# Why the discrete subgroup of $R^n$ has finite balls?

Consider a discrete subgroup $G$ of $R^n$. Then let we have an open ball $B\subset G$ (topology in $G$ is induced from $R^n$). Why is $|B|<\infty$?

For example I can take $A=\{\tfrac{1}{1},\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4},\cdots, -1\}$, which is discrete set. Naturally I can find an open ball in A which is not finite, however $A$ is not a group. So I guess being group is necessary.

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Hint: infinite bounded subsets of $\mathbb{R}^n$ have an accumulation point. – t.b. Jul 17 '11 at 21:01
It is discrete: please note it does not contain $0$. And for every $\frac{1}{k}$ we can find small enough ball which does not touch $\frac{1}{k-1}$ and $\frac{1}{k+1}$. – Yosif Jul 17 '11 at 21:09
Ah, I misread $0$ for $-1$. Note that subgroups are homogeneous: The map $g \mapsto g+h$ is a homeomorphism of $G$ onto itself. – t.b. Jul 17 '11 at 21:11
So if we can find any point which is has an open ball that does not touch any other point, therefore such ball with its center in any other point does not touch other points. And therefore it has to be finite. Is that right? (In other words in discrete subgroups the radius of separating ball has to be good for all points). – Yosif Jul 17 '11 at 21:20
Yes you can do it this way, that's even better than the argument I had in mind initially. – t.b. Jul 17 '11 at 21:24

Here are two key facts:

1) We are working inside $\mathbb{R}^n$, in which a subset is compact iff it is closed and bounded (Heine-Borel).

2) A discrete subgroup $G$ of any Hausdorff topological group is closed (a proof of this was given recently on this site). More generally, any locally compact subgroup of a Hausdorff group is closed.

So let $G$ be a discrete subgroup of $\mathbb{R}^n$. We want to prove that for every ball $B$ of finite radius, $B \cap G$ is finite. We easily reduce to the following special case: for all $r \geq 0$, the intersection of $G$ with the closed ball $B_r$ of radius $r$ centered at $0$ is finite.

Now let us apply the above facts: since $B_r$ is closed and bounded in $\mathbb{R}^n$, it is compact. Since $G$ is closed in $\mathbb{R}^n$, $G \cap B_r$ is closed in $B_r$, so $G \cap B_r$ is compact. Moreover, $G \cap B_r$ is a subspace of the discrete space $G$, so it is discrete. But a topological space which is both compact and discrete is necessarily finite.

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You probably mean this thread in 2). – t.b. Jul 27 '11 at 23:43
@Theo: of course, you're right. The next time this happens, if you just go ahead and edit the link into my answer I'll take it as a big favor. :) – Pete L. Clark Jul 28 '11 at 2:34
Okay, I'll do that :) – t.b. Jul 28 '11 at 2:36