# Prove that if X is second-countable and every compact subset of X is closed, then X is Hausdorff.

Question: Prove that if $X$ is second-countable and every compact subset of $X$ is closed, then $X$ is Hausdorff.

I know that the second-countability of $X$ is what will make the proof work at some point, since if you remove that from the hypothesis you can take $X$ to be an uncountable set with the cocountable topology as a counterexample. I'm just having a lot of trouble seeing how to tie it into the proof.

I apologize for not being able to condense the subsequent ramblings into a shorter notation, I'm still a bit shaky with LaTeX. So far I have attempted using the following facts in the proof:

• $X$ being second-countable implies that $X$ has a countable dense subset, hence separable, so there must be some sequence which has an element in every open subset of $X$;

• $X$ being second-countable implies that there exists some collection of open subsets of $X$ such that any open subset of $X$ can be expressed as the union of some subfamily of the aforementioned collection of open subsets;

• $X$ being second-countable implies that every open cover of $X$ has a countable subcover (Lindelof);

• $X$ is KC (every compact subset of $X$ is closed);

• the basic ways to show that $X$ is Hausdorff (for each pair of points there exists disjoint neighborhoods, diagonal is closed, every singleton is equal to the intersection of all closed neighborhoods of that singleton, etc).

I just can't seem be able to comfortably use the second-countability to complete the proof without getting the nagging feeling that my logic is wrong. If anyone could give a full formal proof or give any amount of insight I'll be very, very happy. Please and thank you!

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Thanks for the edit @EricStucky, that run on list in paragraph form was dreadful. –  WayMoreQuestionsThanAnswers Apr 26 '13 at 11:06

Let $x$ and $y$ be distinct points of $X$. Since $X$ is second countable, it is certainly first countable, and there are local bases $\{B_x(n):n\in\Bbb N\}$ and $\{B_y(n):n\in\Bbb N\}$ at $x$ and $y$, respectively. We may further assume that $B_x(n)\supseteq B_x(n+1)$ and $B_y(n)\supseteq B_y(n+1)$ for each $n\in\Bbb N$.
Suppose that $x$ and $y$ cannot be separated by disjoint open sets. Then for each $n\in\Bbb N$ we can choose a point $x_n\in B_x(n)\cap B_y(n)$. Let $K=\{x_n:n\in\Bbb N\}\cup\{x\}$. Now show that $K$ is compact but not closed.
@whyquestions: Let $\mathscr{U}$ be any open cover of $K$; some $U\in\mathscr{U}$ contains $x$, and there is an $n\in\Bbb N$ such that $x\in B_x(n)\subseteq U$, so $U\supseteq\{x\}\cup\{x_k:k\ge n\}$, since the sets $B_x(k)$ are nested. Thus, $U$ contains all but finitely many points of $K$. It takes only finitely many more members of $\mathscr{U}$ to cover those points, so $\mathscr{U}$ has a finite subfamily covering $K$, and $K$ is compact. On the other hand, every open nbhd of $y$ contains one of the $B_y(n)$ and therefore contains a point of $K$, so $y\in\operatorname{cl}K\setminus K$. –  Brian M. Scott Apr 26 '13 at 13:07