I kindly ask your expertise on the following point:
Let $P$ be an unbounded polyhedron of $R^d$ defined by linear inequalities as
$$P = \{x\in R^d : \ell_1 (x)< \alpha_1,\ \ \ell_2 (x)\leq \alpha_2, \ \ \ldots, \ell_t (x)\leq \alpha_t, \}$$
where only the first inequality is strict and the others are weaker.
My question is: Does there exists some $\varepsilon >0$ such that the closed polyhedron $P_\varepsilon$
$$P_\varepsilon = \{x\in R^d : \ell_1 (x)\leq \alpha_1 - \varepsilon,\ \ \ell_2 (x)\leq \alpha_2, \ \ \ldots, \ell_t (x)\leq \alpha_t, \}$$
contains the same integer points of $P$, i.e.
$$P_\varepsilon \cap Z^d = P \cap Z^d$$
For $d=1$ and for polytopes, this is true.
Thank you in advance for your help.
For a bounded polyhedron, the answer would be "yes"; for an unbounded one, not necessarily so.
Indeed, for any $\varepsilon>0$ there are integer points $(n,m)$ that satisfy $n-\sqrt2m>0$, but not $n-\sqrt2m\geqslant\varepsilon$. Ditto for any other irrational slope.
To make the answer always "yes", you need a stricter condition, like rational coefficients in $\ell_1 (x)$ or the boundedness of the polyhedron's face corresponding to it.
So it goes.