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Problem: Let $c_1, \dots , c_n$ be points of $\mathbb{R}^d$ in a ball of radius $1$ so that the distance between any two of them is strictly greater than $\sqrt{2}$. Show that $n \le d+1$.

I need this result for a different problem, but my understanding of packings is somewhat weak. What would an effective proof look like?

My one idea is to assume that $n > d+1$, which would mean that the points $c_1 , \dots , c_n$ are affinely dependent. Maybe given this, and if we represent each point as a ball of radius $\sqrt{2}/2$, if these points are in the unit ball then at least two of the balls must intersect.

Wesley
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  • Just to make sure we aren't dealing with an XY-problem, may we know what this "different problem" of yours is? – Arthur May 15 '19 at 17:37
  • The problem is if you can fit some number of points in the unit sphere with this condition whether they are affinely dependent or independent. I first need to know whether the statement of this proposition is actually true or not. – Wesley May 15 '19 at 17:54
  • So when you say "Show that $n\le d+1$", what you really mean is "Is it necessarily true that $n\le d+1$?", right? The way you put it, it looks like you somehow know that it's true, but can't prove it. Which is rather misleading. – TonyK May 15 '19 at 18:43
  • To me it seems like it's necessarily true. I want to see what a proof would like it – Wesley May 15 '19 at 18:55
  • Your post doesn't really support your feeling that it's necessarily true. Do you have any other grounds for this belief? If so, they might provide the starting point for a proof. – TonyK May 15 '19 at 19:21

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"where every point is $\sqrt2$ apart" probably means that the pairwise distance of points of this set is of this size. Thus it needs to be the vertex-set of a regular simplex. And from the given embeding dimension you clearly have at most $n \le d+1$.

OTOH. the circumradius of the regular simplex with edge size $\sqrt2$ happens to be $\sqrt\frac{d}{d+1}$, which indeed is lesser than $1$. So your statement not only is true, but all allowed values $n$ from that relation indeed can be packed accordingly.

--- rk

Dr. Richard Klitzing
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