Operations on Sets. Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) August 6, 2012

Transcription

1 on Sets Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) August 6, 2012 Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

2 Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

3 Often it is convenient to visualize various relations between sets. We use for that A = {1, a, {2, b}} Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

4 Intersection The intersection of sets A and B, denoted by A B, is the set that contains those elements both in A and B A B = {x x A x B} {1, 3, 5, 7} {2, 3, 4, 5, 6} = {3, 5} {Jan, Feb, Dec} {Jan, Feb, Mar} = {Jan, Feb} {x y(x = 2y)} {x y(x = 3y)} = {x y(x = 6y)} Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

5 Union The union of sets A and B, denoted by A B, is the set that contains those elements that are either in A or in B A B = {x x A x B} {1, 3, 5, 7} {2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6, 7} {Jan, Feb, Dec} {Jan, Feb, Mar} = {Jan, Feb, Mar, Dec} {x y(x = 2y)} {x y(x = 3y)} = {x y(x = 2y x = 3y)} Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

6 Disjoint Sets A and B are said to be disjoint if A B = {1, 3, 5, 7} {2, 4, 6} = {Jan, Feb, Dec} {May, Aug} = Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

7 Principle of Inclusion-exclusion Principle of inclusion-exclusion For any finite sets A and B A B = A + B A B To count the elements in A B, we first count the elements of A, then the elements of B. Elements of A B are counted twice so we subtract the number of such elements. If A and B are disjoint, then A B = A + B Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

8 Symmetric Difference The symmetric difference of sets A and B denoted by A B, is the set that contains those elements that are either in A or in B, but not in both. A B = {x x A x B} {1, 3, 5, 7} {2, 3, 4, 5, 6} = {1, 2, 4, 6, 7} {Jan, Feb, May} {May, Feb, Aug} = {Jan, Aug} Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

9 Disjoint Sets and Symmetric Difference Theorem Sets A and B are disjoint if and only if A B = A B Proof. Notice first that A B A B. Suppose that A and B are disjoint. To prove the equality, it suffices to show that A B A B. Take x A B. It belongs to A or B, but x / A B, as the intersection is empty. Therefore, x A B. Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

10 Complement Let A be a set and U a universe, A U. The complement of A denoted by A, is the set that contains all elements in U that do not belong to A. A = {x x U x / A} = {x x / A} Let the universe be the set of all integers, and A = {x y x = 2y}. Then A is the set of all odd integers Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

11 Difference The difference of sets A and B denoted by A B, is the set containing those elements in A, but not in B. A B = {x x A x / B} {1, 3, 5} {1, 2, 3} = {5} Clearly, A = U A Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

12 Connection with logic connectives Similar to logic connectives and formulas, expressions built from set operations and sets also satisfy some laws There is a tight connection between set operations and logic connectives corresponds to complement X corresponds to union corresponds to intersection corresponds to symmetric difference 0 corresponds to 1 corresponds to U Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

13 Double Complement law : (A) = A DeMorgan s Laws (A B) = A B (A B) = A B Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

14 (cntd) Commutative laws A B = B A A B = B A Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

15 (cntd) Commutative laws Associative laws A B = B A A B = B A A (B C) = (A B) C A (B C) = (A B) C Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

16 (cntd) Commutative laws Associative laws Distributive laws A B = B A A B = B A A (B C) = (A B) C A (B C) = (A B) C A (B C) = (A B) (A C) A (B C) = (A B) (A C) Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

17 (cntd) Commutative laws Associative laws Distributive laws Idempotent laws A B = B A A B = B A A (B C) = (A B) C A (B C) = (A B) C A (B C) = (A B) (A C) A (B C) = (A B) (A C) A A = A A A = A Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

18 (cntd) Identity laws A = A A U = A Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

19 (cntd) Identity laws Complement laws A = A A U = A A A = U A A = Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

20 (cntd) Identity laws Complement laws Domination laws A = A A U = A A A = U A A = A U = U A = Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

21 (cntd) Identity laws Complement laws Domination laws Absorption laws A = A A U = A A A = U A A = A U = U A = A (A B) = A A (A B) = A Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

22 (cntd) Theorem A B = A B Proof. We will show that A B A B and A B A B. (1) We show that A B A B. Take x A B. By the definition, x / A B. Therefore, x / A or x / B. Hence x A or x B. Thus x A B. (2) We show that A B A B. Take x A B. By definition, x A or x B. Therefore, x / A or x / B. This implies x / A B. And, finally, x A B. Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets

23 Another proof Theorem A B = A B Another way to prove equalities in sets is to use the set builder construction and some logic Proof. A B = {x x / A B} by definition of complement = {x (x A B)} by definition of does not belong = {x (x A x B)} by definition of intersection = {x (x A) (x B)} by DeMorgan s Law = {x (x / A) (x / B)} by definition of does not belong = {x (x A) (x B)} by definition of complement = {x x A B} by definition of union = A B Gazihan Alankuş (Based on original slides by Brahim Hnich et al.) on Sets