Two metric spaces are quasi-isometric if and only if each contains a separated net and the two nets are bilipschitz equivalent

Let $X$ and $Y$ be quasi-isometric, and by [L2] choose a quasi-isometric embedding $f:X\to Y$ with coarsely dense image. Fix constants $L\ge 1$, $C\ge 0$ and $R\ge 0$ such that $f$ is $(L,C)$-quasi-isometric and $f[X]$ is $R$-coarsely dense. Put $\delta :=2LC+1$. Consider the poset of $\delta$-separated subsets of $X$, ordered by inclusion. It is nonempty because $\mathrm{\varnothing }$ is $\delta$-separated, and the union of a chain of $\delta$-separated subsets is again $\delta$-separated; so Zorn's lemma gives a maximal $\delta$-separated subset $A\subseteq X$. If some $x\in X$ satisfied $d(x,A)\ge \delta$, then $A\cup \{x\}$ would still be $\delta$-separated, contradicting maximality. Hence every point of $X$ lies within distance $\delta$ of $A$, so $A$ is a separated net in $X$.

For distinct $a,a^{\mathrm{\prime }}\in A$ one has $d_Y(f(a),f(a^{\mathrm{\prime }}))\ge L^{-1}{d}_X(a,a^{\mathrm{\prime }})-C\ge (L^{-1}-\frac{C}{\delta })d_X(a,a^{\mathrm{\prime }})$ and $d_Y(f(a),f(a^{\mathrm{\prime }}))\le (L+\frac{C}{\delta })d_X(a,a^{\mathrm{\prime }}),$ so the restriction $f{\mathrm{\mid }}_A:A\to f[A]$ is bilipschitz. Also, if $y\in Y$, choose $x\in X$ with $d_Y(y,f(x))\le R$ and then choose $a\in A$ with $d_X(x,a)<\delta$; then $d_Y(y,f(a))\le d_Y(y,f(x))+d_Y(f(x),f(a))\le R+L\delta +C\mathrm{.}$ So $f[A]$ is a separated net in $Y$, bilipschitz equivalent to $A$.

Conversely, let $A\subseteq X$ and $B\subseteq Y$ be separated nets, and let $\varphi :A\to B$ be a bilipschitz equivalence. By [L7] the inclusions $A\hookrightarrow X$ and $B\hookrightarrow Y$ are isometric embeddings, and because $A$ and $B$ are nets those inclusions have coarsely dense image; hence [L2] makes them quasi-isometries. By [L5] the map $\varphi$ is a quasi-isometry. Transitivity of quasi-isometry now gives $X\simeq A\simeq B\simeq Y$, so $X$ and $Y$ are quasi-isometric.