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It’s known that every closed subset of a compact topological space is also compact. However, it’s not always true that compact subspaces are closed(by taking the cofinite topology on $\mathbb{Z}$ the set of positive integers $\mathbb{Z}+$ is compact but is not closed). The following is my attempt to replace “closed” with a weaker condition so that the converse actually holds.

Let $(X, \mathcal{T})$ be a compact space. The idea for the following arguments is that in the proof of “closed subset $A$ of compact space is compact”, we show the complement $X-A$ can be removed so that the rest of the open sets also form a cover:

For a subset $A$ of $X$, define $$K(A)=\bigcap\{U\in \mathcal{T}: U\subset A\}.$$ If there exists a closed set $C$ in $X$ such that $A\subset C\subset K(A)$, we say $A$ is $K$-closed. Apparently every closed set is also $K$-closed.

Now if $A$ is $K$-closed, then by taking $X-C$ together with an open cover of $A$, we have an open cover, thus a finite subcover $\{U_i\}$of $X$. It can then be shown that if $X-C=U_i$ for some $U_i$, then it is removable so that the rest of the finite cover also covers $A$.

Now I wonder if the converse holds:

If $A\subset X$ is compact, is it true that $A$ must be $K$-closed?

Is there an alternate way to modify the result so that the converse holds?

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$\mathbb{Z}+$ is compact in $\mathbb{Z}$ with the cofinite topology, but $K(\mathbb{Z}+)=\mathbb{Z}+$ so it’s not $K$-closed. My attempt does not solve the problem. Does anyone have a better idea?

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  • $\begingroup$ $K(A) = \bigcap \{O \in \mathcal{T}: A \subseteq O\}$ of course. $\endgroup$ – Henno Brandsma Dec 12 '18 at 23:03
  • $\begingroup$ Thank you for the comment. I have re-edited the question. $\endgroup$ – William Sun Dec 12 '18 at 23:11
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No. For instance, let $X=\{0,1\}$ with $\{0\}$ open but $\{1\}$ not open. Then $A=\{0\}$ is compact but $K(A)=A=\{0\}$ is not closed so $A$ is not $K$-closed.

Ultimately, the issue is that the "reason" $A$ is compact could be totally unrelated to the "reason" $X$ is compact. In particular, $X$ might have a point $x$ (like $1$ in the example above) whose only neighborhood is $X$ itself, which automatically guarantees that $X$ is compact. If $x\not\in A$, there's never going to be any nice way to relate open covers of $A$ to open covers of $X$ to learn anything interesting using the compactness of $A$, since open covers of $X$ are all trivial ($X$ itself must always be one of the sets).

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