Does anyone know if it has been proved what the maximum number of points in$ n$-dimensional space, for any two points with equal distance？ Does anyone know if it has been proved what the maximum number of points in$ n$-dimensional space, for any two  points with equal distance.
case when $N=1$,it is  the maximum is $2$
case $N=2$, it is clear the maximum is $3$,in other words, Three vertices of an equilateral triangle
case $N=3$,it is clear the maximum is $4$,that's mean is four vertices of a positive tetrahedral
so for  General,I conjecture the maximum if $N=n$, it's $n+1?$
 A: Yes, the general result is $n+1$ and the points must be the vertices of a regular simplex. (Adding the "... must be regular simplex part" makes it easy to show this by induction).
A: Let $X \subseteq  \Bbb{R}^n$ be such that $d(v, w) = d > 0$ whenever $v, w \in X$ with $v \neq w$. Assume without loss of generality that $0\in X$ and let $v_0 = 0$. If $n = 0$, we are done: $\Bbb{R}^0 = X = \{v_0\}$. Otherwise, $X$ has at least two elements and we can pick $v_1 \in X \backslash \{v_0\}$. Then $v_1$ spans a $1$-dimensional subspace $V_1$ of $\Bbb{R}^n$, and (by induction) $V_1 \cap X = \{v_0, v_1\}$. Continuing inductively for $i \le n$ we can find $v_1, v_2, \ldots v_i \in X$ spanning an $i$-dimensional subspace $V_i$ such that $V_i \cap X = \{v_0, v_1, \ldots, v_i\}$. But then $V_n = \Bbb{R}^n$ and $X = V_n \cap X = \{v_0, v_1, \ldots, v_n\}$. As Hagen von Eitzen points out each set $V_i$ will comprise the vertices of a regular $i$-simplex, but that fact is irrelevant to the induction, so I would be interested to know why Hagen thinks that simplifies the proof.
