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The taks is: Show that in every infinite compact space there is a countable subset that is not closed.

At first I read that it should be closed and I had an idea to take a point $x_1 \in X$ and an open set $U_1$ containing $x_1$. If $X/U_1$ is finite then we're done. If not, then it is still compact and we take $x_2 \in X\setminus U_1$ and an open set $U_2$ which should be distinct from $U_1$ and so on... And we would get a countable and closed set.

But I don't know how to do this task. Could somebody tell me how to do it?

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  • $\begingroup$ Do you know that in a compact space, every infinite subset has a limit point? $\endgroup$ – Daniel Fischer May 28 '15 at 19:06
  • $\begingroup$ I don't know it yet... $\endgroup$ – nilcorc May 28 '15 at 19:08
  • $\begingroup$ Your argument doesn't work for $\mathbb{N}$ with the trivial (indiscrete) topology. It is compact and infinite, and for any $x\in\mathbb{N}$, $\mathbb{N}$ is the only open set containing $x$, so $\mathbb{N}\setminus\mathbb{N}=\emptyset$ is finite. But of course $\mathbb{N}$ is not the set we're looking for, since it is closed. $\endgroup$ – Moya May 28 '15 at 19:17
  • $\begingroup$ What is your definition of "compact"? $\endgroup$ – Omnomnomnom May 28 '15 at 19:36
  • $\begingroup$ A space is compact iff from every open cover we can choose a finite subcover. $\endgroup$ – nilcorc May 28 '15 at 19:39
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Lemma: In compact spaces all infinite subsets have a limit point. (But they don't necessarily contain it.)

(Proof sketch: Take, by contradiction, an infinite subset $A$ with no limit points. $A$ is closed as $A'=\emptyset$ and so $A=\bar{A}$, closed subsets of compact spaces are compact so $A$ is compact, but every point in $A$ is isolated which gives us an infinite cover by open sets [the singletons] with no finite subcover, contradicting $A$ compact.)

Now take a countably infinite subset of your compact space, by the lemma it has at least 1 limit point. If it does not contain this limit point, it is not closed and countably infinite. If it does contain the limit point, take that limit point out and the set you're left with is not closed and still countably infinite.

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