Discrete subgroups of $(\mathbb{R},+)$ I  have a simple question on discrete subgroups. 
We say that a subset $C$ of $\mathbb{R}$ is discrete if every point of $C$ is isolated in the topology inherited from $\mathbb{R}$.
Is it true that every discrete subgroup of $(\mathbb{R},+)$ is of the form $r\mathbb{Z}$ with $r \in \mathbb{R}$? And if so, how can i prove it?
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My attempt: I don't really know how to prove it, i mean, i cannot think of an example of a discrete subset of $\mathbb{R}$ which is not, somehow, related to $\mathbb{Z}$...
Thanks in advance!
 A: Yes, it is true. If $G$ is such a group, then let $r=\inf\bigl(G\cap(0,\infty)\bigr)$. Then $r\in G$ (since $G$ is discrete) and $G=r\mathbb Z$, since:

*

*if $m\in\Bbb Z$, $mr\in G$, which proves that $r\Bbb Z\subset G$;

*is $g\in G\setminus r\Bbb Z$, then there is some $m\in\Bbb Z$ such that $0<mr-g<r$, but then $mr-g\in G$, which is impossible, since $r=\inf\bigl(G\cap(0,\infty)\bigr)$.

A: Even though this is an old question, I really like the following solution:
Let $G\subset \mathbb{R}$ denote a discrete subgroup. Then by classical covering space theory, $\mathbb{R}\to\mathbb{R}/G$ is a covering and $\mathbb{R}/G$ is a connected $1$-manifold. Furthermore by the classification of $1$-manifolds, it has to be compact (since the only connected non-compact $1$-manifold $\mathbb{R}$ is simply-connected), hence again by the classification it is $S^1$. By elementary covering space theory $\pi_1(S^1)\cong \mathbb{Z}\cong G$.
Since a map from $\mathbb{Z}$ to $\mathbb{R}$ is specified by the value of $1$, it has the required form.
