faces of a simple polytope Question: Why is the face of a simple polytope a simple polytope?
My reasoning thus far:
I only have to show that this is true for any facet $F$ of $P$. Now let's take $v$ a vertices of $F$ and we show that $v$ is in $d-1$ facets of $F$.
Now take $F'$ a facet of $F$ containing $v$. Then by the diamond property, there exists a unique facet $G$ of $P$ containing $v$ such that $F'$ is in $G$ and $F$. Therefore the number of facets of $F$ containing $v$ is smaller or equal to $d-1$ (since there are only $d-1$ facets of $P$ containing $v$ other than $F$).
However I don't see how I can show that it cannot be strictly smaller than $d-1$

My answer to my question:
Two results that I use:
1) If $P$ is simple, $Fig^v(P)$ is a simplex
2) The order set $(F^v,\subset)$ is isomorphic to the face lattice of $Fig^v(P)$ ($F^v$ is the set of faces of $P$ containing $v$)
By 2) we have that the face lattice of $F$ in the polytope $F^v$ is isomorphic to the face lattice of a facet of $Fig^v(P)$. By 1) it is the face lattice of a simplex. Therefore $F$ in $F^v$ is a simplex and it has d-1 facets who all include $v$. However it is possible that there are over facets of $F$ in the polytope $P$ who contain $v$. So we know that there are at least $d-1$ facets of $F$ who contain $v$.
