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

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.

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?

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

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.

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.

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.

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.

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?

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).

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.

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.

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.

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

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.

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