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I'm studying the Topology-Munkres book and there is a Theorem that states that Every compact subspace of a Hausdorff space is closed and I was wondering if there is any example where the "Hausdorff" condition is not needed, I mean could you give an example where the statement is fulfill only by Every compact subspace of a space $X$ is closed

Is just that I'm really bad finding examples and I want to know this, thank you.

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  • $\begingroup$ If you have "Every compact subspace of a space X is closed"... every point is closed (since is compact). So, you find that X is Hausdorff. $\endgroup$ – user99126 Oct 8 '13 at 11:18
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    $\begingroup$ @user99126: You are almost right, except that closed points means T1, not Hausdorff. $\endgroup$ – Moishe Kohan Oct 8 '13 at 11:58
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Let $X$ be an uncountable set with the cocountable topology, i.e., say that $A\subset X$ is closed iff $A=X$ or $A$ is countable. Certainly $X$ is not Hausdorff.

Which subsets $A\subset X$ are compact? Certainly $A$ is compact if it is finite. If $A$ is infinite, let $\{a_1,a_2,\dots\}$ be a countably infinite subset. Then the open sets $X\backslash\{a_i,a_{i+1},\dots\}$ cover $A$ but have no finite subcover, so $A$ is not compact. Since every finite set is closed, every compact subspace of $X$ is closed.

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If every subspace is closed (like in the discrete topology) then the statement is trivially fulfilled.

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    $\begingroup$ Such space is automatically Hausdorff. $\endgroup$ – Moishe Kohan Oct 8 '13 at 12:11

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