# Discrete Structures Chapter 3 Set Theory Nurul Amelina Nasharuddin Multimedia Department.

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Discrete Structures Chapter 3 Set Theory Nurul Amelina Nasharuddin Multimedia Department

Basics of Set Theory Sets are used to group objects together Set notation: {1, 2, 3}, {{1, 2}, {3}, {1, 2, 3}}, {1, 2, 3, …}, , {x  R | -3 < x < 6} Set A is called a subset of set B, written as A  B, when  x, x  A  x  B (A is contained in B and B contains A)  is a subset for every set Any set is a subset of itself Visual representation of the sets Distinction between  and 

Basics of Set Theory A is a proper subset of B, written as A  B, when A is a subset of B and  x  B and x  A (every element A is in B but there is at least one element B not in A) When A is a proper subset of B, A  B Eg: The set of all men is a proper subset of the set of all people. {1, 3}  {1, 2, 3, 4} {1, 2, 3, 4}  {1, 2, 3, 4}

Example Let U = {1, 2, 3, 4, 5, 6, x, y, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}} Then |U| = 11. (a)If A = {1, 2, 3, 4} then |A| = 4 and A  U A  U A  U {A}  U {A}  U{A}  U (b) Let B = {5, 6, x, y, A} = {5, 6, x, y, {1, 2, 3, 4}}. Then |B| = 5, not 8. And A  B {A}  B {A}  B {A}  B A  B (that is, A is not a subset of B) A  B (that is, A is not a proper subset of B).

Cardinality of Sets If A is a finite set, then cardinality of A, denoted by n(A) or |A|, the number of elements A contains Eg: A = {x, y, z}. n(A) = 3 Eg: Let S be the set of odd positive integers less than 10 Then |S| = 5 We see that A  B  |A|  |B|, and A  B  |A|  |B|.

Example For U = {1, 2, 3, 4, 5}, A = {1, 2}, and B = {1, 2}, we see that A is a subset of B (that is, A  B), but it is not a proper subset of B (or, A  B). A  B  |A|  |B|, 2  2 so it is true that A  B A  B  |A|  |B|, 2  2 false so it is false that A  B.

Set Operations Set A equals set B, written A = B, iff every element of set A is in set B and vice versa (A  B and B  A) Proof technique for showing sets equality Eg: Let A = {n  Z | n = 2p, for some integer p}, B = {k  Z | k = 3r + 1, for some integer r}. Is A = B? Ans: No. 2  A since 2 = 2(1); but 2  B. 3r + 1= 2 3r = 1 r= 1/3 (not an integer), so 2  B There is an element in A that not in B, A  B.

Set Operations Union of two sets (A  B) is a set of all elements that belong to at least one of the sets Intersection of two sets (A  B) is a set of all elements that belong to both sets Difference of two sets (B – A) is a set of elements in one set, but not the other Complement of a set (A c or ) is a difference between universal set and a given set Symmetric difference of two sets (A  B) is the set that contains the elements which is in A but not in B and in B but not in A

Set Operations Symbolically: A  B = {x  U | x  A or x  B} A  B = {x  U | x  A and x  B} B – A = {x  U | x  B and x  A} A c = {x  U | x  A} A  B = {x | x  A  B but x  A  B} Eg: Let U = {a, b, c, d, e, f, g}, A = {a, c, e, g}, B = {d, e, f, g} A  B = A c = A  B =A  B = {a, c, d, f} B – A =

Venn Diagrams Union Intersection Difference ( A – B)Complement Difference

Empty Set Empty set (null set) is the set containing no elements Denoted by  or {} Eg: {1, 3}  {2, 5} =  Eg: Describe the set D = {x  R | 3  x  2}  D =  Empty set is a subset of any set There is exactly one empty set

Disjoint Set Two sets are disjoint, iff they have no elements in common A and B are disjoint, A  B =  Eg: Are A = {1, 3, 5} and B = {2, 4, 6} disjoint? Yes. {1, 3, 5}  {2, 4, 6} =  Sets A 1, A 2, …, A n are mutually disjoint, iff no two sets A i and A j with distinct subscripts have any elements in common A i  A j =  whenever i  j Eg: Are B 1 = {2, 4, 6}, B 2 = {3, 7}, B 3 = {4, 5} mutually disjoint? No. B 1 and B 3 both contain 4.

Power Set Power set of A, denoted by P(A), is the set of all subsets of A Theorem: If A  B, then P(A)  P(B) Theorem: If set X has n elements, then P(X) has 2 n elements Eg: P({x, y}) = { , {x}, {y}, {x, y}}

Power Set Eg: Let C = {1, 2, 3, 4}. What is the power set of C? P(C) = { , {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4},{2, 3, 4}, C}

Cartesian Products Ordered n-tuple is a set of ordered n elements Two ordered n-tuples (x 1, x 2, …, x n ) and (y 1, y 2, …, y n ) are equal iff x 1 = y 1, x 2 = y 2, …, x n = y n Eg: Is (1, 2) = (2, 1)? No. By definition of equality of ordered pair, (1, 2) = (2, 1)  1 = 2 and 2 = 1 Cartesian product of A and B, denoted by A x B, is the set of all ordered pairs (a, b) where a is in A and b is in B. A x B = {(a, b) | a  A and b  B} Eg: Let A = {x, y} and B = {1, 2, 3} A x B = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}

