What is the maximal size of an equal-distance set in $\mathbb{R}^n$? Let $A\subseteq \mathbb{R}^n$ with the casual metric and $c\in\mathbb{R}^+$ be a real positive number, such that for every $a_1, a_2\in A$ if $a_1\neq a_2$ then $d(a_1,a_2)=c$.
What is the maximal size of such a set $A$?
My intuition says that it's $n+1$, i.e. the dimension of the euclidean space we're working with plus one, but I'm not sure on how to prove this - assuming this is correct?
It's most easy to see this in $\mathbb{R}$  - two points on the real line, and in $\mathbb{R}^2$ - a triangle, but I'm not sure it's true for every $n\in \mathbb{N}$.
 A: Yes it's true, and the proof is by induction on the cardinality $k = |A|$. 
By scaling, we can reduce to the case that the equal-distance is $1$.
Now one proves a stronger statement by induction, namely that $k \le n+1$ and there is an isometry $\mathbb{R}^n \mapsto \mathbb{R}^n$ that takes $A$ to an equilateral $k-1$ simplex of side length $1$ contained in $\mathbb{R}^{k-1}$. This is obviously true for $k=1$. 
Assuming it is true for numbers $<k$, let $A'$ be obtained by removing one point of $A$, and so there is a similarity taking $A'$ to the vertex set of an equilaterial $k-2$ simplex $\sigma$ of side length $1$ in $\mathbb{R}^{k-2}$. So we may assume that $A'$ is the vertex set of this simplex. Now take the set of unit radius spheres around the points of $A'$ and intersect them. That set of intersections is a sphere of some dimension whose diameter is strictly less than $1$. Therefore the set $A$ can contain at most one point on that sphere. And any single point on that sphere, union $A'$, is the vertex set of an equilateral $k-1$ simplex, which we can rotate to be in $\mathbb{R}^{k-1}$ by a rotation that fixes $\mathbb{R}^{k-2}$.
