Minkowski's Theorem

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

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

Let $\mu$ be a translation invariant measure on $\R^n$

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

Let the volume of $S$ be greater than $2^n d$.

Then $S$ contains a non-zero point of $L$.

Proof
Let $D$ be any fundamental parallelepiped.

Then by definition:


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

where:
 * $A + \vec x := \left\{{\vec a + \vec x : \vec a \in A}\right\}$

By Intersection Distributes over Union:

Since $\frac 1 2 S \cap \R^n = \frac 1 2 S$ (an element in $A \subset B$ iff it's in $A$ and $B$) and the


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

where:
 * $\frac 1 2 S := \left\{{\frac 1 2 \vec s: \vec s \in S}\right\}$

Consider the intersection of $D + \vec{x}$ and $\frac{1}{2} S$. This is obtained by adding $\vec{x}$ to every point in $D$ then taking those points that are also in $S$. However, this is the same as subtracting $\vec{x}$ from all elements of $\frac{1}{2} S$, taking the elements that are also in $D$ and adding $\vec{x}$ to restore them to their original position.


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

Since, by hypothesis, $\mu$ is translation invariant:

Aiming for a contradiction, suppose the sets $\displaystyle \left\{{\frac 1 2 S - \vec x: \vec x \in L}\right\}$ are pairwise disjoint.

That is:
 * $\displaystyle \forall \vec x, \vec y \in L: \left({\frac 1 2 S - \vec x}\right) \cap \left({\frac 1 2 S - \vec y}\right) \ne \varnothing \iff \vec x \ne \vec y$)

Then:

which is a contradiction.

So $\displaystyle \left\{{\frac 1 2 S - \vec x: \vec x \in L}\right\}$ are not pairwise disjoint.

This means:


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

Therefore there exist $\vec {p_1}, \vec {p_2} \in L$ such that $\vec {p_1} \ne \vec {p_2}$ and:
 * $\displaystyle \frac 1 2 \vec {p_1} - \vec x = \frac 1 2 \vec {p_2} - \vec y$

and therefore:
 * $\displaystyle \frac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) = \vec x - \vec y \in L$

Since $S$ is convex, we have that:
 * $\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \in S$

As $\vec {p_1} \ne \vec {p_2}$ by definition:
 * $\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \ne \vec 0$

Hence the result.