# On infinite iterations of the functor of idempotent probability measures

Kh. F. Kholturaev
Tashkent institute of irrigation and agricultural mechanization engineers

###### Abstract

In this paper we establish that the functor of idempotent probability measures
acting in the category of compacta and their continous mappings is perfect
metrisable.

2010 Mathematics Subject Classification. Primary 54C65, 52A30; Secondary
28A33.

Key words and phrases: metric, metrization of functors, idempotent probability
measures.

###### Contents

  1. 1 Itroduction
  2. 2 Preliminaries
  3. 3 Uniform metrizability of the functor of idempotent probability measures
  4. 4 Perfect metrizability of the functor of idempotent probability measures

##  1 Itroduction

The notion of idempotent measure finds important applications in different
parts of mathematics, mathematical physics, economics, mathematical biology
and others. One can find a row of applications of idempotent mathematics from
[3].

Let $\mathbb{R}$ be the real line. Consider the set
$\mathbb{R}\cup\\{-\infty\\}$ with two algebraic operations: addition $\oplus$
and multiplication $\odot$ defined as $u\oplus v=\max\\{u,\,v\\}$ and $u\odot
v=u+v$. The set $\mathbb{R}\cup\\{-\infty\\}$ forms semifield with respect to
this operations and, the unity $\mathbf{1}=0$ and zero $\mathbf{0}=-\infty$,
i. e.

  * $(i)$

the addition $\oplus$ and the multiplication $\odot$ are associative;

  * $(ii)$

the addition $\oplus$ is commutative;

  * $(iii)$

the multiplication $\odot$ is distributive with respect to the addition
$\oplus$;

  * $(iv)$

each nonzero element $x\in\mathbb{R}\cup\\{-\infty\\}$ is intertible.

It denotes by $\mathbb{R}_{\max}$. It is idempotent, i. e. $x\oplus x=x$ for
all $x\in\mathbb{R}_{\max}$, and commutative, i. e. the multiplication $\odot$
is commutative.

Let $X$ be a compact Hausdorff space, $C(X)$ the algebra of continuous
functions $\varphi\colon X\to\mathbb{R}$ with the usual algebraic operations.
On $C(X)$ the operations $\oplus$ and $\odot$ we define as following:

$\displaystyle\varphi\oplus\psi=\max\\{\varphi,\,\psi\\},\quad\mbox{where}\quad\varphi,\,\psi\in
C(X),$
$\displaystyle\varphi\odot\psi=\varphi+\psi,\quad\mbox{where}\quad\varphi,\,\psi\in
C(X),$
$\displaystyle\lambda\odot\varphi=\varphi+\lambda_{X},\quad\mbox{where}\quad\varphi\in
C(X),\quad\lambda\in\mathbb{R},$

where $\lambda_{X}$ is a contant function.

Recall [15] that a functional $\mu\colon\,C(X)\to\mathbb{R}$ is called to be
an idempotent probability measure on $X$, if:

  * $1)$

$\mu(\lambda_{X})=\lambda$ for each $\lambda\in\mathbb{R}$;

  * $2)$

$\mu(\lambda\odot\varphi)=\mu(\varphi)+\lambda$ for all
$\lambda\in\mathbb{R}$, $\varphi\in C(X)$;

  * $3)$

$\mu(\varphi\oplus\psi)=\mu(\varphi)\oplus\mu(\psi)$ for every
$\varphi,\,\psi\in C(X)$.

For a compact Hausdorff space $X$ a set of all idempotent probability measures
on $X$ we denote by $I(X)$. Consider $I(X)$ as a subspace of
$\mathbb{R}^{C(X)}$. In the induced topology the sets of the view

$\displaystyle\langle\mu;\,\varphi_{1},\,\dots,\,\varphi_{k};\,\varepsilon\rangle=\\{\nu\in
I(X):\,|\mu(\varphi_{i})-\nu(\varphi_{i})|<\varepsilon,\,i=1,\,\dots,\,k\\},$

form a base of neighbourhoods of the idempotent measure $\mu\in I(X)$, where
$\varphi_{i}\in C(X)$, $i=1,\,\dots,\,k$, and $\varepsilon>0$. The topology
generated by this base coincide with point-wise convergence topology on
$I(X)$. The topological space $I(X)$ is compact [15]. For a given map
$f\colon\,X\to Y$ of compact Hausdorff spaces the map $I(f)\colon\,I(X)\to
I(Y)$ defines by the formula $I(f)(\mu)(\varphi)=\mu(\varphi\circ f)$, $\mu\in
I(X)$, where $\varphi\in C(Y)$. The construction $I$ is a covariant functor,
acting in the category of compact Hausdorff spaces and their continuous maps.

Since $I$ is a normal functor, for an arbitrary idempotent measure $\mu\in
I(X)$ we may define the support of $\mu:$ $\mbox{supp}\,\mu=\cap\\{A\subset
X:\,\bar{A}=A,\,\mu\in I(A)\\}$. For a point $x\in X$ by the rule
$\delta_{x}(\varphi)=\varphi(x)$, $\varphi\in C(X)$, we define the Dirac
measure $\delta_{x}$ supported on the singleton $\\{x\\}$, i. e.
$\textrm{supp}\,\delta_{x}=\\{x\\}$.

Let $\mu_{1},\,\mu_{2}\in I(X)$. Put

$\displaystyle\Lambda_{1\,2}=\Lambda(\mu_{1},\,\mu_{2})=\\{\xi\in
I(X^{2}):\,I(\pi_{i})(\xi)=\mu_{i},\,i=1,\,2\\},$

where $\pi_{i}\colon\,X\times X\to X$ is the projection onto $i$-th factor,
$i=1,\,2$. Note that $\Lambda_{1\,2}\neq\varnothing$ (see [6], Page 7).

