I am self-learning about Voronoi diagrams and in the book I am reading about, to prove Theorem 4.6, it assumes that the Voronoi vertex $v$ must be incidient to at least three Voronoi regions.
That made me think, why is it at least three? Why doesn't two work?
A Voronoi region $Vor(p)$ is the intersection of all halfplanes $H(p,q)$, where $q$ is any other site in the point set $S$. Then why wouldn't a Voronoi vertex be the point that is at the midpoint of the halfplane line? I suppose it wouldn't be possible to find the midpoint of a halfplane line if the halfplane line is "infinite" length similar to a perpendicular bisector of a triangle side is.
Let's first consider the setup of the theorem within a 2 dimensional space, where the Voronoi complex is being considered. Then just consider the dual complex, i.e. the Delone (btw. french transscriptions of that russian name rather would write: Delaunay, both of which are being found in literature) complex. That one here would need to have at least 2 dimensional simplices for domains, i.e. triangles. Thus you are done here, when re-dualizing that.
Within higher dimensions the number of vertices of a simplex clearly increases. And simplices still are the smallest possible Delone domains. As the count of vertices of any Delone domain re-dualizes into the number of Voronoi domains, which are incident to a vertex of the Voronoi complex, you will be done here as well after all.
--- rk