Is $ \ \mathcal{P}(A\cap B) = \mathcal{P}(A)\cap \mathcal{P}(B)$ True? I have a feeling it is true, but cannot find a way to prove it.
Help is welcome =)
 A: Yes, this statement is true.  Let $x \in \mathcal{P}(A \cap B)$.  Then, by the definition of the power set, $x \subseteq A \cap B$, which by definition of intersection implies that $x \subseteq A$ and $x \subseteq B$.  So by definition of power set, $x \in \mathcal{P}(A)$ and $x \in \mathcal{P}(B)$.  Hence, $x \in \mathcal{P}(A) \cap \mathcal{P}(B)$.  We have shown that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$.  It's true that $\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)$ by "reversing" the proof of $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$.  Rewrite that to obtain the equivalence of both sets.
A: It's a simple equivalency proof :
$ E \in \mathcal P (A\cap B)  \Leftrightarrow E\subset A\cap B
\Leftrightarrow \left[E\subset A \textrm{ and  } E\subset B\right] \Leftrightarrow \left[E\in\mathcal P (A) \textrm{ and } E\in \mathcal P(B) \right]\Leftrightarrow E\in\mathcal P (A) \cap \mathcal P(B)$
You should maybe show carefully this equivalence :
$$E\subset A\cap B
\Leftrightarrow \left[E\subset A \textrm{ and  } E\subset B\right]$$
going back to the definition : $E\subset F \Leftrightarrow \forall x \in E, x\in F$.
