# Decreasing sequence of closed set in a metric space is convergent?

Let $\{E_n\}$ be a collection of bounded and closed subsets in a metric space $X$ such that $E_{n+1} \subset E_n$ and $lim_{n\to\infty} diam E_n = 0$.

It's a theorem that if $X$ is complete, then $\bigcap_{n\in \mathbb{N}} E_n$ is a singleton.

However, i have proved that $\bigcap_{n\in \mathbb{N}} E_n$ is a singleton where $Int(E_n) ≠\emptyset$ even if $X$ is not complete, but just an arbitrary metric space.

I don't actually believe my proof.. Is it true? Or please give me a counterexample!

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Where is your proof? – Did Nov 12 '12 at 10:48
I have updated my answer a bit. Now the interiors of $E_n$ aren't empty. – Dan Shved Nov 12 '12 at 10:58
@Dan I found a foolish mistake in my argument.. Thank you – Katlus Nov 12 '12 at 11:03

Looks really weird. Suppose $X=\mathbb{R}\setminus\{0\}$ and $E_n=(0,\frac{1}{n}]$. All the $E_n$ are closed in $X$, their diameters converge to $0$, and the intersection is empty.