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Question:

Is it true that for any sets A and B, P(A)∪P(B) = P(A∪B)?

Solution:

Let A = {0, 1} and B = {1, 2}
∴ A∪B = {0, 1, 2}
P(A) = {∅, {0}, {1}, {0, 1}}
P(B) = {∅, {1}, {2}, {1, 2}}
P(A∪B) = {∅, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}}
P(A)∪P(B) = {∅, {0}, {1}, {0, 1}, {2}, {1, 2}}
∴P(A)∪P(B) ≠ P(A∪B)
False