# Proof of equality of the intersection of power sets to the power set of an intersection dependent on associativity?

In Naive Set Theory p. 20, Halmos states the following theorem: $$\bigcap_{X \in \mathscr C}\mathscr P(X) = \mathscr P(\bigcap_{X \in \mathscr C}X)$$

He states that this can be generalized from proofs used in the previous exercise wherein we prove $$\mathscr P(E) \cap \mathscr P(F) = \mathscr P(E \cap F)$$

The way I approached it was to extend the original proof by invoking the associativity of intersection and how this allows us to arbitrarily extend such statements through the idea of using sets to represent other combinations of sets.

Let $$Y \in \mathscr P(E \cap F \cap G \ldots \cap N)$$

$$\therefore Y \subset (E \cap F \cap G \ldots \cap N)$$ $$\therefore (Y \subset E) \land (Y \subset F) \land (Y \subset G) \ldots \land (Y \subset N)$$ $$\therefore (Y \in \mathscr P(E)) \land (Y \in \mathscr P(F)) \land (Y \in \mathscr P(G)) \ldots \land (Y \in \mathscr P(N))$$ $$\therefore Y \in (\mathscr P(E) \cap \mathscr P(F) \cap \mathscr P(G) \ldots \cap \mathscr P(N))$$

I think I'm overlooking something. Is it acceptable to use '$$\ldots$$' in this scenario? Is articulating an intersection of an arbitrarily large collection in this manner legitimate? At this point in my study, I've already established how associativity allows these sorts of arbitrary extensions by letting us use individual sets in an intersection to stand in for other intersections. Even so, is this sufficiently rigorous?

$$\cap$${ P(A) : A in C } = P($$\cap$$C).
X in $$\cap$${ P(A) : A in C } iff
X subset $$\cap$$C iff