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CHAPTER 8 Proofs Involving Sets CHAPTER 8 Proofs Involving Sets tudents in their first advanced mathematics classes are often surprised S by the extensive role that sets play and by the fact that most of the proofs they encounter are proofs about sets. Perhaps you’ve already seen such proofs in your linear algebra course, where a vector space was defined to be a set of objects (called vectors) that obey certain properties. Your text proved many things about vector spaces, such as the fact that the intersection of two vector spaces is also a vector space, and the proofs used ideas from set theory. As you go deeper into mathematics, you will encounter more and more ideas, theorems and proofs that involve sets. The purpose of this chapter is to give you a foundation that will prepare you for this new outlook. We will discuss how to show that an object is an element of a set, how to prove one set is a subset of another and how to prove two sets are equal. As you read this chapter, you may need to occasionally refer back to Chapter 1 to refresh your memory. For your convenience, the main definitions from Chapter 1 are summarized below. If A and B are sets, then: A B ©(x, y): x A, y Bª, £ Æ 2 2 A B ©x :(x A) (x B)ª, [ Æ 2 _ 2 A B ©x :(x A) (x B)ª, \ Æ 2 ^ 2 A B ©x :(x A) (x B)ª, ¡ Æ 2 ^ ∉ A U A. Æ ¡ Recall that A B means that every element of A is also an element of B. ⊆ 8.1 How to Prove a A 2 We will begin with a review of set-builder notation, and then review how to show that a given object a is an element of some set A. 132 Proofs Involving Sets © ª Generally, a set A will be expressed in set-builder notation A x : P(x) , Æ where P(x) is some statement (or open sentence) about x. The set A is understood to have as elements all those things x for which P(x) is true. For example, © ª © ª x : x is an odd integer ..., 5, 3, 1,1,3,5,... Æ ¡ ¡ ¡ © ª A common variation of this notation is to express a set as A x S : P(x) . Æ 2 Here it is understood that A consists of all elements x of the (predetermined) set S for which P(x) is true. Keep in mind that, depending on context, x could be any kind of object (integer, ordered pair, set, function, etc.). There is also nothing special about the particular variable x; any reasonable symbol x, y, k, etc., would do. Some examples follow. © ª © ª n Z : n is odd ..., 5, 3, 1,1,3,5,... 2 Æ ¡ ¡ ¡ ©x N : 6 xª ©6,12,18,24,30,...ª 2 j Æ ©(a, b) Z Z : b a 5ª ©...,( 2,3),( 1,4),(0,5),(1,6),...ª 2 £ Æ Å Æ ¡ ¡ ©X P(Z): X 1ª ©...,© 1ª,©0ª,©1ª,©2ª,©3ª,©4ª,...ª 2 j j Æ Æ ¡ Now it should be clear how to prove that an object a belongs to a set © ª © ª x : P(x) . Since x : P(x) consists of all things x for which P(x) is true, to © ª show that a x : P(x) we just need to show that P(a) is true. Likewise, to © 2 ª show a x S : P(x) , we need to confirm that a S and that P(a) is true. 2 2 2 These ideas are summarized below. However, you should not memorize these methods, you should understand them. With contemplation and practice, using them becomes natural and intuitive. How to show a ©x : P(x)ª How to show a ©x S : P(x)ª 2 2 2 Show that P(a) is true. 1. Verify that a S. 2 2. Show that P(a) is true. © ª Example 8.1 Let’s investigate elements of A x : x N and 7 x . This set © ª Æ 2 j has form A x : P(x) where P(x) is the open sentence (x N) (7 x). Thus Æ 2 ^ j 21 A because P(21) is true. Similarly, 7,14,28,35, etc., are all elements of 2 A. But 8 A (for example) because P(8) is false. Likewise 14 A because ∉ ¡ ∉ P( 14) is false. ¡ © ª Example 8.2 Consider the set A X P(N): X 3 . We know that © ª © ª Æ 2 ¯© j ªj¯ Æ © ª 4,13,45 