Let $(X,\,\rho)$ be a metric compact space. A function $\rho_{I}\colon
I(X)\times I(X)\to\mathbb{R}$ defined as

$\rho_{I}(\mu_{1},\,\mu_{2})=\inf\\{\sup\\{\xi(\rho)\oplus\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\xi\\}:\,\xi\in\Lambda_{1\,2}\\}.$
(1)

is a metric on $I(X)$ which is an extension of the metric $\rho$ (Theorem 1
[6]). Besides, the metric $\rho_{I}$ generates point-wise converging topology
on $I(X)$ (Theorem 2 [6]). Note in [5, 14] where considered another metric.

Note that a function

$\rho_{I}(\mu_{1},\,\mu_{2})=\inf\\{\sup\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\xi\\}:\,\xi\in\Lambda_{1\,2}\\},$
(2)

suggested by A. Zaitov, also is a metric on $I(X)$. It is obvious that the
equalities (1) and (2) define the same metric.

##  2 Preliminaries

All of the concepts and results in this section we take from [1]. Under a
functor we mean a covariant functor acting in the category $\mathfrak{Top}$ of
topological spaces and their continuous maps, and some subcategories of
$\mathfrak{Top}$.

###### Definition 2.1

A functor $\mathcal{F}$, acting in the category $\mathfrak{Comp}$ of compact
Hausdorff spaces and their continuous maps, is said to be seminormal if it
satisfies the following conditions:

  1. $1)$

$\mathcal{F}$ preserves empty set and singleton, i. e.
$\mathcal{F}(\varnothing)=\varnothing$ and
$\mathcal{F}(\mathbf{1})=\mathbf{1}$ take places, where $\mathbf{1}$ is a
singleton.

  2. $2)$

$\mathcal{F}$ preserves intersections, i. e.
$\mathcal{F}(\bigcap\limits_{F\in\mathcal{B}}F)=\bigcap\limits_{F\in\mathcal{B}}\mathcal{F}(F)$
for a given compact Hausdorff space $X$ and for every family $\mathcal{B}$ of
closed subsets of $X$;

  3. $3)$

$\mathcal{F}$ is monomorphic, i. e.
$\mathcal{F}(i)\colon\,\mathcal{F}(A)\to\mathcal{F}(X)$ is an embedding for
every given embedding $i\colon\,A\to X$;

  4. $4)$

$\mathcal{F}$ is continuous, i. e.
$\mathcal{F}(\lim\mathcal{S})=\lim(\mathcal{F}(\mathcal{S}))$ for each
spectrum $\mathcal{S}=\\{X_{\alpha},\,\pi^{\beta}_{\alpha};\,A\\}$ of compact
Hausdorff spaces and their corresponding projections.

If a functor $\mathcal{F}$ is seminormal then there exists a unique natural
transformation $\eta^{\mathcal{F}}=\eta\colon\,\mbox{Id}\to\mathcal{F}$ of
identity functor Id into the functor $\mathcal{F}$. Moreover, this
transformation is monomorphism, i. e. for each compact Hausdorff space $X$ the
map $\eta_{X}\colon\,X\to\mathcal{F}(X)$ is an embedding.

###### Definition 2.2

An acting in the category $\mathfrak{MComp}$ of metrisable compact spaces and
their continuous maps seminormal functor $\mathcal{F}$ is said to be
metrisable if for any metrisable compact $X$ and for each metric $d=d_{X}$ on
$X$ it is possible to assign a metric $d_{\mathcal{F}(X)}$ on the compact
$\mathcal{F}(X)$ such that the following conditions hold:

  1. $P1)$

if $i\colon\,(X_{1},\,d^{1})\to(X_{2},\,d^{2})$ is an isometric embedding then

$\displaystyle\mathcal{F}(i)\colon\,(\mathcal{F}(X_{1}),\,d^{1}_{\mathcal{F}(X_{1})})\to(\mathcal{F}(X_{2}),\,d^{2}_{\mathcal{F}(X_{2})})$

also is an isometric embedding;

  2. $P2)$

the embedding $\eta_{X}\colon(X,\,d)\to(\mathcal{F}(X),\,d_{\mathcal{F}(X)})$
is an isometry;

  3. $P3)$

$\mbox{diam}(\mathcal{F}(X),\,d_{\mathcal{F}(X)})=\mbox{diam}(X,\,d)$.

Fix a seminormal functor $\mathcal{F}$ and a compact Hausdorff space $X$, and
put
$\eta_{n-1,n}=\eta_{\mathcal{F}^{n-1}(X)}:\mathcal{F}^{n-1}(X)\rightarrow\mathcal{F}^{n}(X)$.
For positive integers $n<m$ put

$\displaystyle\eta_{n,m}=\eta_{m-1,m}\circ...\circ\eta_{n+1,n+2}\circ\eta_{n,n+1}.$

A direct sequence

$\displaystyle\setcounter{MaxMatrixCols}{11}\begin{CD}X@>{\eta_{0,\,1}}>{}>\mathcal{F}(X)@>{\eta_{1,\,2}}>{}>\dots
@>{\eta_{n-1,\,n}}>{}>\mathcal{F}^{n}(X)@>{\eta_{n,\,n+1}}>{}>\mathcal{F}^{n+1}(X)@>{\eta_{n+1,\,n+2}}>{}>\dots\,\,.\end{CD}$
(3)

arises.

