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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.2011-07-17
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    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 $\frac{1}{k+1}$.2011-07-17
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    Ah, I misread $0$ for $-1$. Note that subgroups are *homogeneous*: The map $g \mapsto g+h$ is a homeomorphism of $G$ onto itself.2011-07-17
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    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).2011-07-17
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    Yes you can do it this way, that's even better than the argument I had in mind initially.2011-07-17
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    Mod note: cleaned up some comments made obsolete by subsequence edits to the question.2011-11-10

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