If $X$ is not Hausdorff, then does $X$ have a compact subset that is not closed?

Asked Jan 15 '17 at 18:56

Active Jan 15 '17 at 19:12

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This is something of a converse to the theorem which says if $X$ is Hausdorff and $C$ is a compact subset of $X$, then $C$ is closed in $X$. But I do not know if true or not.

Question: If $X$ is T$_1$ but not Hausdorff, then does $X$ necessarily have a compact subset that is not closed?

edited Jan 15 '17 at 19:12

asked Jan 15 '17 at 18:56

Forever Mozart

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no edit needed ;) – Forever Mozart Jan 15 '17 at 19:02
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Relevant: http://math.stackexchange.com/questions/955382/spaces-where-all-compact-subsets-are-closed – J126 Jan 15 '17 at 19:07
@JoeJohnson126 ooh thank you.
in my head I was sort of thinking that $X$ is T$_1$. If single points are not closed then the space is not very interesting to me.
– Forever Mozart Jan 15 '17 at 19:09
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@ForeverMozart, then add that condition to the body of the question. – Mariano Suárez-Álvarez Jan 15 '17 at 19:11
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Related: https://math.stackexchange.com/questions/328725 – Watson Jan 15 '17 at 19:37

If $X$ is not Hausdorff, then does $X$ have a compact subset that is not closed?

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