Fix a metric $d$ on a compactum $X$ and a metrization of the functor
$\mathcal{F}$. A metric on the compactum $\mathcal{F}^{n}(X)$, generated by
this metrization we denote by $d_{n}$. Then every of the maps

$\displaystyle\eta_{n,m}:(\mathcal{F}^{n}(X),d_{n})\rightarrow(\mathcal{F}^{m}(X),d_{m})$

is an isometric embedding. The limit of sequence (3) in the category metric
spaces and their isometric maps we denote by $(\mathcal{F}^{+}(X),\,d_{+})$.
We give more detail definition of the metric $d_{+}$. Considering so far
$\mathcal{F}^{+}(X)$ as a limit of (3) in the set category, a limit of the
embeddings $\eta_{n,m}:\mathcal{F}^{n}(X)\rightarrow\mathcal{F}^{m}(X)$ as
$m\rightarrow\infty$ we denote by
$\eta_{n}:\mathcal{F}^{n}(X)\rightarrow\mathcal{F}^{+}(X)$. Then

$\displaystyle\mathcal{F}^{+}(X)=\bigcup\\{\eta_{n}(\mathcal{F}^{n}(X)):n\in\omega\\},$

and a metric $d_{+}$ defines by the metrics $d_{n}$ on the summands
$\eta_{n}(\mathcal{F}^{n}(X))$, i. e. for $x,\,y\in\mathcal{F}^{+}(X)$ we have

$\displaystyle\begin{CD}d_{+}(x,\,y)=d_{n}(a,\,b)\end{CD}$ (4)

where $\eta_{n}(a)=x$, $\eta_{n}(b)=y$. The definition of the metric $d_{+}$
by equality (4) is correct, since at $n<m$ the maps $\eta_{n,\,m}$ are
isometric embeddings.

If $f\colon\,X\to Y$ is a continuous map, then it is possible to determine a
map $\mathcal{F}^{+}(f)\colon\,\mathcal{F}^{+}(X)\to\mathcal{F}^{+}(Y)$. It
becomes as follows. For $x\in\mathcal{F}^{+}(X)$ there exist $n$ and
$a\in\mathcal{F}^{n}(X)$ such that $x=\eta_{n}(a)$. We put
$\mathcal{F}^{+}(f)(x)=\eta_{n}\mathcal{F}^{n}(f)(a)$. The correctness of this
definition follows from what $\eta_{n,\,m}$ is a natural transformation of the
functor $\mathcal{F}^{n}$ to the functor $\mathcal{F}^{m}$.

###### Definition 2.3

A metrisable functor $\mathcal{F}$ is said to be uniformly metrisable, if its
some metrization has the following property

  1. $P4)$

for any continuous map $f\colon\,(X_{1},\,d^{1})\to(X_{2},\,d^{2})$ the map

$\displaystyle\mathcal{F}^{+}(f)\colon\,(\mathcal{F}^{+}(X_{1}),\,d^{1}_{+})\to(\mathcal{F}^{+}(X_{2}),\,d^{2}_{+})$

is uniformly continuous.

A metrization of a functor, satisfying property $P4)$, is called uniformly
continuous.

For a uniformly metrisable functor $\mathcal{F}$ the operation
$\mathcal{F}^{+}$ is a functor, acting from the category
$\mathfrak{MComp}=\mathfrak{Metr}\cap\mathfrak{Comp}$ of compacta into the
category of (sigma-compact) metrisable spaces. Moreover Definition 2.3
directly follows

###### Proposition 2.1

If $f\colon\,(X_{1},\,d^{1})\to(X_{2},\,d^{2})$ is a homeomorphism of compacta
then for a uniformly metrisable functor $\mathcal{F}$ the map
$\mathcal{F}^{+}(f)\colon\,(\mathcal{F}^{+}(X_{1}),\,d_{+}^{1})\to(\mathcal{F}^{+}(X_{2}),\,d_{+}^{2})$
is a uniformly homeomorphism.

Therefore a metric space $(\mathcal{F}^{+}(X),\,d_{+})$ topologically does not
depend of the choose a metric $d$ on the compactum $X$. Consequently, the
operation $\mathcal{F}^{+}$ may be consider as a functor, acting from the
category compacta into the category metrisable spaces and uniformly continuous
maps.

Each compactum $(X,\,d)$ is assigned a completion
$(\mathcal{F}^{++}(X),\,d_{++})$ of the space $(\mathcal{F}^{+}(X),\,d_{+})$,
and each continuous map $f\colon\,(X_{1},\,d^{1})\to(X_{1},\,d^{1})$ is
assigned a map
$\mathcal{F}^{++}(f)\colon\,(\mathcal{F}^{++}(X_{1}))\to(\mathcal{F}^{++}(X_{2}))$
which is an extension of the map $\mathcal{F}^{+}(f)$ on the completions of
the spaces $(\mathcal{F}^{+}(X_{1}),\,d_{+}^{1})$ and
$(\mathcal{F}^{+}(X_{2}),\,d_{+}^{2})$. Thereby it is defined a functor
$\mathcal{F}^{++}$, acting from the category metric compact spaces into
complete metric spaces and uniformly continuous maps. The functor
$\mathcal{F}^{++}$ is as topological invariant as the functor
$\mathcal{F}^{+}$, and it is possible to consider $\mathcal{F}^{++}$ as a
functor, acting from the category compacta into the Polish spaces.

Let for a seminormal functor $\mathcal{F}$ except the natural transformation
$\eta\colon\,\mbox{Id}\to\mathcal{F}$, it defines still a natural
transformation $\psi^{\mathcal{F}}\colon\,\mathcal{F}^{2}\to\mathcal{F}$. If
at the same time for every compact $X$ the equalities

$\displaystyle\psi_{X}\circ\mathcal{F}(\eta_{X})=\mbox{id}_{\mathcal{F}(X)},$
(5)
$\displaystyle\psi_{X}\circ\eta_{\mathcal{F}(X)}=\mbox{id}_{\mathcal{F}(X)},$

are executed, then the functor $\mathcal{F}$ is said to be semimonadic. And if
the equality

$\displaystyle\psi_{X}\circ\mathcal{F}(\psi_{X})=\psi_{X}\circ\psi_{\mathcal{F}(X)}$

is still carried out, then $\mathcal{F}$ is said to be monadic.

