Here is my attempt to prove that $\mathbb Q$ is not locally compact. (My questions are below the proof.)

Suppose $\mathbb Q$ is locally compact. Then it is locally compact at every point. Let $x\in \mathbb Q$. By the local compactness at $x$, there is a compact subset $K$ of $\mathbb Q$ and a neighborhood $U$ of $x$ in $\mathbb Q$ such that $x\in U\subseteq K\subseteq Q$. (The neighborhood $U$ is of the form $(a,b)\cap \mathbb Q$ for some $a,b\in\mathbb Q$; here $x\in (a,b)\subset \mathbb R$.) Consider a sequence of points of $U$ in the topological space $K$ that converges to an irrational point. Then the irrational limit lies in $\overline K$. But since $K$ is a compact subspace of the Hausdorff space $\mathbb Q$, it is closed in $\mathbb Q$, so $\overline K=K$. So the irrational limit lies in $K$, which is a contradiction to the fact that $K\subseteq Q$.

The only moment that I'm not sure about is the existence of the sequence. Is that part true at all? How to prove it? I do know that given an irrational number in $\mathbb R$, one can take a sequence of shrinking neighborhoods of that irrational number, and that will give a sequence of rational points converging to that irrational number, but in our case the sequence consists of points of $K$, and $K$ is a subset of $\mathbb Q$.

And otherwise does the proof look okay?

  • 2
    $\begingroup$ Your answer has error. Irrational point doesn't belongs to X=K . Hence its is not limit point. $\endgroup$
    – Cloud JR K
    May 27, 2020 at 19:58

1 Answer 1


Yes, your proof is fine. I would just mention why you can suppose without loss of generality that $U$ is the intersection of an open interval $(a,b)$ and $\mathbb{Q}$.

As for the sequence, yes you may indeed find such a sequence. For this, pick any irrational number $r\in (a,b)$. Then for any $n\in \mathbb{N}$, there exists a rational number $x_n$ such that $$ r < x_n < r+\frac{1}{n}. $$ This gives you a sequence of rational numbers converging to $r$. Notice also that for all $n$ sufficiently large, $x_n \in (a,b)$. Therefore, let $(y_n)$ be a subsequence of $(x_n)$ such that $y_n\in (a,b)$ for all $n\in\mathbb{N}$. So we see that $(y_n)$ is in fact a sequence in $(a,b)\cap\mathbb{Q} = U$ and therefore also in $K$ and $y_n$ converges to the irrational number $r$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.