A because 4,13,45 P(N) and ¯ 4,13,45 ¯ 3. Also 1,2,3 A, © ª2 © 2 ª ¯© Æ ª¯ 2 10,854,3 A, etc. However 1,2,3,4 A because ¯ 1,2,3,4 ¯ 3. Further, © ª 2 © ª ∉ 6Æ 1,2,3 A because 1,2,3 P(N). ¡ ∉ ¡ ∉ How to Prove A B 133 ⊆ © ª Example 8.3 Consider the set B (x, y) Z Z : x y (mod 5) . Notice Æ 2 £ ´ (8,23) B because (8,23) Z Z and 8 23 (mod 5). Likewise, (100,75) B, 2 2 £ ´ 2 (102,77) B, etc., but (6,10) B. 2 ∉ Now suppose n Z and consider the ordered pair (4n 3,9n 2). Does 2 Å ¡ this ordered pair belong to B? To answer this, we first observe that (4n 3,9n 2) Z Z. Next, we observe that (4n 3) (9n 2) 5n 5 5(1 n), Å¡ ¡ 2 £ ¢ Å ¡ ¡ Æ¡ Å Æ ¡ so 5 (4n 3) (9n 2) , which means (4n 3) (9n 2) (mod 5). Therefore we j Å ¡ ¡ Å ´ ¡ have established that (4n 3,9n 2) meets the requirements for belonging Å ¡ to B, so (4n 3,9n 2) B for every n Z. Å ¡ 2 2 Example 8.4 This illustrates another common way of defining a set. © 3 ª Consider the set C 3x 2 : x Z . Elements of this set consist of all the Æ Å 2 values 3x3 2 where x is an integer. Thus 22 C because 22 3( 2)3 2. Å ¡ 2 ¡ Æ ¡ Å You can confirm 1 C and 5 C, etc. Also 0 C and 1 C, etc. ¡ 2 2 ∉ 2 ∉ 8.2 How to Prove A B ⊆ In this course (and more importantly, beyond it) you will encounter many circumstances where it is necessary to prove that one set is a subset of an- other. This section explains how to do this. The methods we discuss should improve your skills in both writing your own proofs and in comprehending the proofs that you read. Recall (Definition 1.3) that if A and B are sets, then A B means that ⊆ every element of A is also an element of B. In other words, it means if a A, then a B. Therefore to prove that A B, we just need to prove that 2 2 ⊆ the conditional statement “If a A, then a B” 2 2 is true. This can be proved directly, by assuming a A and deducing a B. 2 2 The contrapositive approach is another option: Assume a B and deduce ∉ a A. Each of these two approaches is outlined below. ∉ How to Prove A B How to Prove A B ⊆ ⊆ (Direct approach) (Contrapositive approach) Proof. Suppose a A. Proof. Suppose a B. 2 . ∉ . Therefore a B. Therefore a A. 2 ∉ Thus a A implies a B, Thus a B implies a A, 2 2 ∉ ∉ so it follows that A B. so it follows that A B. ⊆ ■ ⊆ ■ 134 Proofs Involving Sets In practice, the direct approach usually results in the most straight- forward and easy proof, though occasionally the contrapositive is the most expedient. (You can even prove A B by contradiction: Assume ⊆ (a A) (a B), and deduce a contradiction.) The remainder of this section 2 ^ ∉ consists of examples with occasional commentary. Unless stated otherwise, we will use the direct approach in all proofs; pay special attention to how the above outline for the direct approach is used. © ª © ª Example 8.5 Prove that x Z : 18 x x Z : 6 x . 2 j ⊆ 2 j © ª Proof. Suppose a x Z : 18 x . 2 2 j This means that a Z and 18 a. 2 j By definition of divisibility, there is an integer c for which a 18c. Æ Consequently a 6(3c), and from this we deduce that 6 a. Æ j© ª Therefore a is one of the integers that 6 divides, so a x Z : 6 x . © ª © 2ª 2 j We’ve shown a x Z : 18 x implies a n Z : 6 x , so it follows that © ª © 2 2ª j 2 2 j x Z : 18 x x Z : 6 x . 2 j ⊆ 2 j ■ © ª © ª © ª Example 8.6 Prove that x Z : 2 x x Z : 9 x x Z : 6 x . 2 j \ 2 j ⊆ 2 j © ª © ª Proof. Suppose a x Z : 2 x x Z : 9 x . 2 2 j \ 2 j © ª © ª By definition of intersection, this means a x Z : 2 x and a x Z : 9 x . © ª 2 2 j 2 2 j Since a x Z : 2 x we know 2 a, so a 2c for some c Z. Thus a is even. 2 © 2 j ª j Æ 2 Since a x Z : 9 x we know 9 a, so a 9d for some d Z.
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