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Given sets are A = {a,b} and B = {a,b,c}.

Clearly, every element of A is the element of B, therefore A⊄BA∪ B = {a, b, c].

(i) X ∪ Y = {1, 2, 3, 5} X ∩ Y = {1, 3} 

(ii) A ∪ B = {a, b, c, e, i, o, u} A ∩ B = {a} 

(iii) A = {3, 6, 9, 12,…….. } B = { 1, 2, 3, 4, 5, 6} 

∴ A ∪ B = {l, 2, 3, 4, 5, 6, 9, 12, ………….. } 

(iv) A = {2, 3, 4, 5, 6} B = {7, 8, 9} 

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} 

A ∩ B = {2, 3, 4, 5, 6, 7, 8, 9}

(v) A ∪ B = {1, 2, 3} = A 

A ∩ B = φ = B

X∪Y= {Ram, Geeta, Akbar, Divid, Ashok} 

This is the set of students from Class XI who are in the hockey team or the foot ball team or both.

A ∪ B = {a, c, i, o, u} = A

∴ A ∪ B = A

A ∪ B = {2, 4, 6, 8, 10, 12}.

If A and B are two sets such that A∩B =φ, then A and B are called disjoint sets.

For any sets A and B we have De morgans law 

(A ∪B)’ =A’ ∩B’ , (A ∩ B)’ = A’ ∪ B’ 

=A – (A–B) 

= A ∩ (A–B)’ 

= A∩(A∩B’)’ 

= A∩(A’∪ B’)’) 

= A ∩ (A’∪B) 

= (A ∩ A’) ∪ (A ∩ B) 

= ϕ ∪ (A ∩ B) 

= (A ∩ B)

Expanding 

(A ∩ A’) ∪ (A∩ B) 

(A ∩ A’) =ϕ 

⇒ ϕ ∪ (A∩ B) 

(A ∪ ϕ =A) 

⇒ (A∩ B)

(i) To show: A ∪ (A ∩ B) = A
We know that
A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A … (1)
Also, A ⊂ A ∪ (A ∩ B) … (2)
∴ From (1) and (2), A ∪ (A ∩ B) = A
(ii) To show: A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
= A {from (1)}

Let x ∈ A ∴ (X) ∈ P(A)

= {x} ∈ P(B) ∵ P(A) = P(B)

∴ x ∈ B.

∴ A ⊂ B

Let y∈ B ∴ (y)∈ P(B)

(y) ∈ P(A) ∵ P(A) = P(B)

y ∈ A

B ⊂ A

From(1) and (2),we get A = B

To show, A’ – B’ = B – A 

We need to show 

A’ – B’ ⊆ B – A 

B – A ⊆ A’ – B’ 

Let, x ϵ A’ – B.’ 

⇒ x ϵ A’ and x ∉ B.’ 

⇒ x ∉ A and x ϵ B 

⇒ x ϵ B – A 

It is true for all x x ϵ A’ – B’ 

∴ A’ – B’ = B – A 

Hence Proved

To show: A = (A ∩ B) ∪ (A – B)
Let x ∈ A
We have to show that x ∈ (A ∩ B) ∪ (A – B)

Case I

 x ∈ A ∩ B
Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B) 

Case II

 x ∉ A ∩ B
⇒ x ∉ A or x ∉ B
∴ x ∉ B [x ∉ A]
∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)
It is clear that
A ∩ B ⊂ A and (A – B) ⊂ A
∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)
From (1) and (2), we obtain
A = (A ∩ B) ∪ (A – B)
To prove: A ∪ (B – A) ⊂ A ∪ B
Let x ∈ A ∪ (B – A)

⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or (x ∈ B and x ∉ A)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)
⇒ x ∈ (A ∪ B)
∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)
Next, we show that (A ∪ B) ⊂ A ∪ (B – A).
Let y ∈ A ∪ B
⇒ y ∈ A or y ∈ B
⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)
⇒ y ∈ A or (y ∈ B and y ∉ A)
⇒ y ∈ A ∪ (B – A)
∴ A ∪ B ⊂ A ∪ (B – A) … (4)
Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.

