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It is trivial to see that:

If $X$ is Hausdorff, then every compact subset of $X$ is closed.

I am asking under what condition does the converse hold, i.e. when does

If every compact subset of $X$ is closed, then $X$ is Hausdorff

hold.

EDIT:

Sorry for whoever just read the question, I corrected the title and the whole question. It should be clear now.

Alex Ravsky
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king_S
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    The question in the title is not asking when the converse holds. The converse is if every compact subset of $X$ is closed, then $X$ is Hausdorff. Did you actually want to ask when this is true? – Brian M. Scott Apr 20 '13 at 23:37
  • @BrianM.Scott Sorry about the confusion, I already edited the question – king_S Apr 20 '13 at 23:43
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    But what you say is not correct: the converse of If $X$ is Hausdorff, then every compact subset of $X$ is closed is If every compact subset of $X$ is closed, then $X$ is Hausdorff, and that’s not what you’re asking about. If you really mean the question in the title and the last sentence of the body, then you should remove the incorrect remark about the converse. – Brian M. Scott Apr 20 '13 at 23:46
  • @BrianM.Scott Totally my bad. I just corrected my title and rewrote the body paragraph. Very sorry for the confusion caused. – king_S Apr 21 '13 at 00:08
  • That’s okay; I just wanted to make sure that we were trying to answer the right question! – Brian M. Scott Apr 21 '13 at 00:09
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    Related: If every compact set is closed, then is the space Hausdorff?. – Martin Sleziak Apr 21 '13 at 06:53
  • @MartinSleziak I am aware of the thread you gave, and I already looked that up since that was answered by Brian too. I am not looking for counterexample, but the conditions which it holds. – king_S Apr 21 '13 at 13:25

1 Answers1

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As for me, this questions sounds unnaturally, but seems to be not so trivial. It seems the following. Let $X$ be a space such that every compact subset of $X$ is closed. Then $X$ is $T_1$, because each one-point subset is compact.

Proposition 1. If $X^2$ is a sequential space, then $X$ is Hausdorff.

Proof. Suppose the opposite. Then the diagonal $\Delta=\{(x,x)\in X^2:x\in X\}$ is not closed in $X^2$. Since $X^2$ is a sequential space then there exists a sequence $\{(x_n,x_n)\}$ of points of $\Delta$, converging to a point $(x,y)\in X^2\backslash\Delta$. Without loss of generality we may suppose that $x_n\not=y$ for each $n$. The sequence $\{x_n\}$ converges to $x$. Therefore a set $X_0=\{x\}\cup\{x_n\}$ is compact. Hence $X\backslash X_0$ is an open neighborhood of $y$. Since the sequence $\{x_n\}$ converges to $y$, there exists a number $n$ such that $x_n\in X\backslash X_0,$ a contradiction. $\square$

Example 1. There exists a non-Hausdorff space $X$ such that each compact subset of $X$ is closed.

Put $X=\omega\cup\{\alpha\}\cup\{\alpha’\}$, where $\alpha\not=\alpha’$ and $\{\alpha, \alpha’\}\cap\omega=\varnothing$. Let $\mathcal F$ be a free ultrafilter on the set $\omega$. Define a topology $\tau$ on the $X$ as follows. A subset $U$ of $X$ belongs to $\tau$ iff $U\subset\omega$ or $U\cap\omega\in\mathcal F$. Since $\mathcal F$ is an ultrafilter, we can easily check that each compact subset of $X$ is finite and, hence, closed in $T_1$-space $X$.

The following questions are already answered by Martin. For the sake of future advances in this direction I formulate the following questions.

Exists a non-Hausdorff space $X$ such that each compact subset of $X$ is closed, provided:

  1. $X$ is compact?

  2. $X$ is locally compact?

  3. $X$ is a "$k$-space"?

  4. $X$ is a "$k_\omega$-space"?

  5. $X$ is a Fréchet-Urysohn space?

  6. $X$ is a sequential space?

Alex Ravsky
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    Another simple example of a space in which compact sets are closed is the cocountable topology on an uncountable space (compact sets are finite). I can answer three of your questions: 1. and 5. Yes: the one-point compactification of $\mathbb{Q}$. 2. Probably depends on the notion of local compactness in non-Hausdorff setting. The answer is no if every point has a base of compact (hence closed) neighborhoods because then it is regular and $T_1$. – Martin Apr 21 '13 at 09:45