Not a duplicate of
Prove that if $I ≠ \emptyset$ then $\bigcap_{i \in I}A_{i} \in \bigcap_{i \in I} \mathscr P (A_{i})$
To Prove $ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $
This is exercise $3.3.15$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\{A_i | i ∈ I\}$ is an indexed family of sets and $I \neq \emptyset$. Prove that $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\mathscr P(A_i)$.
Here is my proof:
Let $A$ be an arbitrary element of $\bigcap_{i\in I}\mathscr P(A_i)$. Let $x$ be an arbitrary element of $A$. Since $I\neq \emptyset$, let $i$ be an arbitrary element of $I$. From $\bigcap_{i\in I}\mathscr P(A_i)$ and $i\in I$, $A\in\mathscr P(A_i)$ and so $A\subseteq A_i$. From $A\subseteq A_i$ and $x\in A$, $x\in A_i$. Thus if $i \in I$ then $x\in A_i$. Since $i$ was arbitrary, $\forall i\Bigr(i\in I\rightarrow x\in A_i\Bigr)$ and so $x\in\bigcap_{i\in I}A_i$. Thus if $x\in A$ then $x\in\bigcap_{i\in I}A_i$. Since $x$ was arbitrary, $\forall x\Bigr(x\in A\rightarrow x\in\bigcap_{i\in I}A_i\Bigr)$ and so $A\subseteq\bigcap_{i\in I}A_i$ and ergo $A\in\mathscr P(\bigcap_{i\in I}A_i)$. Therefore if $A\in\bigcap_{i\in I}\mathscr P(A_i)$ then $A\in\mathscr P(\bigcap_{i\in I}A_i)$. Since $A$ was arbitrary, $\forall A\Bigr(A\in\bigcap_{i\in I}\mathscr P(A_i)\rightarrow A\in\mathscr P(\bigcap_{i\in I}A_i)\Bigr)$ and so $\bigcap_{i\in I}\mathscr P(A_i)\subseteq\mathscr P(\bigcap_{i\in I}A_i)$.
Let $A$ be an arbitrary element of $\mathscr P(\bigcap_{i\in I}A_i)$. This means $A\subseteq\bigcap_{i\in I}A_i$. Since $I\neq\emptyset$, let $i$ be an arbitrary element of $I$. Let $x$ be an arbitrary element of $A$. From $A\subseteq\bigcap_{i\in I}A_i$ and $x\in A$, $x\in \bigcap_{i\in I}A_i$. From $x\in \bigcap_{i\in I}A_i$ and $i\in I$, $x\in A_i$. Thus if $x\in A$ then $x\in A_i$. Since $x$ was arbitrary, $\forall x\Bigr(x\in A\rightarrow x\in A_i\Bigr)$ and so $A\subseteq A_i$ and ergo $A\in\mathscr P(A_i)$. Thus if $i\in I$ then $A\in \mathscr P(A_i)$. Since $i$ was arbitrary, $\forall i\Bigr(i\in I\rightarrow A\in \mathscr P(A_i)\Bigr)$ and so $A\in\bigcap_{i\in I}\mathscr P(A_i)$. Therefore if $A\in\mathscr P(\bigcap_{i\in I}A_i)$ then $A\in \bigcap_{i\in I}\mathscr P(A_i)$. Since $A$ was arbitrary, $\forall A\Bigr(A\in\mathscr P(\bigcap_{i\in I}A_i)\rightarrow A\in\bigcap_{i\in I}\mathscr P(A_i)\Bigr)$ and so $\mathscr P(\bigcap_{i\in I}A_i)\subseteq \bigcap_{i\in I}\mathscr P(A_i)$.
Since $\bigcap_{i\in I}\mathscr P(A_i)\subseteq\mathscr P(\bigcap_{i\in I}A_i)$ and $\mathscr P(\bigcap_{i\in I}A_i)\subseteq \bigcap_{i\in I}\mathscr P(A_i)$, then $\mathscr P(\bigcap_{i\in I}A_i)= \bigcap_{i\in I}\mathscr P(A_i)$. Therefore we can rewrite $\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\mathscr P(A_i)$ as $\bigcap_{i\in I}A_i\in\mathscr P(\bigcap_{i\in I}A_i)$ which is equivalent to $\bigcap_{i\in I}A_i\subseteq \bigcap_{i\in I}A_i$ which is by definition true. $Q.E.D.$
Is my proof valid$?$
Thanks for your attention.
Best Answer
Your proof is valid but way longer than it needs to be.
Really, let $i_0\in I$. Since $\cap_{i\in I} A_i\subseteq A_{i_0},$ you get that $\cap_{i\in I} A_i\in \mathscr{P}(A_{i_0})$. Since $i_0$ was arbitrary, we get that $\cap_{i\in I} A_i\in \cap_{i\in I} \mathscr{P}(A_i)$.