Prove that if $I ≠ \emptyset$ then $\bigcap_{i \in I}A_{i} \in \bigcap_{i \in I} \mathscr P (A_{i})$

Question

Suppose $\{A_{i} | i \in I\}$ is an indexed family of sets and $ I ≠ \emptyset$. Prove that $\bigcap_{i \in I}A_{i} \in \bigcap_{i \in I} \mathscr P (A_{i})$

My attempt:

Let $i_{0}$ be arbitrary index such that $i_{0} \in I$. Since $A_{i_{0}} \in \{A_{i} | i \in I\}$, it follows that $\bigcap_{i \in I}A_{i} \subseteq A_{i_{0}}$. By definition of power set, we concude that $\bigcap_{i \in I}A_{i} \in \mathscr P(A_{i_{0}})$. Since $i_{0}$ was arbitrary, we conclude that $\forall i \in I (\bigcap_{i \in I}A_{i} \in \mathscr P(A_{i}))$, and therefore $\bigcap_{i \in I}A_{i} \in \bigcap_{i \in I} \mathscr P (A_{i}) $

Is it correct?

Note that the assumption that $I\neq\varnothing$ isn't really necessary. — Asaf Karagila, Jul 30 '19 at 16:09
@AsafKaragila The author of the book included this condition, I just copied the whole statement. — Stokolos Ilya, Jul 30 '19 at 16:24
I guessed, I am just saying that this is something to observe. — Asaf Karagila, Jul 30 '19 at 16:24
@AsafKaragila But wait, suppose $I = \emptyset$, then $\bigcap_{i \in I}A_{i} = \emptyset$. But what about $\bigcap_{i \in I} \mathscr P(A_{i})$? Is it $\emptyset$ or ${\emptyset}$? — Stokolos Ilya, Jul 30 '19 at 16:30
Why does it hold that $\bigcap_{i\in\varnothing}A_i=\varnothing$? — Asaf Karagila, Jul 30 '19 at 16:31
@AsafKaragila Well, my guess was that if $I = \emptyset$ then ${A_{i} | i \in I } = \emptyset$, and hence $\bigcap_{i \in I}A_{i} = \emptyset$. If it is not correct, then what is $\bigcap_{i \in I}A_{i}$? And why theorem above holds even when $I = \emptyset?$ — Stokolos Ilya, Jul 30 '19 at 18:13
Apply the definition of $\bigcap_{i\in\varnothing}A_i$ to find out, then apply it to $\bigcap_{i\in\varnothing}\mathcal P(A_i)$ as well, and see what you get. — Asaf Karagila, Jul 30 '19 at 18:16
@AsafKaragila. Not empty I is useful to avoid the technical hassles of $\cap$empty set. — William Elliot, Jul 30 '19 at 22:54
@William: Yes, I figured that part out. I was making a small remark to the OP that this assumption is not really necessary and is never truly used. — Asaf Karagila, Jul 30 '19 at 22:56

Prove that if $I ≠ \emptyset$ then $\bigcap_{i \in I}A_{i} \in \bigcap_{i \in I} \mathscr P (A_{i})$

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