# Characterizing Lipschitz images of injective metric spaces

Taras Banakh, Judyta Bąk, Joanna Garbulińska-Węgrzyn,
Magdalena Nowak, Michał Popławski  T. Banakh (ORCID 0000-0001-6710-4611): Ivan
Franko National University of Lviv (Ukraine), and Mathematics Department, Jan
Kochanowski University in Kielce, Uniwersytecka 7, 25-406 Kielce, Poland
<EMAIL_ADDRESS>J. Bąk: (ORCID 0000-0001-8027-7226): Mathematics
Department, Jan Kochanowski University in Kielce, Uniwersytecka 7, 25-406
Kielce, Poland<EMAIL_ADDRESS>J. Garbulińska-Węgrzyn (ORCID
0000-0001-7217-2002): Mathematics Department, Jan Kochanowski University in
Kielce, Uniwersytecka 7, 25-406 Kielce, Poland<EMAIL_ADDRESS>M.
Nowak (ORCID 0000-0003-1915-0001): Mathematics Department, Jan Kochanowski
University in Kielce, Uniwersytecka 7, 25-406 Kielce, Poland
<EMAIL_ADDRESS>M. Popławski (ORCID 0000-0002-2725-9675):
Mathematics Department, Jan Kochanowski University in Kielce, Uniwersytecka 7,
25-406 Kielce, Poland<EMAIL_ADDRESS>
###### Abstract.

A metric space $X$ is injective if every non-expanding map $f:B\to X$ defined
on a subspace $B$ of a metric space $A$ can be extended to a non-expanding map
$\bar{f}:A\to X$. We prove that a metric space $X$ is a Lipschitz image of an
injective metric space if and only if $X$ is Lipschitz connected in the sense
that for every points $x,y\in X$, there exists a Lipschitz map $f:[0,1]\to X$
such that $f(0)=x$ and $f(1)=y$. In this case the metric space $X$ carries a
well-defined intrinsic metric. A metric space $X$ is a Lipschitz image of a
compact injective metric space if and only if $X$ is compact, Lipschitz
connected and its intrinsic metric is totally bounded. A metric space $X$ is a
Lipschitz image of a separable injective metric space if and only if $X$ is a
Lipschitz image of the Urysohn universal metric space if and only if $X$ is
analytic, Lipschitz connected and its intrinsic metric is separable.

###### Key words and phrases:

Injective metric space, Lipschitz map, Lipschitz image, Urysohn metric space,
Lipschitz connected metric space

###### 2010 Mathematics Subject Classification:

Primary:54E35, 54E40; Secondary: 51F30, 54C55, 54E45, 54E50, 54F15

##  1\. Introduction

By the classical Hahn–Mazurkiewicz Theorem [11, 16], a Hausdorff space $X$ is
a continuous image of the unit segment $[0,1]$ if and only if $X$ is a Peano
continuum, i.e., $X$ is compact, metrizable, connected and locally connected.
A metric counterpart of the Hahn–Mazurkiewicz Theorem is due to Ważewski [19]
and Eilenberg, Harrold [7] who proved that a metric space $X$ is the image of
the closed interval $[0,1]$ under a Lipschitz map if and only if $X$ has
finite length. The unit interval and the real line are the most important
examples of injective metric spaces. So, it is natural to ask which metric
spaces are Lipschitz images of injective metric spaces. In this paper we
address this problem and characterize metric spaces which are Lipschitz images
of (separable or totally bounded) injective metric spaces.

For a metric space $X$, its metric will be denoted by $d_{X}$. For a map
$f:X\to Y$, its Lipschitz constant is defined as a (finite or infinite) number

$\operatorname{Lip}(f)\coloneqq\sup\limits_{{x,x^{\prime}\in X}\atop{x\neq
x^{\prime}}}\frac{d_{Y}(f(x),f(x^{\prime}))}{d_{X}(x,x^{\prime})}\in[0,\infty].$

In this definition we assume that $\sup\varnothing=0$.

A map $f:X\to Y$ between metric spaces $X,Y$ is called

  * •

Lipschitz if $\operatorname{Lip}(f)<\infty$;

  * •

$L$-Lipschitz for a real constant $L$, if $\operatorname{Lip}(f)\leq L$;

  * •

non-expanding if it is 1-Lipschitz.

A metric space $X$ is Lipschitz connected if for every points $x,y\in X$ there
exists a Lipschitz map $f:[0,1]\to X$ such that $f(0)=x$ and $f(1)=y$. Such a
map $f$ is called a Lipschitz path connecting the points $x,y$. It is clear
that every Lipschitz connected metric space is a path connected topological
space. Every Lipschitz connected metric space $X$ carries the intrinsic metric
$d_{IX}:X\times X\to\mathbb{R}$ defined by

$d_{IX}(x,y)\coloneqq\inf\\{\operatorname{Lip}(f):f\in
X^{[0,1]}\;\wedge\;f(0)=x\;\wedge\;f(1)=y\\}.$

for any points $x,y\in X$. In the above formula, $X^{[0,1]}$ stands for the
set of all functions from $[0,1]$ to $X$.

A metric space $X$ is called injective if every $1$-Lipschitz map $f:B\to X$
defined on a subspace $B$ of a metric space $A$ extends to a $1$-Lipschitz map
$\bar{f}:A\to X$.

Injective metric spaces form a well-known and well-studied class of complete
metric spaces, [4], [10, p. 47]. Such spaces are known under two alternative
names: hyperconvex metric spaces [1, 9, 14] and $1$-Lipschitz Absolute
Retracts [5, Ch.I.1.]. The following (known) characterization of injective
metric spaces can be deduced from [1, 5].

######  Theorem 1.1.

For a metric space $X$, the following conditions are equivalent:

  1. (1)

$X$ is an injective metric space;

  2. (2)

$X$ is $1$-Lipschitz Absolute Retract, which means that for every metric space
$Y$ containing $X$, there exists a $1$-Lipschitz map $f:Y\to X$ such that
$f(x)=x$ for all $x\in X$;

  3. (3)

$X$ is hyperconvex in the sense that for any indexed family of closed balls
$\big{(}B[x_{i},r_{i}]\big{)}_{i\in I}$ in $X$ with $d_{X}(x_{i},x_{j})\leq
r_{i}+r_{j}$ for all $i,j\in I$, the intersection $\bigcap_{i\in
I}B[x_{i},r_{i}]$ is not empty.

  4. (4)

$X$ is metrically convex (for every $a,b\in X$ and $t\in[0,1]$ there exists
$x\in X$ such that $d_{X}(a,x)=t\cdot d_{X}(a,b)$ and $d_{X}(x,b)=(1-t)\cdot
d_{X}(a,b)$) and has a binary intersection property (every collection of
mutually intersecting closed balls in $X$ has a common point).

  5. (5)

for every metric space $Y$ containing $X$ and for every metric space $Z$,
every Lipschitz map $X\to Z$ has an extension $Y\to Z$ with the same Lipschitz
constant.

###### Proof.

The implications $(5)\Leftrightarrow(1)\Leftrightarrow(2)$ are proved in [5,
Proposition 1.2]; $(3)\Rightarrow(2)$ in [1, Theorem 8], and
$(2)\Leftrightarrow(4)$ in [5, Proposition 1.4]. To see that
$(4)\Rightarrow(3)$, take any indexed family
$\big{(}B[x_{i},r_{i}]\big{)}_{i\in I}$ of closed balls in $X$ such that
$d_{X}(x_{i},x_{j})\leq r_{i}+r_{j}$ for every $i,j\in I$. The metric
convexity of $X$ ensures that for any distinct indices $i,j\in I$, the closed
balls $B[x_{i},r_{i}]$ and $B[x_{j},r_{j}]$ have a common point. The binary
intersection property guarantees that $\bigcap_{i\in
I}B[x_{i},r_{i}]\neq\emptyset$. ∎

Another important source of injective metric spaces are complete
$\mathbb{R}$-trees, see [6, 14]. A metric space $X$ is called an
$\mathbb{R}$-tree if for every points $x,y\in X$ the metric segment

$[x,y]\coloneqq\\{z\in X:d_{X}(x,y)=d_{X}(x,z)+d_{X}(z,y)\\}$

is isometric to the interval $[0,d_{X}(x,y)]$ on the real line, and for every
points $x,y,z\in X$ with $[x,y]\cap[y,z]=\\{y\\}$ we have
$d_{X}(x,z)=d_{X}(x,y)+d_{X}(y,z)$. By [14], every complete $\mathbb{R}$-tree
is an injective metric space. Nevertheless, classes of $\mathbb{R}$-trees and
injective metric spaces are distinct. For instance $(0,1)$ (with the euclidean
metric) is $\mathbb{R}$-tree, which is not injective, since injective metric
spaces are complete. On the other hand, $[0,1]^{2}$ with the maximum metric is
an injective metric space, which is not an $\mathbb{R}$-tree (the metric
segment $[(0,0),(1,0)]$ is not isometric to $[0,1]$).

A metric space $Y$ is called a Lipschitz image (resp. $1$-Lipschitz image) of
a metric space $X$ if there exists a surjective Lipschitz map $f:X\to Y$ (with
$\operatorname{Lip}(f)\leq 1$).

One can easily reformulate Hahn-Mazurkiewicz theorem [11, 16] to the following
version$\colon$ a Hausdorff space $X$ is a continuous image of the unit
segment $[0,1]$ if and only if $X$ is a continuous image of some compact
injective metric space $Y$. However, the Ważewski result in [19] implies that
the class of Lipschitz images of the unit interval $[0,1]$ is strictly smaller
than the class of Lipschitz images of compact injective metric spaces: a
counterexample is the injective metric space $[0,1]^{2}$ which is not the
Lipschitz image of the unit segment $[0,1]$.

