Think about the map for Minkowski’s theorem. It cuts a plane into two by two squares and stacks the squares on each other. A plane is a surface that stretches on forever and doesn’t have thickness.
Create the formulation. Have L represent the lattice of determinant. Determinants are objects used to analyze and solve linear equations. Have S represent the convex subset of Rn. In this case, if x is S, then –x is also S. A convex subset is a situation where the line AB is inside S.
Understand that f is not injective, meaning that each “A” has its own unique and matching member in “B.” If it were, it would not overlap with anything else and would be area-preserving for all of S. Since this is not true, f is not injective. Therefore, f(p1) = f(p2) for points p1,p2 in S.
Consider that S is symmetric about the origin and –p1 is also a point for S. S is convex, so the line segment between –p1 and p2 are completely in S. Therefore, if the convex subset’s volume is greater than 2n d(L), S has at least one lattice point beyond the origin.