As the title suggests, I am trying to prove:
Prove that if any subset of the vertex set of a polytope defines a face, then the polytope is a simplex.
For polytope $P$ with $n$ vertices $\{v_1,...,v_n\}$. I guess this could be proved by arguing that the face lattice of $P$ is isomorphic to the Boolean lattice.
However, I was thinking about proving it by contradiction:
- Suppose polytope $P$ is not a simplex and $dim(P)\leq n-1$. Then there is proper face $F$ of $P$, which is not a simplex. Let $dim(F)=m$ and $F=conv(V)$ where $V=\{v_1,...,v_{m+2}\}$. So, we get $v_1,...,v_{m+2}$ are affinely dependent. Now let, $F'=conv(V \cup v_{m+3})$ for $v_{m+3}$ not in affine hull of $F$.
I was thinking of getting some contradiction here by showing that $F'$ cannot be a face, but I'm unsure how to proceed.
- I considered showing $dim(P)\geq n-1$ in another approach. But I still don't know how to do it.