To Prove $ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $ $$ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i)  $$ , $ I \neq \phi $
MY ATTEMPT
I use proof by Contradiction.  Assume $ \bigcap_{i \in I} A_i \notin \bigcap_{i \in I} P(A_i)  $$
Let $ x \in  \bigcap_{i \in I} A_i $
i.e  $ \{ x \}  \notin ( P(A_1 ) \land P(A_2 \land... \land P(A_n)) )$
So $ \{x\} \not \subseteq A_i  \forall i \in I$
So $ x \notin A_i \forall i \in I$
So $x \notin \bigcap_{i \in I} A_i  $
Hence we arrive at contradiction
 A: When you start by letting $x\in\bigcap_{i\in I}A_i$ and supposing that $x\notin\bigcap_{i\in I}\wp(A_i)$, you’re already getting off on the wrong foot: that would be a reasonable start if you were trying to show that $\bigcap_{i\in I}A_i$ was a subset of $\bigcap_{i\in I}\wp(A_i)$, but that isn’t what you want to show. You need to show that $\bigcap_{i\in I}A_i$ is an element of $\bigcap_{i\in I}\wp(A_i)$. In symbols, you’re setting out to try to prove that $\bigcap_{i\in I}A_i\color{red}{\subseteq}\bigcap_{i\in I}\wp(A_i)$, but what you need to prove is that $\bigcap_{i\in I}A_i\color{red}{\in}\bigcap_{i\in I}\wp(A_i)$.
Let’s back up for a minute and take a good look at the objects involved. In fact, let’s look at a very simple example. Suppose that $I=\{1,2,3\}$, so that you have sets $A_1,A_2$, and $A_3$. To be absolutely definite, let’s suppose that $A_1=\{1,2,4,5\}$, $A_2=\{2,3,4,5\}$, and $A_3=\{4,5,6\}$. Then
$$\begin{align*}
\bigcap_{i\in I}A_i&=A_1\cap A_2\cap A_3\\
&=\{1,2,4,5\}\cap\{2,3,4,5\}\cap\{4,5,6\}\\
&=\{4,5\}\;.
\end{align*}$$
This a subset of each of the sets $A_1,A_2$, and $A_3$, so like $A_1,A_2$, and $A_3$, it is a set of integers. Now what is $\bigcap_{i\in I}\wp(A_i)$?
$$\begin{align*}
\wp(A_1)=&\big\{\varnothing,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\\
&\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\big\}\;,
\end{align*}$$
$$\begin{align*}
\wp(A_2)=&\big\{\varnothing,\{2\},\{3\},\{4\},\{5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\},\\
&\{2,3,4\},\{2,3,5\},\{2,4,5\},\{3,4,5\},\{2,3,4,5\}\big\}\;,
\end{align*}$$
and
$$\wp(A_3)=\big\{\varnothing,\{4\},\{5\},\{6\},\{4,5\},\{4,6\},\{5,6\},\{4,5,6\}\big\}\;,$$
and the intersection of these three sets is
$$\big\{\varnothing,\{4\},\{5\},\{4,5\}\big\}\;:$$
$\varnothing,\{4\},\{5\}$, and $\{4,5\}$ are the only sets of integers that are elements of all three power sets.
$\bigcap_{i\in I}A_i=\{4,5\}$ cannot possibly be a subset of $\bigcap_{i\in I}\wp(A_i)$: it’s the wrong kind of object. If it were a subset of $\bigcap_{i\in I}\wp(A_i)$, its elements would also be elements of $\bigcap_{i\in I}\wp(A_i)$. But the elements of $\{4,5\}$ are integers, while the elements of $\bigcap_{i\in I}\wp(A_i)$ are sets of integers.
$\{4,5\}$ can, however, be an element of $\bigcap_{i\in I}\wp(A_i)$, and indeed we see that it is:
$$\bigcap_{i\in I}\wp(A_i)=\big\{\varnothing,\{4\},\{5\},\color{red}{\{4,5\}}\big\}\;.$$
Now let’s go back and consider how to prove the result. You don’t need a proof by contradiction: you can show directly that $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\wp(A_i)$. For each $i\in I$, $\bigcap_{i\in I}A_i$ is a subset of $A_i$: $\bigcap_{i\in I}A_i\subseteq A_i$. By definition this means that $\bigcap_{i\in I}A_i\in\wp(A_i)$. Thus, $\bigcap_{i\in I}A_i\in\wp(A_i)$ for each $i\in I$, and that by definition means that $\bigcap_{i\in I}A_i$ is in the intersection of those power sets: $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\wp(A_i)$. And that’s what we wanted to prove.
Once you understand what’s going on here, you might try to prove the stronger result that
$$\bigcap_{i\in I}\wp(A_i)=\wp\left(\bigcap_{i\in I}A_i\right)\;.$$
A: Let $A=\bigcap_{i\in I}A_i$. By definition, $A\subseteq A_j$ for each $j\in I$ and thus $j\in I\implies A\in\mathcal P(A_j)$. $$\therefore A\in\bigcap_{i\in I}\mathcal P(A_i)$$
A: Why not a direct proof?
$$\bigcap_{i \in I}A_i \in \bigcap_{i \in I}P(A_i) \iff \forall k \in I: \bigcap_{i \in I} A_i \in P(A_k)$$
$$\iff \forall k \in I: \bigcap_{i \in I} A_i \subseteq A_k$$
and the latter statement is obvious.
