Prove that there are $n$-dimensional polytopes that don't have any diagonal for any positive integer $n\geq2$.

Question

My friend happened to find a question on Combinatorial Geometry. We know that in $2$ dimensional Euclidean Plane triangles have no diagonal; and, in $3$ dimensional space, Tetrahedrons have no diagonal. So, naturally, a question comes in mind about higher dimensions.

The question is to prove that there exists $n$-dimensional polytopes that don't have any diagonal for any positive integer $n\geq2$.

Answer 1

1. A triangle has no diagonal. - Obvious.
2. A tetrahedron is nothing but the pyramid on the base of a triangle. Thus the height (as a potential diagonal wrt. the top vertex) obviously hits the bottom facet (here: triangle face) perpendicularily. Putting this together with the fact that the tetrahedron is regular (and thus esp. both vertex- and facet- (here: triangle face) transitive), you'd get this assumption here too.
3. Finally note that the arguments provided in step 2. will apply for all simplices of any further dimension as well.

--- rk