# Cyclic $d$-polytope -- does every facet border every other facet?

A cyclic $$d$$-polytope is "neighborly," where every set of vertices sized $$d/2$$ together make up a face. What I'm interested in: in a cyclic $$d$$-polytope with $$N$$ facet ($$d-1$$ dimensional faces), does every facet "border" every other facet (i.e., share a $$d-2$$ dimensional face)?

Why I think it might: we can use the upper bound theorem in dual form (https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node12.html) to find the total number of $$d-2$$ dimensional faces for a $$d$$-polytope with $$N$$ facets. If you use the formula in the provided link, we find that there are $$N(N-1)/2$$ such $$d-2$$ dimensional faces. Since every such face borders $$2$$ facets, this total number of $$d-2$$ dimensional faces suggests that the property holds.

This unfortunately fails in even the simplest case: when $$d = 2$$.
Consider the cyclic polytope $$C_{2}(T) \subset \mathbb{R}^{2}$$ where $$T = \{ t_{1}, t_{2}, t_{3}, t_{4} \}_{<} \subset \mathbb{Z}$$. It has facets $$F_{i, i + 1} = \mathrm{conv}\{(t_{i}, t_{i}^{2}), (t_{i + 1}, t_{i + 1}^{2}) \}$$, and it's not hard to see that facets $$F_{i, j}, F_{k, \ell}$$ border each other if and only if $$\{i, j \} \cap \{ k, \ell \}$$ is a singleton set.
Consider the facets $$F_{1, 4}, F_{2, 3}$$. They do not meet at a vertex of $$C_{2}(T)$$ since their indexing sets are disjoint.