Apparently the following theorem;
Theorem: In $\Bbb R^2 $ we can be sure of a covering of at least $\lceil A \rceil$ lattice points where is a measurable region of any shape and area A.
Can be used to prove Minkowski's theorem.
Let $\mathcal R$ be a region in the plane which is.
(i) Convex
(ii) Symmetric about the origin
(iii) Area > 4
Does anyone know the general process to doing such a thing and does it generalize to the larger dimensional cases?
My only real thought on how to do this is to look at a square of area 4 it contains only the origin then inscribe other convex entities inside it and make its area slightly bigger than 4. it is clearly intuitive that any other shape is less area efficient to lattice point containing than a square so if a square with more than area 4 does then everything should.
As I understand the question, the goal is to prove Minkowski's Theorem using the above result. It works in any dimension $n\geq 2$. We will use:
Theorem: for any measurable set $A\subset \mathbb{R}^n$, there exists $v \in \mathbb{R}^n$ such that the intersection of the lattice with the translate $(v+A)\cap \mathbb{Z}^n$ contains at least $\lceil\mathcal{L}(A)\rceil$ points, where $\mathcal{L}$ stands for the Lebesgue measure.
Let $\mathcal{R}\subset \mathbb{R}^n$ be a convex region, symmetric, of area $>2^n$. Consider the set $A=\mathcal{R}/2$ (the image under homothety of ratio 1/2). So the area of $A$ is $>1$. So there are two points $x\neq y \in \mathbb{Z}^n\cap(v+A)$. Then $x,y$ can be written: $$x=v+a/2, \; y=v+b/2,$$ where $a,b \in \mathcal{R}$. Since $\mathcal{R}$ is symmetric, $-b$ is also in $\mathcal{R}$. So $$x-y=\frac{a+(-b)}2,$$ is a nonzero point with integer coordinates (since $x\neq y$), and the barycenter of the two points $a,-b$ of $\mathcal{R}$, so also in $\mathcal{R}$ by convexity, as required.