# Facets shared by two points on a convex polytope

I have a convex polytope of arbitrary dimension. Let $\mathcal{F} (A)$ denote the set of facets that vertex $A$ belongs to. If two vertices share an edge, is it true that the disunion of $\mathcal{F} (A)$ and $\mathcal{F} (B)$ has a single element only, i.e. the share all but a single supporting facet? If so, where can I find the proof of this? Many thanks.

• What does "disunion" mean? Nov 3, 2015 at 17:57

Assuming that "disunion" means the symmetric difference (the set of facets containing either the vertex $A$, or the vertex $B$, but not both), then no. E.G. take a cube, and two vertices $A$ and $B$ connected by an edge. There is a face containing $A$ but not $B$, and a face containing $B$ but not $A$, so there are two facets in the symmetric difference $\mathcal{F}(A) \mathbin{\triangle}\mathcal{F}(B)$. There will always be at least two (otherwise one vertex would be in all the facets containing another vertex, which is impossible.)