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.
 A: 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.  
