The power set of the intersection of two sets equals the intersection of the power sets of each set It would be great if someone could verify this proof.
Theorem:
$\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$
Proof:
First I prove that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Take any subset $X \subseteq A \cap B$. Then $X \in \mathcal{P}(A \cap B)$. Also, $X \subseteq A \wedge X \subseteq B$, meaning that $X \in \mathcal{P}(A) \wedge X \in \mathcal{P}(B)$. This, again, means that $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, and this proves the statement.
Then I prove that $\mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(A \cap B)$. Again,  take any subset $X \subseteq A \cap B$. Using just the same arguments as above, this is also a truth.
The theorem follows.

In the fullest generality, consider a nonempty set $I$ and a family $A$ of sets indexed by $I$ (such that we will write $A_i$ for the set corresponding to index $i \in I$).
We then have:
$$X \in \bigcap_{i \in I} \mathscr{P}(A_i) \Longleftrightarrow (\forall i)(i \in I \Rightarrow X \subseteq A_i) \Longleftrightarrow X \subseteq \bigcap_{i \in I} A_i \Longleftrightarrow X \in \mathscr{P}\left(\bigcap_{i \in I}A_i\right)$$
and the equality
$$\bigcap_{i \in I}\mathscr{P}(A_i)=\mathscr{P}\left(\bigcap_{i \in I}A_i\right)$$
follows from the axiom of extensionality.
