How to prove if $P(A) = P(B) \cap P(C)$ then $A = B \cap C$ Given three sets $A,B,C$
please help me to prove that if $P(A) = P(B) \cap P(C)$ then $A = B \cap C$
 A: Following is an elementary proof.
\begin{align}
&x\in B\cap C\\
\implies& x\in B \ \ \&\ \ x\in C\\
\implies& \{x\}\in P(B)\ \ \&\ \ \{x\}\in P(C)\\
\implies& \{x\}\in P(B)\cap P(C)\\
\implies& \{x\}\in P(A)\\
\implies& x\in A
\end{align}
Hence, $B\cap C\subset A$
The other side is exactly similar.
A: Here's a pathway to a solution. First prove the following two facts:


*

*If $X$ is a set, then $\bigcup \mathcal{P}(X) = X$.

*If $X$ and $Y$ are sets, then $\mathcal{P}(X) \cap \mathcal{P}(Y) = \mathcal{P}(X \cap Y)$
Then piece these two facts together to obtain the proof of the result in your question.
[In case the notation is unfamiliar to you: we have $\bigcup \mathcal{F} = \bigcup_{U \in \mathcal{F}} U = \{ x \mid x \in U \text{ for some } U \in \mathcal{F} \}$.]
A: Try to prove the following elementary facts:
1) for any sets $A, B$ one has $\mathscr{P}(A) \cap \mathscr{P}(B)=\mathscr{P}(A \cap B)$.
2) for any sets $A, B$ one has
$$A \subseteq B \Longleftrightarrow \mathscr{P}(A) \subseteq \mathscr{P}(B)$$
and consequently 
$$A=B \Longleftrightarrow \mathscr{P}(A)=\mathscr{P}(B)$$ 
which follows straight away from the axiom of extensionality.
With this you will have completely solved the problem.
Keep in mind the more general relation
$$\bigcap_{i \in I}\mathscr{P}(A_i)=\mathscr{P}(\bigcap_{i \in I}A_i)$$
valid for any family $A$ of sets indexed by an arbitrary nonempty index set $I$.
