Question

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

Answer 1

Let $XϵP(A\cap B)$.
Then each element of $X$ is an element of $A$ and $B$, hence $X$ is also in $P\left(A\right)$ and $P\left(B\right)$ $\Rightarrow XϵP\left(A\right)\cap P\left(B\right)$.
Now
Let $YϵP\left(A\right)\cap P\left(B\right)$.
Then $YϵP\left(A\right)$ and $YϵP\left(B\right)$. Therefore each element of $Y$ is an element of $A$ and $B$. Hence each element of $Y$ is in $A\cap B\Rightarrow YϵP(A\cap B)$.
$X$ and $Y$ are arbitrary, hence we have shown that any set in $P(A\cap B)$ is in $P\left(A\right)\cap P\left(B\right)$ and vice versa.
From this we can conclude that the two sets have identical composition and are thus equal.