Dimension of polyhedron

Question

Let $A \in R^{m×d}$, $b \in R^m$. $P=[x \in R^d: Ax \leq b]$ is a polyhedron. Suppose there is some $\bar x \in R^d$ such that $A\bar x<b$, that is, all inequalities are satisfied strictly. Now how can I show $dim(P) = d$?

Someone suggested me to use

1. Results of affine independence

2. $\exists \epsilon_i>0$ such that $A(\bar x + \epsilon_i*e_i)<b$, where $e_i$ is a d-dimentional vector where the i-th entry is $1$ and others are $0$.

However, I still don't get how to utilize these hints. Please help me.

Answer 1

$H = \{x \in \mathbb R^d \mid Ax \lt b\}$ is open as it is the intersection of $m$ open subspaces of $\mathbb R^d$. Therefore if $\overline{x}$ belongs to $H$, it exists an open ball $B(\overline{x},\epsilon)$ centered on $\overline{x}$ of radius $\epsilon$ included in $H$.

As any open ball contains $d+1$ points that are affinely independent, we get the desired result $\dim P = d$.