# Are compact subsets closed in the one-point compactification of $\mathbb{Q}$?

Let $\omega\mathbb{Q}$ denote the one-point compactification of $\mathbb{Q}$, where $\omega$ is a point not in $\mathbb{Q}$, and define a topology $\tau$ on $\omega\mathbb{Q}$ by $$\tau=\{U\subseteq \mathbb{Q}\ |\ U\text{ is open in }\mathbb{Q}\}\cup \{\omega\mathbb{Q}\backslash K\ |\ K\text{ is a compact subset of }\mathbb{Q}\}.$$

Is it true that any compact subset of $\omega\mathbb{Q}$ is then closed?

If $C\subseteq \omega\mathbb{Q}$ is compact and $\omega\notin C$, I think that $C$ will then be compact in $\mathbb{Q}$, so the complement of $C$ is an element of $\tau$, hence $C$ is closed. However, if $\omega\in C$, I'm not sure what to do. I think I'd need to show the complement of $C$ is an open set in $\mathbb{Q}$, which I believe to be true, but I can't prove this explicitely.

• The compactification of Q is Hausdorff. In any Hausdorf space compact sets are closed – Kavi Rama Murthy Nov 17 '17 at 6:13
• @KaviRamaMurthy It is not Hausdorff, as $\mathbb{Q}$ is not locally compact. That's the whole point of the question. – Henno Brandsma Nov 17 '17 at 9:18

Let’s continue. If $\omega\in C$ and $\Bbb Q\setminus C$ is not open in $\Bbb Q$ then $\Bbb Q\cap C$ is not closed in $\Bbb Q$, so there exists a point $x\in \Bbb Q\setminus C$ and a sequence $\{x_n\}$ of points of $\Bbb Q\cap C$ converging to the point $x$. A set $K=\{x\}\cup \{x_n\}$ is a compact subset of $\Bbb Q$. So a family $\{\omega\Bbb Q\setminus K\}\cup\{(\Bbb Q\setminus K)\cup \{x_n\} \}$ is an open cover of the compact $C$, containing no finite subcover, a contradiction.
• Is $\{x_n\}$ open in $C$? – Henno Brandsma Nov 17 '17 at 11:49
• @HennoBrandsma Thanks, corrected. A bug in my head had turned a switch and I thought that $\Bbb Q$ has a discrete subspace topology. :-) – Alex Ravsky Nov 17 '17 at 12:23