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？

Question

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 $n+1?$

I finally understood the question. "Maximum number of points in $n$-dimensional space such that any two (different) points are at the same distance". How people are able to interpret this from the text in the question baffles me. — Git Gud, Oct 08 '15 at 18:29
@dafinguzman Using the formulation in my comment above, a circumference won't do it. I suppose this is what it is intended. — Git Gud, Oct 08 '15 at 18:30
@GitGud: I am risking being flamed horribly, but I think native English speakers are supposed to be adequately fluent in the international language called "English as a Foreign Language" to decode the OP's question (particularly using the title, which is hard to interpret any other way). — Rob Arthan, Oct 08 '15 at 19:47

score 7 · Answer 1 · answered Oct 08 '15 at 15:28

7

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).

answered Oct 08 '15 at 15:28

Hagen von Eitzen

374,180

(+1). I like the point that the stronger assertion has the more direct proof. – martini Oct 08 '15 at 15:30
why it easy to show this by induction? – Oct 08 '15 at 15:32
@abandon: the answer to your comment was a bit long to fit in a comment, so see my answer. – Rob Arthan Oct 08 '15 at 16:05
@martini: I wasn't able to make any use of the strengthening of the inductive hypothesis. How does it lead to a more direct proof? – Rob Arthan Oct 08 '15 at 19:55

score 1 · Answer 2 · edited Oct 09 '15 at 04:05

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.

Apologies to Hagen for misspelling his name. – Rob Arthan Oct 12 '15 at 21:41 — Rob Arthan, Oct 12 '15 at 21:41

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？

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