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It seems plausible that any real tree or ${\mathbb{R}}$-tree in the sense of the definition in https://en.wikipedia.org/wiki/Real_tree admits an isometric embedding into the Banach space $\ell_1(\Gamma)$ for a set $\Gamma$ of sufficiently large cardinality (the space $\ell_1(\Gamma)$ is defined as the space of real-valued functions on $\Gamma$ with countable support and norm $||f||=\sum_{\gamma\in\hbox{supp} f}|f(\gamma)|$). My question is:

Is this known? If "Yes", I would appreciate references.

P.S. I like the proof of this fact suggested by Yves de Cornulier (see below). I also agree with Bill Johnson that the fact looks like "well known" and most probably has been published somewhere. If someone knows a reference, I would be very thankful for it (I cannot accept the second answer, but I would be happy to upvote it.)

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  • $\begingroup$ What do you mean by an isometric embedding? $\endgroup$ Dec 14, 2015 at 8:29
  • $\begingroup$ Isometric embedding means map preserving the distance. $\endgroup$
    – YCor
    Dec 14, 2015 at 10:39
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    $\begingroup$ I guess you haven't you read my paper with Lindenstrauss, Preiss, and Schechtman on $\ell_1$-trees. Embed any interval or half line of the tree into a one dimensional space. Take any branch point and add a new $\ell_1$ dimension and move an interval or ray branching off in this new direction. Continue transfinitely. $\endgroup$ Dec 14, 2015 at 16:46
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    $\begingroup$ @BillJohnson There are 3 papers with the same 4-tuple of authors; I guess you refer to this one sciencedirect.com/science/article/pii/S0022123602939248 (JLPS, Lipschitz Quotients from Metric Trees and from Banach Spaces Containing l1, J. Funct. Anal. 194 (2002), no. 2, 332–346). Where exactly do you prove this? I indeed see the ingredients of the proof at some places but can't single out the precise statement. $\endgroup$
    – YCor
    Dec 14, 2015 at 17:18
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    $\begingroup$ @Bill I have not noticed this result stated explicitly in your paper. I also sketched a proof of the fact obtained by transfinite enumeration of all geodesics (which is more complicated than the proof by Yves below). The point is that I needed the fact that all trees are embeddable into RNP spaces to abandon some direction of thought, and I wanted to be sure that I did not do so prematurely. I asked the question to become sure in this. $\endgroup$ Dec 14, 2015 at 17:37

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It's true: there's an isometric embedding of every real tree $T$ on some $\ell^1$-space, i.e., a set $A$ and a map $f:T\to \ell^1(A)$ that is an isometric embedding, that is, satisfies $\|f(x)-f(x')\|=d(x,x')$ for all $x\in T$.

I don't know a reference; here's a proof. Let $(x_t)_{t<\alpha}$ be an enumeration of points on the real tree $T$ by an ordinal $\alpha$, assume that the convex hull of them is $T$.

Let $T_\beta$ be the closed convex hull of $\{x_t:t<\beta\}$ (so $T_\alpha=T$). Let us construct by induction an isometric embedding $f_\beta:T_\beta\to\ell^1(\beta)\subset\ell^1(\alpha)$, so that whenever $\beta\le\gamma$ then $f_\gamma$ extends $f_\beta$.

For $\beta=1$, $T_\beta=\{x_0\}$, and we just prescribe $f_\beta(x_0)=0$. For $\beta=\gamma+1>1$ a successor ordinal, $T_\beta$ is the convex hull of $T_\gamma\cup\{x_\gamma\}$. Let $y_\gamma$ be the projection of $x_\gamma$ on $T_\gamma$. So $$T_\beta=T_\gamma\cup [y_\gamma,x_\gamma].$$ We define $f_\beta$ so as to extend $f_\gamma$ on $T_\gamma$, and prescribe $f_\beta(y)=f_\gamma(y_\gamma)+d(y,y_\gamma)e_\beta$ for every $y\in [y_\gamma,x_\gamma]$, where $(e_t)$ is the canonical basis of $\ell^1(\alpha)$ and $d$ is the distance in the tree. This is well-defined (note that $f_\gamma(y_\gamma)$ is defined twice) and defines an isometric embedding. Finally, for $\beta$ a limit ordinal, then $T_\beta$ is the closure of $T'_\beta=\bigcup_{\gamma<\beta}T_\gamma$, we just define $f_\beta$ on $T'_\beta$ by saying its graph is the union of graphs of all $f_\gamma$ for $\gamma<\beta$; this is an isometric embedding into a complete metric space, so extends uniquely to an isometric embedding of the closure $T_\beta$ into $\ell^1(\beta)$.

So $f_\alpha$ is the required isometric embedding.

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  • $\begingroup$ @Yves. Thank you. I also produced something like this, but less organized than your argument and I was thinking that probably better-organized arguments are known and published. $\endgroup$ Dec 14, 2015 at 17:19

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