Proof request: every polytope has a facet When I was reading Ziegler's book "Lectures on Polytopes" this statement appeared to be never proven formally.

Question: does every (convex, bounded, non-empty) polytope have a facet?

Here I assume that the polytope $P\subset\Bbb R^d$ has dimensions $d$, and a facet is a face of dimension $d-1$.
Depending on the amount of theory developed to this point there are many ways to prove this. But it seems some element in the chain towards such a proof it missing in the book. E.g, proving any of the following statements might already be sufficient:

*

*the face lattice is graded of length $d+1$ (stated but never proven in Ziegler)

*the face lattice is co-atomic (proven using duality and the next unproven fact)

*given a $\delta$-face $f$ of $P$, its dual face in $P^\circ$ is of dimension $d-1-\delta$ (not proven as far as I can tell)

Maybe I am just overlooking the proof in the book, but an elementary self-contained proof of the existence of a facet is welcome anyway.
 A: I am not familiar with Ziegler's book and will be using Grunbaum as a reference. That is, I define a face of a convex set to be its intersection with a supporting hyperplane, and a facet to be a maximal proper face.
Let us prove that every polyhedron $P$ has a facet. Let $\mathcal{H}:=\{H_i\}_{i=1}^n$ be an irredundant family of hyperplanes such that $P:=\cap_{i=1}^nH_i^+$, where $H_i^+$ is the positive closed half-space defined by $H_i$. Then every $H_i$ is a supporting hyperplane for $P$. By definition, $P\cap H_1$ is a face of $P$. Additionally, since $\mathcal{H}$ is irredundant, $P\cap H_1$ has codimension $1$, i.e. $aff(P\cap H_1)=H_1$.
Let $P\cap H$ be another proper face of $P$. Suppose that $P\cap H_1\subseteq P\cap H$. Then, $$P\cap H_1\subseteq H\Rightarrow aff(P\cap H_1)\subseteq H\Rightarrow H_1\subseteq H\Rightarrow H_1=H\Rightarrow P\cap H_1=P\cap H.$$ Therefore, $P\cap H_1$ is a facet.
A: Depending on the amount of convex geometry and topology available at this point, the following would be a possible reasoning.
If $P$ is full-dimensional, then every $x\in\partial P$ is contained in some proper face of $P$. This follows from the existence of a supporting hyperplane at $x$, whose intersection with $P$ is a face.
If we assume that $P$ has no facet, then $\partial P$ is the union of convex sets of dimension $\le d-2$. But the boundary of a $d$-dimensional compact convex set is homeomorphic to the $(d-1)$-sphere, which cannot be embedded in such a union.
