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- 1. D i s c re teS 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-intersection05/26/12
- 6. Exercise-union05/26/12
- 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 LawA∪B = B∪A Commutative (A∪B)c = Ac∩Bc De Morgan’s LawA∩B = B∩A Law (A∩B)c = Ac∪BcA∪(B∪C) A∩(B∪C) == (A∪B)∪C Associative (A∩B)∪(A∩C) Distributive LawA∩(B∩C) Law A∪(B∩C) == (A∩B)∩C (A∪B)∩(A∪C)A∪(A∩B) = A Absorption A ∪ Ac = U Complement LawA∩(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 ∪ BcProof: 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

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