I am working on the proof of the following theorem and I have some questions about it:
Let $H$ be a lattice in $\mathbb{R}^n$, $S$ a measurable subset of $\mathbb{R}^n$ which is convex, compact and symmetric with respect to $0$. If $\mu(S) \ge 2^nvol(H)$ then $H$ contains a nonzero point which is in $S$.
If we have $\mu(S) \gt 2^nvol(H)$ then we define a map $\phi: \cfrac{1}{2}S \to \mathbb{R}^n /H$ then it can not be injective and we can find two points $x,y$ so that $\phi(x)=\phi(y)$ which gives us $\phi(x)-\phi(y)=0$ $ \iff \phi(x-y)=0 (How? ) \iff x-y \neq 0 \in H \cap S$. So we should consider the other case when $\mu(S) = 2^nvol(H)$
My questions:
1) If we set $S_\epsilon=S(1+\epsilon)$, then each $S_\epsilon$ contains $S$. $S_\epsilon \cap H$ is nonempty for any $\epsilon > 0$ by first case. Then how can we say that we can construct a sequence converging to a point $x$ which is limit point of $S$?
2)Why $S_\epsilon$ is compact for any $\epsilon$?
3)What do we mean by volume of a lattice $H$? What exactly is $vol(\mathbb{R^n}/H)$ or $vol(H)$ which is given by determinants, not by integrals?
Thank you.
Take a sequence $\epsilon_n\to0$, for instance $\epsilon n=1/n$. For each $n$ take $x_n\in H\cap (1+\epsilon_n)S$. The sequence $(x_n)$ is bounded: it has a convergent subsequence (Bolzano-Weierstrass). As $H$ is discrete, this convergent subsequence is eventually constant. Its limit is in $S$.
As $S$ is compact, $(1+\newcommand{\ep}{\epsilon}\ep)S$ is compact since $x\to (1+\ep)x$ is a homeomorphism.
The volume of a lattice, is really the volume of a fundamental domain of a lattice: that is of $$\{t_1 v_1+\cdots+t_nv_n:0\le t_i<1\}$$ where $v_1,\ldots,v_n$ form a basis of the lattice. As this is a parallelotope there is an easy determinant formula for the volume (form a matrix from the lattice basis, and then take the absolute value of its determinant).