$L\subset \mathbb{R}^{n}$ lattice of max. rank
$\Leftrightarrow \mathbb{R}^{n}/L$ compact in quotient topology
$\Leftrightarrow \exists$ bounded subset $B\subset\mathbb{R}^{n}$ s.t. $L+B = \mathbb{R}^{n}$
I just managed to do the first implication:
$L = \mathbb{Z} \omega_{1} + ... + \mathbb{Z} \omega_{n}$.
$P:=\{\lambda_{1}\omega_{1} + ...+ \lambda_{n}\omega_{n} | 0 \leq \lambda_{i} \leq 1\}.$
Then there exists an equivalence relation on $\mathbb{R}^{n}$, s.t. two numbers $x,x'$ are equivalent modulo $L$ $\Leftrightarrow$ $x-x'\in L$. $\pi:\mathbb{R}^{n}\rightarrow\mathbb{R}^n/L$ with $x\mapsto x+L$. We can find for all $x\in\mathbb{R}^{n}$ a $x'\in P$ s.t. $x-x'\in L$.
$\Rightarrow \pi(x)=\pi(x')$ $\Rightarrow \pi\big|_{P} \rightarrow \mathbb{R}^n$/L bijective.
Hence $\pi(P)=\mathbb{R}^n/L$ and since $P$ is compact, so is $\mathbb{R}^n/L$
For the next implication I have no idea yet.
I guess I have to use that $\mathbb{R}^n = \bigcup\limits_{i=1}^n U_{i}$, but I don't know how to get to $L+B = \mathbb{R}^{n}$. Maybe I must also use that $L$ is discrete? But do I only know that from the lattice, or does my proof for the first implication show that as well?
Hope someone feels like helping :)
best, Kathrin