Let x ϵ A ∪ (B – A) 

⇒ x ϵ A or x ϵ (B–A) 

⇒ X ϵ A or x ϵ B or x ∉ A 

⇒ x ϵ A or x ϵ B 

⇒ x ϵ (A ∪ B) 

∴ A ∪ (B – A) ⊂ (A ∪ B)…….(1) 

Let and x ϵ (A ∪ B) 

⇒ x ϵ A or x ϵ B 

⇒ x ϵ A or x ϵ B and x ∉ A 

⇒ x ϵ A or x ϵ B–A 

⇒ x ϵ A ∪ (B–A) 

∴ (A ∪ B) ⊂A ∪ (B – A)…….(2) 

From (1) and(2), we get 

A ∪ (B – A) = A ∪ B.

False
Let A = {0, 1} and B = {1, 2}
∴ A ∪ B = {0, 1, 2}
P(A) = {Φ, {0}, {1}, {0, 1}}
P(B) = {Φ, {1}, {2}, {1, 2}}
P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}
P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}
∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Let A = {φ{a}}, and B = {b}

∴ A∪B = {a, b}

P(A) = {φ,{a}}, P(B) = {φ, {b}}

and P(A∪B) = {φ,{a},{b} {a,b}}………………. (1)

and P(A)∪P(B) = {φ,{a},{b}} ………………. (2)

From (1) and (2), we get, P(A∪B) ≠ P(A)∪P{B)

No. consider the following sets A, B and C :

A = {1, 2, 3}

B = {2, 3, 5}

C = {4, 5, 6}

Now

(A ∩ B) ∪ C = ({1, 2, 3} ∩ {2, 3, 5}) ∪ {4, 5, 6}

= {2, 3} ∪ {4, 5, 6}

= {2, 3, 4, 5, 6}

And

A ∩ (B ∪ C) = {1, 2, 3} ∩ [{2, 3, 5} ∪ {4, 5, 6}

= {1, 2, 3} ∩ {2, 3, 4, 5, 6}

= {2, 3}

Therefore, (A ∩ B) ∪ C ≠ A ∩ (B ∪ C)

We know that (A ∩ B) ⊂ A and (A – B) ⊂ A 

⇒(A ∩ B) ∩ (A – B)⊂ A….(1) 

Let and x ϵ (A ∩ B) ∩ (A – B) 

⇒ x ϵ (A ∩ B) and x ϵ (A–B) 

⇒ x ϵ A and x ϵ B and x ϵ A and x ∉ B 

⇒ x ϵ A and x ϵ A [∵ x ϵ B and x ∉ B are not possible simultaneously] 

→ x ϵ A 

∴ (A ∩ B) ∩ (A – B)⊂ A…(2) 

From (1) and (2), we get 

A = (A ∩ B) ∩ (A – B)

It is a False statement. 

Let, A = {3} and B = {4} 

Then, 

P(A) = {ϕ , {3}} 

And P(B) = {ϕ , {4}} 

∴ P(A) ∪ P (B) = {ϕ , {1}, {2}} 

Now, 

A ∪ B = {1, 2} 

And P(A ∪ B) = {ϕ , {1}, {2}, {1, 2}} 

Hence, 

P(A) ∪ P(B) ≠ P (A ∪ B)

We have B’⊂ A’ 

To Show: A ⊂ B 

Let, x ϵ A 

⇒ x∉ A’ [∵ A ∩ A’ = ϕ ] 

⇒ x ∉ B’ [ ∵ B’ ⊂ A’ ] 

⇒ x ϵ B [∵ B ∩ B’ = ϕ] 

Thus, 

x ϵ A ⇒ x ϵ B 

This is true for all x ϵ A 

∴ A ⊂ B