16 CHAPTER 1 PRELIMINARIES n 13. Prove thatn<2 for alln N. 2 n 14. Prove that 2 <n! for alln 4,n N. 2 n 2 15. Prove that 2n 3 2 for alln 5,n N. 2 16. Find all natural numbersn such thatn 2 <2 n. Prove your assertion. 3 17. Find the largest natural numberm such thatn n is divisible bym for alln N. Prove your assertion. 2 ffiffiffi ffiffiffi ffiffiffi ffiffiffi 18. Prove that 1=p1 1= p2 1= pn > pn for alln N,n> 1. þ þ þ 2 k 19. LetS be a subset ofN such that (a) 2 S for allk N, and (b) ifk S andk 2, then 2 2 2 k 1 S. Prove thatS N. 2 ¼ 1 20. Let the numbersx n be defined as follows:x 1 : 1,x 2 : 2, andx n 2 : xn 1 x n for all ¼ ¼ þ ¼ 2 ðþ þ Þ n N. Use the Principle of Strong Induction (1.2.5) to show that 1 x n 2 for alln N. 2 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.