The following three characterizations proved in Sections 5, 6, 7 are the main
results of this paper.

######  Theorem 1.2.

For a metric space $X$, the following conditions are equivalent:

  1. 1)

$X$ is a Lipschitz image of an injective metric space;

  2. 2)

$X$ is a Lipschitz image of a complete $\mathbb{R}$-tree;

  3. 3)

$X$ is Lipschitz connected.

######  Theorem 1.3.

For a metric space $X$, the following conditions are equivalent:

  1. 1)

$X$ is a Lipschitz image of a totally bounded injective metric space;

  2. 2)

$X$ is a Lipschitz image of a compact $\mathbb{R}$-tree;

  3. 3)

$X$ is compact, Lipschitz connected, and its intrinsic metric is totally
bounded.

######  Theorem 1.4.

For a metric space $X$, the following conditions are equivalent:

  1. 1)

$X$ is a Lipschitz image of a separable injective metric space;

  2. 2)

$X$ is a Lipschitz image of a separable complete $\mathbb{R}$-tree;

  3. 3)

$X$ is a $1$-Lipschitz image of the Urysohn universal space;

  4. 4)

$X$ is analytic, Lipschitz connected, and its intrinsic metric is separable.

Let us recall that a topological space $X$ is analytic if it is a continuous
image of a separable complete metric space. The Urysohn universal metric space
is a complete separable metric space $\mathbb{U}$ such that any isometric
embedding of a subset $A$ of a finite metric space $B$ admits an extension to
an isometric embedding of $B$ in $\mathbb{U}$. It is well-known [18] that a
Urysohn universal metric space exists and is unique up to an isometry.

##  2\. $\mathbb{R}$-convex metric spaces and $\mathbb{R}$-trees

######  Definition 2.1.

A metric space $X$ is called $\mathbb{R}$-convex if for every points $x,z\in
X$, the metric segment

$[x,z]\coloneqq\\{y\in X:d_{X}(x,y)+d_{X}(y,z)=d_{X}(x,z)\\}$

is isometric to the segment $[0,d_{X}(x,z)]$ of the real line.

######  Proposition 2.2.

For a metric space $X$, the following conditions are equivalent:

  1. 1)

$X$ is $\mathbb{R}$-convex;

  2. 2)

for every points $x,y\in X$, there exists a unique set $I\subseteq X$ such
that $x,y\in I$ and $I$ is isometric to the segment $[0,d_{X}(x,y)]$ on the
real line;

  3. 3)

for every points $x,y\in X$ there exists a unique $1$-Lipschitz map
$f:[0,d_{X}(x,y)]\to X$ such that $f(0)=x$ and $f(d_{X}(x,y))=y$.

###### Proof.

The implications $(1)\Rightarrow(2)\Rightarrow(3)$ are obvious. To prove that
$(3)\Rightarrow(1)$, assume that a metric space $X$ satisfies the condition
(3). Given any points $x,y\in X$, we need to show that the metric segment
$[x,y]$ is isometric to the interval $[0,d_{X}(x,y)]$ in the real line. By the
condition (3), there exists a unique $1$-Lipschitz map $f:[0,d_{X}(x,y)]\to X$
such that $f(0)=x$ and $f(d_{X}(x,y))=y$. The triangle inequality implies that
the image $I\coloneqq f\big{[}[0,d_{X}(x,y)]\big{]}$ is a subset of the metric
segment $[x,y]$. Assuming that $[x,y]\neq I$, we can find a point
$z\in[x,y]\setminus I$. Then $d_{X}(x,y)=d_{X}(x,z)+d_{X}(z,y)$. By condition
(3), there exist $1$-Lipschitz maps $\alpha:[0,d_{X}(x,z)]\to X$ and
$\beta:[0,d_{X}(z,y)]\to X$ such that $\alpha(0)=x$,
$\alpha(d_{X}(x,z))=z=\beta(0)$ and $\beta(d_{X}(z,y))=y$. The $1$-Lipschitz
maps $\alpha,\beta$ compose a $1$-Lipschitz map $\gamma:[0,d_{X}(x,y)]\to X$
defined by

$\gamma(t)=\begin{cases}\alpha(t)&\mbox{if $t\in[0,d_{X}(x,z)]$};\\\
\beta(t-d_{X}(x,z))&\mbox{if $t\in[d_{X}(x,z),d_{X}(x,y)]$}.\end{cases}$

Since $\gamma(0)=x=f(0)$, $\gamma(d_{X}(x,y))=y=f(d_{X}(x,y))$, and
$\gamma\neq f$, we obtain a contradiction with the uniqueness of the
$1$-Lipschitz path $f$. This contradiction shows that $[x,y]=I$. The triangle
inequality implies that the $1$-Lipschitz map $f:[0,d_{X}(x,y)]\to[x,y]$ is an
isometry, witnessing that the metric space $X$ is $\mathbb{R}$-convex. Indeed,
for every $u,z$ such that $0\leq u\leq z\leq d_{X}(x,y)$ we have $|u-z|\geq
d_{X}(f(u),f(z))\geq d_{X}(x,y)-d_{X}(x,f(u))-d_{X}(f(z),y)\geq
d_{X}(x,y)-|0-u|-|z-d_{X}(x,y)|=|u-z|.$ ∎

######  Proposition 2.3.

If a metric space $X$ is $\mathbb{R}$-convex, then for every points $x,y,z\in
X$, the intersection $[x,y]\cap[x,z]$ is equal to the metric segment $[x,u]$
for a unique point $u\in[x,y]\cap[x,z]$.

###### Proof.

Since the metric space $X$ is $\mathbb{R}$-convex, there exist isometries
$\alpha:[0,d_{X}(x,y)]\to[x,y]$ and $\beta:[0,d_{X}(x,z)]\to[x,z]$ such that
$\alpha(0)=x=\beta(0)$, $\alpha(d_{X}(x,y))=y$, and $\beta(d_{X}(x,z))=z$. By
the compactness of $[x,z]$, the closed set $\alpha^{-1}[x,z]$ in
$[0,d_{X}(x,y)]$ has the largest element $s$. Since $\alpha$ is an isometry,
$s=d_{X}(\alpha(0),\alpha(s))=d_{X}(x,\alpha(s))$. Let $u=\alpha(s)$ and
$t=\beta^{-1}(u)$. Since $\beta$ is an isometry,
$t=d_{X}(\beta(0),\beta(t))=d_{X}(x,u)=s$. Observe that
$\alpha^{\prime}\coloneqq\alpha{\restriction}_{[0,s]}:[0,s]\to X$ and
$\beta^{\prime}\coloneqq\beta{\restriction}_{[0,t]}:[0,t]\to X$ are two
$1$-Lipschitz maps from the interval $[0,s]=[0,t]$ to $X$ such that such that
$\alpha^{\prime}(0)=x=\beta^{\prime}(0)$ and
$\alpha^{\prime}(s)=u=\beta^{\prime}(t)$. Proposition 2.2 ensures that
$\alpha^{\prime}=\beta^{\prime}$ and hence
$[x,y]\cap[x,z]=\alpha[[0,s]]=[x,u]$. The uniqueness of the point $u$ implies
from the uniqueness of the end-points of the segment $[x,u]=[x,y]\cap[x,z]$. ∎

We observe that a metric space $X$ is $\mathbb{R}$-tree if $X$ is an
$\mathbb{R}$-convex metric space such that for every points $x,y,z\in X$ with
$[x,y]\cap[y,z]=\\{y\\}$, we have $d_{X}(x,z)=d_{X}(x,y)+d_{X}(y,z)$.

The following important fact can be found in [14].

######  Proposition 2.4.

Every complete $\mathbb{R}$-tree is an injective metric space.

##  3\. Lipschitz connected metric spaces

In this section we study Lipschitz connected and $L$-Lipschitz connected
metric spaces in more details.

######  Definition 3.1.

Let $L$ be a real number. A metric space $X$ is called $L$-Lipschitz connected
if for every points $x,y\in X$, there exists an $L$-Lipschitz path
$f:[0,d_{X}(x,y)]\to X$ such that $f(0)=x$ and $f(d_{X}(x,y))=y$.

If an $L$-Lipschitz connected metric space contains more than one point, then
$L\geq 1$. Therefore, the $1$-Lipschitz connectedness is the strongest notion
among the $L$-Lipschitz connectedness properties.

######  Example 3.2.

Every injective metric space $X$ is $1$-Lipschitz connected.

###### Proof.

Given any point $x,y\in X$, consider the doubleton $\\{0,d_{X}(x,y)\\}$ in the
real line and the $1$-Lipschitz map $f:\\{0,d_{X}(x,y)\\}\to X$ defined by
$f(0)=x$ and $f(d_{X}(x,y))=y$. By the injectivity of $X$, this $1$-Lipschitz
map admits a $1$-Lipschitz extension $\bar{f}:[0,d_{X}(x,y)]\to X$, which is a
$1$-Lipschitz path witnessing that $X$ is $1$-Lipschitz connected. ∎

As we have mentioned in the introduction, every Lipschitz connected metric
space $X$ carries the intrinsic metric $d_{IX}:X\times X\to\mathbb{R}$ defined
by

$\displaystyle d_{IX}(x,y)$ $\displaystyle=\inf\\{\operatorname{Lip}(f):f\in
X^{[0,1]}\;\wedge\;f(0)=x\;\wedge\;f(1)=y\\}$
$\displaystyle=\inf\\{C\in[0,\infty):\exists f\in
X^{[0,C]}\;(f(0)=x\;\wedge\;f(C)=y\;\wedge\;\operatorname{Lip}(f)\leq 1)\\}$

for any points $x,y\in X$. The second equality in the definition of the
intrinsic metric holds because a function $f\in X^{[0,1]}$ is $L$-Lipschitz if
and only if the function $f_{L}:[0,L]\to X$, $f_{L}:t\mapsto f(t/L)$, is
$1$-Lipschitz.

