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1 Set Theory
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Notation S={a, b, c} refers to the set whose elements are a, b and c. a S means “a is an element of set S”. d S means “d is not an element of set S”. {x S | P(x)} is the set of all those x from S such that P(x) is true. E.g., T={x Z | 0<x<10}. Notes: 1) {a,b,c}, {b,a,c}, {c,b,a,b,b,c} all represent the same set. 2) Sets can themselves be elements of other sets, e.g., S={ {Mary, John}, {Tim, Ann}, …}
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3 Relations between sets Definition: Suppose A and B are sets. Then A is called a subset of B: A B iff every element of A is also an element of B. Symbolically, A B x, if x A then x B. A B x such that x A and x B. B A A B B A A BA B
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4 Relations between sets Definition: Suppose A and B are sets. Then A equals B: A = B iff every element of A is in B and every element of B is in A. Symbolically, A=B A B and B A. Example: Let A = {m Z | m=2k+3 for some integer k}; B = the set of all odd integers. Then A=B.
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Operations on Sets Definition: Let A and B be subsets of a set U. 1. Union of A and B: A B = {x U | x A or x B} 2. Intersection of A and B: A B = {x U | x A and x B} 3. Difference of B minus A: B A = {x U | x B and x A} 4. Complement of A: A c = {x U | x A} Ex.: Let U=R, A={x R | 3<x<5}, B ={x R| 4<x<9}. Then 1) A B = {x R | 3<x<9}. 2) A B = {x R | 4<x<5}. 3) B A = {x R | 5 ≤x<9}, A B = {x R | 3<x ≤4}. 4) A c = {x R | x ≤3 or x≥5}, B c = {x R | x ≤4 or x≥9}
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6 Properties of Sets Theorem 1 (Some subset relations): 1) A B A 2) A A B 3) If A B and B C, then A C. To prove that A B use the “element argument”: 1. suppose that x is a particular but arbitrarily chosen element of A, 2. show that x is an element of B.
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7 Proving a Set Property Theorem 2 (Distributive Law): For any sets A,B and C: A (B C) = (A B) (A C). Proof: We need to show that (I) A (B C) (A B) (A C) and (II) (A B) (A C) A (B C). Let’s show (I). Suppose x A (B C) (1) We want to show that x (A B) (A C) (2)
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8 Proving a Set Property Proof (cont.): x A (B C) x A or x B C. (a) Let x A. Then x A B and x A C x (A B) (A C) (b) Let x B C. Then x B and x C. Thus, (2) is true, and we have shown (I). (II) is shown similarly (left as exercise).■
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