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.
"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