As in this question, let $\tau$ be the standard topology on $\mathbb{R}$ and let $\tau'$ be the compact complement topology on $\mathbb{R}$: $$\tau' = \{\mathbb{R} \setminus K \mid K \text{ is $\tau$-compact}\} \cup \{\varnothing, \mathbb{R}\}$$

One of the commenters makes the claim that the $\tau'$-compact subsets are precisely the $\tau$-closed subsets, but I'm having trouble filling in the details. Here's what I have so far:

Suppose $A$ is $\tau$-closed and $\mathcal{U}$ is a $\tau'$-open cover of $A$. If $\mathbb{R} \in \mathcal{U}$, then $\{\mathbb{R}\}$ is a finite subcover. Otherwise, choose any nonempty $U \in \mathcal{U}$. By the definition of $\tau'$, $\mathbb{R} \setminus U$ is $\tau$-compact, so $A \setminus U=A \cap (\mathbb{R} \setminus U)$ is $\tau$-compact. Since $\tau' \subseteq \tau$, we have $\mathcal{U}$ is a $\tau$-open cover of $A \setminus U$. Let $\mathcal{V} \subseteq \mathcal{\mathcal{U}}$ be a finite subcover of $A \setminus U$. Then $\mathcal{V} \cup \{U\}$ is a finite subcover of $A$, so $A$ is $\tau'$-compact.

I don't have a proof of the converse. Given a $\tau'$-compact set $A$, we want to show that it is $\tau$-closed. The comment suggests supposing it is not $\tau$-closed. Then you could find a ($\tau$- ?) compact set $K$ such that the intersection $A \cap K$ is not ($\tau$- ?) closed (how?). We could then conclude that $A \cap K$ is $\tau'$-compact, but I don't see how that helps us.

Have I made any errors? Can anyone finish the proof or give a counterexample to the converse?


1 Answer 1


Given a $\tau'$-compact set $A$, we want to show that it is $\tau$-closed.

Assume that $A$ is not $\tau$-closed. Pick any point $x\in\overline{A}^\tau\setminus A$ and consider a $\tau'$-open cover $$\mathcal U=\{\Bbb R\setminus [x-1/n,x+1/n]:n\in\mathbb N\}$$ of the set $A$. It has no finite subcover, contradicting the $\tau'$-compactness of $A$.

From the other hand, by the definition of $\tau'$, a subset $A$ of $\Bbb R$ is $\tau'$-closed iff its complement $\Bbb R\setminus A$ is $\tau'$-open iff the complement $\Bbb R\setminus(\Bbb R\setminus A)=A$ of the complement is $\tau$-compact or $\Bbb R$.

  • $\begingroup$ Just to clarify, $\mathcal{U}$ is a $\tau'$-open cover because $x \notin A$ and it doesn't have a finite subcover because $x \in \overline{A}^\tau$ and so $[x-1/n,x+1/n] \cap A$ is always nonempty? $\endgroup$
    – ngenisis
    May 8, 2017 at 21:51
  • 1
    $\begingroup$ @ngenisis The second part is true, the family $\mathcal U$ is a cover of $A$ because $x\not\in A$ and $\mathcal U$ is $\tau'$-open because each set of the form $\{\Bbb R\setminus [x-1/n,x+1/n]:n\in\mathbb N\}$ is $\tau'$-open directly by the definition of topology $\tau'$. $\endgroup$ May 8, 2017 at 21:55

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