Question

(6). Assume that P(A) = P(B). Show that A = B.

(7). It is true for any sets A and B, P(A) U P(B) = P(A U B) ? Justify your answer.​

Answer 1

(6). Every set is a member of power set so that, A ∈ P(A) 

Given that P (A) = P (B) So that 

A ∈ P(B) 

A is a element of power set of B so that, 

A ⊂ B ... (1) 

Similarly we can prove that 

B ⊂ A ... (2) 

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

(7). Let A = {1, 2} and B = {2, 3}

A ∪ B = {1, 2} ∪ {2, 3} = {1, 2, 3}

P(A) = {Φ, {1}, {2}, {1, 2}}

P(B) = {Φ, {2}, {3}, {2, 3}}

P(A ∪ B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}}

We can observe that P(A) ∪ P(B) ≠ P(A ∪ B).

Hence given statement is false.

Answer 2

$\huge\mathbb\red{Answer:-}$

(6). Every set is a member of power set so that, A ∈ P(A) 

Given that P (A) = P (B) So that 

A ∈ P(B) 

A is a element of power set of B so that, 

A ⊂ B ... (1) 

Similarly we can prove that 

B ⊂ A ... (2) 

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

_________________________

(7). Let A = {1, 2} and B = {2, 3}

A ∪ B = {1, 2} ∪ {2, 3} = {1, 2, 3}

P(A) = {Φ, {1}, {2}, {1, 2}}

P(B) = {Φ, {2}, {3}, {2, 3}}

P(A ∪ B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

P(A) ∪ P(B) = {Φ, {1}, {2}, {3}, {1, 2}, {2, 3}}

We can observe that P(A) ∪ P(B) ≠ P(A ∪ B).

Hence given statement is false.