1 Set Theory Second Part

2 Disjoint Set let A and B be a set. the two sets are called disjoint if their intersection is an empty set. Intersection of set:  A={1,3,5,7,9}  B={2,4,6,8,10} A  B= { } = , thus A and B are disjoint

3 Cartesian Product Ordered n-tuples has a 1 as its first element, a 2 as its second element, …and a n as its nth element. (a 1,a 2,…,a n ). 2 ordered n-tuples are equal if and only if each corresponding pair of their elements is equal.  (a 1,a 2,…,a n ) = (b 1,b 2,…,b n ) if & only if a i = b i for i = 1,2,…,n. 2-tuples are called ordered pairs:  ordered pairs (a,b) & (c,d) are equal if a = c & b = d.  ordered pairs (a,b) & (b,a) are equal if a = b.

4 Cartesian Product Is (1,2) = (2,1)? No. Is (3,(-2) 2,1/2) = (√9,4,3/6)? Yes. Let A and B be sets. The Cartesian product of A and B, denoted by A x B, is the set of all ordered pairs (a, b) where a  A  b  B. Hence, A x B = { (a,b) | a  A  b  B}.

5 Cartesian Product E.g. What is the Cartesian product of A = {1,2} and B = (a,b,c} ? A x B = {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c)}. The Cartesian product of the set A 1, A 2, …, A n, denoted by A 1 x A 2 x … x A n is the set of ordered n-tupples (a 1, a 2,…,a n ), where a i belongs to A i for i=1,2,…,n. In other words, A 1 x A 2 x … x A n = {(a 1, a 2,…,a n ) | a i  A i for i =1,2,…,n}

6 Set Theory Objectives On completion of this chapter, student should be able to: Identify element and subset of a set Perform set operations Establish set properties (equalities) using element arguments.

7 Properties of Set Subset relations 1.Inclusion of Intersection: For all sets A and B, (a) (A  B)  A and (b) (A  B)  B 2. Inclusion in Union (a) A  A  B and (b) B  A  B 3. Transitive of Subsets if A  B and B  C, then A  C (One set is a subset of another)

8 Element Argument The basic method for proving that one set is a subset of another. Let sets X and Y be given. To prove that X  Y 1. Suppose that x is a particular but arbitrarily chosen element of x 2. Show that x is an element of Y.

9 Example Prove that for all sets A and B, A  B  A Proof: Suppose x is any element of A  B. Then x  A and x  B by definition of intersection. In particular x  A.

10 Set Identities

11 Set Identities

12 Set Identities 11. U c =  and  c =U Complements of U &  12. A – B = A  B c Set Difference Law

13 Proving Set Identities Basic Method Let sets X and Y be given. To prove that X = Y: 1.Prove that X  Y 2.Prove that Y  X.

14 Distributive Law Prove that A  (B  C) = (A  B)  (A  C) Suppose that x  A  (B  C). Then x  A or x  (B  C). Case 1: (x  A) Since x  A then x  A  B and also x  A  C by definition of union Hence x  (A  B)  (A  C) Case 2: (x  B  C) Since (x  B  C), then x  B and x  C. Since x  B, x  A  B and since x  C, x  A  C. Hence x  (A  B)  (A  C)

15 Show that (A  B)  (A  C)  A  (B  C) Suppose that x  (A  B)  (A  C) Then x  (A  B) and x  (A  C). Case 1: (x  A) Since x  A we can conclude that A  (B  C) Case 2: (x  A ) Since x  A, then x  B  C. It follows that x  A  (B  C). Hence (A  B)  (A  C)  A  (B  C) Therefore, A  (B  C) = (A  B)  (A  C)

16 Question ???

17 De Morgan’s Law Prove that (A  B) c = A c  B c Show that (A  B) c  A c  B c x  (A  B) c then x  (A  B). It means x  A and x  B by DeMorgan’s Law. Hence x  A c and x  B c or x  A c  B c So (A  B) c  A c  B c Next is to show that A c  B c  (A  B) c Do it yourself!

18 Algebraic Proofs of Set Identities Can use set identities to derive new set identities or to simplify a complicated set expression A  B) c  A c  B c Prove that (A  B)–C = (A–C)  (B–C) (A  B)–C = (A  B)  C c by the set difference law = C c  (A  B) by the comm. law = (C c  A)  (C c  B) by the distributive law = (A  C c )  (B  C c ) comm. law = (A-C)  (B-C) by the set difference law

19 Question ???

20 Another example Prove that A-(A  B)=(A-B) A-(A  B)= A  (A  B) c by the set difference law = A  (A c  B c ) by the De Morgan’s law = (A  A c )  (A  B c ) the distributive law =   (A  B c ) the complement law = (A  B c ) the identity law = A-B by the set difference law

21 Activity Use set identities to: Show that (A  (B  C)) c = (C c  B c )  A c Simplify the expression [((A  B)  C) c  B c )] c

22 Summary Covered: LO: Identify element and subset of a set Basic definitions: e.g. element and subset of a set, empty set, power set, cartesian product LO: Perform set operations Union, intesection, set difference, complement, symmetric difference. LO: Establish set properties (equalities) using element arguments. Proving set identities using element argument

23 TERIMA KASIH