# Characterize the compact subspaces in a topological space.

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?

• $K(A) = \bigcap \{O \in \mathcal{T}: A \subseteq O\}$ of course. – Henno Brandsma Dec 12 '18 at 23:03
• Thank you for the comment. I have re-edited the question. – William Sun Dec 12 '18 at 23:11

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).