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

Question

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.

Answer 1

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 $\cap_{i\in I} A_i\in \cap_{i\in I} \mathscr{P}(A_i)$.

Answer 2

Apart from the fact that you argue directly using elements, there is another reason your proof is longer than WoolierThanThou's. You actually prove the stronger statement that $$\bigcap_{i\in I}\mathscr P(A_i) = \mathscr P\left(\bigcap_{i\in I}A_i\right).$$ That is, you prove that every element of the set on the left belongs to the set on the right, and vice versa. But you are only asked to prove that one specific element of the set on the right belongs to the set on the left.

If you do want to prove the equality above, here is a way to do it with sets. Let $B$ be an arbitrary set. Then $$B \in \bigcap_{i\in I}\mathscr P(A_i) \quad \text{ iff } \quad \forall i \in I \ \ B \in \mathscr P(A_i) \quad \text{ iff } \quad \forall i \in I \ \ B \subseteq A_i.$$ At the same time, $$B \in \mathscr P\left(\bigcap_{i\in I}A_i\right) \quad \text{ iff } \quad B \subseteq \bigcap_{i\in I}A_i \quad \text{ iff } \quad \forall i \in I \ \ B \subseteq A_i.$$

Since the sets $\bigcap_{i\in I}\mathscr P(A_i)$ and $\mathscr P(\bigcap_{i\in I}A_i)$ have the same elements, they are equal.

Everything here is just a matter of definitions except for the last equivalence, which is easy to check. It amounts to permuting the universal quantifiers in $\forall x \in B \ \forall i \in I \ \ x \in A_i$.