# by using properties prove that(i) a U b = a intersection b imlies a=b(ii) for any sets a and b show that p(a intersection b) = p(a) itersection p (b)(iii) if a,b And c be the sets such that aUb = aUc and a intersection b = a itersection c show that b = c(iv) a intersection (aUb) = a and aU(a intersection b) = a

(i)

To prove: P (A ∪ B) = A ∩ B ⇔ A = B

Proof: Firstly, Let A = B Then,

A ∪ B = A and A ∩ B = A

⇒ A ∪ B = A ∩ B

Then, A = B ⇒ (A ∪ B) = (A ∩ B)  ...  (1)

Conversely, let A ∪ B = A ∩ B. Then we have to prove that

A = B For this, let From (2) and (3), we get A = B

Thus, A ∪ B = A ∩ B ⇒ A = B  ... (4)

From (1) and (4), we have

A ∪ B = A ∩ B ⇔ A = B

(ii)

In order to prove that P (A ∩ B) = P (A) ∩ P (B), it is sufficient to prove that P (A ∩ B) ⊂ P (A) ∩ P (B) and P (A) ∩ P (B) ⊂ P (A ∩ B)

First let X ∈ P (A ∩ B)

⇒ X ⊂ A ∩ B

⇒ X ⊂ A  and X ⊂ B

⇒ X ∈ P (A)  and X ∈ P (B)

⇒ X ∈ P (A) ∩ P (B)

∴ P (A ∩ B)  ⊂ P (A) ∩ P (B)  ...  (1)

Now, let

y ∈ P (A)  ∩ P (B). Then,

y ∈ P (A)  ∩ P (B)

y ∈ P (A) and  y ∈ P (B)

y ⊂ A  and y ⊂ B

y ⊂ A ∩ B

y ∈ P (A ∩ B)

∴ P (A) ∩ P (B)  ⊂ P (A ∩ B)  ...  (2)

From (1) and (2), we get

P (A ∩ B)  = P (A) ∩ P (B)

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