Set Theory Supartha Podder uOttawa Set Theory A set is an - PowerPoint PPT Presentation

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} 4/19

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} T {∅} ⊂ {∅} 4/19

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} T {∅} ⊂ {∅} F {∅} ∈ {{∅}} 4/19

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} T {∅} ⊂ {∅} F {∅} ∈ {{∅}} T {{∅}} ∈ {∅ , {∅}} 4/19

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} T {∅} ⊂ {∅} F {∅} ∈ {{∅}} T {{∅}} ∈ {∅ , {∅}} {{∅}} ⊂ {{∅} , {∅}} F 4/19

Subsets Power set of A is the set of all possible subsets of A . P ( A ) = { B : B ⊆ A } Example: Let A = { 1 , 2 } then P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} . Statement T/F Statement T/F ∅ ∈ {∅} T ∅ ∈ {∅ , {∅}} T {∅} ∈ {∅} F {∅} ⊆ {∅} T {∅} ⊂ {∅} F {∅} ∈ {{∅}} T {{∅}} ∈ {∅ , {∅}} {{∅}} ⊂ {{∅} , {∅}} F F 4/19

Venn Diagrams 5/19

Venn Diagrams V = { a , e , i , o , u } . 5/19

Venn Diagrams V = { a , e , i , o , u } . 5/19

Venn Diagrams V = { a , e , i , o , u } . V = Complement of V . 5/19

Venn Diagrams V = { a , e , i , o , u } . V = Complement of V . 5/19

Venn Diagrams Set Intersection: A ∩ B = ∀ x { x ∈ A ∧ x ∈ B } . 6/19

Venn Diagrams Set Intersection: A ∩ B = ∀ x { x ∈ A ∧ x ∈ B } . 6/19

Venn Diagrams Set Intersection: A ∩ B = ∀ x { x ∈ A ∧ x ∈ B } . Set Union: A ∪ B = ∀ x { x ∈ A ∨ x ∈ B } . 6/19

Venn Diagrams Set Intersection: A ∩ B = ∀ x { x ∈ A ∧ x ∈ B } . Set Union: A ∪ B = ∀ x { x ∈ A ∨ x ∈ B } . 6/19

Venn Diagrams Set Difference: A \ B = ∀ x { x ∈ A ∧ x / ∈ B } . 7/19

Venn Diagrams Set Difference: A \ B = ∀ x { x ∈ A ∧ x / ∈ B } . 7/19

Venn Diagrams Set Difference: A \ B = ∀ x { x ∈ A ∧ x / ∈ B } . Symmetric Difference: A △ B = { ( A \ B ) ∪ ( B \ A ) } . 7/19

Venn Diagrams Set Difference: A \ B = ∀ x { x ∈ A ∧ x / ∈ B } . Symmetric Difference: A △ B = { ( A \ B ) ∪ ( B \ A ) } . 7/19

Example Let A = { 1 , 2 , 3 , 5 } , B = { 2 , 4 , 8 } . 8/19

Example Let A = { 1 , 2 , 3 , 5 } , B = { 2 , 4 , 8 } . A ∩ B = 8/19

Example Let A = { 1 , 2 , 3 , 5 } , B = { 2 , 4 , 8 } . A ∩ B = { 2 } . A \ B = 8/19

Example Let A = { 1 , 2 , 3 , 5 } , B = { 2 , 4 , 8 } . A ∩ B = { 2 } . A \ B = { 1 , 3 , 5 } A ∪ B = 8/19

Example Let A = { 1 , 2 , 3 , 5 } , B = { 2 , 4 , 8 } . A ∩ B = { 2 } . A \ B = { 1 , 3 , 5 } A ∪ B = { 1 , 2 , 3 , 4 , 5 , 8 } . 8/19

