# Bound for the cardinality of maximal $r$-separable subsets contained in a ball of radius $R$ in $\mathbb R^d$

Let $$B$$ be a closed ball in $$\mathbb R^d$$ of radius $$R$$ and let $$N=N_R(r)$$ denote the maximal cardinality of the $$r$$-separated sets (meaning any two points in this set have distance at least $$r$$) that can be contained in $$B$$.

If $$r>2R$$ then $$N_R(r)=1$$.

If $$r\le 2R$$, I expect $$N$$ can be bounded by something like $$C(\frac{2R}{r})^{d+\delta}$$ where $$C,\delta$$ are absolute constants (I may be wrong here. Let me know the correct form).

I wonder what are naive/classical/latest results on this estimation.

• Hint: compare volumes. This is naive/classical and not so bad. You get the bound $(\frac{2R}{r}+1)^d$. – Mikael de la Salle Oct 8 at 7:39
• The key observation is that balls of radius $r/2$ centered in the separated set cannot intersect and have to lie completely in a ball of radius $R+r/2$. Conversely any such balls of radius $r/2$ in $B_{R+r/2}$ have a center in $B_R$ and if they do not intersect their centers have distance at least $r$, so your problem is equivalent to a sphere packing problem, for which there should be a lot of advanced results. – mlk Oct 8 at 8:05