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

Suppose first that $X$ and $Y$ are quasi-isometric. If $X$ is empty, a quasi-inverse $Y\to X$ forces $Y$ empty, and the empty subsets are separated nets linked by their unique bilipschitz equivalence. Otherwise choose a quasi-isometry $f:X\to Y$ with coarse Lipschitz quasi-inverse $g:Y\to X$. Let $g$ have constants $L_g\ge 1,C_g\ge 0$ and let $d_X(gf(x),x)\le D_X$ and $d_Y(fg(y),y)\le D_Y$. Then $d_X(x,x^{\mathrm{\prime }})\le L_g{d}_Y(fx,f{x}^{\mathrm{\prime }})+C_g+2D_X$, so $f$ has a quasi-isometric lower bound; its coarse Lipschitz bound supplies the upper bound. Fix common constants $L\ge 1,C\ge 0$ for these bounds. The second composite estimate says $f[X]$ is $D_Y$-coarsely dense.

Put $\delta :=2LC+1$. Zorn's lemma applied to the poset of $\delta$-separated subsets of $X$ gives a maximal one $A$: a union of a chain is still separated, and the empty set starts the poset. For each $x\in X$ there must be an $a\in A$ with $d_X(x,a)<\delta$, since otherwise adjoining $x$ would contradict maximality. Thus $A$ is a $\delta$-separated, $\delta$-net; in particular it is nonempty. This quantified argument also covers the initially empty candidate subset without using $d(x,\mathrm{\varnothing })$.

For distinct $a,a^{\mathrm{\prime }}\in A$, separation absorbs the additive error: $d_Y(fa,f{a}^{\mathrm{\prime }})\ge (L^{-1}-C\mathrm{/}\delta )d_X(a,a^{\mathrm{\prime }})$ and $d_Y(fa,f{a}^{\mathrm{\prime }})\le (L+C\mathrm{/}\delta )d_X(a,a^{\mathrm{\prime }})$. The first coefficient is positive because $\delta >LC$. Hence $f{\mathrm{\mid }}_A$ is a bilipschitz equivalence onto $f[A]$, and $f[A]$ is separated. Given $y\in Y$, choose $x\in X$ with $d_Y(y,fx)\le D_Y$ and then $a\in A$ with $d_X(x,a)<\delta$; one has $d_Y(y,fa)<D_Y+L\delta +C$. Thus $f[A]$ is also a net in $Y$.

Conversely, suppose separated nets $A\subseteq X$ and $B\subseteq Y$ are linked by a bilipschitz equivalence $\varphi :A\to B$. If $A$ is empty, its net property forces $X$ empty, while bijectivity forces $B$ and hence $Y$ empty; their unique maps are quasi-inverses. Otherwise choose net radii $R_X,R_Y$ and, using choice, select maps $p_X:X\to A$ and $p_Y:Y\to B$ within those radii, with $p_X(a)=a$ and $p_Y(b)=b$ on the nets. The triangle inequality gives $d_X(p_{X}x,p_X{x}^{\mathrm{\prime }})\le d_X(x,x^{\mathrm{\prime }})+2R_X$, and likewise for $p_Y$, so both maps are coarse Lipschitz. Define $F:X\to Y$ by $F=\varphi p_X$ and $G:Y\to X$ by $G={\varphi }^{-1}{p}_Y$, viewing the net values in their ambient spaces. Bilipschitz bounds make both maps coarse Lipschitz. Since $p_X$ and $p_Y$ fix their nets, $GF=p_X$ and $FG=p_Y$; these are within $R_X$ and $R_Y$ of the respective identities. Hence $F$ and $G$ are quasi-inverses and $X,Y$ are quasi-isometric. Together with step 3.1, this proves both directions.