Minkowski's Theorem

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

Let $d$ be the covolume 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:


 * $\ds \R^n = \coprod \limits_{\vec x \mathop \in L} \paren {D + \vec x}$

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

By Intersection with Subset is Subset:
 * $\dfrac 1 2 S \cap \R^n = \dfrac 1 2 S \iff \dfrac 1 2 S \subseteq \R^n$

Hence by Intersection Distributes over Union:


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

where:
 * $\dfrac 1 2 S := \set {\dfrac 1 2 \vec s: \vec s \in S}$

Consider the intersection of $D + \vec x$ and $\dfrac 1 2 S$.

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 $\dfrac 1 2 S$, taking the elements that are also in $D$ and adding $\vec x$ to restore them to their original position.

Thus:
 * $\ds \frac 1 2 S \cap \paren {D + \vec x} = \paren {\paren {\frac 1 2 S - \vec x} \cap D} + \vec x$

Since,, $\mu$ is translation invariant:

the sets $\ds \set {\frac 1 2 S - \vec x: \vec x \in L}$ are pairwise disjoint.

That is:
 * $\ds \forall \vec x, \vec y \in L: \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O \iff \vec x \ne \vec y$

Then:

which is a contradiction.

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

This means:


 * $\ds \exists \vec x, \vec y \in L: \vec x \ne \vec y, \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O$

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

and therefore:
 * $\ds \frac 1 2 \paren {\vec {p_1} - \vec {p_2} } = \vec x - \vec y \in L$

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

As $\vec {p_1} \ne \vec {p_2}$ by definition:
 * $\dfrac 1 2 \paren {\vec {p_1} - \vec {p_2} } \ne \vec 0$

Hence the result.