How do you prove that an object is an element of a set? I am stuck on this particular problem:
Suppose $\{A_i \mid i \in I \}$ is an indexed family of sets and $I ≠ \varnothing$. Prove that
$\bigcap_{i\in I}A_i \in \bigcap_{i\in I}P(A_i)$. I understand the notation and the problem itself does not seem that hard however i have no idea how to prove this rigorously enough for a lack of a better word.
Any advice how to approach this type of problem?
 A: To prove $x \in \bigcap_i B_i$ for some $x$, by the definition of intersection, you have to prove $x \in B_i$ for all $i$. In your case, $B_i = P(A_i)$, hence $x\in B_i$ means $x \subseteq A_i$. With $x = \bigcap_{i \in I} A_i$, you have to show $\bigcap_{i \in I} A_i \subseteq A_i$ for all $i \in I$. Which is true by the definition of intersection.
A: You have
$$ \cap_{i \in I} A_i \subset A_i \quad \forall i \in I, $$
thus
$$ \cap_{i \in I} A_i \in \mathcal{P}(A_i) \quad \forall i \in I, $$
therefore
$$ \cap_{i \in I} A_i \in \cap_{i \in I} \mathcal{P}(A_i) \quad \forall i \in I. $$
A: If "$I = \varnothing$" is not a typo, you are looking at an intersection of an empty family of sets. It is standard to define the intersection of an empty family of subsets of $A$ as being $A$any element of $A$ is in each $A_i$ if $I = \varnothing$. But there has to be some understanding of what set $A$ all the $A_i$ are (in that context) to be thought of as being "natural" subsets of. In a situation like this, the $\mathcal{P}(A_i)$ would probably be thought of as being natural subsets of $\mathcal{P}(A)$. So  $\bigcap_{i\in I}A_i$ would be $A$, whereas  $\bigcap_{i\in I}\mathcal{P}(A_i)$ would be $\mathcal{P}(A)$. And $A \in \mathcal{P}(A)$.
