Check that a set is locally compact in $\mathbb{R}$

Question

Is the subgroup $S=\{m+n\alpha|\;m,n\in \mathbb{Q}\}$, where $\alpha$ is a fixed irrational number, locally compact in $\mathbb{R}$ ?

Approach: I can see that $S$ is dense in $\mathbb{R}$. But I am not able to proceed after that.

Answer 1

It is not. It suffices to show that $0$ does not have any open neighborhood in $S$ with compact closure.

Take any such neighborhood $U$. Then there is an open interval $I \subseteq \mathbb{R}$ around $0$ such that $I \cap S \subseteq U$. Then $U$ is dense in $I$, but $I \not \subseteq U$ since $U$ is countable, so there is some $a \in I \setminus U$. Now take any sequence $u_n$ of elements of $U$ converging to $a$. It is a sequence of elements of $\mathrm{cl}_S(U)$ without a subsequence convergent to a point in $\mathrm{cl}_S(U)$, hence $\mathrm{cl}_S(U)$ is not compact.