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Properties of Sets Lecture 26 Section 5.2 Tue, Mar 6, 2007.

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1 Properties of Sets Lecture 26 Section 5.2 Tue, Mar 6, 2007

2 Proving Basic Properties Theorem: Let A and B be sets. Then A  B  A. Proof: Let x  A  B. Then x  A and x  B. Therefore, x  A. So A  B  A.

3 Comments The proof uses the logic that S  T if and only if x  S  x  T. A Venn diagram alone does not constitute a proof. This theorem is suggestive of the “specialization” argument form (p. 40) p  q  p What is the connection?

4 Identity Laws Theorem: Let A be any set. Then A  U = A, A   = A.

5 DeMorgan’s Laws Theorem: Let A and B be any two sets. Then (A  B) c = A c  B c, (A  B) c = A c  B c.

6 Set-Theoretic Proofs Theorem: Let A, B, and C be sets. Then (A – B)  (C – B) = (A  C) – B. Proof: Let x  (A – B)  (C – B). Then x  A – B and x  C – B,  x  A, and x  C, and x  B,  x  A  C and x  B,  x  (A  C) – B.

7 Set-Theoretic Proof It then follows that (A – B)  (C – B)  (A  C) – B. The second half of the proof will show that (A  C) – B  (A – B)  (C – B). However, the logic is exactly the reverse of the first half. We may handle that by saying “and conversely” at the end of the first half.

8 Comment The preceding theorem is equivalent to the logical equivalence (p  q)  (r  q) = (p  r)   q which is not hard (at all!) to prove. Proof: (p  q)  (r  q) = p  q  r  q = p  r  q = (p  r)   q.

9 Question Why was that so much easier than the original proof? Because we know a lot about the operators , , and . We could use the “algebra” of , , and . Is there an algebra of , , and complement?

10 Algebraic Properties of Sets See Theorem 1.1.1, p. 14. Commutative Laws: A  B = B  A. A  B = B  A. Associative Laws: (A  B)  C = A  (B  C). (A  B)  C = A  (B  C).

11 Algebraic Properties of Sets Distributive Laws: A  (B  C) = (A  B)  (A  C). A  (B  C) = (A  B)  (A  C). Identity Laws: A  U = A. A   = A.

12 Algebraic Properties of Sets Complement Laws: A  A c = U. A  A c = . Double Complement Law: (A c ) c = A. Idempotent Laws: A  A = A. A  A = A.

13 Algebraic Properties of Sets Universal Bound Laws: A  U = U. A   = . DeMorgan’s Laws: (A  B) c = A c  B c. (A  B) c = A c  B c.

14 Algebraic Properties of Sets Absorption Laws: A  (A  B) = A. A  (A  B) = A. Complements of U and . U c = .  c = U.

15 The Proof Revisited Theorem: Let A, B, and C be sets. Then (A – B)  (C – B) = (A  C) – B. Proof: (A – B)  (C – B) = (A  B c )  (C  B c ) = (A  C)  (B c  B c ) = (A  C)  B c = (A  C) – B.

16 The Proof Revisited Theorem: Let A, B, and C be sets. Then (A – B)  (C – B) = (A  C) – B. Proof:


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