Minkowski's Theorem

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

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

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
Consider $f:S \rightarrow \mathbb{R}^{n},(x_{1},x_{2},\dotsc,x_{n})\mapsto (x_{1}\pmod{2},x_{2}\pmod{2},\dotsc,x_{n} \pmod{2})$. Intuitively, this function cuts $\mathbb{R}^{n}$ into hypercubes with a side length of 2 and then squishes them together into one hypercube with a side length of 2. Let

$$\operatorname{Vol}:\mathscr{P}(\mathbb{R}^{n}) \rightarrow \mathbb{R},X \mapsto \underset{X}{\int \int \dotsb \int} \mathrm{d}x_{1}\mathrm{d}x_{2} \dotsb \mathrm{d}x_{n}$$

be the volume function (where $\mathscr{P}(X)$ is the powerset of a $X$). This definition gives us

$$ \operatorname{Vol}(A \cup B) = \underset{A \cup B}{\int \int \dotsb \int} \mathrm{d}x_{1}\mathrm{d}x_{2} \dotsb \mathrm{d}x_{n} = \underset{A}{\int \int \dotsb \int} \mathrm{d}x_{1}\mathrm{d}x_{2} \dotsb \mathrm{d}x_{n} + \underset{B}{\int \int \dotsb \int} \mathrm{d}x_{1}\mathrm{d}x_{2} \dotsb \mathrm{d}x_{n} = \operatorname{Vol}(A)+\operatorname{Vol}(B)$$

This will be useful later. Moving on, since $f(S)=\operatorname{im}(f) \subseteq \left[ 0,2 \right]^{n}$ we have $ \operatorname{Vol}(f(S)) \leqslant 2^{n}$. Assume $f$ is injective. Intuitively, this means that the hypercubes were squished in a way so that the pieces of $S$ cut out by the hypercubes don't overlap. If $f$ were preserved the volume of $S$, then $f(S)=S > 2^{n}$ which contradicts the fact that the $f(S) \subseteq \left[ 0,2 \right]^{n}$. Notice that the restriction of $f$ to $\left[ 2a_{1}, 2b_{1} \right]\times\left[ 2a_{2}, 2b_{2} \right] \times \dotsb \times \left[ 2a_{n}, 2b_{n} \right]$ is bijective for any sequences $(a_{i}),(b_{i}) \in \mathbb{Z}^{\mathbb{N}}$. This means that $f$ is locally area preserving, and thus preserves the area of all the the pieces cut out of $S$ by the hypercubes. Since $f$ is injective, we have that the pieces of $S$ cut out by the hypercubes, don't overlap which means that $f$ preserves the sum of the areas of those pieces. Clearly, the pieces partition $S$ which means that the sum of their volumes is the sum of $S$. Thus

\begin{align*} \operatorname{Vol}(f(S)) &=\operatorname{Vol} (\bigcup_{(a_{i}),(b_{i}) \in \mathbb{Z}^{\mathbb{N}}} f(S) \cap \left[ 2a_{1}, 2b_{1} \right]\times\left[ 2a_{2}, 2b_{2} \right] \times \dotsb \times \left[ 2a_{n}, 2b_{n} \right]) \\ &= \sum\limits_{(a_{i}),(b_{i}) \in \mathbb{Z}^{\mathbb{N}}} \operatorname{Vol}( f(S) \cap \left[ 2a_{1}, 2b_{1} \right]\times\left[ 2a_{2}, 2b_{2} \right] \times \dotsb \times \left[ 2a_{n}, 2b_{n} \right])\\ &= \sum\limits_{(a_{i}),(b_{i}) \in \mathbb{Z}^{\mathbb{N}}} \operatorname{Vol}( S \cap \left[ 2a_{1}, 2b_{1} \right]\times\left[ 2a_{2}, 2b_{2} \right] \times \dotsb \times \left[ 2a_{n}, 2b_{n} \right]) \\ &=\operatorname{Vol} (\bigcup_{(a_{i}),(b_{i}) \in \mathbb{Z}^{\mathbb{N}}} S \cap \left[ 2a_{1}, 2b_{1} \right]\times\left[ 2a_{2}, 2b_{2} \right] \times \dotsb \times \left[ 2a_{n}, 2b_{n} \right]) \\ &=\operatorname{Vol}(S) \end{align*}

So $f$ is not injective. By definition, this means that $\exists p_{1},p_{2} \in S: p_{1} \neq p_{2}, f(p_{1})=f(p_{2})$. So $\exists (c_{i}) \in \mathbb{Z}^{\mathbb{N}} : p_{1} = p_{2}+(2c_{1},2c_{2},\dotsc,2c_{n}) $ Since $S$ is symmetric about the origin, $-p_{1} \in S$. Also, since $S$ is convex the line between $-p_{1}$ and $-p_{2}$ is entirely contained in $S$. Specifically this means that the midpoint is contained in $S$. In otherwords, $$(c_{1},c_{2},\dotsc,c_{n}) = \frac{1}{2} ((2c_{1},2c_{2},\dotsc,2c_{n})+p_{1}-p_{1}) = \frac{1}{2} (p_{2}-p_{1}) \in S $$