I am trying to demonstrate that the set $T \subset \mathbb{R}^N$ of translations leaving some discrete set $L \subset \mathbb{R}^N$ invariant (i.e. $T = \{ t \in \mathbb{R}^N : l \in L \to l+t\in L \})$ necessarily possesses some "basis" $t_1,\ldots,t_n, n\leq N $ that "spans" $T$, (i.e. $\forall t \in T , \exists m_1,\ldots,m_n \in \mathbb{Z} : t = \sum_i^n m_i t_i$). I am not sure if this theorem is even true.
My thinking is to set $t_1$ equal to the smallest element in $T$. If $t_1$ spans $T$ then we are done. Otherwise set $t_2$ equal to the smallest element of $T$ outside the span of $t_1$. If $t_1$ and $t_2$ together span $T$, we are done. Otherwise, set $t_3=\ldots$
We continue until we get to the $N^{th}$ basis vector $t_N$. From here we must show that $T$ is the span of $t_1,\ldots,t_N$. For brevity we denote this span by $S$.
We attempt proof by contradiction. Letting $t$ be some vector outside $S$, we define $t'$ to be the shortest vector from $t$ to any point in $S$. $t'$ then lies in the Wigner-Seitz cell of $S$.
Here I run out of steam. For $N<4$ and a cubic $S$ you can show that $t'$ is necessarily shorter than $t_N$ and produce the desired contradiction. This proof fails for $N>4$.