# Rational numbers are not locally compact

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?

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.

• "Consequently, the set $[-\epsilon,\epsilon]\cap \mathbb Q$ is a closed subset of $K$". Does the argument below prove this? $[-\epsilon,\epsilon]\cap \mathbb Q$ is closed in $\mathbb Q$ if it equals $K\cap V$ where $V$ is closed in $\mathbb Q$. Let $V=[-\epsilon,\epsilon]\cap \mathbb Q$. Then $K\cap V=V$, and $V$ is closed in $\mathbb Q$ because $[-\epsilon,\epsilon]$ is closed in $\mathbb Q$. Sep 29, 2018 at 2:43
• Rather than directly answer your question, I will point out that if you are unsure about this sort of subspace topology argument then I would recommend working out what it means in terms of more familiar definitions of being closed (e.g. the definition involving sequences) which will greatly clarify the situation (instead of a just a "yes" or "no" answer from me). Sep 29, 2018 at 2:49

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.

• See math.stackexchange.com/questions/3031986/… for more details. Dec 10, 2018 at 1:32
• I don't find your argument convincing. Limit point of subset depends on the Topological space. Here you take X as K. Therefore that irrational point doesn't belongs to X . So it's not making sense to talk about it being a limit point of A in X. But l think since compact space are invariant under subspace topology. You can modify your argument as X=R. May 27, 2020 at 19:54

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.

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.