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.
 A: "Again,  take any subset $X \subseteq A \cap B$."
You might want to change the wording here. "Take any set $X$ such that $X \subseteq A$ and $X \subseteq B$. Then $X \subseteq A \cap B$."
As written it sounds like you're assuming what you want to prove. Otherwise good...
A: For the first half, you want to prove that any member of $\mathcal{P}(A \cap B)$ is a member of  $\mathcal{P}(A) \cap \mathcal{P}(B)$. So the proof should start "Suppose $X \in \mathcal{P}(A \cap  B)$". Then you derive $X \subseteq A \cap B$, and the rest goes as you state. (Except, don't use "means" for "implies".)
For the second half, you aren't going to use "just the same arguments" or you'd end up showing that $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A \cap B)$ which isn't what you meant! 
A: $
\newcommand{pow}[1]{\mathcal P(#1)}
$For comparison, here is another way to prove this: start with the most complex side, investigate which elements are in that set by expanding the definitions, and then try to simplify.  So we calculate, for any $\;V\;$,
\begin{align}
& V \in \pow A \cap \pow B \\
\equiv & \qquad \text{"definition of $\;\cap\;$"} \\
& V \in \pow A \;\land\; V \in \pow B \\
\equiv & \qquad \text{"definition of $\;\pow{\quad}\;$, twice"} \\
& V \subseteq A \;\land V \subseteq B \\
\equiv & \qquad \text{"set theory: property of $\;\cap\;$"} \\
& V \subseteq A \cap B \\
\equiv & \qquad \text{"definition of $\;\pow{\quad}\;$"} \\
& V \in \pow{A \cap B} \\
\end{align}
By set extensionality, this proves $\;\pow A \cap \pow B \;=\; \pow{A \cap B}\;$.
A: I was looking to prove a similar statement, but about an arbitrary intersection. This example seemed a good foundation to base the more general proof:
   
Let $\textbf{X}$ be a nonempty collection of sets. Prove that 
$\mathcal{P}(\bigcap \textbf{X}) = \bigcap \{\mathcal{P}(A) \mid A \in \textbf{X} \}$.  
Proof. Let $\textbf{X} = \{X_1, X_2, \dots, X_{n-1}, X_n\}$ be a nonempty 
collection of sets and let $x \in \mathcal{P}(\bigcap \textbf{X})$. Then
 
      
$\begin{aligned}
    x \in \mathcal{P}(\bigcap \textbf{X}) &\iff x \in \mathcal{P}(X_1 \cap X_2 \cap \dots \cap X_{n-1} \cap X_n) \\
    &\iff x \subseteq (X_1 \cap X_2 \cap \dots \cap X_{n-1} \cap X_n) \\
    &\iff x \subseteq X_1 \wedge x \subseteq X_2 \wedge \dots \wedge x \subseteq X_{n-1} \wedge x \subseteq X_n \\
    &\iff x \in \mathcal{P}(X_1) \wedge x \in \mathcal{P}(X_2) \wedge \dots \wedge x \in \mathcal{P}(X_{n-1}) \wedge x \in \mathcal{P}(X_n) \\
    &\iff x \in \mathcal{P}(X_1) \cap \mathcal{P}(X_2) \cap \dots \cap \mathcal{P}(X_{n-1}) \cap \mathcal{P}(X_n) \\
    &\iff x \in \bigcap \{\mathcal{P}(A) \mid A \in \textbf{X} \}.
    \end{aligned}$
 
Since $x$ was an arbitrary element of $\mathcal{P}(\bigcap \textbf{X})$, we have shown that
${\forall x \big( x \in \mathcal{P}(\bigcap \textbf{X}) \iff x \in \bigcap \{\mathcal{P}(A) \mid A \in \textbf{X} \} \big)}$, 
or $\mathcal{P}(\bigcap \textbf{X}) = \bigcap \{\mathcal{P}(A) \mid A \in \textbf{X} \}$. $\square$
 
 
However, the follow-up problem leaves me with a question. I mechanically changed intersections to unions and ands to ors from the proof above, but I don't know how to incorporate the condition "if
and only if $\bigcup \textbf{X} \in \textbf{X}$". I believe one of the steps below does not actually hold unless I use this condition, and I think it's step three, but I'm not sure of the justification:
 
Let $\textbf{X}$ be a nonempty collection of sets. Prove that 
$\mathcal{P}(\bigcup \textbf{X}) = \bigcup \{\mathcal{P}(A) \mid A \in \textbf{X} \}$ if
and only if $\bigcup \textbf{X} \in \textbf{X}$.
 
Proof. Let $\textbf{X} = \{X_1, X_2, \dots, X_{n-1}, X_n\}$ be a nonempty 
collection of sets and let $x \in \mathcal{P}(\bigcup \textbf{X})$. Then
 
      
$\begin{aligned}
    x \in \mathcal{P}(\bigcup \textbf{X}) &\iff x \in \mathcal{P}(X_1 \cup X_2 \cup \dots \cup X_{n-1} \cup X_n) \\
    &\iff x \subseteq (X_1 \cup X_2 \cup \dots \cup X_{n-1} \cup X_n) \\
    &\iff x \subseteq X_1 \vee x \subseteq X_2 \vee \dots \vee x \subseteq X_{n-1} \vee x \subseteq X_n \\
    &\iff x \in \mathcal{P}(X_1) \vee x \in \mathcal{P}(X_2) \vee \dots \vee x \in \mathcal{P}(X_{n-1}) \vee x \in \mathcal{P}(X_n) \\
    &\iff x \in \mathcal{P}(X_1) \cup \mathcal{P}(X_2) \cup \dots \cup \mathcal{P}(X_{n-1}) \cup \mathcal{P}(X_n) \\
    &\iff x \in \bigcup \{\mathcal{P}(A) \mid A \in \textbf{X} \}.
    \end{aligned}$
 
Since $x$ was an arbitrary element of $\mathcal{P}(\bigcup \textbf{X})$, we have shown that
${\forall x \big( x \in \mathcal{P}(\bigcup \textbf{X}) \iff x \in \bigcup \{\mathcal{P}(A) \mid A \in \textbf{X} \} \big)}$, 
or $\mathcal{P}(\bigcup \textbf{X}) = \bigcup \{\mathcal{P}(A) \mid A \in \textbf{X} \}$.
$\square$ 
A: 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.
A: 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 :)
