Show that for any Sets $A$ and $B$, $\mathcal{P}(A) \cap \mathcal{P}(B) = \mathcal{P}(A\cap B)$ Question: show that for any sets $A$ and $B$, $\mathcal{P}(A) \cap \mathcal{P}(B) = \mathcal{P}(A \cap B).$
I want to prove it. Consider the following attempted proof.
$$(1)\:\:\:A \cap B \in \mathcal{P}(A \cap B) $$
$$(2)\:\:\:A \in \mathcal{P}(A) $$
$$(3)\:\:\:B \in \mathcal{P}(B)  $$
We can get $(4)$ from $(2)$ and $(3):$
$$(4)\:\:\:A \cap B \in \mathcal{P}(A) \cap \mathcal{P}(B)$$
So, for any set $A$ and $B$, we can get from $(1)$ and $(4):$
$$\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$$
Is this correct?
 A: Let $C\in P(A)\cap P(B)$ then $C\in P(A)$ and $C\in P(B)$ implying $C\subseteq A$ and $C\subseteq B$ implying $C\subseteq A\cap B$ implying $C\in P(A\cap B)$.
Thus, $P(A)\cap P(B)\subseteq P(A\cap B)$.
Now let $D\in P(A\cap B)$ then $D\subseteq A\cap B$ implying $D\subseteq A$ and $D\subseteq B$ implying $D\in P(A)$ and $D\in P(B)$ implying $D\in P(A)\cap P(B)$.
Thus, $P(A\cap B)\subseteq P(A)\cap P(B)$.
The above shows $P(A\cap B)=P(A)\cap P(B)$.
A: You haven't justified how one uses $(2)$ and $(3)$ to get $(4),$ yet. From $(4),$ you can fairly directly conclude that $$\mathcal P(A\cap B)\subseteq\mathcal P(A)\cap\mathcal P(B),\tag{$\star$}$$ but $(1)$ isn't sufficient to give you the reverse inclusion on its own.
It's important in a proof not to just say things like "we can get" or "it follows that." While it may be true, it doesn't explain how. Textbooks will often do such things, but this is to give the reader a chance to "fill in the blanks" and increase their own understanding.
I recommend you explain explicitly why $(2)$ and $(3)$ give us $(4),$ and how $(\star)$ follows from $(4).$ That should give you an idea how to finish the proof.
