Is there a countable metric space $U$ such that any countable metric space is bi-Lipschitz equivalent to a subset of $U$? How about $c\_{00}(\mathbb{Q})$ where $\mathbb{Q}$ is the rational numbers? Thanks!



Does there exist a countable metric space which is Lipschitz universal for all countable metric spaces?
The affirmative answer to this problem follows from



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> **Lemma.** For any countable dense subsets $X,Y$ in the half-line $\mathbb R\_+=[0,+\infty)$ there exists a $C^2$-smooth function $f:\mathbb R\_+\to\mathbb R\_+$ such that
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> $\bullet$ $f(X)\subseteq Y\cup\{0\}$;
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> $\bullet$ $f(0)=0$;
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> $\bullet$ $1<f'(x)<2$ for all $x>0$;
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> $\bullet$ $f''(x)<0$ for all $x>0$.
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> 


*Proof.* Such a function $f$ can be found by a standard back-and-forth argument. $\quad\square$


Now take any countable metric space $(X,d)$ and consider the countable subset $d(X\times X)$ of $\mathbb R\_+$. By the above lemma, there exists a function $f:\mathbb R\_+\to\mathbb R\_+$ such that $f(d(X\times X))\subseteq \mathbb Q$, $f(0)=0$, $1< f'(x)< 2$ and $f''(x)<0$ for all $x>0$. These properties of the function $f$ imply that $$f(x+y)\le f(x)+f(y)\quad\mbox{and}\quad x<f(x)< 2x$$for all $x,y\in\mathbb R\_+$, and hence the functionn
$$\rho:X\times X\to\mathbb R\_+,\quad \rho:(x,y)\mapsto f(d(x,y))$$is a metric on $X$, which is bi-Lipschitz equivalent to the metric $d$.


Thus the metric space $(X,d)$ is bi-Lipschitz equivalent to the metric space $(X,\rho)$ whose metric takes its values in the set of rationals. The latter space is isometric to a subspace of some (canonical) dense subset of the universal Urysohn space.


