Is it true that for any sets A and B, P(A)∪P(B)=P(A∪B)? Justify your answer
Let A={0,1} and B={1,2}
∴A∪B={0,1,2}
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}}
P(A)∪P(B)={ϕ,{0},{1},{0,1},{2},{1,2}}
∴P(A)∪P(B)≠P(A∪B)
False