2.3 Algebraic properties of set operations

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1 CHAPTER 2. SETS Algebraic properties of set operations When a mathematician thinks of algebraic properties what they mean is this: we have two things and we combine them to make a third; what sorts of properties does this combination satisfy? When the two things are real numbers, we get properties like the commutative law, the associative law, etc. In the previous section, the two things would be two sets, and we combine them in four di erent ways: A [ B, A \ B, A B, A B. What sorts of properties do these combinations satisfy? We can start to get an idea of what to look for by looking at the algebraic properties of the real numbers and asking if there are analogous properties here. Example 1. Consider the following property of real numbers: a + b = b + a where a and b can be real numbers. (a) What statements do you get if you replace a and b with sets A and B, and + with [ or \ or or? (b) Two of your new statemtents are true, and two are false. Prove the true ones and find counter examples for the false ones. This is where we ended on Wednesday, October 3

2 CHAPTER 2. SETS 39 Solution: (a) The new statements are: A [ B = B [ A A \ B = B \ A A B = B A A B = B A (b) The first two statements in part (a) are true, and the second two are false. We prove A [ B = B [ A. Let x 2 A [ B. Thenx 2 A or x 2 B. Thenx 2 B or x 2 A. Then x 2 B [ B. Let x 2 B [ B. Thenx 2 B or x 2 A. Thenx 2 A or x 2 B. Thenx 2 A [ B. We prove next that A \ B = B \ A. Let x 2 A \ B. Thenx 2 A and x 2 B. Thenx 2 B and x 2 A. Thenx 2 B \ B. Let x 2 B \ B. Thenx 2 B and x 2 A. Thenx 2 A and x 2 B. Thenx 2 A \ B. Now we show that A B = B A is false. Let A = {1} and B = ;. Then A B = {1} and B A = ;. Note that {1} 6= ;. Finally, we show that A B = B A is false. Let A = {1} and B = {2}. ThenA B = {(1, 2)} and B A = {(2, 1)}. Note that (1, 2) 6= (2, 1) and so {(1, 2)} 6= {(2, 1)}. This is where we ended on Monday, February 12

3 CHAPTER 2. SETS 40 Example 2. Consider the following property of real numbers: a (b + c) =(a b)+(a c) where a, b and c can be real numbers. (a) What statements do you get if you replace a, b and c with sets A, B and C, and and + with [ or \ or or? (b) Some of the above statements are true and some are false. Make some conjectures about which are true and which are false. Prove at least one true one and find a counter example for at least one falso one. Solution: (a) We are given four set operations, and are choosing two of them, so there will be 6 choices, but each choice can be put in two orders, so there will be 12 new statements: A [ (B \ C) =(A [ B) \ (A [ C) A \ (B [ C) =(A \ B) [ (A \ C) A [ (B C) =(A [ B) (A [ C) A (B [ C) =(A B) [ (A C) A \ (B C) =(A \ B) (A \ C) A (B \ C) =(A B) \ (A C) A [ (B C) =(A [ B) (A [ C) A (B [ C) =(A B) [ (A C) A \ (B C) =(A \ B) (A \ C) A (B \ C) =(A B) \ (A C) A (B C) =(A B) (A C) A (B C) =(A B) (A C) (b) For most of the true statements we (finally) get lazy and prove both directions of the set containment at the same time. We prove that A [ (B \ C) =(A [ B) \ (A [ C). For any x we have x 2 A [ (B \ C) () x 2 A or x 2 B \ C () x 2 A or (x 2 B and x 2 C) () (x 2 A or x 2 B) and (x 2 A or x 2 C) () (x 2 A [ B) and (x 2 A [ C) () x 2 (A [ B) \ (A [ B). distribute and/or We prove that A \ (B [ C) =(A \ B) [ (A \ C). For any x we have x 2 A \ (B [ C) () x 2 A or x 2 B [ C () x 2 A or (x 2 B or x 2 C) () (x 2 A or x 2 B) or (x 2 A or x 2 C) () (x 2 A \ B) or (x 2 A \ C) () x 2 (A \ B) [ (A \ B). distribute and/or We prove that A [ (B C) =(A [ B) (A [ C) is false.let A = {1}, B = {1}, and C = ;. Then B C = {1}, A [ (B C) ={1},

4 CHAPTER 2. SETS 41 A [ B = {1}, A [ C = {1}, (A [ B) (A [ C) ={1} {1} Since {1} 6= ; we have that the two sides are not equal. I think (but have not triple checked) A (B [ C) =(A B) [ (A C) is false: the [ should be changed to \. I think (but have not triple checked) A \ (B C) =(A \ B) (A \ C) is true (although Wikipedia writes it di erently). I think (but have not triple checked) A (B \ C) =(A B) \ (A C) is false: the \ should be changed to a [. I think (but have not triple checked) A [ (B C) =(A [ B) (A [ C) is false. I think (but have not triple checked) A (B [ C) =(A B) [ (A C) istrue. I think (but have not triple checked)a \ (B C) =(A \ B) (A \ C) istrue I think (but have not triple checked) A (B \ C) =(A B) \ (A C) istrue. I think (but have not triple checked) A (B C) =(A B) (A C) is false. Ithink(buthavenottriplechecked)A (B C) =(A B) (A C) istrue. Theorems about Set Operations Theorem Let A, B and C be any sets and let U be a universal set. The following hold. 1. A [ B = B [ A and A \ B = B \ A (Commutative Laws). 2. A \ (B \ C) =(A \ B) \ C and A [ (B [ C) =(A [ B) [ C (Associative Laws). 3. A \ (B [ C) =(A \ B) [ (A \ C) and A [ (B \ C) =(A [ B) \ (A [ C) (Distributive Laws). 4. A [;= A and A \ U = A (identities). 5. A [ U = U and A \;= ; (domination laws). = ;. Theorem If A, B and C are any sets in a universal set U, then (A c ) c = A A B = A \ B c (A [ B) c = A c \ B c and (A \ B) c = A c [ B c (De Morgan s Laws for sets). Theorem Let A, B and C be any sets. 1. If A C and B C then A [ B C. 2. If C A and C B, thenc A \ B. 3. A B if and only if B c A c Theorem Let A, B and C be any sets. The following hold: 1. (A [ B) C =(A C) [ (B C) 2. (A \ B) C =(A C) \ (B C) 3. (A B) C =(A C) (B C) 4. Suppose A and B are nonempty. Then A B = B A () A = B. 5. ; A = ;. 6. If A 1 A and B 1 B then A 1 B 1 A B. 7. If A and B each have at least two elements, then there are subsets of A B that cannot be written the form A 1 B 1 for any subsets A 1 A and B 1 B. Theorem Let A and B be any finite sets. 1. A + B = A [ B + A \ B. 2. If A \ B = ;, then A [ B = A + B. 3. A B = A B.

5 CHAPTER 2. SETS 42 Theorem (Cardinality of Finite Power Set). If A is a finite set, then P(A) =2 A. In other words, the number of subsets of A is equal to 2 n where n is the number of elements of A. Some books use the notation 2 A instead of P(A). In this notation the theorem becomes 2 A =2 A. This is where we ended on Friday, October 5

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