Minkowski's Theorem

Theorem
Let $L$ be a lattice in $\mathbb{R}^{n}$.

Let $d$ be the determinant of $L$.

Let $\mu$ be a measure on $\mathbb{R}^{n}$

Let $S$ be a convex subset of $\mathbb{R}^{n}$ that is symmetric about the origin, i.e. such that:
 * $\forall p \in S : -p \in S$

Then if the volume of $S$ is greater than $2^{n}d$, then $S$ contains a non-zero point of $L$.

Proof
Let $D$ be any fundamental parallelepiped. Then obviously


 * $\displaystyle \mathbb{R}^{n}=\coprod\limits_{\vec{x} \mathop \in L} \left( D+\vec{x} \right)$

Thus


 * $\displaystyle \frac{1}{2} S = \coprod\limits_{\vec{x} \mathop \in L} \left( \frac{1}{2} S \cap \left( D+\vec{x} \right) \right)$

Clearly,


 * $\frac{1}{2} S \cap \left(D+\vec{x} \right) = \left( \left( \frac{1}{2} S -\vec{x} \right) \cap D\right) - \vec{x}$

However,translation is invariant under a measure, so


 * $\mu\left(\frac{1}{2}S \cap \left( D + \vec{x} \right) \right) = \mu\left( \left( \left(\frac{1}{2} S - \vec{x} \right) \cap D \right) - \vec{x} \right) = \mu\left( \left( \frac{1}{2} S - \vec{x} \right) \cap D \right)$

Assume $\left\lbrace \frac{1}{2} S - \vec{x} \vert \vec{x} \in L \right\rbrace$ is pairwise disjoint (i.e. $\forall \vec{x},\vec{y} \in L$, $\left( \frac{1}{2} S - \vec{x} \right) \cap \left( \frac{1}{2} S - \vec{y} \right) \neq \emptyset \iff \vec{x} \neq \vec{y}$). Then

Which is a contradiction.

So $\left\lbrace \frac{1}{2} S - \vec{x} \vert \vec{x} \in L \right\rbrace$ is not pairwise disjoint, which means


 * $\exists \vec{x}$,$\vec{y} \in L: \vec{x} \neq \vec{y}$, $\left( \frac{1}{2} S - \vec{x} \right) \cap \left( \frac{1}{2} S - \vec{y} \right) \neq \emptyset$

Thus


 * $\exists \vec{p}_{1}$, $\vec{p}_{2} \in L:\vec{p}_{1} \neq \vec{p}_{2}$, $\frac{1}{2} \vec{p}_{1} - \vec{x} = \frac{1}{2} \vec{p}_{2} - \vec{y} \implies \frac{1}{2} \left( \vec{p}_{1} - \vec{p}_{2} \right) = \vec{x} - \vec{y} \in L$

Since $\frac{1}{2} \left( \vec{p}_{1} - \vec{p}_{2} \right)$ is the midpoint of the line between $\vec{p}_{1}$ and $\vec{p}_{2}$ and $S$ is convex, we have that $\frac{1}{2} \left( \vec{p}_{1} - \vec{p}_{2} \right) \in S$