Set Properties and Laws of Set Theory Inclusion of Intersection: A  B  A and A  B  B Inclusion in Union: A  A  B and B  A  B Transitivity of Inclusion: (A  B  B  C)  A  C Set Definitions: i.x  X  Y  x  X or x  Y ii.x  X  Y  x  X and x  Y iii.x  X – Y  x  X and x  Y iv.x  X c  x  X v.(x, y)  X  Y  x  X and y  Y

RulesName of rule A  B = A  B A  B = B  A Commutative Laws (A  B)  C = A  (B  C) (A  B)  C = A  (B  C) Associative Laws A  (B  C) = (A  B)  (A  C) A  (B  C) = (A  B)  (A  C) Distributive Laws A  U = A A   = A Identity Law A  A c =  A  A c = U Complement Law (A c ) c = ADouble Complement Law A  A = A A  A = A Idempotent Laws A   =  A  U = U Universal Bound Laws (A  B) c = A c  B c (A  B) c = A c  B c De Morgan’s Laws A  (A  B) = A A  (A  B) = A Absorption Laws A – B = A  B c Set Difference Law U c =   c = U Complements of U and  If A  B, then A  B = A and A  B = B Intersection and Union with a subset

Formal Proving for Set Identities Basic method for proving that sets are equal: Let sets X and Y be given. To prove that X = Y: 1.Prove that X  Y 2.Prove that Y  X.

Proving: Distributive Law Proof by cases Prove that A  (B  C) = (A  B)  (A  C) We must show that A  (B  C)  (A  B)  (A  C) and (A  B)  (A  C)  A  (B  C) For 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) by definition of intersection

Case 2: (x  B  C) Since (x  B  C), then x  B and x  C Since x  B, then x  A  B and since x  C, then x  A  C by definition of union Hence x  (A  B)  (A  C) In both cases, x  (A  B)  (A  C) Hence A  (B  C)  (A  B)  (A  C) by definition of subset Do for (A  B)  (A  C)  A  (B  C)! Proving: Distributive Law Proof by cases

Prove that A  (B  C) = (A  B)  (A  C) A  (B  C) = {x | (x  A)  (x  (B  C))} = {x | (x  A)  (x  B  x  C)} = {x | ((x  A)  (x  B))  ((x  A)  (x  C))} = {x | (x  A  B)  (x  A  C)} = (A  B)  (A  C) Proving: Distributive Law Proof by cases

Prove that (A  B) c = A c  B c (A  B) c = {x | x  (A  B)} = {x | ~ (x  A  x  B)} = {x | ~ (x  A)  ~ (x  B)} = {x | x  A  x  B} = {x | x  A c  x  B c } = {x | x  (A c  B c )} = A c  B c Proving: De Morgan’s Law Proof by cases

Algebraic Proofs of Set Identities Can use set identities to derive new set identities or to simplify a complicated set expression, eg. (A  B) c  A c  B c Eg: 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

Example of Simplification Using laws of set theory, simplify the expression: De Morgan’s Law Double Complement Law Associative Law Commutative Law Associative Law Absorption Law

Membership table To establish set equalities We observe that for sets A, B  U, an element x  U satisfies exactly one of the following four situations: a)x  A, x  B b)x  A, x  B c)x  A, x  B d)x  A, x  B When x is an element of a given set, we write a 1(T); when x is not in the set, we enter a 0 (F). 3 rd row of the table shows that x  A, but x  B. So, x  A  B but x is in A  B. ABAcAc A  BA  B 00100 01101 10001 11011

Membership Table Using membership table, prove the A  (B  C) = (A  B)  (A  C) ABC B  CA  (B  C)A  BA  C(A  B)  (A  C) 00000000 00100010 01000100 01111111 10001111 10101111 11001111 11111111 Since these columns are identical, A  (B  C) = (A  B)  (A  C)

Duality The dual of s, denoted s d is obtained from a statement, s by replacing –  by U and U by  and –  by  and  by  Eg: s: s d :

Counting and Venn Diagrams By using Venn diagrams, show that For example, Set = regions 1, 3, 4, 6; while set = regions 1, 2, 4, 7. Therefore, = regions 1, 4. Set C = regions 4, 6, 7, 8; therefore set = regions 1, 2, 3, 5. = regions 1, 2, 3, 4, 5. ?

Counting and Venn Diagrams If A and B are finite sets, then n(A  B) = n(A) + n(B) - n(A  B). In particular, if A and B are disjoint then n(A  B) = n(A) + n(B). If A and B are finite sets, then n(AB) = n(A)*n(B)

Example In a class of 50 college freshmen, 30 are studying C++, 25 are studying Java, and 10 are studying both languages. How many freshmen are studying either computer language? We let U be the class of 50 freshmen, A is the subset of those students studying C++, and B is the subset of those studying Java. n(A) = 30, n(B) = 25, n(A  B) = 10 n(A  B) = n(A) + n(B) - n(A  B) = 30 + 25 – 10 = 45

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