Suppose $P_1$ and $P_2$ are two $n$-dimensional convex polytopes. Does $\partial P_1 \subseteq\partial P_2$ imply that $P_1 = P_2$?

Question

Given two convex polytopes $P_1$ and $P_2$ with the same dimension, I want to know if the boundary of $P_1$ (denoted $\partial P_1$) being contained in the boundary of $P_2$ (denoted $\partial P_2$) is enough to conclude that $P_1$ and $P_2$ are the same polytope.

From $\partial P_1 \subseteq\partial P_2$ we immediately have that $\mathcal V_1 \subseteq \mathcal V_2$ (where $\mathcal V_i$ is the vertex set of $P_i$), which also implies that $P_1 \subseteq P_2$.

My rough idea of how to proceed is as follows: Suppose that there exists $v^\dagger$ such that $v^\dagger \in \mathcal V_2 \setminus \mathcal V_1$; if there are many such $v^\dagger$, pick it so that there exists a vertex, $v$, adjacent to it (in $P_2$) such that $v \in \mathcal V_1$. Now, there must be a vertex $v^*$ in $\mathcal V_1$ such that $v^*$ is adjacent to $v$ in $P_1$ but not in $P_2$ (not sure how to show this formally, but basically the idea is that in $P_2$, $v^\dagger$ is ''in-between'' $v$ and $v^*$...). Then, the convex hull of $\{v,v^*\} \subseteq \partial P_1$ but $\{v,v^*\} \not\subseteq \partial P_2$, a contradiction. Therefore, we have $\mathcal V_1 = \mathcal V_2$, giving us that $P_1 = P_2$.

Appreciate any ideas about how to formalize the step of showing that such a $v^*$ must exist, or showing that the claim is false (or, if it is correct, an alternate proof!).

I know that the $v^*$ existence step must necessarily leverage that the polytopes have the same dimension, but it's not clear to me how to do this exactly...

Answer 1

This has a very quick proof using a bit of topological machinery. The boundary of an $n$-dimensional convex polytope is homeomorphic to $S^{n-1}$. The only subset of $S^{n-1}$ which is homeomorphic to $S^{n-1}$ is $S^{n-1}$ itself (this follows, for instance, from invariance of domain, which implies such a subset must be open, but it must also be closed by compactness, so it is all of $S^{n-1}$ by connectedness).

So, $\partial P_1\subseteq \partial P_2$ implies $\partial P_1$ is actually all of $\partial P_2$. In particular, $P_1$ contains all the vertices of $P_2$ so they are equal.

Answer 2

Here is a proof without topology.

As you already mentioned, $P_1\subseteq P_2$. So, if we choose a point $x$ from the interior of $P_1$ it is also in the interior of $P_2$. For every point $y\in\partial P_2$, consider the ray $r$ from $x$ through $y$. Because of convexity, the ray $r$ hits $\partial P_1$ and $\partial P_2$ exactly once. But since $\partial P_1\subseteq \partial P_2$, this intersection must be the same point, namely $y$. Thus $y\in\partial P_1$. Since this holds for all $y\in\partial P_2$, we have $\partial P_2\subseteq\partial P_1\implies\partial P_1=\partial P_2\implies P_1=P_2$.