This seems a very basic problem so it's probably been tackled multiple times, but I am wondering how many equidistant* points can there be at an $n$-dimensional euclidean space (let's say $\mathbb{R}^n$ with the euclidean distance)
My conjecture is that, for $n$ dimensions, there can be exactly $n+1$ of those points. This is obvious for $n=1$. For $n=2$ we have the triangle and, for $n=3$, the tetrahedron. I cannot think of a way to arrange more points for $n \leq 3$ I think an extra dimension will always allow us to set a new point along that new axis, in the "center" of the previous figure (like going form a segment to a triangle and from a triangle to a tetrahedron)
So, am I right about my intuitions or do they break at some point (no pun intended)? Also, is it possible to have more than $n+1$ equidistant points?
So, am I right there?
*a set of points all separated the same distance from each other
Best Answer
In fact those n+1 vertex polytopes of n-dimensional space with equal pairwise distances are called simplices (singular: simplex). This is, as those are the most simple polytopes within each dimension. For obvious reasons there will be no polytope within n-dimensional space with fewer vertices. Moreover those simplices all are regular polytopes.
The described construction by @AsuraPath is that of continued pyramidisation: take a vertex atop the simplex of one dimension less.
--- rk