The definition of the intrinsic metric implies that $d_{X}\leq d_{IX}$ for
every Lipschitz connected metric space $X$. Also any $1$-Lipschitz map $f:X\to
Y$ between Lipschitz connected metric spaces remains $1$-Lipschitz with
respect to the intrinsic metrics on $X$ and $Y$, which means that

$d_{IY}(f(x),f(x^{\prime}))\leq d_{IX}(x,x^{\prime})$

for every points $x,x^{\prime}\in X$.

The following lemma describes an interplay between the completions of a
Lipschitz connected metric space with respect to the intrinsic metric and the
original metric.

######  Lemma 3.3.

Let $X$ be a Lipschitz connected metric space, $\overline{X}$ be the
completion of $X$, $\overline{X}_{I}$ be the completion of the metric space
$X_{I}\coloneqq(X,d_{IX})$, and
$\overline{\operatorname{id}}:\overline{X}_{I}\to\overline{X}$ be the unique
continuous extension of the identity map $\operatorname{id}:X_{I}\to X$. Then
$\overline{\operatorname{id}}^{-1}[X]=X_{I}$. In other words, every Cauchy
sequence $(x_{n})_{n\in\mathbb{N}}$ in $(X,d_{IX})$, which is convergent to a
point $x$ in $(X,d_{X})$ is convergent to $x$ in $(X,d_{IX}),$ too.

###### Proof.

Given any point $x\in\overline{X}_{I}$ with
$\overline{\operatorname{id}}(x)\in X$, we need to prove that $x\in X_{I}$.
Since $x\in\overline{X}_{I}$, there exists a sequence
$(x_{n})_{n\in\mathbb{N}}$ in $X_{I}$ such that
$d_{IX}(x,x_{n})<\frac{1}{2^{n+2}}$ for all $n\in\mathbb{N}$. For every
$n\in\mathbb{N}$ we have

$d_{IX}(x_{n},x_{n+1})\leq
d_{IX}(x_{n},x)+d_{IX}(x,x_{n+1})<\tfrac{1}{2^{n+2}}+\tfrac{1}{2^{n+3}}<\tfrac{1}{2^{n+1}}.$

So, we can find a 1-Lipschitz path
$\gamma_{n}:[\tfrac{1}{2^{n+1}},\tfrac{1}{2^{n}}]\to X$ such that
$\gamma_{n}(\tfrac{1}{2^{n+1}})=x_{n+1}$ and
$\gamma_{n}(\tfrac{1}{2^{n}})=x_{n}$. The paths $\gamma_{n}$,
$n\in\mathbb{N}$, compose a 1-Lipschitz path $\gamma:[0,1]\to X$ defined by
$\gamma(0)=\overline{\operatorname{id}}(x)$ and
$\gamma{\restriction}_{[\tfrac{1}{2^{n+1}},\tfrac{1}{2^{n}}]}=\gamma_{n}$ for
all $n\in\mathbb{N}$. The definition of $\gamma$ guarantees that

$\lim_{n\to\infty}d_{IX}(x,x_{n})=0=\lim_{n\to\infty}d_{IX}(x_{n},\operatorname{id}^{-1}(\overline{\operatorname{id}}(x)))$

and hence
$x=\operatorname{id}^{-1}(\overline{\operatorname{id}}(x))\in\operatorname{id}^{-1}[X]=X_{I}$.
∎

A metric $d$ on a set $X$ will be called compact (resp. complete, separable,
analytic) if so is the metric space $(X,d)$.

######  Lemma 3.4.

Let $(X,d_{X})$ be a Lipschitz connected metric space.

  1. (1)

If the metric $d_{X}$ is complete, then so is the intrinsic metric $d_{IX}$.

  2. (2)

The metric $d_{IX}$ is analytic if and only if $d_{IX}$ is separable and the
metric $d_{X}$ is analytic.

  3. (3)

If the metric $d_{IX}$ is separable (resp. totally bounded, compact), then so
is the metric $d_{X}$.

###### Proof.

(1) Suppose that the metric $d_{X}$ is complete and take a Cauchy sequence
$(x_{n})_{n\in\mathbb{N}}$ in the metric space $(X,d_{IX}).$ Since $d_{X}\leq
d_{IX}$, the sequence $(x_{n})_{n\in\mathbb{N}}$ remains Cauchy in the metric
space $(X,d_{X})$. Since $(X,d_{X})$ is complete, this sequence converges to
some point $x\in X$ in the metric space $(X,d_{X})$. Lemma 3.3 guarantees that
$\lim_{n\to\infty}x_{n}=x$ in $(X,d_{IX})$ too, witnessing that the intrinsic
metric $d_{IX}$ is complete.

(2) If the metric $d_{IX}$ is analytic, then $d_{IX}$ is separable and $d_{X}$
is analytic, because of the continuity of the identity map
$(X,d_{IX})\to(X,d_{X})$.

Next, assume that the metric $d_{IX}$ is separable and $d_{X}$ is analytic.
Let $\overline{X}$ be the completion of the metric space $(X,d_{X})$ and
$\overline{X}_{I}$ be the completion of the metric space
$X_{I}\coloneqq(X,d_{IX})$. Let $\bar{I}:\overline{X}_{I}\to\overline{X}$ be
the continuous extension of the (1-Lipschitz) identity map $I:X_{I}\to X$. By
Lemma 3.3, $X_{I}=\bar{I}^{-1}[X]$. Since the spaces $\overline{X}_{I}$ and
$\overline{X}$ are Polish, the preimage $X_{I}=\bar{I}^{-1}[X]$ is an analytic
subspace of $\overline{X}_{I}$, by [13, 14.4].

The statement (3) follows from the $1$-Lipschitz property of the identity map
$(X,d_{IX})\to(X,d_{IX})$. ∎

The following simple example shows that the implications of Lemma 3.4 cannot
be reversed in a general case.

######  Example 3.5.

Let $L$ be any line in the complex plane $\mathbb{C}$ and
$o\in\mathbb{C}\setminus L$ be any point. For every uncountable compact zero-
dimensional set $K\subseteq L$, the cone

$X:=\bigcup_{x\in K}[o,x]$

over $K$ with vertex $o$ is a Lipschitz connected compact (by [8, 3.2.11])
metric space with respect to the metric $d_{X}$ inherited from the complex
plane. Thus $(X,d_{X})$ is also separable and totally bounded. Since
$d_{IX}(x,y)\geq 2\cdot\operatorname{dist}(o,L)$ for any distinct points
$x,y\in K$, the metric space $(X,d_{IX})$ is not separable (and hence not
totally bounded and not compact).

If the compact space $K$ has no isolated points, then for every countable
dense set $D\subseteq K$, the Lipschitz connected metric subspace
$Y=\bigcup_{x\in D}[o,x]$ of $X$ is not complete (by the Baire Theorem)
whereas the intrinsic metric $d_{IY}$ is $\sigma$-compact and complete.

######  Example 3.6.

The metric subspace $X=\\{(x,\sin(\frac{1}{x}))\colon x>0\\}$ of the Euclidean
plane is Lipschitz connected and not complete whereas the intrinsic metric on
$X$ is complete.

##  4\. The $\mathbb{R}$-trees generated by weighted trees

In this section we describe a general construction of a (complete)
$\mathbb{R}$-tree, determined by a weighted tree.

######  Definition 4.1.

A forest is a partially ordered set $(T,\preceq)$ such that for every $t\in T$
the set

${\downarrow}t\coloneqq\\{x\in T:x\preceq t\\}$

is finite and linearly ordered.

For any elements $s,t\in T$, the largest element of the linearly ordered set
${\downarrow}s\cap{\downarrow}t$ is called the infimum of $s$ and $t$ and is
denoted by $s\wedge t$.

For an element $t$ of a forest $T$, the set
$\mathrm{succ}(t)\coloneqq\min\\{s\in T:t\prec s\\}$ is called the set of
immediate successors of $t$ in the forest.

A subset $B$ of a forest $T$ is called a branch of $T$ if $B$ is a maximal
linearly ordered subset of $T$. The boundary $\partial T$ of a forest $T$ is
the set of all infinite branches in $T$.

For a forest $T$ by $\min T$ denote the set of minimal elements of $T$. For a
nonempty forest $T$, the set $\min T$ is not empty. A forest $T$ is a tree if
the set $\min T$ is a singleton, consisting of the smallest element of $T$,
called the root of the tree $T$ and denoted by $r_{T}$.

Observe that for every element $r$ of a forest $T$, the set

${\uparrow}r\coloneqq\\{x\in T:r\preceq x\\}$

is a tree with root $r$.

######  Definition 4.2.

A weighted tree is a pair $(T,w)$ consisting of a tree $T$ and a function
$w:T\to[0,+\infty)$ assigning to every vertex $t\in T$ its weight $w_{t}=w(t)$
so that $w_{t}=0$ if and only if $t$ is the root of $T$.

