# A Banach space C(K) reading the dimension of K

Damian Głodkowski  Institute of Mathematics of the Polish Academy of Sciences,
ul. Śniadeckich 8, 00-656 Warszawa, Poland  Faculty of Mathematics,
Informatics, and Mechanics, University of Warsaw, ul. Banacha 2, 02-097
Warszawa, Poland<EMAIL_ADDRESS>
###### Abstract.

Assuming Jensen’s diamond principle ($\diamondsuit$) we construct for every
natural number $n>0$ a compact Hausdorff space $K$ such that whenever the
Banach spaces $C(K)$ and $C(L)$ are isomorphic for some compact Hausdorff $L$,
then the covering dimension of $L$ is equal to $n$. The constructed space $K$
is separable and connected, and the Banach space $C(K)$ has few operators i.e.
every bounded linear operator $T:C(K)\rightarrow C(K)$ is of the form
$T(f)=fg+S(f)$, where $g\in C(K)$ and $S$ is weakly compact.

###### Key words and phrases:

Banach spaces of continuous functions, Banach spaces with few operators,
covering dimension, diamond principle

###### 2020 Mathematics Subject Classification:

03E35, 46E15, 47L10, 54F45

This research was funded in part by the NCN (National Science Centre, Poland)
research grant no. 2021/41/N/ST1/03682. For the purpose of Open Access, the
author has applied a CC-BY public copyright licence to any Author Accepted
Manuscript (AAM) version arising from this submission.

##  1\. Introduction

In [19] Koszmider showed that there is a compact Hausdorff space $K$ such that
whenever $L$ is compact Hausdorff and the Banach spaces $C(K)$ and $C(L)$ are
isomorphic, the dimension of $L$ is greater than zero. In the light of this
result Pełczyński asked, whether there is a compact space $K$ with $\dim(K)=k$
for given $k\in\omega\backslash\\{0\\}$, such that if $C(K)\sim C(L)$, then
$\dim(L)\geq k$ ([21, Problem 4]). We show that the answer to this question is
positive, if we assume Jensen’s diamond principle ($\diamondsuit$). Namely, we
prove the following:

###### Theorem 6.9.

Assume $\diamondsuit$. Then for every $k\in\omega\cup\\{\infty\\}$ there is a
compact Hausdorff space $K$ such that $\dim(K)=k$ and whenever $C(K)\sim
C(L)$, $\dim(L)=k$.

Note that typically dimension of $K$ is not an invariant of the Banach space
$C(K)$ under isomorphisms. For instance, the classical result by Miljutin says
that if $K,L$ are compact metrizable uncountable spaces, then the Banach
spaces $C(K)$ and $C(L)$ are isomorphic ([26]). This also shows that $C(K)$
with the desired property cannot admit any complemented copy of $C(L)$ where
$L$ is compact, metrizable and uncountable (indeed, if $C(K)\sim X\oplus
C(L)$, then $C(K)\sim X\oplus C(L)\oplus C([0,1]^{n})\sim C(K)\oplus
C([0,1]^{n})$ for any $n\in\omega$). Another result by Pełczyński says that if
$G$ is an infinite compact topological group of weight $\kappa$, then $C(G)$
is isomorphic to $C(\\{0,1\\}^{\kappa})$ ([29]).

On the other hand the space $C(K)$ remembers many topological and set-
theoretic properties of $K$. For example Cengiz showed that if $C(K)\sim
C(L)$, then $K$ and $L$ have the same cardinalities ([4]). If $K$ is
scattered, then by Pełczyński-Semadeni theorem $L$ is scattered as well
([30]). In this case both spaces must be zero-dimensional. If $K$ is an
Eberlein compact, then $L$ is also Eberlein ([27]). If $K$ is a Corson compact
and $L$ is homogeneous, then $L$ is Corson ([33]).

Although the isomorphic structure of $C(K)$ does not remember the dimension of
$K$, the metric structure of $C(K)$ contains such information, since by the
Banach-Stone theorem $K$ and $L$ are homeomorphic, whenever $C(K)$ and $C(L)$
are isometric. Similar results were obtained by Gelfand, Kolmogorov and
Kaplansky in the category of rings of functions on compact spaces and in the
category of Banach lattices ([15, 18]). It is also worth to mention that the
covering dimension of $K$ is an invariant for the space $C_{p}(K)$ of
continuous functions on $K$ with the pointwise topology ([31]).

The key property of the space $K$ that we construct to prove Theorem 6.9 is
the fact that the Banach space $C(K)$ has few operators i.e. every bounded
operator $T:C(K)\rightarrow C(K)$ is of the from $T=gI+S$, where $g\in C(K)$
and $S$ is weakly compact. Schlackow showed that if the Banach space $C(K)$
has few operators, $C(K)\sim C(L)$ and both spaces $K,L$ are perfect, then $K$
and $L$ are homeomorphic ([36]). We improve this result under the assumption
that $K$ is separable and connected.

