Consider \begin{align}\min \sqrt2 x_1 – x_2 \\ 1 \leq x_2 \leq \sqrt2 x_1 \\ x_1,x_2\text{ integer.} \end{align}
How do you show that $conv\{x \in \mathbb{Z}^2: 1 \leq x_2 \leq \sqrt2 x_1\} $ is not a polyhedron?
I know by Meyer's Theorem that it is a polyhedron if it can be represented with a system $Ax\leq b$ with rational $A,b$, but the converse it not true in general.
Using Minkowski-Weyl (a polyhedron is the Minkowski sum of the convex hull $Q$ of a finite number of points, say $v^1,…,v^p$, and the finitely generated cone C with generators $r^1,…,r^q$), I find that it has extreme ray $(1,0)$. I expect that there are infinitely points $v^1,…,v^p$, which results in $conv\{x \in \mathbb{Z}^2: 1 \leq x_2 \leq \sqrt2 x_1\} $ not being a polyhedron, but how to prove this exactely?
Best Answer
Update: As mr-e-man has pointed out in his comment both of my previous suggestions for answering this question were wrong. I can't delete this post until it has been unaccepted. In the meantime, mr-e-man's comments effectively give a proper answer.