Intersection of infinitely many closed nested sets in a complete space $(X,d)$ is a complete metric space. It needs to be proved, that if $A_1\supset A_2\supset ...$ are non-empty closed sets and $A_n$ is a finite sum of closed sets with diameter $\leq 1/n$ then $\bigcap_{n=1}^{\infty} A_n$ is nonempty and compact.
The problem is I don't know how to start and I'm not quite sure how should I try to obtain the compactness of this intersection. As the closed sets are mentioned, I think it may have to do with Baire's category theorem, but I don't see nowhere dense sets here. Then again, it'd be easier if $diam\ A_n \longrightarrow 0$, but that doesn't necessarily have to be the case.
 A: Suppose $A_1 = \cup_{k=1}^{N} F_k^1$ is a disjoint union of closed sets. Then $A_n = \cup_{k=1}^{N} F_k^n$ where $F_k^n \subseteq F_k^1$. For simplicity we replace $F_k^n$ by $A_n$. Now we would like to show $\cap_n A_n$ is a singleton.
If $(A_n)$ is nonempty, pick $x_n \in A_n$.
For $n \geq m$, $x_n \in A_n \subseteq A_m$ and $x_m \in A_m$ implies $d(x_n, x_m) \leq diam(A_m) \leq \frac{1}{m}$.
Hence $(x_n)$ is Cauchy and convergent by completeness. Let $x = \lim_{n} x_n$.
We then see $x \in \cap_n A_n$. For $n \geq 1$, for $m \geq n$, $x_m \in A_n$ and therefore $x = \lim_m x_m \in A_n$ by the fact that $A_n$ is closed.
Finally, we see that $\cap_n A_n = \{x\}$ is a singleton. Assume $y \in \cap_n A_n \setminus \{x\}$. Then $d(y, x) > 0$.
Pick $m$ such that $diam(A_m) \leq \frac{1}{m} \leq \frac{d(y, x)}{2}$. Then $x, y \in A_m \subseteq \cap_n A_n$ and $d(y, x) > diam(A_m)$. Contradiction.
This shows that, in the original setting, $\cap_n A_n$ is a finite set since each $F_k^n$ is either empty or a singleton.
A: Let $\bigcap_{n=1}^{\infty} A_n = A$.
For each n let $\mathscr C_n$ be a finite set of closed, non-empty sets with diameter $< 1/n$ and $\cup \mathscr C_n = A_n$. Consider finite sequences of the form $<C_1, ...,C_n>$ with each $C_n \in \mathscr C_n$ and $C_1 \cap ... \cap C_n$ non-empty. Every member of every $\mathscr C_n$ is in at least one such sequence so there are infinitely many of them. But there are only finitely many possibilities for the first member so there is a $B_1 \in \mathscr C_1$ which is in infinitely many sequences. Similarly there is $B_2 \in \mathscr C_2$ which is second in infinitely many sequences starting with $B_1$. Inductively we get an infinite sequence $<B_n>$, every initial sequence of which is of the specified form. Any sequence $<x_n>$ with each $x_n \in B_n$ must be Cauchy and so has a limit, which will be a point in A.
Now, suppose $<F_n>$ is any sequence of closed subsets of A with the finite intersection property. Replace $A_n$ by $F_1\cap...\cap F_n$. This is still a union of finitely many closed sets of diameter less than $1/n$ so the same argument shows that $\bigcap_{n=1}^{\infty} F_n$ is non-empty. Thus A is countably compact.
Let $\mathscr S = \{A \setminus C \mid C \in \mathscr C_n, n \in \Bbb N \}$. This is a sub-basis for the topology on A, showing it is second countable. Together with countably compact that implies it is compact.