For a positive integer $n$ we put

$\displaystyle\psi_{n+1,\,n}=\psi_{\mathcal{F}^{n-1}(X)}\colon\,\mathcal{F}^{n+1}(X)\to\mathcal{F}^{n}(X)$

and for $n<m$:

$\displaystyle\psi_{m,\,n}=\psi_{n+1,\,n}\circ\psi_{n+2,\,n+1}\circ\dots\circ\psi_{m-1,\,m-2}\circ\psi_{m,\,m-1}.$

For each semimonadic functor the following inverse sequence arises

$\displaystyle\setcounter{MaxMatrixCols}{11}\begin{CD}\mathcal{F}(X)@<{\psi_{2,\,1}}<{}<\mathcal{F}^{2}(X)@<{\psi_{3,\,2}}<{}<\dots
@<{\psi_{n,\,n-1}}<{}<\mathcal{F}^{n}(X)@<{\psi_{n+1,\,n}}<{}<\mathcal{F}^{n+1}(X)@<{\psi_{n+2,\,n+1}}<{}<\dots\,\,.\end{CD}$
(6)

Denote by $\mathcal{F}^{\omega}$ the limit of the sequence (6). Since
$\psi_{n+1,\,n}$ are natural transformations the operation
$\mathcal{F}^{\omega}$ is functorial. The functor $\mathcal{F}^{\omega}$ acts
as in the category $\mathfrak{Comp}$ as in its subcategory $\mathfrak{MComp}$.

For $1\leq n<m$ put $q_{n,\,m}=\mathcal{F}(\eta_{n-1,\,m-1})$. Equality (5)
yields

$\displaystyle\psi_{m,\,n}\circ q_{n,\,m}=\mbox{id}_{\mathcal{F}^{n}(X)}.$

###### Definition 2.4

A uniformly metrisable semimonadic functor $\mathcal{F}$ is said to be perfect
metrisable, if some of its metrization along with properties $P1)\,-\,P4)$ has
the following properties

  1. $P5)$

$\psi_{2,1}:(\mathcal{F}^{2}(X),\,d_{2})\rightarrow(\mathcal{F}(X),\,d_{1})$
is a non-expending map;

  2. $P6)$

for every pair of $a\in\mathcal{F}^{2}(X)$ and $x\in X$ we have

$\displaystyle d_{1}(\psi_{2,1}(a),\,\eta_{0,1}(x))=d_{2}(a,\,\eta_{0,2}(x)).$

Further, we denote by $\mathcal{F}$ a perfect metrisable functor. By
$\theta_{n}\colon\,\mathcal{F}^{+}(X)\to\mathcal{F}^{n}(X)$ we denote a
natural projection, which on the summand $\eta_{m}(\mathcal{F}^{m}(X))$ is
defined as follows:

$\displaystyle\theta_{n}=\psi_{m,n}\circ\eta_{m}^{-1}.$

The map $\theta_{n}$ non-expanding with respect to (4) and $P5)$. Thence as
uniformly continuous map $\theta_{n}$ extends on the completion
$\mathcal{F}^{++}(X)$ of the space $\mathcal{F}^{+}(X)$. This extension we
denote by $\overline{\theta}_{n}$.

At $n_{1}<n_{2}$ the following equality holds

$\displaystyle\theta_{n_{1}}=\psi_{n_{2},n_{1}}\circ\eta_{n_{2}}.$ (7)

Really, for $x\in\eta_{m}(\mathcal{F}^{m}(X))$ at $m>n_{2}$ we have

$\displaystyle\theta_{n_{1}}(x)=\psi_{m,\,n_{1}}\circ\eta_{m}^{-1}(x)=\psi_{n_{2},\,n_{1}}\circ\psi_{m,\,n_{2}}\circ\eta_{m}^{-1}(x)=\psi_{n_{2},\,n_{1}}\circ\theta_{n_{2}}(x).$

Since a continuous map into a Hausdorff space is defined in a unique way with
itself values on everywhere dense subspace, (7) implies

$\displaystyle\overline{\theta}_{n_{1}}=\psi_{n_{2},n_{1}}\circ\overline{\theta}_{n_{2}}.$

Therefore there exists unique map
$\theta\colon\,\mathcal{F}^{++}(X)\to\mathcal{F}^{\omega}(X)$ (a limit of maps
$\overline{\theta}_{n}\colon\,\mathcal{F}^{++}(X)\to\mathcal{F}^{n}(X)$) such
that for each $n$ one has

$\displaystyle\overline{\theta}_{n}=\psi_{n}\circ\theta,$

where $\psi_{n}\colon\,\mathcal{F}^{\omega}(X)\to\mathcal{F}^{n}(X)$ is a
through projection of the inverse sequence (6)

###### Proposition 2.2

Maps

$\displaystyle\eta_{n}\circ\overline{\psi}_{n}\colon\,\mathcal{F}^{++}(X)\to\mathcal{F}^{n}(X)\to\mathcal{F}^{+}(X)\hookrightarrow\mathcal{F}^{++}(X)$

converge to the identity map uniformly on compact sets.

