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

Set Theory

Supartha Podder

uOttawa

Set Theory

A set is an unordered collection of objects called elements.

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · },

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’.

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’. d / ∈ V : ‘d is not a member of V or d is not in V ’.

Question

Are {1, 3, 5}, {3, 1, 5} equal?

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’. d / ∈ V : ‘d is not a member of V or d is not in V ’.

Question

Are {1, 3, 5}, {3, 1, 5} equal? Answer - YES.

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’. d / ∈ V : ‘d is not a member of V or d is not in V ’.

Question

Are {1, 3, 5}, {3, 1, 5} equal? Answer - YES. Are {1, 3, 5}, {1, 1, 3, 3, 3, 5} equal?

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’. d / ∈ V : ‘d is not a member of V or d is not in V ’.

Question

Are {1, 3, 5}, {3, 1, 5} equal? Answer - YES. Are {1, 3, 5}, {1, 1, 3, 3, 3, 5} equal?Answer - YES.

1/19

Set Theory

A set is an unordered collection of objects called elements. Example: V= {a,e,i,o,u}, N = {1, 2, 3, · · · }, Z = {· · · , −2, −1, 0, 1, 2, · · · }, a ∈ V : ‘a is a member of V or a is in V ’. d / ∈ V : ‘d is not a member of V or d is not in V ’.

Question

Are {1, 3, 5}, {3, 1, 5} equal? Answer - YES. Are {1, 3, 5}, {1, 1, 3, 3, 3, 5} equal?Answer - YES. Two sets A and B are equal iff they have the same elements. A = B iff ∀x(x ∈ A ↔ x ∈ B}

1/19

Set Definitions

R+ = {x | x ∈ R, x > 0}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}. (a, b] = {x | a < x ≤ b}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}. (a, b] = {x | a < x ≤ b}. [a, b) =? Set Notations: List Notation: e.g., S = {12, 23, −4, · · · }.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}. (a, b] = {x | a < x ≤ b}. [a, b) =? Set Notations: List Notation: e.g., S = {12, 23, −4, · · · }. Set builder notation e.g., S = {∀i | i ∈ Z, i ≥ 50}.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}. (a, b] = {x | a < x ≤ b}. [a, b) =? Set Notations: List Notation: e.g., S = {12, 23, −4, · · · }. Set builder notation e.g., S = {∀i | i ∈ Z, i ≥ 50}. empty set {} or ∅.

2/19

Set Definitions

R+ = {x | x ∈ R, x > 0}. {a, b} = {x | x = a or x = b}. [a, b] = {x | a ≤ x ≤ b}. (a, b) = {x | a < x < b}. (a, b] = {x | a < x ≤ b}. [a, b) =? Set Notations: List Notation: e.g., S = {12, 23, −4, · · · }. Set builder notation e.g., S = {∀i | i ∈ Z, i ≥ 50}. empty set {} or ∅. A set may contain different kind of elements. For example: A = {1, {2}}, B = {1, {2, {3}}, {{{5}}}}.

2/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) .

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A)

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO Cardinality of a finite set A is |A|= number of distinct elements of A.

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO Cardinality of a finite set A is |A|= number of distinct elements of A. Example: A = {1, 2, 2, 2, 2, 4}. What is |A|?

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO Cardinality of a finite set A is |A|= number of distinct elements of A. Example: A = {1, 2, 2, 2, 2, 4}. What is |A|?|A| = 3.

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO Cardinality of a finite set A is |A|= number of distinct elements of A. Example: A = {1, 2, 2, 2, 2, 4}. What is |A|?|A| = 3. Power set of A is the set of all possible subsets of A. P(A) = {B : B ⊆ A}

3/19

Subsets

Subet A ⊆ B if and only if ∀x(x ∈ A → x ∈ B) . Proper subset A ⊂ B or A B: ∀x(x ∈ A → x ∈ B) ∧ ∃y(y ∈ B ∧ y / ∈ A) Example: {4, 5, 2} ⊆ {4, 5, 5, 2, 1}?: YES Example: {4, 5, 2} ⊂ {4, 5, 5, 2}?: NO Cardinality of a finite set A is |A|= number of distinct elements of A. Example: A = {1, 2, 2, 2, 2, 4}. What is |A|?|A| = 3. Power set of A is the set of all possible subsets of A. P(A) = {B : B ⊆ A}

3/19

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}}.

