# Countable and not closed subset of infinite compact space

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?

• Do you know that in a compact space, every infinite subset has a limit point? – Daniel Fischer May 28 '15 at 19:06
• I don't know it yet... – nilcorc May 28 '15 at 19:08
• 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. – Moya May 28 '15 at 19:17
• What is your definition of "compact"? – Omnomnomnom May 28 '15 at 19:36
• A space is compact iff from every open cover we can choose a finite subcover. – nilcorc May 28 '15 at 19:39

(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.)