Answer:
This statement is true
Step-by-step explanation:
Remember that subset F of a metric (or topological) space X is said to be closed if X-F is open according to the metric (topology) of F.
Let F⊆X. For the "if" part, suppose that [tex]F=F_1\cap F_2\cap \cdots \cap F_n[/tex] where [tex]F_k[/tex] is a closed set for all k. Then by De Morgan's law, [tex]X-F=(X-F_1)\cup(X-F_2)\cup \cdots\cup(X-F_n)[/tex].
Now, since Fk is closed for all k, then X-Fk is open. In every metric (topological) space, the union of an arbitrary family of open sets open sets is open, thus X-F is open, that is, F is closed.
For the "only if" implication, suppose that F is closed. We always have that F=F∩F (y∈F if and only if y∈F and y∈F if and only if y∈F∩F). then F is a finite intersection of closed sets (F and F).
