Is it true that for any sets A and B, P(A) ∪ P(B) = P (A ∪ B)? Justify your answer.
AnswerGiven statement is false.
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.
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