# face of polytope $P'=$ conv($P\cup \{v\}$)

I'm reading Grunbaum's book on convexe polytopes and property 5.2.1 is perturbing me.

So we have a polytope $P$ and another polytope $P'=$ conv($P\cup \{v\}$).

He states that it is "obvious" that a face of $P'$ is either a face of $P$, or the convex hull of the union of a face of $P$ and $v$.

On a drawing it is indeed obvious. And this is how I proceed:

$F'$ a face of $P'$ ($F' = P'\cap H$ for some valid hyperplane $H$). If $v\not\in F'$ then somehow you have to prove that actually no other point in $P'$ not in $P$ can be in $F'$. (How do you prove this?) Therefore $P'\cap H = P\cap H$ and since $H$ remains valid for $P$, we get that $F'$ is also a face of $P$.

If $v\in F'$ we have that $F = P\cap H\subset P'\cap H = F'$ so $F\cup \{v\}\subset F'$ and conv($F\cup \{v\}$)$\subset$ conv($F') = F'$ But how do you get the other inclusion?

Let $F'$ be a face of $P'$ having $v$ as one vertex. It is the intersection of a supporting hyperplane $H$ of $P'$ with $P'$ (so $P'$ is within one closed half-space defined by $H$). Then $H$ is also a supporting hyperplane for $P$ and $F=H\cap P$ is a face of $P$ (I'm taking the empty set to be a face.)
We need to show that $F'$ is the convex hull of $F\cup\{v\}$. Certainly $F'$ contains this convex hull. But each point $x$ of $F'$ is a point of $P'$ and so lies in a segment $[uv]$ from a point $u\in P$ to $v$. Assume $x\ne v$. Then the whole segment $[uv]$ lies in $H$. Therefore $u\in P\cap H=F$. So $x$ lies in the convex hull of $F\cup \{v\}$
• Ok. And how do you show that if $v\not\in F'$ then the facet is included in $P$? by contradiction I suppose but I don't see how – tomak Apr 21 '17 at 13:24
• @tomak The case $v\notin F'$ is the easy case. If $H$ is a supporting hyperplane for $P'$ with $H\cap P'=F'$ then $H$ is also a supporting hyperplane for $P$ and $H\cap P=F'$ too as all vertices of $F'$ are in $P$. – Lord Shark the Unknown Apr 21 '17 at 13:46