5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The compactification of Q is Hausdorff. In any Hausdorf space compact sets are closed $\endgroup$ – Kavi Rama Murthy Nov 17 '17 at 6:13
  • 1
    $\begingroup$ @KaviRamaMurthy It is not Hausdorff, as $\mathbb{Q}$ is not locally compact. That's the whole point of the question. $\endgroup$ – Henno Brandsma Nov 17 '17 at 9:18
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Is $\{x_n\}$ open in $C$? $\endgroup$ – Henno Brandsma Nov 17 '17 at 11:49
  • $\begingroup$ @HennoBrandsma Thanks, corrected. A bug in my head had turned a switch and I thought that $\Bbb Q$ has a discrete subspace topology. :-) $\endgroup$ – Alex Ravsky Nov 17 '17 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.