thatH 1 is an injection ofS ð Þ ¼ 2fð Þ¼ g intoN, note that ifs,t S andn st : H 1 s H 1 t , thens H(n st) t. 2 ¼ ð Þ¼ ð Þ ¼ ¼ (c) (a) IfH 1 is an injection ofS intoN, then it is a bijection ofS ontoH 1 S N. By ) ð Þ Theorem 1.3.9(a),H 1(S) is countable, whence the setS is countable. Q.E.D.
1.3.11 Theorem The setQ of all rational numbers is denumerable.
Proof. The idea of the proof is to observe that the setQ þ of positive rational numbers is contained in the enumeration:
1 1 2 1 2 3 1 1 ; 2 ; 1 ; 3 ; 2 ; 1 ; 4 ;...; 20 CHAPTER 1 PRELIMINARIES which is another ‘‘diagonal mapping’’ (see Figure 1.3.2). However, this mapping is not an 1 2 injection, since the different fractions 2 and 4 represent the same rational number.
Figure 1.3.2 The setQ þ
To proceed more formally, note that sinceN N is countable (by Theorem 1.3.8), it follows from Theorem 1.3.10(b) that there exists a surjectionf ofN ontoN N. Ifg: N N Q þ is the mapping that sends the ordered pair (m,n) into the rational number ! having a representationm=n, theng is a surjection ontoQ þ. Therefore, the compositiong f is a surjection ofN ontoQ þ, and Theorem 1.3.10 implies thatQ þ is a countable set. Similarly, the setQ of all negative rational numbers is countable. It follows as in Example 1.3.7(b) that the setQ Q 0 Q þ is countable. SinceQ containsN, it ¼ [ f g [ must be a denumerable set. Q.E.D.
The next result is concerned with unions of sets. In view of Theorem 1.3.10, we need not be worried about possible overlapping of the sets. Also, we do not have to construct a bijection. S 1.3.12 Theorem If Am is a countable set for each m N, then the union A: m1 1 Am is countable. 2 ¼ ¼
Proof. For eachm N, letw m be a surjection ofN ontoA m. We defineb:N N A by 2 ! b m;n : w n : ð Þ ¼ mð Þ We claim thatb is a surjection. Indeed, ifa A, then there exists a leastm N such that 2 2 a A m, whence there exists a leastn N such thata w m n . Therefore,a b(m,n). 2 2 ¼ ð Þ ¼ SinceN N is countable, it follows from Theorem 1.3.10 that there exists a surjection f:N N N whenceb f is a surjection ofN ontoA. Now apply Theorem 1.3.10 again ! to conclude thatA is countable. Q.E.D.
Remark A less formal (but more intuitive) way to see the truth of Theorem 1.3.12 is to enumerate the elements ofA m,m N, as: 2 A1 a 11;a 12;a 13;... ; ¼ f g A2 a 21;a 22;a 23;... ; ¼ f g A3 a 31;a 32;a 33;... ; ¼ f: g
1.3 FINITE AND INFINITE SETS 21
We then enumerate this array using the ‘‘diagonal procedure’’:
a11;a 12;a 21;a 13;a 22;a 31;a 14;...; as was displayed in Figure 1.3.1.
Georg Cantor Georg Cantor (1845–1918) was born in St. Petersburg, Russia. His father, a Danish businessman working in Russia, moved the family to Germany several years later. Cantor studied briefly at Zurich, then went to the University of Berlin, the best in mathematics at the time. He received his doctorate in 1869, and accepted a position at the University of Halle, where he worked alone on his research, but would occasionally travel the seventy miles to Berlin to visit colleagues. Cantor is known as the founder of modern set theory and he was thefirst to study the concept of infinite set in rigorous detail. In 1874 he proved thatQ is countable and, in contrast, thatR is uncountable (see Section 2.5), exhibiting two kinds of infinity. In a series of papers he developed a general theory of infinite sets, including some surprising results. In 1877 he proved that the two-dimensional unit square in the plane could be put into one-one correspondence with the unit interval on the line, a result he sent in a letter to his colleague Richard Dedekind in Berlin, writing ‘‘I see it, but I do not believe it.’’ Cantor’s Theorem on sets of subsets shows there are many different orders of infinity and this led him to create a theory of ‘‘transfinite’’ numbers that he published in 1895 and 1897. His work generated considerable controversy among mathematicians of that era, but in 1904, London’s Royal Society awarded Cantor the Sylvester Medal, its highest honor. Beginning in 1884, he suffered from episodes of depression that increased in severity as the years passed. He was hospitalized several times for nervous breakdowns in the Halle Nervenklinik and spent the last seven months of his life there.
