A lattice subgroup in $\Bbb R^n$ is discrete Let $v_1,\dots,v_n \in \Bbb R^n$ be linearly independent vectors and let $L = \{\sum_{i=1}^n m_i v_i \mid m_i \in \Bbb Z\}$, which is a free abelian group. I want to prove that it is discrete, so I would like to show that
$$B(\vec 0,r) \cap L = \{\overrightarrow 0\}$$
where $r = \min\limits_{1≤i≤n} \|v_i\|$ (with the usual euclidean norm and balls).
Assume that $x = \sum_{i=1}^n m_i v_i \in L$ is non-zero and has norm $\|x\|<r$. If we assume that the $v_i$'s are orthonormal, then $\|x\|^2=\sum_{i=1}^n |m_i|^2<r=1 \implies m_i=0$ for all $i$ (since we have integers $m_i$).
How do I manage the other cases, where the $v_i$'s are not necessarily orthonormal?
Thank you!
 A: $\mathbb R^n$ is a finite dimensional vector space. Hence all norms are equivalent in $\mathbb R^n$.
For a vector $\sum_{i=1}^n \alpha_i v_i$,
$$\left\Vert \sum_{i=1}^n \alpha_i v_i \right\Vert_\infty = \sup\limits_i \vert \alpha_i \vert$$ is a norm. For $\sum_{i=1}^n m_i v_i \in L \setminus \{0\}$, you have
$$\left\Vert \sum_{i=1}^n m_i v_i \right\Vert_\infty \ge 1$$
And as $\Vert \cdot \Vert_\infty$ is equivalent to $\Vert \cdot \Vert$, it exists $c > 0$ such that 
$$\Vert x \Vert \ge c \Vert x \Vert_\infty$$ for all $x \in \mathbb R^n \setminus \{0\}$. Hence for all $\sum_{i=1}^n m_i v_i \in L \setminus \{0\}$
$$\left\Vert \sum_{i=1}^n m_i v_i \right\Vert \ge c\left\Vert \sum_{i=1}^n m_i v_i \right\Vert_\infty \ge c >0$$ which allows to conclude.
A: Suppose by contradiction that for every $k\in\mathbb N$ there are non zero vectors $x^k\in\mathbb{R}^n$ and $m^k\in\mathbb{N}^n$ such that
$$x^k=\sum_{i=1}^n m_i^ke_i$$
and $|x^k|<\frac1k$. Then
$$\left|\frac{x^k}{|m^k|}\right|\leq|x^k|<\frac1k.$$
Now by compactness of the unit Euclidean sphere in $\mathbb{R}^n$ there is an accumulation point $m$ of $\frac{m^k}{|m^k|}$ and we obtain in the limit that
$$\sum_{i=1}^n m_ie_i=0$$
contradicting linear independence of the $e_i$'s.
PS The way you are trying to prove it is too strong (think of two very close vectors; then their difference is very small in norm). 
