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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!

Eric Stucky
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WayMoreQuestionsThanAnswers
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2 Answers2

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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 $n\in\Bbb N$. Finally, compact sets in $X$ are closed, so all singleton subsets of $X$ are closed, $X$ is therefore $T_1$, and we may assume that $x\notin B_y(0)$ and $y\notin B_x(0)$.

Now 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 $K=\{x_n:n\in\Bbb N\}\cup\{x\}$, and show that $K$ is compact but not closed.

Brian M. Scott
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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
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    @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 ? –  Apr 26 '13 at 11:40
  • @BrianM.Scott Since you're much more proficient at both topology and using LaTeX, could you help whyquestions out? – WayMoreQuestionsThanAnswers Apr 26 '13 at 13:04
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    @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
  • @WayMoreQuestionsThanAnswers: Thanks for the ping; for some reason I didn’t get one when whyquestions posted the comment. – Brian M. Scott Apr 26 '13 at 13:08
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    @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
  • @WayMoreQuestionsThanAnswers: Yes, that would explain it. You’re welcome! – Brian M. Scott Apr 26 '13 at 13:23
  • Note: the argument above assumes $X$ is $T_1$. This is no loss of generality since KC spaces are always $T_1$. – PatrickR May 19 '20 at 22:06
  • @PatrickR: It would be more accurate to say that it silently uses the obvious fact that a KC space is $T_1$: it’s a silent inference from the hypothesis, not an extra assumption. – Brian M. Scott May 19 '20 at 22:10
  • @BrianM.Scott what happen if $x_n = y$ for all $n$ natural number or for $n$ after a $n_0$? – James R. Jul 23 '23 at 18:34
  • @JamesR.: In that case ${y}$ is a compact set that isn’t closed. But to clean up that case a little more neatly I’ll modify the answer slightly to observe that if compact sets are closed, the space is necessarily $T_1$, and we can choose the local bases at $x$ and $y$ so that $y\notin B_x(0)$ and $x\notin B_y(0)$. – Brian M. Scott Jul 23 '23 at 21:46
  • @BrianM.Scott thanks – James R. Jul 24 '23 at 00:08
  • @JamesR.: You’re welcome. – Brian M. Scott Jul 24 '23 at 01:06
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For kicks I looked this up on the pi-Base to see if the result is obtainable there: https://topology.pi-base.org/spaces?q=Second%20Countable%2Bkc%2B~%24T_2%24 As it turns out, first countable and US is sufficient to guarantee Hausdorff. This was shown in this post: If every convergent sequence has a unique limit point in space $X$, then is $X$ Hausdorff?

Steven Clontz
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