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?
Best Answer
Claim. For all $\epsilon>0$, the set $[-\epsilon,\epsilon]\cap\mathbb Q$ is non-compact.
Proof. Fix an irrational number $\alpha\in (-\epsilon,\epsilon)$ and let $\{x_n\}_{n\in\mathbb N}$ be a sequence of rational numbers in $[-\epsilon,\epsilon]$ converging (in $\mathbb R)$ to $\alpha$. Then $\{x_n\}$ is an infinite sequence with no $\mathbb Q$-convergent subsequence, and therefore $[-\epsilon,\epsilon]\cap\mathbb Q$ is non-compact. $\square$
To show that $0$ has no neighborhood contained in a compact set, suppose for contradiction that there was some $0\in U\subseteq K\subseteq \mathbb Q$ with $U$ open and $K$ compact. Then for some $\epsilon>0$ we have $(-\epsilon,\epsilon)\cap\mathbb Q\subseteq U$. Consequently, the set $[-\epsilon,\epsilon]\cap \mathbb Q$ is a closed subset of the compact set $K$, and must therefore be compact, contradicting the claim. This proves the result.