This Presentation is a Courtesy of bilalAmjad

1. D i s c re te
S tru c tu re s
(Discrete Mathematics)
Topic: Set Operations
©bilalAmjad
bilalamjad78633@yahoo.com

2. Set operations: Union
 Formal definition for the union of two sets:
A U B = { x | x ∈ A or x ∈ B } or
A U B = { x ∈ U| x ∈ A or x ∈ B }
 Further examples
 {1, 2, 3} ∪ {3, 4, 5} = {1, 2, 3, 4, 5}
 {a, b} ∪ {3, 4} = {a, b, 3, 4}
 {1, 2} ∪ ∅ = {1, 2}
 Properties of the union operation
 A∪∅=A Identity law
 A∪U=U Domination law
 A∪A=A Idempotent law
 A∪B=B∪A Commutative law
 A ∪ (B ∪ C) = (A ∪ B) ∪ C Associative law

3. 05/26/12

4. Set operations: Intersection
 Formal definition for the intersection of two sets:
A ∩ B = { x | x ∈ A and x ∈ B }
 Examples
 {1, 2, 3} ∩ {3, 4, 5} = {3}
 {a, b} ∩ {3, 4} = ∅
 {1, 2} ∩ ∅ = ∅
 Properties of the intersection operation
 A∩U=A Identity law
 A∩∅=∅ Domination law
 A∩A=A Idempotent law
 A∩B=B∩A Commutative law
 A ∩ (B ∩ C) = (A ∩ B) ∩ C Associative law

5. Exercise-intersection
05/26/12

6. Exercise-union
05/26/12

7. Disjoint sets
 Formal definition for disjoint sets:
two sets are disjoint if their intersection is the
empty set
 Further examples
 {1, 2, 3} and {3, 4, 5} are not disjoint
 {a, b} and {3, 4} are disjoint
 {1, 2} and ∅ are disjoint
• Their intersection is the empty set
 ∅ and ∅ are disjoint!
• Because their intersection is the empty set

8. Set operations: Difference
 Formal definition for the difference of two sets:
A - B = { x | x ∈ A and x ∉ B }
 Further examples
 {1, 2, 3} - {3, 4, 5} = {1, 2}
 {a, b} - {3, 4} = {a, b}
 {1, 2} - ∅ = {1, 2}
• The difference of any set S with the empty set will be the
set S

9. Complement sets
 Formal definition for the complement of a set:
A = { x | x ∉ A } = Ac
 Or U – A, where U is the universal set
 Further examples (assuming U = Z)
 {1, 2, 3}c = { …, -2, -1, 0, 4, 5, 6, … }
 {a, b}c = Z
 Properties of complement sets
 (Ac)c = A Complementation law
 A ∪ Ac = U Complement law
 A ∩ Ac = ∅ Complement law

10. Set identities
A∪∅ = A A∪U = U
Identity Law Domination law
A∩U = A A∩∅ = ∅
A∪A = A Idempotent Complementation
(Ac)c = A
A∩A = A Law Law
A∪B = B∪A Commutative (A∪B)c = Ac∩Bc
De Morgan’s Law
A∩B = B∩A Law (A∩B)c = Ac∪Bc
A∪(B∪C) A∩(B∪C) =
= (A∪B)∪C Associative (A∩B)∪(A∩C)
Distributive Law
A∩(B∩C) Law A∪(B∩C) =
= (A∩B)∩C (A∪B)∩(A∪C)
A∪(A∩B) =
A Absorption A ∪ Ac = U
Complement Law
A∩(A∪B) = Law A ∩ Ac = ∅
A

11. How to prove a set identity
 For example: A∩B=B-(B-A)
 Four methods:
 Use the basic set identities
 Use membership tables
 Prove each set is a subset of each other
 Use set builder notation and logical equivalences

12. What we are going to prove…
A∩B=B-(B-A)
A B
B-(B-A)
B-A
A∩B

13. Proof by Set Identities
 A ∩ B = A - (A - B) = B – (B – A)
Proof: A - (A - B) = A - (A ∩ Bc)
= A ∩ (A ∩ Bc)c
= A ∩ (Ac ∪ B)
= (A ∩ Ac) ∪ (A ∩ B)
= ∅ ∪ (A ∩ B)
=A∩B

14. Showing each is a subset of the others
 (A ∩ B)c = Ac ∪ Bc
Proof: Want to prove that
(A ∩ B)c ⊆ Ac ∪ Bc and Ac ∪ Bc ⊆ (A ∩ B)c
(i) x ∈ (A ∩ B)c
⇒ x ∉ (A ∩ B)
⇒ ¬ (x ∈ A ∩ B)
⇒ ¬ (x ∈ A ∧ x ∈ B)
⇒ ¬ (x ∈ A) ∨ ¬ (x ∈ B)
⇒x∉A∨x∉B
⇒ x ∈ Ac ∨ x ∈ B c
⇒ x ∈ Ac ∪ B c
(ii) Similarly we show that Ac ∪ Bc ⊆ (A ∩ B)c

15. Exercise