Let $\Omega\subset \mathbb{R}^m$ be a bounded Lipschitz domain. Let $D$ be a countable dense subset of $\Omega$, denoted as $D = \{p\_1,p\_2,p\_3\ldots \}$. Define the minimum seperation distance among first $n$ points of $D$ as
$$h(n) = \min\_{1\le i,j\le n,i\ne j}\|p\_i-p\_j\|\_2 .$$


I'd like to know the convergence rate of $h(n)$ as $n\to\infty$. Surely $h(n)\to 0$, but at what rate. Has it got anything to do with the dimension of the space. Is it $$h(n) = O(n^{-\frac{1}{m}})$$ os it simply $h(n) = O(1/n)?$


I feel there are some notions that I am not aware, which is the reason this question is puzzling me.



What is the convergence rate of the minimum separation distance?
If $\Omega$ is any bounded domain in $\mathbb{R}^m$, Let $B\_0$ be a ball which contains it, and let $B$ be a ball with the same center but with its radius increased by $1$. Say $R$ and $V$ are the radius and volume of $B$.


Fix $r < {\rm min}(\frac{1}{2}h(n), 1)$. Then the $r$-balls about the points $p\_1$, $\ldots$, $p\_n$ are disjoint, and as they are all contained in $B$ we have $nv \leq V$, where $v$ is the volume of a ball of radius $r$. Thus $nr^m \leq R^m$, so $r \leq Rn^{-\frac{1}{m}}$. As this is true of every $r < \frac{1}{2}h(n)$, we get $h(n) \leq 2Rn^{-\frac{1}{m}}$.


To see that $O(n^{-\frac{1}{m}})$ is best possible, let $\Omega$ be the unit cube in $\mathbb{R}^m$ and form $D$ as the union, in order, of the sets $D\_k = 2^{-k}L \cap \Omega \setminus (D\_1 \cup \cdots \cup D\_{k-1})$, where $L$ is the integer lattice in $\mathbb{R}^m$. Then $|D\_1 \cup \cdots \cup D\_k| = |2^{-k}L \cap \Omega| = 2^{km}$, so $h(2^{km}) \geq 2^{-k}$ (the minimum distance between points in $2^{-k}L \cap \Omega$). That is, for $n = 2^{km}$ we have $h(n) \geq n^{-\frac{1}{m}}$ in this example.


