Discrete subgroups of $(\mathbb{R},+)$

Question

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!

José Carlos Santos · Accepted Answer · 2020-10-30T09:13:03.313

5

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 $r=\inf\bigl(G\cap(0,\infty)\bigr)$.

edited Oct 30 '20 at 09:13

answered Apr 06 '18 at 09:38

José Carlos Santos

427,504

4

It may helpful to expand a bit on why $G = r\mathbb Z$. The inclusion $r \mathbb Z \subset G$ follows almost directly from $r\in G$. For the reverse, suppose that $x\in G\setminus r\mathbb Z$. Then there is some $n$ so that $0 < nr - x < r$. But $nr-x \in G$, contradicting the minimality of $r$. – Jalex Stark Apr 06 '18 at 09:45
2

@JalexStark In fact. But I would write that kind of stuff only if the OP asked me to do it. – José Carlos Santos Apr 06 '18 at 09:46
One small question, you mean $r = \inf (G \cap (0,+\infty) )$, right? – El.Gon.Zalo Apr 06 '18 at 09:52
@El.Gon.Zalo Yes. I've edited my answer. I hope that everything is clear now. – José Carlos Santos Apr 06 '18 at 09:53
@JoséCarlosSantos That makes sense. I'll leave my comment up for any readers who want the details. – Jalex Stark Apr 06 '18 at 10:06
Thank you so much! – El.Gon.Zalo Apr 06 '18 at 12:09

score 5 · Answer 2 · answered Oct 30 '20 at 08:26

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.

This is a really beautiful application of convering theory and classification of dimension 1 manifolds. — Didier, Oct 30 '20 at 13:51

Discrete subgroups of $(\mathbb{R},+)$

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