Why is $\Bbb R\setminus\{\frac1n\mid n\in\Bbb N\}$ not locally compact?

Question

I have a question: if I take in $(\mathbb{R},|.|)$ the set $A=\left\{\frac1n, n\in \mathbb{N}\right\}$ and I consider the set $B=\mathbb{R}\setminus A$

I want to prove that $B$ is not locally compact, we have that $0\in B$ is there a compact which contains the open $]-\varepsilon,\varepsilon[,\forall\varepsilon>0$ ?

Thank you

Answer 1

If $B$ were locally compact, then $0\in B$ would have a relative neighborhood in $B$, that is, a set $N\cap B$ where $N$ is a neighborhood of $0$ in $X$, such that $N\cap B$ is compact. Now $\Bbb R$ is Hausdorff, so this $N\cap B$ would have to be closed. Can you show that no such set is closed?