I was reading some notes on polyhedral analysis and encountered a proof that confused me. The proof is in the book named "Integer and Combinatorial Optimization" by Wolsey and Nemhauser.
Let $M = \{l, 2, ... , m\}, M^{=} = \{i \in M: a_ix = b_i, \forall x \in P\}$ and let $M^{\leq} = M \setminus M^{=}$ where $P$ is a rational polyhedron.
$Definition:$ A point $x \in P$ called an inner point of $P$ if $a^ix < b_i , \forall i \in M^{\leq}$.
$Proposition:$ Every nonempty polyhedron has an inner point.
$Proof:$ If $M^{\leq} = \emptyset$, every point of $P$ is inner. Otherwise, for $i \in M^{\leq}$ there exists a point $x^i \in P$ with $a^i x^i < b_i$ . Now $ \hat{x}= (1 / |M^{\leq}|)\sum_{i \in M^{\leq}}x^i \in P$ since $P$ is convex. Since $a^i \hat{x} < b_i, \forall i \in M^{\leq}$, $\hat{x}$ is an inner point.
Doesn't the second statement (Otherwise, for $i \in M^{\leq}$ there exists a point $x^i \in P$ with $a^i x^i < b_i$) imply that we already assume that there exists an inner point? If that is the case, why do we show that there exist $\hat{x}$ satifying the condition of an inner point?
No. There is a different $x^i$ for every $i.$ But we want the same $x$ for every $i.$ The proof constructs such an $x.$ On the other hand, I don't think the construction works, since while $\hat{x} \in P,$ I don't really see why $\hat{x}$ satisfies all the strict inequalities.