We close this section with one of Cantor’s more remarkable theorems.
1.3.13 Cantor’s Theorem If A is any set, then there is no surjection of A onto the set A of all subsets of A. Pð Þ Proof. Suppose thatw:A A is a surjection. Sincew(a) is a subset ofA, eithera belongs tow(a) or it does!Pð not belongÞ to this set. We let
D: a A:a= w a : ¼ 2f2 ð Þ g
SinceD is a subset ofA, ifw is a surjection, thenD wa 0 for somea A. ¼ ð Þ 0 2 We must have eithera D ora = D. Ifa D, then sinceD wa 0 , we must have 0 2 0 2 0 2 ¼ ð Þ a0 wa 0 , contrary to the definition ofD. Similarly, ifa 0 = D, thena0 = wa 0 so that a 2D,ð which Þ is also a contradiction. 2 2 ð Þ 0 2 Therefore,w cannot be a surjection. Q.E.D.
Cantor’s Theorem implies that there is an unending progression of larger and larger sets. In particular, it implies that the collection N of all subsets of the natural numbers Pð Þ N is uncountable. 22 CHAPTER 1 PRELIMINARIES
Exercises for Section 1.3
1. Prove that a nonempty setT 1 isfinite if and only if there is a bijection fromT 1 onto afinite setT 2. 2. Prove parts (b) and (c) of Theorem 1.3.4. 3. LetS: {1, 2} andT: {a,b,c}. (a) Determine¼ the number¼ of different injections fromS intoT. (b) Determine the number of different surjections fromT ontoS. 4. Exhibit a bijection betweenN and the set of all odd integers greater than 13. 5. Give an explicit definition of the bijectionf fromN ontoZ described in Example 1.3.7(b). 6. Exhibit a bijection betweenN and a proper subset of itself.
7. Prove that a setT 1 is denumerable if and only if there is a bijection fromT 1 onto a denumerable setT 2. 8. Give an example of a countable collection offinite sets whose union is notfinite. 9. Prove in detail that ifS andT are denumerable, thenS T is denumerable. [ 10. (a) If (m,n) is the 6th point down the 9th diagonal of the array in Figure 1.3.1, calculate its number according to the counting method given for Theorem 1.3.8. (b) Given thath(m, 3) 19,findm. ¼ 11. Determine the number of elements in S , the collection of all subsets ofS, for each of the following sets: Pð Þ (a)S: {1, 2}, (b)S:¼ {1, 2, 3}, (c)S:¼ {1, 2, 3, 4}. Be sure¼ to include the empty set and the setS itself in S . Pð Þ 12. Use Mathematical Induction to prove that if the setS hasn elements, then S has 2 n elements. Pð Þ 13. Prove that the collection N of allfinite subsets ofN is countable. Fð Þ APPENDIX B FINITE AND COUNTABLE SETS
We will establish the results that were stated in Section 1.3 without proof. The reader should refer to that section for the definitions. Thefirst result is sometimes called the ‘‘Pigeonhole Principle.’’ It may be interpreted as saying that ifm pigeons are put inton pigeonholes and ifm>n, then at least two pigeons must share one of the pigeonholes. This is a frequently used result in combinatorial analysis. It yields many useful consequences.
B.1 Theorem Let m;n N with m>n. Then there does not exist an injection fromN m 2 intoN n.
Proof. We will prove this by induction onn. Ifn 1 and ifg is any map ofN m m>1 intoN 1, then it is clear thatg 1 g m ¼ 1, so thatg is not injective. ð Þ ð Þ ¼¼ ð Þ¼ Assume thatk> 1 is such that ifm>k, there is no injection fromN m intoN k. We will show that ifm>k 1, there is no functionh:N m N k 1 that is an injection. þ ! þ Case 1: If the rangeh N m N k N k 1, then the induction hypothesis implies thath is ð Þ þ not an injection ofN m intoN k, and therefore intoN k 1. þ Case 2: Suppose thath N m is not contained inN k. If more than one element inN m is mapped intok 1, thenð h isÞ not an injection. Therefore, we may assume that a single þ p N m is mapped intok 1 byh. We now defineh 1 :N m 1 N k by 2 þ ! h q ifq 1;...;p 1; h1 q : ð Þ ¼ ð Þ ¼ h q 1 ifq p;...;m 1: ð þ Þ ¼ Since the induction hypothesis implies thath 1 is not an injection intoN k, it is easily seen thath is not an injection intoN k 1. Q.E.D. þ
We now show that afinite set determines a unique number inN.
