Question:
for any set A and B show that P(A intersection B) =P(A) intersection P(B)
Answer:
Let, x ∈ P(A∩ B)
⇒ x ∈ P(A) and x ∈ P(B)
⇒ x ∈ P(A) ∩ P(B)
⇒ P(A∩ B) ⊆ P(A) ∩ P(B)
Again Let, y ∈ P(A) ∩ P(B)
⇒ y ∈ P(A) and x ∈ P(B)
⇒ y ∈ P(A∩ B)
⇒ P(A) ∩ P(B) ⊆ P(A∩ B)
Hence, P(A∩ B) = P(A) ∩ P(B)
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