###### Proposition 2.3

For arbitrary pair of $x\in\mathcal{F}^{n}(X)$ and $a\in\mathcal{F}^{m}(X)$,
$m\geq n+2$, we have

$\displaystyle
d_{n+1}(\psi_{m,\,n+1}(a),\,\eta_{n,\,n+1}(x))=d_{m}(a,\,\eta_{n,m}(x)).$

###### Theorem 2.1

For each compactum $X$ and every perfect metrisable functor $\mathcal{F}$ the
map $\theta:\mathcal{F}^{++}(X)\rightarrow\mathcal{F}^{\omega}(X)$ is an
embedding.

###### Theorem 2.2

Let a perfect metrisable functor $\mathcal{F}$ and a compactum $X$ be such
that

  * $1)$

$\mathcal{F}^{\omega}(X)$ is homeomorphic to $Q$;

  * $2)$

$\mathcal{F}^{n}(X)$ is homeomorphic to $Q$ for all $n$, beginning with some;

  * $3)$

$\eta_{n,\,n+1}(\mathcal{F}^{n}(X))$ is a $Z$-set in $\mathcal{F}^{n+1}(X))$
for all $n$.

Let, besides, the metrization of the functor $\mathcal{F}$ satisfies the
following condition

  1. $P7)$

for an arbitrary $a\in\mathcal{F}(X)$ the inequality
$d_{1}(a,\,\eta_{0,\,1}(X))\geq\varepsilon$ implies the inequality
$d_{i+1}(q_{1,\,i+1}(a),\,\eta_{i,\,i+1}q_{1,\,i}(a))\geq\varepsilon$ for all
$i\geq 2$.

Then the triple
$(\mathcal{F}^{\omega}(X),\,\theta(\mathcal{F}^{++}(X)),\,\theta(\mathcal{F}^{+}(X)))$
is homeomorphic to the triple $(Q,\,s,\,\mbox{rint}\,Q)$.

Remind, that $Q=\prod\\{[-1,\,1]_{n}:\,n=1,\,2,\,\dots\\}$ is the Hilbert
cube, $s$ is the pseudo-interior of $Q$, and
$\mbox{rint}\,Q\,=\\{x=\\{x_{n}\\}\in Q:\,|x_{n}|\leq t<1\mbox{ for all
}n=1,\,2,\,\dots\\}$. It is well-known that $s$ is homeomorphic to Hilbert
space $l_{2}$, and $\mbox{rint}\,Q$ to $\Sigma$, which is a linear span of $Q$
in $l_{2}$.

##  3 Uniform metrizability of the functor of idempotent probability measures

In this section we verify that the functor $I$ satisfies properties $P4)-P7)$.
For this we need the following construction. Since functor $I$ is normal there
exists unique natural transformation $\eta^{I}=\eta\colon\,\mbox{Id}\to I$ of
identity functor Id into the functor $I$. Here the natural transformation
$\eta$ consists of monomorphisms $\delta_{X},\,X\in\mathfrak{Comp}$. More
detail, the last means that for each compact Hausdorff space $X$ the map
$\delta_{X}\colon\,X\to I(X)$, which defines as $\delta_{X}(x)=\delta_{x}$,
$x\in X$, is an embedding. Thus $\eta=\\{\delta_{X}:\,X\in\mathfrak{Comp}\\}$
is the mentioned natural transformation.

Let $X$ be a compactum. Put

$\displaystyle I^{0}(X)=X,\quad I^{k}(X)=I(I^{k-1}(X)),\quad
k=1,\,2,\,\dots,\quad\mbox{ and }$
$\displaystyle\eta_{n-1,\,n}=\eta_{I^{n-1}(X)}\colon\,I^{n-1}(X)\to I^{n}(X).$

Fix a metric $\rho$ on a compactum $X$ and consider the mertrization
$\rho_{1}$ of the functor $I$ by the Zaitov metric $\rho_{1}=\rho_{I}$,
defined by (2). The metric on $I^{n}(X)$ generated by this metrization we
denote as $\rho_{n}$.

###### Lemma 3.1

Let $X$ be a compactum with a metric $\rho$. Then
$\delta_{X}\colon\,(X,\,\rho)\to(I(X),\,\rho_{1})$ is an isometry.

Proof. For every pair of Dirac measures
$\delta_{x}=0\odot\delta_{x},\,\delta_{y}=0\odot\delta_{y}\in I(X)$, $x,\,y\in
X$, the uniqueness of $(0\odot\delta_{x},\,0\odot\delta_{y})$-admissible
measure $0\odot\delta_{(x,\,y)}\in\Lambda(\delta_{x},\,\delta_{y})\subset
I(X^{2})$ implies that

$\displaystyle\rho_{1}(\delta_{x},\,\delta_{y})=0\odot\delta_{(x,\,y)}(\rho)\oplus\rho(x,\,y)=\rho(x,\,y).$

$\scriptstyle\blacksquare$

###### Lemma 3.2

Let $(X_{1},\rho^{1})$, $(X_{2},\,\,\rho^{2})$ be compacta. Then for each
isometric embedding $i\colon\,(X_{1},\rho^{1})\to(X_{2},\,\,\rho^{2})$ the map
$I(i)\colon\,(I(X_{1}),\,\rho_{1}^{1})\to(I(X_{2}),\,\rho_{1}^{2})$ is also an
isometric embedding.