Every weighted tree $(T,w)$ determines an $\mathbb{R}$-tree

$T_{w}=(\\{r_{T}\\}\times\\{0\\})\cup\bigcup_{t\in T}\\{t\\}\times(0,w_{t}],$

endowed with the metric

$d_{w}((t,t^{\prime}),(s,s^{\prime})):=\begin{cases}\;|t^{\prime}-s^{\prime}|&\mbox{if
$s=t$};\\\
t^{\prime}-w_{t}+s^{\prime}-w_{s}+\sum_{x\in{\downarrow}s\triangle{\downarrow}t}w_{x}&\mbox{otherwise},\\\
\end{cases}$

where $A\triangle B$ stands for the symmetric difference $(A\setminus
B)\cup(B\setminus A)$ of the sets $A,B$. It is easy to check that the metric
space $(T_{w},d_{w})$ is indeed an $\mathbb{R}$-tree. The $\mathbb{R}$-tree
$T_{w}$ will be called the $\mathbb{R}$-tree of the weighted tree $(T,w)$.

The $\mathbb{R}$-tree $T_{w}\subset T\times\mathbb{R}$ carries a natural
lexicographic order

$(s,s^{\prime})\leq_{L}(t,t^{\prime})\quad\Leftrightarrow\quad s\prec
t\;\vee\;(s=t\;\wedge\;s^{\prime}\leq t^{\prime}).$

So, for every element $t^{\prime\prime}\in T_{w}$ we can consider its lower
and upper sets

${\downarrow}t^{\prime\prime}\coloneqq\\{s^{\prime\prime}\in
T_{w}:s^{\prime\prime}\leq_{L}t^{\prime\prime}\\}\quad\mbox{and}\quad{\uparrow}t^{\prime\prime}\coloneqq\\{s^{\prime\prime}\in
T_{w}:t^{\prime\prime}\leq_{L}s^{\prime\prime}\\}$

in the partially ordered set $(T_{w},\leq_{L})$. Since $T$ is a tree, the
lower set ${\downarrow}t^{\prime\prime}$ of every element $t^{\prime\prime}\in
T_{w}$ is linearly ordered. For two elements
$a^{\prime\prime},b^{\prime\prime}\in T_{w}$, let $a^{\prime\prime}\wedge
b^{\prime\prime}$ denote the largest element of the linearly ordered set
${\downarrow}a^{\prime\prime}\cap{\downarrow}b^{\prime\prime}$ in the
partially ordered set $(T_{w},\leq_{L})$.

Now we shall characterize weighted trees whose $\mathbb{R}$-trees are totally
bounded or separable.

######  Definition 4.3.

A weighted tree $(T,w)$ is defined to be precompact if for every
$\varepsilon>0$ there exists a finite set $F\subseteq T$ such that for every
finite linearly ordered set $L\subseteq T\setminus F$ we have $\sum_{x\in
L}w_{x}<\varepsilon$.

######  Proposition 4.4.

A weighted tree $(T,w)$ is precompact if and only if the metric space $T_{w}$
is totally bounded.

###### Proof.

To prove the “if” part, assume that the metric space $T_{w}$ is totally
bounded. Then for every $\varepsilon>0$, there exists a finite set
$F^{\prime\prime}\subseteq T_{w}$ such that for every $x^{\prime\prime}\in
T_{w}$, there exists $y^{\prime\prime}\in F^{\prime\prime}$ with
$d_{w}(x^{\prime\prime},y^{\prime\prime})<\varepsilon$. Since
$F^{\prime\prime}$ is finite, there exists a finite set $F\subseteq T$ such
that $F^{\prime\prime}\subseteq\bigcup\\{{\downarrow}(t,w_{t}):t\in
F\\}\subseteq T_{w}$. Replacing $F$ by the finite set
${\downarrow}F\coloneqq\bigcup_{t\in F}{\downarrow}t$, we can assume that
$F={\downarrow}F$. We claim that $\sum_{x\in L}w_{x}<\varepsilon$ for every
finite linearly ordered set $L\subseteq T\setminus F$. Since $L$ is finite and
linearly ordered, there exists an element $u\in L$ such that
$L\subseteq{\downarrow}u$. By the choice of the set $F^{\prime\prime}$, there
exists a point $y^{\prime\prime}\in F^{\prime\prime}$ such that
$d_{w}((u,w_{u}),y^{\prime\prime})<\varepsilon$. Write the element
$y^{\prime\prime}$ as $(y,y^{\prime})$ for some $y\in T$ and
$y^{\prime}\in[0,w_{y}]$. The choice of the set $F$ ensures that $y\in F$ and
hence $L\subseteq{\downarrow}u\setminus
F\subseteq{\downarrow}u\setminus{\downarrow}y$. Then

$\sum_{x\in L}w_{x}\leq\sum_{x\in{\downarrow}u\setminus{\downarrow}y}w_{x}\leq
d_{w}((u,w_{u}),y^{\prime\prime})<\varepsilon,$

witnessing that the weighted tree $(T,w)$ is precompact.

$L$$(u,w_{u})$$(y,w_{y})$$y^{\prime\prime}=(y,y^{\prime})\in
F^{\prime\prime}$$(u\wedge y,w_{u\wedge y})$

To prove the “only if” part, assume that the weighted tree $(T,w)$ is
precompact. To prove that the metric space $T_{w}$ is totally bounded, fix any
$\varepsilon>0$. By the precompactness of the weighted tree $(T,w)$, there
exists a nonempty finite set $F\subseteq T$ such that $\sum_{x\in
L}w_{x}<\varepsilon$ for every finite linearly ordered set $L\subseteq
T\setminus F$. Replacing $F$ with the finite set
${\downarrow}F\coloneqq\bigcup_{t\in F}{\downarrow}t$, we can assume that
$F={\downarrow}F$. Since the set $F$ is finite, the set

$K\coloneqq{\textstyle\bigcup}\\{{\downarrow}(t,w_{t}):t\in
F\\}=(\\{r_{T}\\}\times\\{0\\})\cup\bigcup_{t\in F}(\\{t\\}\times(0,w_{t}])$

is compact and hence totally bounded in the metric space $T_{w}$. Then there
exists a finite set $F^{\prime\prime}\subseteq K$ such that for every
$x^{\prime\prime}\in K$ there exists $y^{\prime\prime}\in F^{\prime\prime}$
with $d_{w}(x^{\prime\prime},y^{\prime\prime})<\varepsilon$. Replacing
$F^{\prime\prime}$ by the set $F^{\prime\prime}\cup\\{(t,w_{t}):t\in F\\}$, we
can assume that $\\{(t,w_{t}):t\in F\\}\subseteq F^{\prime\prime}$. We claim
that for every $x^{\prime\prime}\in T_{w}$, there exists $y^{\prime\prime}\in
F^{\prime\prime}$ such that
$d_{w}(x^{\prime\prime},y^{\prime\prime})<\varepsilon$.

Indeed, if $x^{\prime\prime}\in K$, then the existence of a point
$y^{\prime\prime}\in F^{\prime\prime}$ with
$d_{w}(x^{\prime\prime},y^{\prime\prime})<\varepsilon$ follows from the choice
of the set $F^{\prime\prime}$. So, we assume that $x^{\prime\prime}\notin K$.
In this case, $x^{\prime\prime}=(x,x^{\prime})$ for some $x\in T\setminus F$
and $x^{\prime}\in(0,w_{x}]$. Since $r_{T}\in{\downarrow}F=F$, there exists a
point $y\in F\cap{\downarrow}x$ such that $F\cap{\downarrow}x={\downarrow}y$.
The choice of the set $F^{\prime\prime}$ ensures that it contains the point
$y^{\prime\prime}\coloneqq(y,w_{y})$. The choice of $F$ guarantees that
$\sum_{t\in{\downarrow}x\setminus{\downarrow}y}w_{t}=\sum_{t\in{\downarrow}x\setminus
F}w_{t}<\varepsilon$ and hence
$d_{w}(x^{\prime\prime},y^{\prime\prime})=x^{\prime}-w_{x}+\sum_{t\in{\downarrow}x\setminus{\downarrow}y}w_{t}\leq\sum_{t\in{\downarrow}x\setminus{\downarrow}y}w_{t}<\varepsilon,$
by the definition of the metric $d_{w}$. This shows that the metric space
$T_{w}$ is totally bounded. ∎

######  Proposition 4.5.

For a weighted tree $(T,w)$, the $\mathbb{R}$-tree $T_{w}$ is separable if and
only if the tree $T$ is countable.

###### Proof.

If the tree $T$ is countable, then $T_{w}\cap(T\times\mathbb{Q})$ is a
countable dense set in the $\mathbb{R}$-tree, witnessing that the metric space
$T_{w}$ is separable.

