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If ${E_n}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X,$ if $E_{n+1} \subset E_n,$ and if $\lim_{n\rightarrow \infty}\text{diam}(E_n) =0,$ then $\bigcap_1^\infty E_n$ contains exactly one point.

I am not sure why I need that the sets are bounded.

My proof is as follows: Begin by constructing a sequence $(p_n)$ such that the $n^{th}$ term is an element of $E_n.$ $\lim_{n\rightarrow \infty}\text{diam}(E_n)=0,$ for each $\epsilon>0$ there is an $N$ such that if $n\geq N$, $|\text{diam}(E_n)-0| <\epsilon.$ Fix N. This implies if $p$ and $q$ are in $E_N$, $d(p,q)<\epsilon$. By the way we constructed our sequence, we know that there is some $p_N\in E_N$ and for all $p_{N'}$ and $p_{M'}$ where $N'\geq N',M'\geq N$, $d(p_{N'},p_{M'})<\epsilon$ because $p_{N'}\in E_{N'} \subset E_{N}$ and $p_{M'}\in E_{M'} \subset E_{N}$. Thus this sequence is Cauchy.

We are given that $X$ is complete, thus every Cauchy sequence converges. Let $\lim_{n\rightarrow\infty}p_n=p$. We know every open ball around $p$ must contain all but finitely many elements of the sequence--thus it must contain some element of each $E_n$. Therefore $p$ is a limit point of each $E_n$ since the sets are nested. Since each $E_n$ is closed, $p$ must be an element of each $E_n$. Thus $\bigcap_1^\infty E_n$ is nonempty.

At this point we just need to show that $p$ is the only element in the grand intersection. Assume via contradiction that there exists more than one point $p$ in the grand intersection. Call it $q$. Let $L=d(p,q)$. Choose $\epsilon<L$. Then for each $n$, $p\in E_n$ and $q\in E_n$ and $\text{diam}(E_n)\geq L>\epsilon$. Thus,$\lim_{n\rightarrow \infty}\text{diam}(E_n)\neq 0$--a contradiction. Therefore $\bigcap_1^\infty E_n$ contains only one point, namely $p$.

I didn't use that the sets were bounded anywhere, which leads me to believe this logic is in some way flawed.

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Do you have a notion of diameter for unbounded sets? If we define $\mbox{Diam}(E_n) = \infty$ when $E_n$ is unbounded, then it does seem that the assumption is superfluous, but if you don't define diameter for sets with infinite diameter maybe that part is added so that $\mbox{Diam}(E_n)$ is defined. Whenever the diameter is finite, we should have that the set is bounded, so if $\mbox{Diam}(E_n) \to 0$ we know that the sequence of sets is bounded for $n$ larger than some $N$. – guy Nov 20 '12 at 5:34
Bounded is implied by the fact that the diameter is bounded and the sets are nested. Presumably $\text{diam} A = \sup_{x,y \in A} d(x,y)$. – copper.hat Nov 20 '12 at 5:46
up vote 1 down vote accepted

$\newcommand{\diam}{\operatorname{diam}}$Your proof is fine. The assumption that the sets are bounded is required to ensure that their diameters exist, since diameters are usually understood to be real numbers. If one allows $\infty$ as the diameter of an unbounded set, then the hypothesis that $\langle\diam E_n:n\in\Bbb N\rangle\to 0$ suffices.

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