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

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).

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.

• "is a subset" purports to mean "is isometric to a subset"? – YCor Apr 22 '18 at 22:45
• @YCor Correct! And corrected. – Taras Banakh Apr 23 '18 at 5:19
• A compact metric space $X$ can be isometrically embedded into $\ell_\infty(X)$, and its closed convex envelope there is a compact subset $K$. Isn't any convex compact subset $K$ of $\ell_\infty(X)$ an absolute $1$-Lipschitz retract? – Pietro Majer Apr 23 '18 at 6:54
• @PietroMajer The order convexity is a sufficient but not necessary condition of being 1-Lipschitz AR. In fact, there are many non-equivalent embeddings into $\ell_\infty$. For example, the unit interval $[0,1]$ is isometric to an order-convex set in $\ell_\infty$ but can also be re-embedded as a diagonal of $\ell^\infty$ and then it is not order-convex, moreover, its order-convex hull is not compact. – Taras Banakh Apr 23 '18 at 8:16
• Taras and/or authors Benyamini and Lindenstrauss have written carefully 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. You may have an easier time to read "Linearization..." from Bull.Acad.Polon.Sci.**16** (1968), pp. 189-193. This paper defines/constructs the metric envelope (and proves its properties), and it proves that the metric envelope of a Banach space is the metric Banach envelope. – Wlod AA Apr 24 '18 at 2:34

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.

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:

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

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 $\ r_f.$

• I've been thinking about this paper as this is the main tool in Lindenstrauss' proof of the fact if a Banach space is $(1+\varepsilon)$-injective for all $\varepsilon > 0$, then it is injective (see also mathoverflow.net/questions/132610/…). Is there a chance for such result in the metric case? – Tomek Kania Apr 23 '18 at 19:57
• @Wlod AA Thank you very much for your answer. But why is your space $Y$ hyperconvex? Is there any reference for this important fact? Cause I cannot find mentioning the space $Met_\delta(X)$ in the paper of Aronszajn and Panickpakdi :( – Taras Banakh Apr 23 '18 at 20:54
• @TomekKania, thank you for the link. – Wlod AA Apr 24 '18 at 2:12
• Taras, I have borrowed the consistent metric terminology (metric maps, MET or Met, etc) from my old papers (wh). I will provide a proof of $\ Y\$ being hyper-convex in my answer--you'll see that it's very straightforward. – Wlod AA Apr 24 '18 at 2:16