Delone sets in $\mathbb{R}$ are bi-Lipschitz equivalent

Question

Why are two Delone sets in $\mathbb{R}$ bi-Lipschitz equivalent?

Definition. $A$ is Delone set in $X$ metric space, if $\exists \delta,\epsilon>0$ such that forall $a\neq a'\in A$,\ $d(a,a')\geq \delta$ and $\forall x\in X, \exists a\in A,\ d(x,a)\leq \epsilon$.

It's true that any Delone set in $\mathbb{R}$ es bi-Lipschitz equivalent to $\mathbb{Z}$?

I need $f:A\to \mathbb{Z}$ such that $f$ bijection and $\frac{1}{L}d(a,a')\leq d(f(a),f(a'))\leq Ld(a,a')$ for all $a,a'\in A$.

I thought to take the function $f(a)=[a]$ integer part. The problem is that there could be two $a,a'\in A$ such that $f(a)=f(a')$...

Answer 1

Let $A\subset\Bbb R$ be a Delone set. Define a map $f:A\to\Bbb Z$ as follows. Pick any points $a_0\in A$ and put $f(a_0)=n$. Let $a\in A\setminus\{a_0\}$ be any point. Put $$f(a)=\cases{+1+|\{b\in A:a_0<b<a\}|, \mbox{if $a>a_0$},\\ -1-|\{b\in A:a<b<a_0\}|, \mbox{if $a<a_0$}.}$$ Pick $\delta$ and $\epsilon$ from the definition of a Delone set. It is easy to check that for each $a,b\in A$ we have $$\delta |f(a)-f(b)|\le d(a,b) \le 2\epsilon|f(a)-f(b)|.$$