TITLE: Rational numbers are not locally compact
QUESTION [8 upvotes]: I'm trying to show that $\mathbb Q$ is not locally compact using this definition:

So I need to show that there is some point $x\in \mathbb Q$ such that no matter what neighborhood of $x$ in $\mathbb Q$ I take, no compact subset of $\mathbb Q$ can contain it.
Any neighborhood of $x$ in $\mathbb Q$ is of the form $(a,b)\cap \mathbb Q$ where $x\in (a,b)$. But I think my problem is that I don't understand/feel how compact sets in $\mathbb Q$ look like (except finite sets). If there is a compact subset of $\mathbb Q$ containing $(a,b)\cap \mathbb Q$, what does it contradict to?

REPLY [2 votes]: There is also a short indirect proof of this fact using Baire Category Theorem. 
If $\mathbb{Q}$ were locally compact, since it is also a Hausdorff space, it would be a Baire space by the Baire Category Theorem. But if we enumerate the rationals as $\{q_n\}_{n \in \mathbb{N}}$, then each singleton $\{q_n\}$ is closed and has empty interior in $\mathbb{Q}$, so their union would have empty interior in $\mathbb{Q}$ too, since we would be in a Baire space. This is a contradiction, since the union of all the singletons is $\mathbb{Q}$, which does not have empty interior in itself.