Rational numbers are not locally compact

Question

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?

Answer 1

A compact set of Q is a sequence with its limit or
several sequences with their limits.

If [a,b] $\cap$ Q is a compact nhood, pick an
irrational r from [a,b] and show [a,r] $\cap$ Q
and [r,b] $\cap$ Q are not compact.

Answer 2

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.

Answer 3

Suppose $x\in \mathbb Q$ and let $(a,b)\cap\mathbb Q$ be a neighborhood of $x$ in $\mathbb Q$ (so $x\in (a,b)\subset \mathbb R$). Suppose there is a compact subset $K$ of $\mathbb Q$ with $(a,b)\cap\mathbb Q\subset K$. Recall the following theorem:

Apply the above with $X=K$ and $A=(a,b)\cap \mathbb Q$. Choose a sequence of rational points in $A$ converging to an irrational point of $[a,b]\subset \mathbb R$. Then by the lemma, the limit, which is an irrational number, lies in $\overline K=K$ ($K$ is a compact subset of a Hausdorff space, hence is closed). But we have assumed $K\subset \mathbb Q$, a contradiction.

Answer 4

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.