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I'm trying to show that $\mathbb Q$ is not locally compact using this definition:

enter image description here

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?

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4 Answers 4

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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.

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  • $\begingroup$ "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 $\mathbb Q$. $\endgroup$
    – user557
    Sep 29, 2018 at 2:43
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    $\begingroup$ 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). $\endgroup$
    – pre-kidney
    Sep 29, 2018 at 2:49
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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:

enter image description here

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.

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  • $\begingroup$ See math.stackexchange.com/questions/3031986/… for more details. $\endgroup$
    – user557
    Dec 10, 2018 at 1:32
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    $\begingroup$ 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. $\endgroup$
    – Cloud JR K
    May 27, 2020 at 19:54
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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.

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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.

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