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 $F\cup \{v\}$

  • $\begingroup$ 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 $\endgroup$ – tomak Apr 21 '17 at 13:24
  • 1
    $\begingroup$ @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 $P$. $\endgroup$ – Lord Shark the Unknown Apr 21 '17 at 13:46
  • $\begingroup$ Yep indeed it was obvious :/ $\endgroup$ – tomak Apr 21 '17 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.