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Minkowski's Theorem

Theorem 1   Let $\zeta$ be a convex set in ${\mathbb{R}}^2$, with $\zeta$ centrally symmetric and $area(\zeta)>4$. Then $\zeta$ must contain a lattice point besides $(0,0)$.

Definition 2   A set is convex whenever $x,y\in\zeta$ implies $\overline{xy}\in\zeta$.

Definition 3   An object $\zeta$ is centrally symmetric if whenever $(0,0)\in\zeta$ and $P\in\zeta$ implies $\overline{0P}\in\zeta$ and $-\overline{0P}\in\zeta$.

1. Take $\frac{1}{2}\zeta=\zeta'$

2. Assign a copy of $\zeta'$ to each lattice point in an $n\times n$ grid.

3. From the above figure we can see $area(\zeta')>1$, thus the area covered by the set s is $(n+1)^2A'$.

4. In fact, all objects cover more (go over a bit). All $(n+1)^2$ objects are contained in a square of side $(n+2s)$, where $s$ is the maximum distance from the center $(0,0)$ to any point of $\zeta'$.

5. Now we want to show the inequality

$$(n+1)^2area(\zeta')>(n+2s)^2$$

to be true. CLAIM:

$$(n+1)^2(area(\zeta'))-(n+2s)^2 = (area(\zeta')-1)(n^2+2n)(area(\zeta)-2s)$$

$${} +area(\zeta')-4s^2$$

$$> 0.$$

Since $area(\zeta)>1$ implies the leading coefficient is positive for large $n$, if we choose $n$ large then we're done! Conclusion, the blobs will overlap for some pair of blobs, but by translating $\zeta'$ and another blob $\zeta''$ cross too! (All is being repeated).

6. Let $(a'',b'')$ be in the intersection of $\zeta', \zeta''$. Then

$$(a'',b'')=(a',b')+(p,q),$$   where$$(a',b')\in\zeta'.$$

Note that $(-a',-b')\in\zeta'$ as well.

Consider the midpoint of $(-a',-b')$ and $(a'',b'')$, then $(\frac{p}{2},\frac{q}{2})\in\zeta'\Rightarrow(p,q)\in\zeta$.

QED