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It is known that If $X$ is a Hausdorff space then every compact subspace of $X$ is closed. Hence closure of compact subspace of $X$ is also compact.

My question: is there any a $T_1$ space $X$ such that if $A$ is a compact subspace of $X$ then closure of $A$ is not compact?

Jaxa
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1 Answers1

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If $X$ is a $T_1$ space, then every finite subset of $X$ is both compact and closed. However, it is still possible for $X$ to have an infinite compact subset whose closure is not compact; the set $X\setminus P$ in the space described in this answer is an example.

Community
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Brian M. Scott
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  • M.Scott: Why every finite subset of $X$ is closed? – Jaxa Mar 06 '15 at 08:03
  • @Jaxa: If $X$ is $T_1$, then ${x}$ is closed for each $x\in X$, and every union of finitely many closed sets is closed. – Brian M. Scott Mar 06 '15 at 08:06
  • M.Scott: I do not understand why $X\setminus P$ in your example is compact. What is closure of $X\setminus P$? – Jaxa Mar 06 '15 at 08:13
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    @Jaxa: The closure of$X\setminus P$ is the whole space. $X\setminus P$ is the compact ordinal space $\omega+1$, which is homeomorphic to ${0}\cup\left{\frac1n:n\in\Bbb Z^+\right}$ with its usual topology. – Brian M. Scott Mar 06 '15 at 08:18