Question

Is it true that for any sets A and $B,P\left(A\right)\cup P\left(B\right)=P\left(A\cup B\right)$? Justify your answer.

Answer 1

$$\begin{gathered} \mathrm{False}. \\ \mathrm{Let}X\in P\left(A\right)\cup P\left(B\right) \\ \Rightarrow X\in P\left(A\right)\mathrm{or}X\in P\left(B\right) \\ \Rightarrow X\subset A\mathrm{or}X\subset B \\ \Rightarrow X\subset \left(A\cup B\right) \\ \Rightarrow X\in P\left(A\cap B\right) \\ \therefore P\left(A\right)\cup P\left(B\right)\subset P\left(A\cup B\right)...\left(1\right) \\ \mathrm{Again},\mathrm{let}X\in P\left(A\cup B\right) \\ \mathrm{But}X\notin P\left(A\right)\mathrm{or}x\notin P\left(B\right)\left[\mathrm{For}\mathrm{example}\mathrm{let}A=\left\{2,5\right\}\mathrm{and}B=\left\{1,3,4\right\}\mathrm{and}\mathrm{take}X=\left\{1,2,3,4\right\}\right] \\ \mathrm{So},X\notin P\left(A\right)\cup P\left(B\right) \\ \mathrm{Thus},P\left(A\cup B\right)\mathrm{is}\mathrm{not}\mathrm{necessarily}\mathrm{a}\mathrm{subset}\mathrm{of}P\left(A\right)\cup P\left(B\right). \end{gathered}$$