Question

For any sets $A$ and $B$, show that $P(A\cap B)=P\left(A\right)\cap P\left(B\right)$.

Answer 1

Part 1 :
Let $X\in P(A\cap B)$
$\Rightarrow X\subset A\cap B$
So, $X\subset A$ and $X\subset B$
$\therefore X\in P\left(A\right)$ and $X\in P\left(B\right)$
$\Rightarrow X\in P(A\cap B)$
$\Rightarrow P(A\cap B)\subset P\left(A\right)\cap P\left(B\right)\cdots \left(i\right)$

Part 2 :
Let $Y\in P\left(A\right)\cap P\left(B\right)$
$\Rightarrow Y\in P\left(A\right)$ and $Y\in P\left(B\right)$
$\Rightarrow Y\subset A$ and $Y\subset B$
$\therefore Y\subset A\cap B$
$\Rightarrow Y\in P(A\cap B)$
i.e., $Y\in P\left(A\right)\cap P\left(B\right)\Rightarrow Y\in P(A\cap B)$
So, $P\left(A\right)\cap P\left(B\right)\subset P(A\cap B)\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$
$P(A\cap B)=P\left(A\right)\cap P\left(B\right)$
Hence proved.