Proof. Let $\mu_{1},\,\nu_{1}\in I(X_{1})$ and
$\mu_{2}=I(i)(\mu_{1}),\,\nu_{2}=I(i)(\nu_{1})\in I(X_{2})$. Then
$I(i)(X_{1})\subset I(X_{2})$ and for every
$\zeta\in\Lambda(\mu_{2},\,\nu_{2})$ supports on $(i\times i)\,(X_{1}\times
X_{1})$. That is way we have

$\rho_{1}^{2}(\mu_{2},\,\nu_{2})=\inf\\{\sup\\{\rho^{2}(x_{2},\,y_{2}):\,(x_{2},\,y_{2})\in\mbox{supp}\,\xi_{2}\\}:\,\xi_{2}\in\Lambda(\mu_{2},\,\nu_{2})\\}=\\\
=\inf\\{\sup\\{\rho^{2}(i(x_{1}),\,i(y_{1})):\,(i(x_{1}),\,i(y_{1}))\in\mbox{supp}\,I(i\times
i)(\xi_{1})\\}:\,\xi_{1}\in\Lambda(\mu_{1},\,\nu_{1})\\}=\\\
=\inf\\{\sup\\{\rho^{1}(x_{1},\,y_{1}):\,(x_{1},\,y_{1})\in\mbox{supp}\,\xi_{1}\\}:\,\xi_{1}\in\Lambda(\mu_{1},\,\nu_{1})\\}=\rho_{1}^{1}(\mu_{1},\,\nu_{1}).$

$\scriptstyle\blacksquare$

###### Lemma 3.3

For any metric $\rho$ on the compactum $X$ the following equality holds

$\displaystyle\mbox{diam}(X,\,\rho)=\mbox{diam}(I(X),\,\rho_{1}).$

Proof. The Proof immediately follows from (2).

$\scriptstyle\blacksquare$

###### Lemma 3.4

[6]. For every pair $\mu,\,\nu\in I(X)$ there exists
$\lambda_{\mu\,\nu}\in\Lambda(\mu,\,\nu)$ such that
$\rho_{1}(\mu,\,\nu)=\sup\\{\rho(x,\,y):\,(x,\,y)\in\mbox{supp}\,\lambda_{\mu\,\nu}\\}$.

###### Lemma 3.5

Let $f\colon X\to Y$ be a continuous map of compacta $(X,\,\rho_{X})$,
$(Y,\,\rho_{Y})$, and

$\displaystyle\rho_{X1}(\mu,\,\nu)=\sup\\{\rho_{X}(x,\,y):\,(x,\,y)\in\mbox{supp}\,\lambda_{\mu\,\nu}\\},$

$\mu,\,\nu\in I(X)$. Then

$\displaystyle\rho_{Y1}(I(f)(\mu),\,I(f)(\nu))\leq\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,\lambda_{\mu\,\nu}\\}.$

Proof. We put $\kappa=I(f\times f)(\lambda_{\mu\,\nu})$. We have

$\displaystyle I(q_{1})(\kappa)(\varphi)=I(q_{1})(I(f\times
f)(\lambda_{\mu\,\nu}))(\varphi)=I(f\times f)(\lambda_{\mu\,\nu})(\varphi\circ
q_{1})=$ $\displaystyle=\lambda_{\mu\,\nu}(\varphi\circ q_{1}\circ f\times
f)=\lambda_{\mu\,\nu}(\varphi\circ f\circ p_{1})=\mu(\varphi\circ
f)=I(f)(\mu)(\varphi),$

i. e. $I(q_{1})(\kappa)=I(f)(\mu)$. Here $p_{1}\colon X^{2}\to X$,
$q_{1}\colon Y^{2}\to Y$ are the projections onto the first factors,
respectively, $\varphi\in C(Y)$. We have used the commutativity of the diagram

$\displaystyle\begin{CD}X\times X@>{f\times f}>{}>Y\times Y\\\
@V{p_{1}}V{}V@V{}V{q_{1}}V\\\ X@>{f}>{}>Y,\end{CD}$

i. e. the equality $q_{1}\circ f\times f=f\circ p_{1}$. Similarly, one can
show $I(q_{2})(\kappa)=I(f)(\nu)$, where $q_{2}\colon Y^{2}\to Y$ is the
projection onto the second factor. Thus, $\kappa\in I(Y^{2})$ is a
$(I(f)(\mu),\,I(f)(\nu))$-admissible measure. Hence,

$\rho_{Y1}(I(f)(\mu),\,I(f)(\nu))\leq\sup\\{\rho_{Y}(t,\,z):\,(t,\,z)\in\mbox{supp}\,\kappa\\}=\\\
=\sup\\{\rho_{Y}(t,\,z):\,(t,\,z)\in\mbox{supp}\,I(f\times
f)(\lambda_{\mu\,\nu})\\}=\\\
=\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,\lambda_{\mu\,\nu}\\}.$

$\scriptstyle\blacksquare$

Note, that since the support $\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})$
of ${}^{1}\lambda_{{}^{1}\mu\,^{1}\nu}=\lambda_{\mu\,\nu}\in I(X^{2})$ is a
compact subset of a given compact Hausdorff space $X$ one may write

$\displaystyle\rho_{X1}(^{1}\mu,\,^{1}\nu)=\max\\{\rho_{X}(x,\,y):\,(x,\,y)\in\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})\\}.$

Therefore, there exist
$(x_{{}^{1}\mu},\,y_{{}^{1}\nu})\in\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})$
such that

$\displaystyle\rho_{X1}(^{1}\mu,\,^{1}\nu)=\rho_{X}(x_{{}^{1}\mu},\,y_{{}^{1}\nu}).$
(8)

Let us consider ${}^{2}\mu$, ${}^{2}\nu$ $\in$ $I^{2}(X)$. Proposition 3.5 and
equality (8) consequently give:

$\displaystyle\rho_{X2}(^{2}\mu,\,^{2}\nu)=\sup\\{\rho_{X1}(^{1}\mu,\,^{1}\nu):\,(^{1}\mu,\,^{1}\nu)\in\mbox{supp}\,(^{2}\lambda_{\mu\,\nu})\\},$
(9)

and

$\displaystyle\rho_{X2}(^{2}\mu,\,^{2}\nu)=\rho_{X1}(^{1}\mu_{{}^{2}\mu},\,^{1}\nu_{{}^{2}\nu})=\sup\\{\rho_{X}(x,\,y):\,(x,\,y)\in\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})$
(10)

Similarly to the proof of inequality in Proposition 3.5, one may establish

$\rho_{Y2}(I^{2}(f)(^{2}\mu),\,I^{2}(f)(^{2}\nu))\leq\\\
\leq\sup\\{\rho_{Y1}(I^{1}(f)(^{1}\mu),\,I^{1}(f)(^{1}\nu)):\,(^{1}\mu,\,^{1}\nu)\in\mbox{supp}\,(^{2}\lambda_{{}^{2}\mu\,^{2}\nu})\\}\\\
\leq\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})\\},$

i. e.