Cartesian Product Cartesian Product of A and B : 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? No (not always). 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? No (not always). A 1 × A 2 × · · · × A n = { ( a 1 , a 2 , · · · , a n ) | a i ∈ A i , 1 ≤ i ≤ n } . 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? No (not always). A 1 × A 2 × · · · × A n = { ( a 1 , a 2 , · · · , a n ) | a i ∈ A i , 1 ≤ i ≤ n } . A 2 = A × A . 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? No (not always). A 1 × A 2 × · · · × A n = { ( a 1 , a 2 , · · · , a n ) | a i ∈ A i , 1 ≤ i ≤ n } . A 2 = A × A . A 2 = A × A = { (1 , 1) , (1 , 2) , (2 , 1) , (2 , 2) } . 9/19

Cartesian Product Cartesian Product of A and B : A × B = { ( a , b ) : a ∈ A , b ∈ B } | A × B | = | A | · | B | . Let A = { 1 , 2 } , B = { a , b , c } . A × B = { (1 , a ) , (1 , b ) , (1 , c ) , (2 , a ) , (2 , b ) , (2 , c ) } Is A × B = B × A ? No (not always). A 1 × A 2 × · · · × A n = { ( a 1 , a 2 , · · · , a n ) | a i ∈ A i , 1 ≤ i ≤ n } . A 2 = A × A . A 2 = A × A = { (1 , 1) , (1 , 2) , (2 , 1) , (2 , 2) } . Disjoint: Two sets A and B are disjoint if they have no elements in common. e.g., A = { 1 , 3 , 5 , 9 } , B = { 2 , 13 , 4 , 7 } . 9/19

Table of Set Identities A \ B = A ∩ B 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A A ∪ B = B ∪ A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A A ∪ B = B ∪ A A ∩ B = B ∩ A 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A A ∪ B = B ∪ A A ∩ B = B ∩ A A ∪ ( B ∪ C ) = ( A ∪ B ) ∪ C 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A A ∪ B = B ∪ A A ∩ B = B ∩ A A ∪ ( B ∪ C ) = ( A ∪ B ) ∪ C A ∩ ( B ∩ C ) = ( A ∩ B ) ∩ C 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A 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 ) 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A 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 = A ∩ B 10/19

Table of Set Identities A \ B = A ∩ B A ∪ A = U A ∪ A = ∅ A ∪ ∅ = A A ∩ ∅ = ∅ A ∪ U = U A ∩ U = A A ∪ A = A A ∩ A = A A = A 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 = A ∩ B A ∩ B = A ∪ B 10/19

Proof Using Venn Diagram Prove using Venn diagram that A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ). 11/19

Proof Using Venn Diagram Prove using Venn diagram that A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ). 11/19

Proof Using Venn Diagram Prove using Venn diagram that A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ). 11/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then x 2 − 3 x + 2 = 0 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then x 2 − 3 x + 2 = 0 x 2 − 2 x − x + 2 = 0 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then x 2 − 3 x + 2 = 0 x 2 − 2 x − x + 2 = 0 x ( x − 2) − ( x − 2) = 0 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then x 2 − 3 x + 2 = 0 x 2 − 2 x − x + 2 = 0 x ( x − 2) − ( x − 2) = 0 ( x − 2)( x − 1) = 0 12/19

Proof Let A = { x ∈ R , x 2 − 3 x + 2 = 0 } . Prove that A ⊆ Z . Proof: Let x ∈ A , then x 2 − 3 x + 2 = 0 x 2 − 2 x − x + 2 = 0 x ( x − 2) − ( x − 2) = 0 ( x − 2)( x − 1) = 0 Thus x = { 1 , 2 } . 12/19

Proof Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C 13/19

Proof Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Assume A ⊆ B . 13/19

Proof Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Assume A ⊆ B . Take any x ∈ A ∪ C , 13/19

Proof Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Assume A ⊆ B . Take any x ∈ A ∪ C , so, x ∈ A ∨ x ∈ C . 13/19

Proof Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Assume A ⊆ B . Take any x ∈ A ∪ C , so, x ∈ A ∨ x ∈ C . By assumption, if x ∈ A then x ∈ B . 13/19