Let $(X,||\cdot||)$ be a complete normed space. Let $F_1, F_2, F_3,\ldots\subseteq X$ be closed, non-empty subsets of $X$.

Assume that $F_1 \supseteq F_2\supseteq F_3\supseteq \cdots$

and that $\sup_{x,y\in F_n}||x-y||\to0$.

Show that the intersection of all $F_n$ is non-empty.

All I can possibly think to go on is that we need to show that there is some $f\in X$ such that $f\in F_1,F_2,F_3,\ldots$

Any pointers in the right direction?

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For each $n \in \mathbb{N}$ pick $x_n \in F_n$, then use your assumption to show the sequence is Cauchy and conclude by proving that the limit is in the intersection (notice that if $n \ge m$, then $x_n \in F_m$, and use the fact that $F_m$ is closed). – Joel Cohen Mar 1 '12 at 0:28
This seems like it should work in any complete metric space. Here are two hints: (1) for each $n$, what does the fact that $F_n$ is non-empty tell you? (2) how are we going to use completeness in proving the result? – user16299 Mar 1 '12 at 0:29
Ah, I see Joel beat me to the punch. – user16299 Mar 1 '12 at 0:29

If we let $x_n \in F_n$, then you can check that $x_1,x_2,\ldots$ is a Cauchy sequence converging (since $X$ is complete) to $x \in X$.
Now fix an $i$. If $j>i$, then $x_j \in F_j \subset F_i$, so since $F_i$ is closed, it follows that $x \in F_i$. Since $x \in F_i$ for all $i$, it follows that $x \in \cap_i F_i$.