In this paper (theorem 2), Chepoi & Hagen say

There exists an infinite $CAT(0)$ cube complex $X$ with constant maximum degree which cannot be isometrically embedded into a Cartesian product of a finite number of trees, i.e., the chromatic number of its crossing graph is infinite.

so that you can't hope to isometrically embed it in a $\mathbb{R}^n$ with $l^1$ metric. But what about $l^1(\mathbb{N})$:

**Question 1**: Under what conditions does a $CAT(0)$ cube complex, with the "polyhedral complex" metric induced by the $l^1$ metric on cubes, embed in $l^1(\mathbb N)$ ?

**Question 2**: Same question but only about embedding the $0$ skeleton of said complex.

My reason for the question is the following: It is known that the $0$-skeleton of a $CAT(0)$ cube complex is a median space, and intuitively it makes sense since it looks a lot like a nice subset of $\mathbb{Z}^n$ for some $n$ big enough, which is itself median. I'm therefore wondering if you could prove median-ness of this $0$-skeleton by embedding your complex in $l^1(\mathbb{N})$, and then arguing that the embedding is actually submedian, or whatever allows you to argue median-ness back for the skeleton.

EDIT: added the obvious "countable" condition. ~~Then an obvious embedding would be to enumerate the vertices and send them to the "basis" of $l^1$ … I wonder if/where that would break.~~