# A sequence of rationals converging to an irrational point (proving that $\mathbb Q$ is not locally compact)

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?

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 $$r$$.