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Given the usual topology $\tau$ on $\Bbb{R}$, define the compact complement topology on $\mathbb{R}$ to be $$\tau'=\{A\subseteq \Bbb{R}:A^C\text{ is compact in }(\Bbb{R},\tau)\} \cup \{\emptyset \}.$$ Are compacts sets in $(\Bbb{R},\tau')$ closed (in $(\mathbb{R} , \tau')$)?

After one hour's struggle I could not prove this. I can see that all compact sets in $(\Bbb{R},\tau)$ are still compact in $(\Bbb{R},\tau')$ and they are trivially closed, but this is certainly not equivalent to the claim to prove.

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  • $\begingroup$ Are you sure this is what you were meant to prove? A similar-sounding statement which is, I believe, true here is that every $\tau$-closed set is $\tau'$-compact. Now if you look at a $\tau$-closed set which is unbounded (and not the whole space), then it will not be $\tau'$-closed: every $\tau'$-open (nonempty) set has a bounded complement. $\endgroup$ – tomasz Sep 1 '15 at 4:12
  • $\begingroup$ @tomasz Thank you for your reply. I am trying to prove the last answer posted here math.stackexchange.com/questions/328725/…, in which it states "the compact sets are precisely the closed sets". I was thinking this is trivial but found out not. If the claim in my question does not hold, then the answer provided in that post should be wrong. $\endgroup$ – Tony Sep 1 '15 at 4:15
  • $\begingroup$ At a glance, I think @AustinMohr made a mistake there. I believe $\tau'$-compact sets are precisely $\tau$-closed sets (sketch of proof: if a set is not $\tau$-closed, you can find an intersection with a compact set which is still not closed; this compact set will be $\tau'$-closed, and the rest should be straightforward) as well as the other way around: $\tau'$-closed sets are precisely the $\tau$-compact sets (this is trivial, though). The same should happen in any locally compact space. Then again, I may be the one who made a mistake, as I did not think much. $\endgroup$ – tomasz Sep 1 '15 at 4:22
  • $\begingroup$ @tomasz I remember feeling a little uneasy about that answer when I posted it. Thanks for looking into it. $\endgroup$ – Austin Mohr Sep 2 '15 at 1:39
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You are correct. Clearly $\Bbb Z$ is not $\tau'$-closed, since it’s not $\tau$-compact, but I claim that it is $\tau'$-compact. Let $\mathscr{U}$ be a $\tau'$-open cover of $\Bbb Z$. Fix $U_0\in\mathscr{U}$; then $\Bbb R\setminus U_0$ is $\tau$-compact and hence bounded. But then $\Bbb Z\setminus U_0$ is bounded and therefore finite, so only finitely many members of $\mathscr{U}$ are needed to cover it.

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