It is not true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)
Justification:
Let us assume,
A = {0, 1}
And, B = {1, 2}
A ∪ B = {0, 1, 2}
According to the question,
We have,
P (A) = {ϕ, {0}, {1}, {0, 1}}
P (B) = {ϕ, {1}, {2}, {1, 2}}
P (A ∪ B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}
Also,
P (A) ∪ P (B) = {ϕ, {0}, {1}, {2}, {0, 1}, {1, 2}}
P (A) ∪ P (B ≠ P (A ∪ B)
Hence, the given statement is false
Answered by Pragya Singh | 1 year agoFind the symmetric difference A Δ B, when A = {1, 2, 3} and B = {3, 4, 5}.