1.3.2 Uniqueness Theorem If S is afinite set, then the number of elements in S is a unique number inN.
Proof. IfthesetShasm elements, there exists a bijectionf 1 ofN m ontoS.IfS also hasn elements, there exists a bijectionf 2 ofN n ontoS. Ifm>n, then (by Exercise 21 1 of Section 1.1)f 2 f 1, is a bijection ofN m ontoN n, which contradicts Theorem B.1. 1 Ifn>m, thenf 1 f 2 is a bijection ofN n ontoN m, which contradicts Theorem B.1. Therefore we havem n. Q.E.D. ¼
B.2 Theorem If n N, there does not exist an injection fromN intoN n. 2
Proof. Assume thatf:N N n is an injection, and letm: n 1. Then the restriction of ! ¼ þ f toN m N is also an injection intoN n. But this contradicts Theorem B.1. Q.E.D.
357 358 APPENDIX B FINITE AND COUNTABLE SETS
1.3.3 Theorem The setN of natural numbers is an infinite set.
Proof. IfN is afinite set, there exists somen N and a bijectionf ofN n ontoN. In this 1 2 case the inverse functionf is a bijection (and hence an injection) ofN ontoN n. But this contradicts Theorem B.2. Q.E.D.
We will next establish Theorem 1.3.8. In connection with the array displayed in Figure 1.3.1, the functionh was defined byh m;n c m n 2 m, where c k 1 2 k 1 k k 1 . We now proveð thatÞ ¼ theð functionþ Þ þh is a bijection. ð Þ¼ þ þþ ¼ 2 ð þ Þ 1.3.8 Theorem The setN N is denumerable. Proof. We will show that the functionh is a bijection. (a) Wefirst show thath is injective. If m;n m ;n , then either (i)m n ð Þ 6¼ ð 0 0Þ þ 6¼ m0 n 0, or (ii)m n m 0 n 0 andm m 0. þ þ ¼ þ 6¼ In case (i), we may supposem n
c k p 1
B.3 Theorem If A N and A is infinite, there exists a functionw:N A such that ! w n 1 >w n n for all n N. Moreover,w is a bijection ofN onto A. ð þ Þ ð Þ 2 Proof. SinceA is infinite, it is not empty. We will use the Well-Ordering Property 1.2.1 of N to give a recursive definition ofw. SinceA 0, there is a least element ofA, which we define to bew 1 ; therefore, w 1 1. 6¼ ð Þ ð Þ SinceA is infinite, the setA 1 : A w 1 is not empty, and we definew 2 to be least ¼ nf ð Þg ð Þ element ofA 1. Thereforew 2 >w 1 1, so thatw 2 2. ð Þ ð Þ ð Þ APPENDIX B FINITE AND COUNTABLE SETS 359
Suppose thatw has been defined to satisfyw n 1 >w n n forn 1;...;k 1, whencew k >w k 1 k 1 so thatw k k. Sinceð þ theÞ setð ÞA is infinite,¼ the set ð Þ ð Þ ð Þ Ak : A w 1 ;...;w k ¼ nf ð Þ ð Þg is not empty and we definew k 1 to be the least element inA k. Thereforew k 1 > w k , and sincew k k, we alsoð þ haveÞ w k 1 k 1. Therefore,w is defined onð allþ Þ ð Þ ð Þ ð þ Þ þ ofN. We claim thatw is an injection. Ifm>n, thenm n r for somer N. If r 1, thenw m w n 1 >w n . Suppose thatw n k ¼>wþn ; we will show2 that w¼n k 1 ð >Þw ¼n ð. Indeed,þ Þ ð thisÞ follows from theð þ factÞ thatð Þw n k 1 w n kð þð1 >þw nÞÞ k >ð wÞ n . Sincew m >w n wheneverm>n, it followsð þð þ thatÞÞw ¼is anðð þ injection.Þþ Þ ð þ Þ ð Þ ð Þ ð Þ We claim thatw is a surjection ofN ontoA. If not, the set A~: A w N is not empty, and we letp be the least element in A~. We claim thatp belongs to the¼ setn ðwÞ1 ;...;w p . Indeed, if this is not true, then f ð Þ ð Þg
p A w 1 ;...;w p A p; 2 nf ð Þ ð Þg ¼ so thatw p 1 , being the least element inA p, must satisfyw p 1 p. But this contradictsð þ theÞ fact thatw p 1 >w p p. Therefore A~ is emptyð þ andÞw is a surjection ð þ Þ ð Þ ontoA. Q.E.D.
