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.

I've encountred this question while studying for a test, here's what i got:
Text book answer:
$X∈P(A∩B) \Leftrightarrow X⊆A∩B \Leftrightarrow X⊆A \ and \ X⊆B \Leftrightarrow X∈P(A) \ and \ X∈P(B) \Leftrightarrow X∈P(A)∩P(B)$
Hope it serves you well :)