Next, assume that the tree $T$ is uncountable. The definition of metric
$d_{w}$ ensures that for every node $t\in T$, the set $\\{t\\}\times(0,w_{t})$
is open in the metric space $T_{w}$. Therefore, the metric space $T_{w}$
contains the uncountable family of pairwise disjoint open sets
$\big{\\{}\\{t\\}\times(0,w_{t}):t\in T\big{\\}}$ and hence $T_{w}$ cannot be
separable, according to [8, Corollary 4.1.16]. ∎

Let $(T,w)$ be a weighted tree. In the set $\partial T$ of all infinite
branches of $T$, consider the subset

$\partial T_{w}:=\\{L\in\partial T:|L|=\infty,\;\sum_{t\in L}w_{t}<\infty\\}.$

On the set $\overline{T}_{w}:=T_{w}\cup\partial T_{w}$ we define a metric
$\bar{d}_{w}$, that extends the metric $d_{w}$ on $T_{w}$ as follows:

$\bar{d}_{w}(x^{\prime\prime},y^{\prime\prime}):=\begin{cases}\;|x^{\prime}-y^{\prime}|&\mbox{if
$x^{\prime\prime}=(x,x^{\prime}),y^{\prime\prime}=(y,y^{\prime})\in T_{w}$ and
$x=y$};\\\
x^{\prime}-w_{x}+y^{\prime}-w_{y}+\sum_{t\in{\downarrow}x\triangle{\downarrow}y}w_{t}&\mbox{if
$x^{\prime\prime}=(x,x^{\prime}),y^{\prime\prime}=(y,y^{\prime})\in T_{w}$ and
$x\neq y$};\\\ x^{\prime}-w_{x}+\sum_{t\in{\downarrow}x\triangle
y^{\prime\prime}}w_{t}&\mbox{if $x^{\prime\prime}=(x,x^{\prime})\in T_{w}$,
$y^{\prime\prime}\in\partial T_{w}$ and $x\notin y^{\prime\prime}$ };\\\
{w_{x}}-x^{\prime}+\sum_{t\in{y^{\prime\prime}\setminus{\downarrow}x}}w_{t}&\mbox{if
$x^{\prime\prime}=(x,x^{\prime})\in T_{w}$, $y^{\prime\prime}\in\partial
T_{w}$ and $x\in y^{\prime\prime}$ };\\\ \;\sum_{t\in
x^{\prime\prime}\triangle y^{\prime\prime}}w_{t}&\mbox{if
$x^{\prime\prime},y^{\prime\prime}\in\partial T_{w}$}.\end{cases}$

######  Proposition 4.6.

For every weighted tree $(T,w)$, the metric space $\overline{T}_{w}$ is a
completion of the metric space $T_{w}$.

###### Proof.

The definition of the metric on $\overline{T}_{w}$ implies that the metric
space $T_{w}$ is a dense subspace of the metric space $\overline{T}_{w}$. To
prove that $\overline{T}_{w}$ is a completion of $T_{w}$, it suffices to show
that every Cauchy sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ in the metric
space $T_{w}$ has a limit in the space $\overline{T}_{w}$. If for some element
$t$ of $T_{w}$, the set $\\{n\in\omega:s^{\prime\prime}_{n}=t\\}$ is infinite,
then $t$ is the limit of the Cauchy sequence
$(s^{\prime\prime}_{n})_{n\in\omega}$ and we are done. So, we can assume that
all elements of the Cauchy sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ are
pairwise distinct and moreover all of them are distinct from the root
$(r_{T},0)$ of the $\mathbb{R}$-tree $T_{w}$. For every $n\in\omega$, write
the element $s^{\prime\prime}_{n}$ of $T_{w}$ as a pair
$(s_{n},s_{n}^{\prime})$ where $s_{n}\in T$ and
$s_{n}^{\prime}\in(0,w_{s_{n}}]$. Let
$\check{s}_{n}\coloneqq\max({\downarrow}s_{n}\setminus\\{s_{n}\\})$ be the
immediate predecessor of $s_{n}$ in the tree $T$.

In the tree $T$, consider the set

$\Lambda\coloneqq\\{t\in T:|\\{n\in\omega:t\preceq\check{s}_{n}\\}|=\omega\\}$

and observe that it contains the root $r_{T}$ of the tree $T$ and hence is not
empty. Two cases are possible.

1\. First assume that the set $\Lambda$ has no maximal elements. In this case
$\Lambda$ contains an increasing sequence $(\lambda_{n})_{n\in\omega}$. For
this sequence the set $b\coloneqq\bigcup_{n\in\omega}{\downarrow}\lambda_{n}$
is an infinite branch of the tree $T$ such that for every $t\in b$ the set
$\\{n\in\omega:t\preceq\check{s}_{n}\\}$ is infinite. Write the branch $b$ as
$b=\\{b_{n}\\}_{n\in\omega}$, where $b_{n}$ is a unique element of $b$ such
that $|{\downarrow}b_{n}|=n+1$. We claim that $b\in\partial T_{w}$. Since the
sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ is Cauchy, for every
$\varepsilon>0$, there exists $m\in\mathbb{N}$ such that
$d_{w}(s^{\prime\prime}_{n},s^{\prime\prime}_{m})<\varepsilon$ for every
$n\geq m$.

Since the sequence $(b_{n})_{n\in\omega}$ is unbounded in the tree $T$, there
exists a number $i\in\omega$ such that $b_{i}\not\preceq\check{s}_{m}$. Given
any number $j\geq i$, find a number $n\geq m$ such that
$b_{j}\preceq\check{s}_{n}$. Observe that $b_{i},b_{j},\check{s}_{n}$ and
$\check{s}_{n}\wedge\check{s}_{m}$ are elements of the linearly ordered set
${\downarrow}\check{s}_{n}$. Assuming that
$\check{s}_{m}\wedge\check{s}_{n}\not\preceq b_{i}$, we conclude that
$b_{i}\preceq\check{s}_{m}\wedge\check{s}_{n}\preceq\check{s}_{m}$, which
contradicts the choice of the number $i$. This contradiction shows that
$\check{s}_{m}\wedge\check{s}_{n}\preceq b_{i}$.

In this case, the definition of the metric $d_{w}$ ensures that

$\sum_{k=i+1}^{j}w_{b_{k}}\leq-
w_{s_{n}}-w_{s_{m}}+\sum_{x\in{\downarrow}s_{n}\triangle{\downarrow}s_{m}}w_{x}\leq
d_{w}(s^{\prime\prime}_{m},s^{\prime\prime}_{n})<\varepsilon$

and hence $\sum_{k=i+1}^{\infty}w_{b_{k}}\leq\varepsilon<\infty$, witnessing
that $b\in\partial T_{w}$.

$\check{s}_{n}$$s_{n}$$b_{j}$$b_{j+1}$$\check{s}_{m}\wedge\check{s}_{n}$$b_{i}$$b_{i+1}$$\check{s}_{m}$$s_{m}$

Since the sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ is Cauchy and
$\varepsilon>0$ was arbitrary, the equality
$b=\lim_{n\to\infty}s^{\prime\prime}_{n}$ will follow as soon as we check that
$\overline{d}_{w}(b,s^{\prime\prime}_{n})<2\varepsilon$ for infinitely many
numbers $n$. Indeed, for every number $n\geq m$ in the infinite set
$\\{n\in\omega:b_{i}\preceq\check{s}_{n}\\}$ we have
$\check{s}_{n}\wedge\check{s}_{m}\preceq b_{i}$ and

$\displaystyle\overline{d}_{w}(s^{\prime\prime}_{n},b)$
$\displaystyle=s_{n}^{\prime}-w_{s_{n}}+\sum_{x\in{\downarrow}s_{n}\triangle
b}w_{x}=s_{n}^{\prime}-w_{s_{n}}+\sum_{x\in{\downarrow}s_{n}\setminus
b}w_{x}+\sum_{x\in b\setminus{\downarrow}s_{n}}w_{x}$ $\displaystyle\leq
s^{\prime}_{n}-w_{s_{n}}+\sum_{x\in{\downarrow}s_{n}\setminus{\downarrow}b_{i}}w_{x}+{\sum_{k=i+1}^{\infty}w_{b_{k}}}$
$\displaystyle\leq
s^{\prime}_{n}-w_{s_{n}}+\sum_{x\in{\downarrow}s_{n}\setminus{\downarrow}s_{m}}w_{x}+\sum_{k=i+1}^{\infty}w_{b_{k}}\leq
d_{w}(s^{\prime\prime}_{n},s^{\prime\prime}_{m})+\varepsilon<2\varepsilon.$

2\. Next, assume that the set $\Lambda$ has a maximal element $\lambda$.
Consider the infinite set
$I\coloneqq\\{n\in\omega:\lambda\preceq\check{s}_{n}\\}$. If for some $s\in
T$, the set $J\coloneqq\\{n\in I:s=s_{n}\\}$ is infinite, then the maximality
of $\lambda$ ensures that $\lambda$ is the immediate predecessor of $s$ in the
tree $T$. Then the subsequence $(s_{n}^{\prime\prime})_{n\in J}$ of the
sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ is contained in the compact set
$\\{(\lambda,w_{\lambda})\\}\cup(\\{s\\}\times(0,w_{s}])$ and hence the Cauchy
sequence $(s_{n}^{\prime\prime})_{n\in\omega}$ converges to some points of
this compact set. So, assume that for every $s\in T$, the set $\\{n\in
I:s_{n}=s\\}$ is finite. In this case we shall show that the Cauchy sequence
$(s_{n}^{\prime\prime})_{n\in\omega}$ converges to the element
$\lambda^{\prime\prime}\coloneqq(\lambda,w_{\lambda})$ of the
$\mathbb{R}$-tree $T_{w}$. Since the sequence
$(s_{n}^{\prime\prime})_{n\in\omega}$ is Cauchy and consists of pairwise
distinct elements, it suffices to show that for every $\varepsilon>0$ there
exists $n\in\omega$ such that
$d_{w}(\lambda^{\prime\prime},s_{n}^{\prime\prime})<\varepsilon$. Since the
sequence $(s^{\prime\prime}_{n})_{n\in\omega}$ is Cauchy, there exists
$m\in\omega$ such that
$d_{w}(s_{n}^{\prime\prime},s_{k}^{\prime\prime})<\varepsilon$ for all
$n,k\geq m$. We can assume that $\lambda\preceq\check{s}_{m}\prec s_{m}$. Thus
the element $\lambda$ has an immediate successor $\hat{\lambda}$ in the tree
$T$ such that $\lambda\prec\hat{\lambda}\preceq s_{m}$. The maximality of
$\lambda$ ensures that $\hat{\lambda}\notin\Lambda$ and hence the set
$\\{n\in\omega:\hat{\lambda}\preceq\check{s}_{n}\\}$ is finite. Then we can
choose a number $n\in I$ such that $n\geq m$ and
$\hat{\lambda}\not\preceq\check{s}_{n}$. Then $s_{n}\wedge
s_{m}=\check{s}_{n}=\lambda$ and

$d_{w}(\lambda^{\prime\prime},s^{\prime\prime}_{n})\leq
s^{\prime}_{n}-w_{s_{n}}+s^{\prime}_{m}-w_{s_{m}}+\sum_{x\in{\downarrow}s_{n}\triangle{\downarrow}s_{m}}w_{x}=d_{w}(s_{n}^{\prime\prime},s_{m}^{\prime\prime})<\varepsilon.$

$\hat{s}_{n}{=}\lambda$$\hat{\lambda}$$s_{m}$$s_{n}$

∎

Propositions 2.4, 4.4, 4.5, 4.6 and the definition of the metric
$\overline{d}_{w}$ imply the following corollary describing the property of
the completion $\overline{T}_{w}$ of the $\mathbb{R}$-tree $T_{w}$ of a
weighted tree $(T,w)$.

