# What's the fewest number of sides required to make a polytope in n dimensions?

In 2 dimensions it takes at least 3 sides to make a polygon, the triangle, and in 3 dimensions it takes at least 4 faces (so far as I'm aware) to make a polyhedron. Can this rule be generalized to higher dimensions, so that the minimum number of sides or faces to make a polytope in n-dimensions is equal to n+1? If so, is there a proof for it? And if not, is there a counterexample?

• yes. See Regular Polytopes by Coxeter. Mar 2, 2014 at 5:37

Firstly, $n+1$ is enough, because we have the $n$-simplex formed by taking $x_1 + \ldots + x_n = 1$ with the coordinate hyperperplanes $x_i = 0$.
To show this is the minimum necessary, assume there is a $n-1$-dimensional polygon in $n$ space with $n$ faces of dimension $n-1$ (which I'll call facets from now on) , and look at one of its facets. This is a $n-2$-dimensional polygon in $n-1$ space. It can have at most $n-1$ facets itself, because there are only $n-1$ other facets of the original polygon that interesect with it. Therefore, if we've shown there are no $n-2$ dimensional polygons with $n-1$ faces, this is absurd, so the proof goes through by induction.
The easiest way to prove that $n+1$ faces are necessary is probably to show that given only $n$ vectors — representing the normals to the $n$ faces of the polytope — there's always a non-zero $n$-vector that has either zero or positive dot product with all of them; such a vector will therefore 'go off to infinity' staying within the half-plane defined by all of the faces, and thus witnesses the unboundedness of the polytope. This can be done via a relatively straightforward application of Gaussian Elimination.