B.4 Theorem If A N, then A is countable. Proof. IfA isfinite, then it is countable, so it suffices to consider the case thatA is infinite. In this case, Theorem B.3 implies that there exists a bijectionw ofN ontoA, so thatA is denumerable and, therefore, countable. Q.E.D.
1.3.9 Theorem Suppose that S and T are sets and that T S. (a) If S is a countable set, then T is a countable set. (b) If T is an uncountable set, then S is an uncountable set.
Proof. (a) IfS is afinite set, it follows from Theorem 1.3.5(a) thatT isfinite, and therefore countable. IfS is denumerable, then there exists a bijectionc ofS ontoN. Since c S N, Theorem B.4 implies thatc S is countable. Since the restriction ofc toT is a ð Þ ð Þ bijection ontoc T andc T N is countable, it follows thatT is also countable. ð Þ ð Þ (b) This assertion is the contrapositive of the assertion in (a). Q.E.D. 2.5 INTERVALS 49 then for anye> 0, there exists anm N such that 0 h j b m a m
The Uncountability ofR The concept of a countable set was discussed in Section 1.3 and the countability of the set Q of rational numbers was established there. We will now use the Nested Interval Property to prove that the setR is an uncountable set. The proof was given by Georg Cantor in 1874 in thefirst of his papers on infinite sets. He later published a proof that used decimal representations of real numbers, and that proof will be given later in this section.
2.5.4 Theorem The setR of real numbers is not countable.
Proof. We will prove that the unit intervalI: 0;1 is an uncountable set. This implies ¼ ½ that the setR is an uncountable set, for ifR were countable, then the subsetI would also be countable. (See Theorem 1.3.9(a).) The proof is by contradiction. If we assume thatI is countable, then we can enumerate the set asI x 1;x 2;...;x n;... . Wefirst select a closed subintervalI 1 ofI such that ¼ f g x1 = I1, then select a closed subintervalI 2 ofI 1 such thatx 2 = I2, and so on. In this way, we obtain2 nonempty closed intervals 2
I1 I 2 I n such thatI n I andx n = In for alln. The Nested Intervals Property 2.5.2 implies that there 2 exists a pointj I such thatj I n for alln. Thereforej x n for alln N, so the enumeration of2I is not a complete2 listing of the elements of 6¼I, as claimed. Hence,2 I is an uncountable set. Q.E.D.
The fact that the setR of real numbers is uncountable can be combined with the fact that the setQ of rational numbers is countable to conclude that the setR Q of irrational numbers is uncountable. Indeed, since the union of two countable sets is countablen (see 1.3.7(c)), if R Q is countable, then sinceR Q R Q , we conclude thatR is also a countable set, n ¼ [ n ðÞ which is a contradiction. Therefore, the set of irrational numbersR Q is an uncountable set. n Note: The set of real numbers can also be divided into two subsets of numbers called algebraic numbers and transcendental numbers. A real number is called algebraic if it is a solution of a polynomial equationPx 0 where all the coefficients of the polynomialP are integers. A real number is called transcendentalðÞ ¼ if it is not an algebraic number. It can be proved that the set of algebraic numbers is countably infinite, and consequently the set of transcendental numbers is uncountable. The numbersp ande are transcendental numbers, but the proofs of these facts are very deep. For an introduction to these topics, we refer the interested reader to the book by Ivan Niven listed in the References.
yBinary Representations We will digress briefly to discuss informally the binary (and decimal) representations of real numbers. It will suffice to consider real numbers between 0 and 1, since the representations for other real numbers can then be obtained by adding a positive or negative number.
y The remainder of this section can be omitted on afirst reading.