######  Corollary 4.7.

For every weighted tree $(T,w)$,

  1. (1)

the complete metric space $\overline{T}_{w}$ is an $\mathbb{R}$-tree and hence
is an injective metric space;

  2. (2)

the space $\overline{T}_{w}$ is compact if and only if the weighted tree
$(T,w)$ is precompact;

  3. (3)

the space $\overline{T}_{w}$ is separable if and only if the tree $T$ is
countable.

##  5\. Proof of Theorem 1.2

To prove Theorem 1.2, we need to demonstrate the equivalence of the following
conditions for every metric space $X$:

  1. (1)

$X$ is a Lipschitz image of an injective metric space;

  2. (2)

$X$ is a Lipschitz image of some complete $\mathbb{R}$-tree;

  3. (3)

$X$ is Lipschitz connected.

The implication $(2)\Rightarrow(1)$ follows from Proposition 2.4 (saying that
every complete $\mathbb{R}$-tree is an injective metric space), and the
implication $(1)\Rightarrow(3)$ follows from the Lipschitz connectedness of
injective metric spaces, proved in Example 3.2.

To prove that $(3)\Rightarrow(2)$, assume that a metric space $X$ is Lipschitz
connected. Fix any point $x_{0}\in X$ and for every point $a\in X$ find a real
number $w_{a}$ and a $1$-Lipschitz path $\gamma_{a}\colon[0,w_{a}]\to X$ such
that $\gamma_{a}(0)=x_{0}$ and $\gamma_{a}(w_{a})=a$. If $a=x_{0}$, then will
assume that $w_{a}=0$.

Consider the tree $T\coloneqq X$ endowed with the partial order

$\preceq:=\\{(x,y)\in X\times X:x=y\;\vee\;x=x_{0}\\}$

The function $w:X\to[0,\infty)$, $w:a\mapsto w_{a}$, turns $T$ into a weighted
tree. The weighted tree $(T,w)$ generates the $\mathbb{R}$-tree $T_{w}$. Since
the tree $T$ contains no infinite branches, the $\mathbb{R}$-tree $T_{w}$
coincides with its completion $\overline{T}_{w}=T_{w}\cup\partial T_{w}$ and
hence $T_{w}$ is a complete $\mathbb{R}$-tree.

The $1$-Lipschitz maps $\gamma_{a}$, $a\in X$, compose a surjective
$1$-Lipschitz map $\gamma:T_{w}\to X$ such that $\gamma((r_{T},0))=x_{0}$ and
$\gamma((a,t))=\gamma_{a}(t)$ for $(a,t)\in\bigcup_{a\in
X}\\{a\\}\times(0,w_{a}]$. Indeed for any $(a,t),(b,s)\in T_{w}$ if $a=b$

$d_{X}(\gamma((a,t)),\gamma((b,s)))=d_{X}(\gamma_{a}(t),\gamma_{a}(s))\leq|t-s|,$

because $\gamma_{a}$ is a 1-Lipschitz map. If $a\neq b$

$\displaystyle d_{X}(\gamma((a,t)),\gamma((b,s)))$
$\displaystyle=d_{X}(\gamma_{a}(t),\gamma_{b}(s))\leq
d_{X}(\gamma_{a}(t),x_{0})+d_{X}(x_{0},\gamma_{b}(s))=$
$\displaystyle=d_{X}(\gamma_{a}(t),\gamma_{a}(0))+d_{X}(\gamma_{b}(0),\gamma_{b}(s))\leq
t+s,$

because $\gamma_{a}$, $\gamma_{b}$ are 1-Lipschitz maps. The $1$-Lipschitz map
$\gamma$ witnesses that $X$ is a $1$-Lipschitz image of the complete
$\mathbb{R}$-tree $T_{w}$.

##  6\. Proof of Theorem 1.3

Theorem 1.3 will follow as soon as we check the equivalence of the following
conditions for every metric space $X$:

  1. (1)

$X$ is a Lipschitz image of some compact $\mathbb{R}$-tree;

  2. (2)

$X$ is a Lipschitz image of a totally bounded injective metric space;

  3. (3)

$X$ is a Lipschitz image of a compact $1$-Lipschitz connected metric space;

  4. (4)

$X$ is a Lipschitz image of an $L$-Lipschitz connected compact metric space
for some $L\geq 1$;

  5. (5)

$X$ is compact, Lipschitz connected, and the intrinsic metric $d_{IX}$ is
totally bounded.

The implication $(1)\Rightarrow(2)$ follows from Proposition 2.4 (saying that
every complete $\mathbb{R}$-tree is an injective metric space), and the total
boundedness of compact metric spaces.

The implication $(2)\Rightarrow(3)$ follows from the completeness and
$1$-Lipschitz connectedness of injective metric spaces, and the implication
$(3)\Rightarrow(4)$ is obvious.

$(4)\Rightarrow(5)$ Assume that a metric space $(X,d_{X})$ is the image of
some $L$-Lipschitz connected compact metric space $(Y,d_{Y})$ under a
Lipschitz map $f:Y\to X$. Then $X$ is compact and Lipschitz connected, so the
intrinsic metric $d_{IX}$ on $X$ is well-defined. Scaling the metric of the
metric space $Y$, we can assume that the Lipschitz map $f$ is $1$-Lipschitz.
In this case the definition of the intrinsic metric implies that the function
$f$ remains $1$-Lipschitz with respect to the intrinsic metrics on the metric
spaces $X$ and $Y$. The $L$-Lipschitz connectedness of the metric space $Y$
implies that $d_{Y}\leq d_{IY}\leq L\cdot d_{Y}$, which implies that the
identity map $(Y,d_{Y})\to(Y,d_{IY})$ is Lipschitz and hence the metric space
$(Y,d_{IY})$ is totally bounded and so is its $1$-Lipschitz image
$(X,d_{IX})$.

$(5)\Rightarrow(1)$ Assume that the metric space $(X,d_{X})$ is compact,
Lipschitz connected, and its intrinsic metric $d_{IX}$ is totally bounded. Let
$D\coloneqq\sup\\{d_{IX}(x,y):x,y\in X\\}$ be the diameter of the (totally)
bounded metric space $(X,d_{IX})$.

Since the metric space $(X,d_{IX})$ is totally bounded, there exists a
sequence $(\mathcal{F}_{n})_{n\in\omega}$ of finite partitions of $X$ such
that $\mathcal{F}_{0}=\\{X\\}$ and for every $n\in\omega$, every set
$F\in\mathcal{F}_{n}$ has $d_{IX}$-diameter $\sup\\{d_{IX}(x,y):x,y\in
F\\}<\frac{D+1}{2^{n}}$. Replacing each partition $\mathcal{F}_{n}$ with
$n\in\mathbb{N}$ by the partition
$\mathcal{F}_{n}\wedge\mathcal{F}_{n-1}\coloneqq\\{A\cap
B:A\in\mathcal{F}_{n-1},\;\;B\in\mathcal{F}_{n},\;\;A\cap
B\neq\varnothing\\}$, we can assume that every set $F\in\mathcal{F}_{n}$ is
contained in a unique set $\check{F}\in\mathcal{F}_{n-1}$. For every
$n\in\omega$ and $F\in\mathcal{F}_{n}$, choose a point $c_{F}\in F$.

Consider the tree
$T\coloneqq\bigcup_{n\in\omega}(\\{n\\}\times\mathcal{F}_{n})$ endowed with
the partial order $\preceq$, defined by $(n,F)\preceq(m,E)$ iff $n\leq m$ and
$E\subseteq F$. Next, endow the tree $T$ with the weight function
$w:T\to[0,\infty)$ defined by

$w(n,F)=w_{(n,F)}=\begin{cases}\frac{D+1}{2^{n-1}}&\mbox{ if $n>0$};\\\
0&\mbox{if $n=0$}.\end{cases}$

Figure 1. The weighted tree $(T,w)$.

Let $T_{w}$ be the $\mathbb{R}$-tree of the weighted tree $(T,w)$. Since the
families $\mathcal{F}_{n}$ are finite and $\sum_{t\in b}w_{t}=2(D+1)$ for
every $b\in\partial T$, the weighted tree $(T,w)$ is precompact. By
Proposition 4.4, the metric space $T_{w}$ is totally bounded and hence its
completion $\overline{T}_{w}$ is a compact $\mathbb{R}$-tree.

