# Hausdorff space and Cantor's intersection theorem

$X$ is a Hausdorff space, $C_i$ is a non-empty closed subset of $X$ and $C_{k+1}\subseteq C_k$ , show that $\displaystyle \bigcap_{i\in \mathbb{N}} C_i$ is compact.

I tried to prove by contradiction. Assume it's not compact, then, we make such open cover of $\displaystyle \bigcap_{i\in \mathbb{N}} C_i$, $\{C_0 \backslash C_1,C_1 \backslash C_2,\dots\}$,which doesn't have finite subcover, because all of them are disjoint. I hope it works, but till now I still cannot find a contradiction with Hausdorff space.

Perhaps what I am thinking is wrong. Any suggestion for this question? Thanks!

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The result is false. Let $X=\Bbb R$, and for $n\in\Bbb N$ let $C_n=\left[1-\frac1{2^n},\to\right)=\left\{x\in\Bbb R:x\ge 1-\frac1{2^n}\right\}$; then

$$\bigcap_{n\in\Bbb N}C_n=[1,\to)=\{x\in\Bbb R:x\ge 1\}\;,$$

which is not compact.

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Brian,what if $X$ compact Hausdorff space? –  Cancan Apr 17 '13 at 23:14
Then the result is trivially true: $\bigcap_{n\in\Bbb N}C_n$ is an intersection of closed sets, so it’s closed and therefore compact. What the nesting gives you in addition is that the intersection is also non-empty. –  Brian M. Scott Apr 17 '13 at 23:17
But, Brian, then I don't understand what the point is here to mention Hausdorf space in this question? –  Cancan Apr 17 '13 at 23:19
@Cancan: There really isn’t any reason to require $X$ to be Hausdorff. –  Brian M. Scott Apr 17 '13 at 23:21
Thanks, Brian! then I'll just ask my professor again. He might have made an improper question :) –  Cancan Apr 17 '13 at 23:23

Consider $X=C_n=\Bbb R$, in such case $C_{n+1}=C_n$ a closed subset, and $\bigcap C_n=\Bbb R$ which is not compact.
Asaf, what if $X$ it's compact Hausdorff space? –  Cancan Apr 17 '13 at 23:15