Let $P$ be a polytope in $\mathbb{R}^d$ given as the solution set to a system of linear inequalities $a_ix \le c_i$ for some row vectors $a_i$ and scalars $c_i$, $i = 1,\dots,n$. For each $i$ define $F_i = P \cap (x \in \mathbb{R}^d \mid a_i x = c_i)$. (The braces weren't working, sorry)
I want to show that any proper face of $P$ is an intersection of some of these faces $F_1,\dots,F_n$. Geometrically this is quite clear but I'm stuck finding an actual proof of this. Any help?
I think I got. Since every face is an intersection of facets its enough to show that every facet is one the $F_i$.
Let $F$ be a facet of $P$. Let $y$ be in the relative interior of $F$. Since $P$ is the disjoint union of the relative interiors of its faces, $y$ is not in the interior of $P$. Thus $y$ is on the boundary of $P$. The boundary is easily seen to be $\bigcup_{i=1}^n F_i$, where $F_i$ are defined in the question. So $y$ is in some $F_j$. We know that $F$ is the smallest face of $P$ containing $y$. Thus $F \subseteq F_j$. The dimensions of $F$ and $F_j$ are easily seen to be the same, namely $d-1$. Hence $F = F_j$.