It remains to construct a $1$-Lipschitz surjective map
$\overline{\gamma}:\overline{T}_{w}\to X$. For every $n\in\mathbb{N}$ and
$F\in\mathcal{F}_{n}$, choose a $1$-Lipschitz function
$\gamma_{n,F}:[0,w_{(n,F)}]\to X$ such that $\gamma_{(n,F)}(0)=c_{\check{F}}$
and $\gamma_{(n,F)}(w_{(n,F)})=c_{F}$. The function $\gamma_{(n,F)}$ exists
because $\\{c_{\check{F}},c_{F}\\}\subseteq\check{F}\cup F=\check{F}$ and the
set $\check{F}\in\mathcal{F}_{n-1}$ has $d_{IX}$-diameter
$<\frac{D+1}{2^{n-1}}=w_{(n,F)}$. Next, consider the function $\gamma:T_{w}\to
X$ defined by $\gamma((0,X),0)=c_{X}$ and $\gamma((n,F),t)=\gamma_{n,F}(t)$
for every $n\in\mathbb{N}$, $F\in\mathcal{F}_{n}$ and $t\in(0,w_{(n,F)}]$. The
definition of the metric $d_{w}$ on the $\mathbb{R}$-tree $T_{w}$ ensures that
the map $\gamma:T_{w}\to X$ is $1$-Lipschitz. Observe that the image
$\gamma[T_{w}]$ is dense in $X$ (because it contains the dense subset
$\\{c_{F}:n\in\mathbb{N},\;F\in\mathcal{F}_{n}\\}$ of $X$).

Since the metric space $X$ is complete, the $1$-Lipschitz map $\gamma:T_{w}\to
X$ admits a unique $1$-Lipschitz extension
$\overline{\gamma}:\overline{T}_{w}\to X$. The total boundedness of the
$\mathbb{R}$-tree $T_{w}$ implies the compactness of its completion
$\overline{T}_{w}$. Then the image $\overline{\gamma}[\overline{T}_{w}]$ is a
compact dense subset of $X$, witnessing that the $1$-Lipschitz map
$\overline{\gamma}:\overline{T}_{w}\to X$ is surjective and hence $X$ is a
$1$-Lipschitz image of the compact $\mathbb{R}$-tree $\overline{T}_{w}$.

######  Remark 6.1.

Although compactness of $(X,d_{X})$ follows from any of the conditions
(1)–(5), the compactness of $(X,d_{X})$ is not sufficient to infer these
conditions, which is showed in Example 3.5.

##  7\. Proof of Theorem 1.4

Theorem 1.4 will follow as soon as we check the equivalence of the following
conditions for every metric space $X$:

  1. (1)

$X$ is a Lipschitz image of some separable complete $\mathbb{R}$-tree;

  2. (2)

$X$ is a Lipschitz image of a separable injective metric space;

  3. (3)

$X$ is a $1$-Lipschitz image of the Urysohn universal metric space;

  4. (4)

$X$ is a $1$-Lipschitz image of a $1$-Lipschitz connected separable complete
metric space;

  5. (5)

$X$ is a Lipschitz image of an $L$-Lipschitz connected analytic metric space
for some $L\geq 1$;

  6. (6)

$X$ is analytic, Lipschitz connected, and the intrinsic metric $d_{IX}$ is
separable.

The implication $(1)\Rightarrow(2)$ follows from Proposition 2.4 (saying that
every complete $\mathbb{R}$-tree is an injective metric space).

To prove that $(2)\Rightarrow(3)$, assume that the metric space $X$ is the
image of a separable injective space $Y$ under a Lipschitz map $f:Y\to X$.
Scaling the metric of the metric space $Y$, we can assume that the Lipschitz
map $f$ is $1$-Lipschitz. By [18], the Urysohn universal metric space
$\mathbb{U}$ contains an isometric copy of any separable metric space. So,
there exists an isometric embedding $e:Y\to\mathbb{U}$. By the injectivity of
$Y$, the $1$-Lipschitz map $e^{-1}:e[Y]\to Y$ defined on the subset $e[Y]$ of
$\mathbb{U}$ admits a $1$-Lipschitz extension $g:\mathbb{U}\to Y$. Then
$f\circ g:\mathbb{U}\to X$ is a surjective $1$-Lipschitz map, witnessing that
$X$ is a $1$-Lipschitz image of the Urysohn universal metric space
$\mathbb{U}$.

To see the implication $(3)\Rightarrow(4)$ it suffices to check that the
Urysohn universal metric space $\mathbb{U}$ is $1$-Lipschitz connected. Given
any points $x,y\in\mathbb{U}$, consider the function
$f:\\{0,d(x,y)\\}\to\\{x,y\\}\subseteq\mathbb{U}$ such that $f(0)=x$ and
$f(d(x,y))=y$. By [3, 12] (see also [2, 15, 17]), the isometric embedding
$f:\\{0,d(x,y)\\}\to\mathbb{U}$ extends to a $1$-Lipschitz map
$f:[0,d(x,y)]\to\mathbb{U}$, witnessing that $\mathbb{U}$ is $1$-Lipschitz
connected.

The implication $(4)\Rightarrow(5)$ is trivial.

$(5)\Rightarrow(6)$ Assume that for some $L\geq 1$, a metric space $(X,d_{X})$
is the image of some $L$-Lipschitz connected analytic metric space $(Y,d_{Y})$
under a Lipschitz map $f:Y\to X$. Then $X$ is analytic and Lipschitz
connected, so the intrinsic metric $d_{IX}$ on $X$ is well-defined. Scaling
the metric of the metric space $Y$, we can assume that the Lipschitz map $f$
is $1$-Lipschitz. In this case the definition of the intrinsic metric implies
that the function $f$ remains $1$-Lipschitz with respect to the intrinsic
metrics on the metric spaces $X$ and $Y$. The $L$-Lipschitz connectedness of
the metric space $Y$ implies that $d_{Y}\leq d_{IY}\leq L\cdot d_{Y}$, which
implies that the identity map $(Y,d_{Y})\to(Y,d_{IY})$ is Lipschitz and hence
the metric space $(Y,d_{IY})$ is separable and so is its $1$-Lipschitz image
$(X,d_{IX})$.

$(6)\Rightarrow(1)$ Assume that the metric space $(X,d_{X})$ is analytic,
Lipschitz connected, and its intrinsic metric $d_{IX}$ is separable. By
Proposition 3.4, the metric space $X_{I}\coloneqq(X,d_{IX})$ is analytic and
hence it is the image of a separable complete metric space $(Z,d_{Z})$ under a
surjective continuous map $f:Z\to X_{I}$. Consider the metric $\hat{d}_{Z}$ on
$Z$, defined by the formula
$\hat{d}_{Z}(z,z^{\prime})\coloneqq\max\\{d_{Z}(z,z^{\prime}),d_{IX}(f(z),f(z^{\prime}))\\}$.
The continuity of the function $f$ ensures that the metric $\hat{d}_{Z}$ is
topologically equivalent to the metric $d_{Z}$. Since $d_{Z}\leq\hat{d}_{Z}$,
the completeness of the metric $d_{Z}$ implies the completeness of the metric
$\hat{d}_{Z}$. Replacing the metric $d_{Z}$ by the metric $\hat{d}_{Z}$, we
can assume that the function $f:Z\to X_{I}$ is $1$-Lipschitz.

Since the metric space $Z$ is separable, there exists a sequence
$(\mathcal{F}_{n})_{n\in\omega}$ of countable partitions of $Z$ such that
$\mathcal{F}_{0}=\\{Z\\}$ and for every $n\in\mathbb{N}$, every set
$F\in\mathcal{F}_{n}$ has diameter $\sup\\{d_{Z}(x,y):x,y\in
F\\}<\frac{1}{2^{n}}$. Replacing each partition $\mathcal{F}_{n}$ by the
partition $\mathcal{F}_{n}\wedge\mathcal{F}_{n-1}\coloneqq\\{A\cap
B:A\in\mathcal{F}_{n-1},\;\;B\in\mathcal{F}_{n},\;\;A\cap
B\neq\varnothing\\}$, we can assume that every set $F\in\mathcal{F}_{n}$ is
contained in a unique set $\check{F}\in\mathcal{F}_{n-1}$. For every
$n\in\omega$ and $F\in\mathcal{F}_{n}$, choose a point $c_{F}\in F$.

Consider the countable tree
$T\coloneqq\bigcup_{n\in\omega}(\\{n\\}\times\mathcal{F}_{n})$ endowed with
the partial order $\preceq$, defined by $(n,F)\preceq(m,E)$ iff $n\leq m$ and
$E\subseteq F$. Next, endow the tree $T$ with the weight function
$w:T\to[0,\infty)$ defined by $w_{(0,Z)}=0$ and
$w_{(n,F)}=d_{IX}(f(c_{\check{F}}),f(c_{F}))+\tfrac{1}{2^{n}}$ for every
$n\in\mathbb{N}$ and $F\in\mathcal{F}_{n}$ (in this formula $\check{F}$
denotes the unique set in $\mathcal{F}_{n-1}$ that contains the set $F$).

Let $T_{w}$ be the $\mathbb{R}$-tree of the weighted tree $(T,w)$ and
$\overline{T}_{w}=T_{w}\cup\partial T_{w}$ be its completion. Proposition 4.5
ensures that the $\mathbb{R}$-tree $T_{w}$ is separable and hence its
completion $\overline{T}_{w}$ is a separable complete $\mathbb{R}$-tree.