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 ∅ ∈ {∅}

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 ∅ ∈ {∅, {∅}}

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 {∅} ∈ {∅}

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 {∅} ⊆ {∅}

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). A1 × A2 × · · · × An = {(a1, a2, · · · , an) | ai ∈ Ai, 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). A1 × A2 × · · · × An = {(a1, a2, · · · , an) | ai ∈ Ai, 1 ≤ i ≤ n}. A2 = 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). A1 × A2 × · · · × An = {(a1, a2, · · · , an) | ai ∈ Ai, 1 ≤ i ≤ n}. A2 = A × A. A2 = 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). A1 × A2 × · · · × An = {(a1, a2, · · · , an) | ai ∈ Ai, 1 ≤ i ≤ n}. A2 = A × A. A2 = 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, x2 − 3x + 2 = 0}. Prove that A ⊆ Z.

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then x2 − 3x + 2 = 0

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then x2 − 3x + 2 = 0 x2 − 2x − x + 2 = 0

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then x2 − 3x + 2 = 0 x2 − 2x − x + 2 = 0 x(x − 2) − (x − 2) = 0

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then x2 − 3x + 2 = 0 x2 − 2x − x + 2 = 0 x(x − 2) − (x − 2) = 0 (x − 2)(x − 1) = 0

12/19

Proof

Let A = {x ∈ R, x2 − 3x + 2 = 0}. Prove that A ⊆ Z. Proof: Let x ∈ A, then x2 − 3x + 2 = 0 x2 − 2x − 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

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. Thus x ∈ A ∨ x ∈ C can be written as x ∈ B ∨ 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. Thus x ∈ A ∨ x ∈ C can be written as x ∈ B ∨ x ∈ C, x ∈ (B ∪ C).

13/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B.

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C)

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C) = ((A ∪ C) ∪ B) ∪ C [associativity]

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C) = ((A ∪ C) ∪ B) ∪ C [associativity] = (A ∪ (C ∪ B)) ∪ C [associativity]

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C) = ((A ∪ C) ∪ B) ∪ C [associativity] = (A ∪ (C ∪ B)) ∪ C [associativity] = (A ∪ (B ∪ C)) ∪ C [commutivity]

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C) = ((A ∪ C) ∪ B) ∪ C [associativity] = (A ∪ (C ∪ B)) ∪ C [associativity] = (A ∪ (B ∪ C)) ∪ C [commutivity] = ((A ∪ B) ∪ C) ∪ C [associativity]

14/19

Alternate Proof

Prove that if A ⊆ B then A ∪ C ⊆ B ∪ C Note that showing A ⊆ B is same as A ∪ B = B. Thus, (A ∪ C) ∪ (B ∪ C) = ((A ∪ C) ∪ B) ∪ C [associativity] = (A ∪ (C ∪ B)) ∪ C [associativity] = (A ∪ (B ∪ C)) ∪ C [commutivity] = ((A ∪ B) ∪ C) ∪ C [associativity]

14/19

Proof Continued

= ((A ∪ B) ∪ C) ∪ C

15/19

Proof Continued

= ((A ∪ B) ∪ C) ∪ C = (B ∪ C) ∪ C [by assumption]

15/19

Proof Continued

= ((A ∪ B) ∪ C) ∪ C = (B ∪ C) ∪ C [by assumption] = B ∪ (C ∪ C) [associativity]

15/19

Proof Continued

= ((A ∪ B) ∪ C) ∪ C = (B ∪ C) ∪ C [by assumption] = B ∪ (C ∪ C) [associativity] = B ∪ C

15/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A.

