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Unformatted text preview: Finish Set Theory Nested Quantifiers Margaret M. Fleck 17 February 2010 This lecture does a final couple examples of set theory proofs. It then fills in material on quantifiers, especially nested ones, from sections 1.3 and 1.4. 1 Announcements Schedule preview: today: finish set theory, nested quantifiers Fri and next Monday: functions (Rosen 2.3) discussions next week: exam review midterm next Wednesday (30 September) 79pm, 141 Wohlers . lec ture is optional exam question/answer. No homework due a week from Friday (2 October) 2 Recap Last lecture, we saw several proofs of identities from set theory. The key step in many of these proofs involved showing that a set A was a subset of a set B by picking a random element from A and showing that it must be in B . We also saw that you can prove two sets A and B to be equal by showing that A B and B A . In particular, remember that we proved the following: 1 Claim 1 (Transitivity) For any sets A , B , and C , if A B and B C , then A C . Lets also recall that the powerset of A ( P ( A )) is the set containing all subsets of A . For example P ( { a, b } ) = { , { a } , { b } , { c } , { a, b } , { b, c } , { a, c } , { a, b, c }} 3 A proof using power sets Now, we can prove a claim about power sets: Claim 2 For all sets A and B , A B if and only if P ( A ) P ( B ) . Our claim is an ifandonlyif statement. The normal way to prove such a statement is by proving the two directions of the implication separately. Although its occasionally possible to do the two directions together, this doesnt always work and is often confusing to the reader. Proof ( ): Suppose that A and B are sets and A B . Suppose that S is an element of P ( A ). By the definition of power set, S must be a subset of A . Since S A and A B , we must have that S B (by transitivity). Since S B , the definition of power set implies that S P ( B ). Since weve shown that any element of P ( A ) is also an element of P ( B ), we have that P ( A )...
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 Spring '08
 Erickson

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