Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$X$ 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!

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
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

Your assumptions are not sufficient.

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.

share|cite|improve this answer
Asaf, what if $X$ it's compact Hausdorff space? – Cancan Apr 17 '13 at 23:15
@Cancan: Then the answer is that it is true; but not every Hausdorff space is compact. – Asaf Karagila Apr 17 '13 at 23:16
But could you please tell me what's the point here to mention Hausdorff to prove the conclusion? I couldn't think of anything that can be related to Hausdorff during proving. – Cancan Apr 17 '13 at 23:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.