How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly? That is, how to construct a cover of an arbitrary neighborhood (e.g. $[0, 1] \cap \mathbb{Q}$) that does not have a finite subcover?
[Math] How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly
compactnessgeneral-topology
Best Answer
Let $X:= \mathbb Q \cap [0,1]$. I'll show that if $U \subset X$ then $U$ doesn't have compact closure. Since $X$ is hausdorff this shows that it's not locally compact. It suffices to show that for any epsilon an open ball in X doesn't have compact closure. Given $x \in X$ and some $\epsilon>0$ pick an irrational $y \in B(x,\epsilon)$. Set $a=x-\epsilon$ and $b=x+\epsilon$, then the following is an open cover of $B(x,\epsilon)$ with no finite subcover:
$$\mathcal U := \left\{ \left( \left(a,y-\frac{1}{n}\right) \cup \left(y+\frac{1}{n},b\right) \right) \cap \mathbb Q \mid n \in \mathbb N\right\}.$$
To see this note that $\mathcal U$ is a nested covering and that for any $n$ it doesn't cover $B(x,\epsilon)$.