is there a simple proof for the following lemma:
An unbounded convex polytope (defined by linear constraints) has either zero integer points or infinite many integer points.
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Edit: the following argument is true when the polyhedron $P$ is the intersection of finitely many linear constraints: $a_1 x_1 + \ldots + a_n x_n \leq b$.
Edit: Let me emphasize that I make the (obviously necessary) extra assumption that $P$ has non-empty interior. I am not assuming, however, that the bounding hyperplanes have rational normal directions.
This will (essentially) follow from this sublemma:
If $x$ is in the interior of $P$, then $P$ contains a $\delta$ neighborhood of a ray $x+te$, $t\ge 0$, for some direction $e\in S$.
To see this, consider $x+te$ for fixed $e$ and observe that if this intersects a bounding hyperplane $H$ for some $t_0>0$, then suitable points $x+t(e')e'$ will also be on $H$, with $t(e')\approx t_0$, for all $e'\approx e$ (that could only fail if $n_H\cdot e=0$ for the normal direction of $H$, but then we would have $x\in H$ already). So if every ray $x+te$, $e\in S$ intersected the boundary, then $P$ would be bounded after all by compactness of the unit sphere. Since there are only finitely many bounding hyperplanes, there is a fixed positive distance to all of them from my ray.
If we now have an $x\in \mathbb Z^n$ in the interior of $P$, then this tube $y+te$, $t\ge 0$, $|y-x|<\delta$ will contain infinitely many other integer points because $x+te$ at least gets arbitrarily close to $\mathbb Z^n$ (the details depend on the rational relations among the $e_j$).
If $x$ was on the boundary, a modified version of this still works.