16/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A. Prove that (A ⊆ B) ↔ (A ∩ B = A).

16/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A. Prove that (A ⊆ B) ↔ (A ∩ B = A). This is a proof of if and only if, hence we need to prove both the following:

1 (A ⊆ B) → (A ∩ B = A). 2 (A ∩ B = A) → (A ⊆ B). 16/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A. Prove that (A ⊆ B) ↔ (A ∩ B = A). This is a proof of if and only if, hence we need to prove both the following:

1 (A ⊆ B) → (A ∩ B = A). 2 (A ∩ B = A) → (A ⊆ B).

Proof of 1.

We assume that A ⊆ B.

16/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A. Prove that (A ⊆ B) ↔ (A ∩ B = A). This is a proof of if and only if, hence we need to prove both the following:

1 (A ⊆ B) → (A ∩ B = A). 2 (A ∩ B = A) → (A ⊆ B).

Proof of 1.

We assume that A ⊆ B. Now assumeing this we need to show A ∩ B = A.

16/19

Proving Two Sets Are Equivalent A = B

One way of showing A = B is by showing A ⊆ B and B ⊆ A. Prove that (A ⊆ B) ↔ (A ∩ B = A). This is a proof of if and only if, hence we need to prove both the following:

1 (A ⊆ B) → (A ∩ B = A). 2 (A ∩ B = A) → (A ⊆ B).

Proof of 1.

We assume that A ⊆ B. Now assumeing this we need to show A ∩ B = A.This requires to show (a) A ∩ B ⊆ A and (b) A ⊆ A ∩ B.

16/19

Proof of 1(a)

So we have assumed A ⊆ B and we will show A ∩ B ⊆ A.

17/19

Proof of 1(a)

So we have assumed A ⊆ B and we will show A ∩ B ⊆ A. Take any x ∈ A ∩ B

17/19

Proof of 1(a)

So we have assumed A ⊆ B and we will show A ∩ B ⊆ A. Take any x ∈ A ∩ B then, x ∈ A and x ∈ B.

17/19

Proof of 1(a)

So we have assumed A ⊆ B and we will show A ∩ B ⊆ A. Take any x ∈ A ∩ B then, x ∈ A and x ∈ B. Thus x ∈ A.

17/19

Proof of 1(a)

So we have assumed A ⊆ B and we will show A ∩ B ⊆ A. Take any x ∈ A ∩ B then, x ∈ A and x ∈ B. Thus x ∈ A. (Note that here we haven’t used our assumption at all).

17/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B.

18/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B. Take any x ∈ A.

18/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B. Take any x ∈ A. From the assumption, A ⊆ B.

18/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B. Take any x ∈ A. From the assumption, A ⊆ B. Thus x ∈ A → x ∈ B.

18/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B. Take any x ∈ A. From the assumption, A ⊆ B. Thus x ∈ A → x ∈ B. so x ∈ A and x ∈ B.

18/19

Proof of 1(b)

Assuming A ⊆ B we will show A ⊆ A ∩ B. Take any x ∈ A. From the assumption, A ⊆ B. Thus x ∈ A → x ∈ B. so x ∈ A and x ∈ B. Hence x ∈ A ∩ B.

18/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B.

19/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B. Assume A ∩ B = A

19/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B. Assume A ∩ B = A Take any x ∈ A,

19/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B. Assume A ∩ B = A Take any x ∈ A, since A = A ∩ B, x ∈ A ∩ B.

19/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B. Assume A ∩ B = A Take any x ∈ A, since A = A ∩ B, x ∈ A ∩ B. So x ∈ A and x ∈ B.

19/19

Proof of 2

Now we will show that if A ∩ B = A then A ⊆ B. Assume A ∩ B = A Take any x ∈ A, since A = A ∩ B, x ∈ A ∩ B. So x ∈ A and x ∈ B. Thus x ∈ A implies x ∈ B, showing that A ⊆ B.

19/19

Thank you for your attention!!