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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
up vote 7 down vote accepted

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.

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Wow, that was fast! Thanks for the great answer, I should have thought of assuming that the intersection of the local bases is nonempty then showing that the set consisting of the countable sequence of points in the intersections along with the original points violates the KC condition, I'll remember this general approach from now on. Thanks again! – WayMoreQuestionsThanAnswers Apr 26 '13 at 11:03
@WayMoreQuestionsThanAnswers: You’re welcome! – Brian M. Scott Apr 26 '13 at 11:05
ı couldnt get the rest to show K is compact but not closed, could you help me ? – user74512 Apr 26 '13 at 11:40
@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
@BrianM.Scott I'm guessing that you weren't notified due to him originally submitting his question as an answer to the general problem. I'm assuming that one of the moderators then switched the question from that location to its current spot as a reply to your solution, which bypassed the notification protocol since it registered as an edited post, not a submitted post. Thanks for showing the rest of the solution! – WayMoreQuestionsThanAnswers Apr 26 '13 at 13:22

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