Is each compact metric space a subset of a compact absolute 1-Lipschitz retract?

Question

A metric space $X$ is called an absolute $L$-Lipschitz retract if for any metric space $Y$ containing $X$ there exists a Lipschitz retraction $r:Y\to X$ with Lipschitz constant $Lip(r)\le L$.

Question. Is each compact metric space isometric to a subset of a compact absolute 1-Lipschitz retract?

Remark 1. Using almost isometric embeddings into the Banach space $c_0$, it can be shown that each compact metric space $X$ is a subset of a compact absolute $(1+\varepsilon)$-Lipschitz retract $Y$, where $\varepsilon$ is any positive real number. The space $Y$ is a suitable cube $\prod_{n\in\omega}[-a_n,a_n]$ in $c_0$ with a bit distorted metric.

Remark 2. There exists also a functorial construction of an embedding of compact metric space $X$ into a compact absolute 8-Lipschitz retract $A(X)$. Given a compact metric space $X$, consider the isometric embedding $X\subset\ell_\infty$ identifying each point $x\in X$ with the distance function $d_X(x,\cdot)$. Next, take the closed convex hull $conv(X)$ of $X$ in $\ell_\infty$. Finally, consider the hyperspace $A(X)$ of non-empty convex compact subsets of $conv(X)$, endowed with the Hausdorff metric. By Theorem 1.7 in the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, the compact metric space $A(X)$ is an absolute 8-Lipschitz retract. I do not know if the constant 8 can be replaced by a smallest constant (say 1).

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Added at Edit. Thanks to the comment of @Wlod AA, I have found an answer to my question on page 32 of the book of Benyamini and Lindenstrauss. They write that Isbell in 1964 suggested the construction of the injective envelope of a metric space, which is the smallest 1-Lipschitz AR containing a given metric space. For a compact metric space its injective envelope is compact, too.

Answer 1

The classical paper on the given theme is by Aronszajn and Panitchkpakdi. I am pretty sure that it contains the required result about embedding metric spaces into metric absolute retracts, i.e. in Lip$_1$ category, and perhaps about embeddings compact metric spaces in compact metric absolute retracts. Indeed, these authors have introduced the notion of the hyper-convex metric spaces which is equivalent to metrically injective spaces (i.e. in Lip$_1$ category, which I call simply metric category or Met for short).

This and further results are contained in later papers by John Isbell and (a bit later and independently) by wh.

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Let me provide perhaps the simplest embedding of arbitrary compact metric space $\ \mathbf X:=(X\ d)\ $ into a compact metric absolute retract.

Let $\ Y:=\mbox{Met}_\delta\ (\mathbf X)\ $ be the set of all metric maps $\ f:X\rightarrow[0;\delta],\ $ where $\ \delta\ $ is the diameter of $\ \mathbf X\ $ (metric maps means Lip$_1).\ $ Then $\ Y\ $ in its uniform distance function is compact, it's hyper-convex i.e. it's an injective metric space (metric absolute retract), and the embedding $\ i:X\rightarrow Y\ $ is given by Kuratowski-Wojdysławski formula:

$$ \forall_{s\ t\in X}\ \ (i(s))(t)\ :=\ d(s\ t) $$

(You need to be archeological to hear about this stuff).

Theorem Let $\ \mathbf X:=(X\ d)\ $ be an arbitrary non-empty metric space of an arbitrary finite diameter. Let $\ -\infty\le a\le b\le\infty.\ $

Then space $\ Y\ \:=\ \mbox{Met}(\mathbf X\,\ \mathbb R\!\cap\![a;b])\ $ of all metric functions $\ f:X\rightarrow\mathbb R\cap[a;b]\ $ is hyper-convex.

Proof   Let $\ \emptyset\ne F\subseteq Y $ and radia $\ r:F\rightarrow[0;\infty)\ $ be such that

$$ \forall_{f\ g\in F}\ r_f+r_g\ge |f-g| $$

Define $\ c : X\rightarrow \mathbb R\!\cap\![a;b]\ $ as follows:

$$ \forall_{x\in X}\ \ c(x)\ :=\ \max(a\ \ sup_{_{x\in X}}\ (f(x)-r_f)) $$

Then, by routine applications of the triangle inequality, the function $\ c\ $ has what it takes:

$$ c\ \in Y\cap\bigcap_{f\in F} B(f\ r_f) $$

where $\ B(f\ r_f)\ $ is the ball centered in $\ f,\ $ of radius $\ r_f.$