# Can i get every face of a polytope by taking a facet (of a facet (of a facet (…))) of the polytope?

Let $P$ be a polytope, i.e. a convex subset of a finite-dimensional real vector space with finitely many extreme points. Let $F$ be a proper face of the polytope, i.e. a subset $F \subset P$ such that there exists a closed half-space containing $P$ such that $F = \partial H \cap P$, where $\partial H$ is the hyperplane given by the boundary of $H$.

I wonder if there must be a sequence $P = F_1 \supset F_2 \supset \ldots \supset F_k = F$ of subsets of $P$ such that $F_{i+1}$ is a facet of $F_i$ for every $i \in \{1, \ldots, k-1\}$ (with the definition that a facet of a polytope is a $(d-1)$-dimensional face of a $d$-dimensional polytope).

How can I prove the existence of such a sequence?

-
Notice that it is enough —by induction— to show that every proper face is a face of a facet. Can you de that? –  Mariano Suárez-Alvarez Nov 27 '11 at 4:55
I agree that this is enough. However, I don't see how to prove this. Can you give me another hint? –  Tom Jonathan Nov 27 '11 at 7:56

The idea is that you take the supporting hyperplane $\partial H$ and rotate it about the face $F$ until it meets the polytope $P$ in a larger face $F'$ (which necessarily includes $F$.) If $F$ is a facet, then its affine hull is all of $\partial H$ and so $\delta H$ cannot be rotated while still containing $F$. But if $F$ is a $(d-i)$-face of the $d$-dimensional polytope $P$, then the affine hull of $F$ is only $(d-i)$-dimensional, leaving $(i-1)$ degrees of freedom in which to rotate $\partial H$. This lets you show that you can repeat the process until you get to a facet.
It is a little tricky to formalize this, but hopefully you understand the idea. One example of a formal argument using this idea is on page 33 of Grünbaum's Convex Polytopes, in the proof of Theorem 3.5 (the somewhat different statement that a face of a face of $P$ is also a face of $P$.) Here's a picture from the text!