It remains to construct a $1$-Lipschitz surjective map
$\overline{\gamma}:\overline{T}_{w}\to X$. For every $n\in\mathbb{N}$ and
$F\in\mathcal{F}_{n}$, choose a $1$-Lipschitz function
$\gamma_{n,F}:[0,w_{(n,F)}]\to X$ such that $\gamma_{n,F}(0)=f(c_{\check{F}})$
and $\gamma_{n,F}(w_{(n,F)})=f(c_{F})$. The function $\gamma_{n,F}$ exists
because $d_{IX}(f(c_{\check{F}}),f(c_{F}))<w_{(n,F)}$. Next, consider the
function $\gamma:T_{w}\to X$ defined by $\gamma((0,X),0)=f(c_{Z})$ and
$\gamma((n,F),t)=\gamma_{n,F}(t)$ for every $n\in\mathbb{N}$,
$F\in\mathcal{F}_{n}$ and $t\in(0,w_{(n,F)}]$. The definition of the metric
$d_{w}$ on the $\mathbb{R}$-tree $T_{w}$ ensures that the map $\gamma:T_{w}\to
X$ is $1$-Lipschitz.

Consider the set $\partial T$ of all infinite branches of the tree $T$ and
observe that every branch $b\in\partial T$ is equal to the set
$\\{(n,F_{n}):n\in\omega\\}$ for some decreasing sequence
$(F_{n})_{n\in\omega}\in\prod_{n\in\omega}\mathcal{F}_{n}$. Observe that for
every $n\geq 2$, the set $\\{c_{\check{F}_{n}},c_{F_{n}}\\}$ is a subset of
the set $\check{F}_{n}\in\mathcal{F}_{n-1}$ and hence
$d_{Z}(c_{F_{n-1}},c_{F_{n}})=d_{Z}(c_{\check{F}_{n}},c_{F_{n}})<\frac{1}{2^{n-1}}$.
Then the sequence $(c_{F_{n}})_{n\in\omega}$ is Cauchy in the complete metric
space $Z$ and has a limit $\lambda(b)\coloneqq\lim_{n\to\infty}c_{F_{n}}\in
Z$, which coincides with the unique point of the intersection
$\bigcap_{n\in\omega}\overline{F_{n}}$. The correspondence $\lambda:\partial
T\mapsto\lambda(b)$, determines a function $\lambda:\partial T\to Z$. Observe
that for every point $z\in Z$ there exists a unique branch
$b=\\{(n,F_{n})\\}_{n\in\omega}\in\partial T$ such that
$z\in\bigcap_{n\in\omega}F_{n}=\bigcap_{n\in\omega}\overline{F}_{n}=\\{\lambda(b)\\}$,
witnessing that the function $\lambda:\partial T\to Z$ is surjective. Taking
into account that diameters of the sets in the partition $\mathcal{F}_{n}$
tend to zero as $n$ tends to infinity, one can show that the map
$\lambda:\partial T\to Z$ is continuous with respect to the topology on
$\partial T$ inherited from the Tychonoff product
$\prod_{n\in\omega}(\\{n\\}\times\mathcal{F}_{n})$ of discrete spaces
$\\{n\\}\times\mathcal{F}_{n}$, $n\in\omega$.

Observe that for every branch $b=\\{(n,F_{n})\\}_{n\in\omega}$ of the tree $T$
and every $n\in\mathbb{N}$ we have
$w_{(n,F_{n})}=d_{IX}(f(c_{F_{n-1}}),f(c_{F_{n}}))+\frac{1}{2^{n}}\leq
d_{Z}(c_{F_{n-1}},c_{F_{n}})+\frac{1}{2^{n}}<\frac{1}{2^{n-1}}+\tfrac{1}{2^{n}}$
and hence $\sum_{t\in b}w_{t}=\sum_{n\in\omega}w_{(n,F_{n})}<\infty$,
witnessing that $\partial T_{w}=\partial T$.

Since the map $\gamma:T_{w}\to X$ is $1$-Lipschitz, there exists a unique
$1$-Lipschitz extension $\overline{\gamma}:\overline{T}_{w}\to\overline{X}$ to
the completion of the space $X$. We claim that
$\overline{\gamma}[\overline{T}_{w}]=X$.

The inclusion $\overline{\gamma}[\overline{T}_{w}]\subseteq X$ will follow as
soon as we check that $\overline{\gamma}[\partial T]\subseteq X$. Fix any
element $b\in\partial T$ and find a unique decreasing sequence of sets
$(F_{n})_{n\in\omega}\in\prod_{n\in\omega}\mathcal{F}_{n}$ such that
$b=\\{(n,F_{n})\\}_{n\in\omega}$. Then
$\lambda(b)=\lim_{n\to\infty}c_{F_{n}}$. The definition of the metric on the
$\mathbb{R}$-tree $\overline{T}_{w}$ ensures that

$b=\lim_{n\to\infty}((n,F_{n}),w_{(n,F_{n})})$

in $\overline{T}_{w}$ and hence
$\overline{\gamma}(b)=\lim_{n\to\infty}\gamma((n,F_{n}),w_{(n,F_{n})})=\lim_{n\to\infty}\gamma_{n,F}(w_{(n,F_{n})})=\lim_{n\to\infty}f(c_{F_{n}})=f(\lim_{n\to\infty}c_{F_{n}})=f(\lambda(b))\in
f[Z]=X$.

Next, we show that $X\subseteq\overline{\gamma}[\overline{T}_{w}]$. Given any
point $x\in X$, choose any point $z\in f^{-1}(x)\subseteq Z$ and find a unique
branch $b=\\{(n,F_{n})\\}_{n\in\omega}\in\partial T$ such that
$z\in\bigcap_{n\in\omega}F_{n}$. Then $z=\lambda(b)$ and
$x=f(\lambda(b))=\overline{\gamma}(b)$.

Therefore, $\overline{\gamma}:\overline{T}_{w}\to X$ is a $1$-Lipschitz
surjective map witnessing that $X$ is a $1$-Lipschitz image of the separable
complete $\mathbb{R}$-tree $\overline{T}_{w}$.

######  Remark 7.1.

Although analyticity and separability of $(X,d_{X})$ follows from each of the
conditions (1)–(6), the analyticity of $(X,d_{X})$ is not sufficient to infer
the equivalent conditions of Theorem 1.4, which is showed in Example 3.5.

## Acknowledgement

This work was co-financed by the Minister of Science (Poland) under the
“Regional Excellence Initiative” program (project no.: RID/SP/0015/2024/01).

## References

  * [1] N. Aronszajn, P. Panitchpakdi, _Extension of uniformly continuous transformations and hyperconvex metric spaces_ , Pacific J. Math. 6:3 (1956), 405–439.
  * [2] T. Banakh, F. Strobin, _Embedding topological fractals in universal spaces_ , J. Fractal Geom. 2 (2015), no. 4, pp. 377–388.
  * [3] S. A. Bogatyi, _Compact homogeneity of Uryson’s universal metric space_ (Russian), Russ. Math. Surv. 55, 332, 2000.
  * [4] B. H. Bowditch, _Median and injective metric spaces_ , Mathematical Proceedings of the Cambridge Philosophical Society 168:1 (2020) 43–55.
  * [5] Y. Beniamini, J. Lindenstrauss, _Geometric Nonlinear Functional Analysis_ , American Mathematical Society Colloquium Publications, v. 48, 2000.
  * [6] I. Chiswell, _Introduction to $\Lambda$-trees_, World Scientific Publishing Co. Pte. Ltd., 2001.
  * [7] S. Eilenberg, O.G. Harrold, Jr. Continua of finite linear measure. I, Amer. J. Math. 65 (1943), 137–146.
  * [8] R. Engelking, _General topology_ , Helderman Verlag Berlin, Sigma series in pure mathematics, vol. 6, 1989.
  * [9] M. Espinola, M.A. Khamsi, _Introduction to hyperconvex spaces_ In: Kirk, W.A., Sims, B. (eds) Handbook of Metric Fixed Point Theory. Springer, Dordrecht. DOI:10.1007/978-94-017-1748-9_13.
  * [10] K. Goebel and W. A. Kirk, _Topics in Metric Fixed Point Theory_ , Cambridge Univ. Press, Cambridge, 1990.
  * [11] H. Hahn, _Mengentheoretische Charakterisierung der stetigen Kurven_ , Sitzungsber. Akad. Wiss. Wien 123 (1914), 2433–2489.
  * [12] G. E. Huhunaišvili, _On a property of Uryson’s universal metric space_ (Russian), Soviet Math. Dokl., 101 (1955), 607–610.
  * [13] A.Kechris, Classical Descriptive Set Theory, Springer, 1995.
  * [14] W. Kirk, _Hyperconvexity of $\mathbb{R}$-trees_, Fund. Math. 156 (1998), 67–72.
  * [15] W. Kubiś, M. Rubin, _Extension theorems and reconstruction theorems for the Urysohn universal space_ , Czechoslovak Math. J. 60:1 (2010), 1–29.
  * [16] S. Mazurkiewicz, _Sur les lignes de Jordan_ , Fund. Math. 1 (1920), 166–209.
  * [17] J. Melleray, _On the geometry of Urysohn’s universal metric space_ , Topology Appl. 154:2 (2007), 384–403.
  * [18] P. S. Urysohn, Sur un espace métrique iuniversel, Bull. Sci. Math 51 (1927) 43–64 and 74–96.
  * [19] T. Ważewski, Kontinua prostowalne w zwiazku z funkcjami i odwzorowaniami absolutnie ciągłymi, Annales de la Société Polonaise de Mathématique, supplement, vol. 3 (1937), pp. 9–49.