$A$ is a subset of $B$ if and only if $P(A) \subset P(B)$ I had to prove the following for a trial calculus exam:
$A\subset B$ if and only if $P(A) \subset P(B)$ where $P(A)$ is the set of all subsets of $A$.
Can someone tell me if my approach is correct and please give the correct proof otherwise?
$PROOF$:
$\Big(\Longrightarrow\Big)$ assume $A\subset B$ is true. Then $\forall$ $a\in A$, $a\in B$
Then for $\forall$ A, the elements $a_1, a_2,$ ... , $a_n$ in A are also in B.
Hence $P(A)\subset P(B)$
$\Big(\Longleftarrow\Big) $ assume $P(A) \subset P(B)$ is true. We prove this by contradiction so assume $A\not\subset B$
Then there is a set $A$ with an element $a$ in it, $a\notin$ B.
Hence $P(A) \not\subset P(B)$
But we assumed $P(A) = P(B)$ is true.
We reached a contradiction.
Hence if $P(A) = P(B)$ then $A\subset B$.

I proved it both sides now, please improve me if I did something wrong :-)

$(\Rightarrow)$ Given any $x\in P(A)$ then $x\subset A$. So, by hypothesis, $x\subset B$ and so $x\in P(B)$.
The counter part is easier, as cited by @Asaf, below.
