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)

 

Due to paucity of time it would not be possible for us to provide the answers to all your queries. We are providing solutions to some your good queries. Try solving rest of the questions yourself and if you face any difficulty then do get back to us.

 

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