$\displaystyle\rho_{Y2}(I^{2}(f)(^{2}\mu),\,I^{2}(f)(^{2}\nu))\leq\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,(^{1}\lambda_{{}^{1}\mu\,^{1}\nu})\\}.$
(11)

We denote $\lambda_{2}=^{1}\lambda_{{}^{1}\mu\,^{1}\nu}$. Then (10) and (11)
has the following view

$\displaystyle\rho_{X2}(^{2}\mu,\,^{2}\nu)=\sup\\{\rho_{X}(x,\,y):\,(x,\,y)\in\mbox{supp}\,\lambda_{2}\\},$
(12)

$\displaystyle\rho_{Y2}(I^{2}(f)(^{2}\mu),\,I^{2}(f)(^{2}\nu))\leq\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,\lambda_{2}\\}.$
(13)

###### Lemma 3.6

Let $f\colon X\to Y$ be a continuous map of compacta $(X,\,\rho_{X})$,
$(Y,\,\rho_{Y})$, and ${}^{k}\mu,\,^{k}\nu\in I^{k}(X)$. Then for every $k\geq
1$ there exists a $\lambda_{k}\in I(X^{2})$ such that

$\displaystyle\rho_{X\,k}(^{k}\mu,\,^{k}\nu)=\sup\\{\rho_{X}(x,\,y):\,(x,\,y)\in\mbox{supp}\,\lambda_{k}\\},$

and

$\displaystyle\rho_{Y\,k}(I^{k}(f)(^{k}\mu),\,I^{k}(f)(^{k}\nu))\leq\sup\\{\rho_{Y}(f(x),\,f(y)):\,(x,\,y)\in\mbox{supp}\,\lambda_{k}\\}.$

Proof. The proof immediately follows from Lemma 3.5 and formulas (12 – 13) by
induction.

$\scriptstyle\blacksquare$

###### Lemma 3.7

If $f\colon\,X\to Y$ is a $(\varepsilon,\,\delta)$-uniformly continuous map of
compacta, then for each $k\geq 1$ the map $I^{k}(f)\colon\,I^{k}(X)\to
I^{k}(Y)$ is $(\varepsilon,\,\delta)$-uniformly continuous.

Proof. The proof immediately follows from the inequality in Lemma 3.6.

$\scriptstyle\blacksquare$

So, summarizing Lemmas 3.1 – 3.7, according to Definition 2.3 we have the main
result of the section.

###### Theorem 3.1

Functor $I$ of idempotent probability measures is uniformly metrisable. More
exactly, for the metrization, introduced by (2), and for every continuous map
$f\colon\,(X_{1},\,d^{1})\to(X_{2},\,d^{2})$ the map

$\displaystyle
I^{+}(f)\colon\,(I^{+}(X_{1}),\,d^{1}_{+})\to(I^{+}(X_{2}),\,d^{2}_{+})$

is uniformly continuous.

##  4 Perfect metrizability of the functor of idempotent probability measures

This section we begin the following statement.

###### Lemma 4.1

[6] Every $\mu\in I(X)$ nay be represented as $\mu=\bigoplus_{x\in
X}\lambda(x)\odot\delta_{x}\in I(X)$ in a unique way, where
$\lambda\colon\,X\to\mathbb{R}$ is an upper semicontinuous map. For each
$x_{0}\in X$ we have $x_{0}\in\mbox{supp}\,\mu$ if and of only if
$\lambda(x_{0})>-\infty$.

Consider now a system $\psi$. The system $\psi$ consists of all mappings
$\psi_{X}\colon\,I^{2}(X)\to I(X)$, acting as the following. Given $M\in
I^{2}(X)$ put $\psi_{X}(M)(\varphi)=M(\overline{\varphi})$, where for any
function $\varphi\in C(X)$ the function
$\overline{\varphi}\colon\,I(X)\to\mathbb{R}$ defines by the formula
$\overline{\varphi}(\mu)=\mu(\varphi)$. Fix a compactum $X$ and for a positive
integer $n$ put $\psi_{n+1,\,n}=\psi_{I^{n-1}(X)}\colon\,I^{n+1}(X)\to
I^{n}(X)$. Note that $\psi_{n+1,\,n}\circ\eta_{n,\,n+1}=Id_{I^{n}(X)}$.

###### Lemma 4.2

$\psi_{1,\,0}\colon\,(I^{2}(X),\,\rho_{2})\to(I(X),\,\rho_{1})$ is a non-
expanding map.

Proof. Let $M$, $N$ $\in$ $I^{2}(X)$. It is required to show that

$\displaystyle\rho_{1}(\psi_{1,\,0}(M),\,\psi_{1,\,0}(N))\leq\rho_{2}(M,\,N).$
(14)

By definition and equation (9) there exists a $(M,\,N)$-admissible measure
$\lambda_{MN}\in I(I(X)\times I(X))$ such that

$\displaystyle\rho_{2}(M,\,N)=\sup\\{\rho_{1}(\mu,\,\nu):\,(\mu,\,\nu)\in\mbox{supp}\,\lambda_{MN}\\},$

For every $\varphi\in C(X)$ we have

$\displaystyle
I(\pi_{I\,1})(\lambda_{MN})(\overline{\varphi})=\lambda_{MN}(\overline{\varphi}\circ\pi_{I\,1})=M(\overline{\varphi})=\psi_{1,\,0}(M)(\varphi)=\overline{\varphi}(\psi_{1,\,0}(M)),$

i. e.
$I(\pi_{I\,1})(\lambda_{MN})(\overline{\varphi})=\overline{\varphi}(\psi_{1,\,0}(M))$.
Analogously,
$I(\pi_{I\,2})(\lambda_{MN})(\overline{\varphi})=\overline{\varphi}(\psi_{1,\,0}(N))$.
Here $\pi_{I\,i}\colon\,I(X)\times I(X)\to I(X)$ is the projection onto $i$-th
factor. The last two equality imply
$(\psi_{1,\,0}(M),\,\psi_{1,\,0}(N))\in\mbox{supp}\,\lambda_{MN}$. That is why

$\displaystyle\rho_{2}(M,\,N)=\sup\\{\rho_{1}(\mu,\,\nu):\,(\mu,\,\nu)\in\mbox{supp}\,\lambda_{MN}\\}\geq\rho_{1}(\psi_{1,\,0}(M),\,\psi_{1,\,0}(N)).$

$\scriptstyle\blacksquare$

###### Lemma 4.3

For each $N\in\psi^{-1}_{1,\,0}(\mu)$ we have
$\rho_{1}(\mu,\,\delta_{x_{0}})=\rho_{2}(\delta_{\delta_{x_{0}}},\,N)$.

Proof. The measure $\mu_{x_{0}}$ is a unique
$(\mu,\,\delta_{x_{0}})$-admissible measure, where
$\mbox{supp}\,\mu_{x_{0}}=\\{(x,\,x_{0}):\,x\in\mbox{supp}\,\mu\\}$ and
measures $\mu_{x_{0}}$ and $\mu$ act the same rule. So, we have

$\displaystyle\rho_{1}(\mu,\,\delta_{x_{0}})=\sup\\{\rho(x,\,x_{0}):\,(x,\,x_{0})\in\mbox{supp}\,\mu_{x_{0}}\\}=\sup\\{\rho(x,\,x_{0}):\,x\in\mbox{supp}\,\mu\\}.$

Similarly, the measure $N_{\delta_{x_{0}}}$ is a unique
$(N,\,\delta_{\delta_{x_{0}}})$-admissible measure, and

$\displaystyle\rho_{2}(\delta_{\delta_{x_{0}}},\,N)=\sup\\{\rho_{1}(\nu,\,\delta_{x_{0}}):\,\nu\in\mbox{supp}\,N\\}.$

Note that if $\psi_{1,\,0}(N)=\mu$ then by definition
$N(\overline{\varphi})=\mu(\varphi)$ for all $\varphi\in C(X)$, from here
$\mu\in\mbox{supp}\,N$. Therefore,

$\displaystyle\rho_{2}(\delta_{\delta_{x_{0}}},\,N)\geq\rho_{1}(\mu,\,\delta_{x_{0}}).$

Finally, inequality (14) finishes the proof.

$\scriptstyle\blacksquare$

###### Lemma 4.4

If $\rho_{1}(\mu,\,\eta_{0,\,1}(X))\geq\varepsilon$ then
$\rho_{2}(I(\eta_{0,\,1})(\mu),\,\,\eta_{1,\,2}(I(X)))\geq\varepsilon$.

Proof. It is clear that

$\displaystyle\rho_{1}(\mu,\,\eta_{0,\,1}(X))=\rho_{1}\left(\mu,\,\bigoplus\limits_{x\in
X}0\odot\delta_{x}\right)$

and

$\displaystyle\rho_{2}(I(\eta_{0,\,1})(\mu),\,\,\eta_{1,\,2}(I(X)))=\rho_{2}\left(\delta_{\mu},\,\bigoplus\limits_{\nu\in
I(X)}0\odot\delta_{\nu}\right).$

For every $\varphi\in C(X)$ we have
$\delta_{\mu}(\overline{\varphi})=\overline{\varphi}(\mu)=\mu(\varphi)$ and

$\displaystyle\bigoplus\limits_{\nu\in
I(X)}0\odot\delta_{\nu}(\overline{\varphi})=\bigoplus\limits_{\nu\in
I(X)}0\odot\nu(\varphi)=\bigoplus\limits_{x\in
X}0\odot\varphi(x)=\bigoplus\limits_{x\in X}0\odot\delta_{x}(\varphi).$

Consequently, $\bigoplus\limits_{x\in
X}0\odot\delta_{x}\in\mbox{supp}\,\bigoplus\limits_{\nu\in I(X)}0\odot\nu$,
which completes the proof.

$\scriptstyle\blacksquare$

Thus, we have established the following result.

###### Theorem 4.1

The functor $I$ is perfect metrisable.

From above established results and results from works [1], [10] and [11] we
have the main statement of the paper.

###### Theorem 4.2

For every compact Hausdorff space $X$ containing more than one point the
triple $(I^{\omega}(X),\,\theta(I^{++}(X)),\,\theta(I^{+}(X)))$ is
homeomorphic to the triple $(Q,\,s,\,\mbox{rint}\,Q)$.

## Acknowledgements

The author would like to thank to professor Adilbek Zaitov the head of the
Department of Mathematics and Natural Disciplines of Tashkent Institute of
Architecture and Civil Engineering for comprehensive support and attention.

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