###### Theorem 4.19.

Suppose that $K$ is a separable connected compact Hausdorff space such that
$C(K)$ has few operators and $L$ is a compact Hausdorff space such that
$C(K)\sim C(L)$.

Then $K$ and $L$ are homeomorphic modulo finite set i.e. there are open
subsets $U\subseteq K,V\subseteq L$ and finite sets $E\subseteq K,F\subseteq
L$ such that $U,V$ are homeomorphic and $K=U\cup E,L=V\cup F$.

The first example (under the continuum hypothesis) of a Banach space $C(K)$
with few operators appeared in the work of Koszmider ([19]). Later, Plebanek
showed how to remove the use of CH from such constructions ([32]). Considered
spaces have many interesting properties (cf. [21, Theorem 13]) e.g. $C(K)$ is
indecomposable Banach space, it is not isomorphic to any of its proper
subspaces nor any proper quotient, it is a Grothendieck space, $K$ is strongly
rigid (i.e. identity and constant functions are the only continuous functions
on $K$) and does not include non-trivial convergent sequences. For more
examples and properties of Banach spaces $C(K)$ with few operators see [3, 11,
20, 21, 22, 23].

In the further part of the paper we show how to construct a Banach space
$C(K)$ with few operators, where $K$ has arbitrarily given dimension. Theorem
6.9 is an almost immediate consequence of Theorem 4.19 and the following
theorem.

###### Theorem 6.8.

Assume $\diamondsuit$. For each $k>0$ there is a compact Hausdorff, separable,
connected space $K$ such that $C(K)$ has few operators and $\dim K=k$.

Our construction is a modification of one of the spaces $K$ from [19, Theorem
6.1], which is a separable connected compact space such that $C(K)$ has few
operators. The original space is constructed as an inverse limit of metrizable
compact spaces $(K_{\alpha})_{\alpha<\omega_{1}}$, where on intermediate steps
we add suprema to countable families of functions in the lattice
$C(K_{\alpha})$ for $\alpha<\omega_{1}$, using the notion of strong extension.
However, the considered families of functions are very general, which leads to
the problem that described operation may rise the dimension of given space and
the final space is infinite-dimensional. We show that under $\diamondsuit$ we
are able to limit the choice of functions in the way that we can control the
dimension of the spaces at each step. In order to control the dimension we
introduce the notion of essential-preserving maps. Similar ideas were studied
in Fedorchuk’s work ([12, 13, 14]). For instance, Fedorchuk considered maps
that are ring-like, monotonic and surjective, which implies that they are
essential-preserving (however, those notions are much stronger and are not
applicable in our context).

One may also consider other notions of dimension such as small or large
inductive dimension. However, since Theorem 3.12 does not work if we replace
the covering dimension with one of the inductive dimensions, we do not know if
the spaces we constructed have finite inductive dimensions.

The structure of the paper is as follows. Section 2 concerns basic
terminology. Section 3 contains necessary results about covering dimension. In
section 4 we prove Theorem 4.19 characterizing properties of spaces $C(K)$
with few operators preserved under isomorphisms. In Section 5 we develop tools
for controlling dimension in some inverse limits of systems of compact spaces.
Section 6 contains the description of the construction leading to the main
theorem of the paper. The last section includes remarks and open questions.

##  2\. Notation and terminology

Most of notation that we use should be standard. For unexplained terminology
check [9, 10, 17]. $\omega$ denotes the set of non-negative integers, which is
also the smallest infinite ordinal number. $\omega_{1}$ is the smallest
uncountable ordinal. $Lim$ stands for the class of all limit ordinals.
$\operatorname{Odd}$ and $\operatorname{Even}$ stand for the classes of odd
and even ordinals respectively. If $f$ is a function, then $f|A$ denotes the
restriction of $f$ to $A$. $\sum_{n\in\omega}f_{n}$ will always denote the
pointwise sum of functions $f_{n}$ (if the sum exists). $[A]^{<\omega}$ is the
family of all finite subsets of $A$. For a topological space $X$, $\dim X$
denotes the covering dimension (also known as Lebesgue covering dimension or
topological dimension, cf. [9, Definition 1.6.7]) of $X$. $X^{\prime}$ stands
for the subset of $X$ consisting of non-isolated points in $X$. A sequence
$(x_{n})_{n\in\omega}$ is said to be non-trivial, if it is not eventually
constant. We say that a topological space $X$ is c.c.c. if every family of
pairwise disjoint open subsets of $X$ is countable. By basic open subset of
$[0,1]^{\omega_{1}}$ we mean a product $\prod_{\alpha<\omega_{1}}U_{\alpha}$
where each $U_{\alpha}\subseteq[0,1]$ is a relatively open interval with
rational endpoints and $U_{\alpha}=[0,1]$ for all but finitely many
$\alpha$’s. A subset $S\subseteq\omega_{1}$ is called stationary, if it has
non-empty intersection with every closed and unbounded subsets of
$\omega_{1}$.

All considered topological spaces are Hausdorff. We work with Banach spaces of
the form $C(K)$ consisting of real-valued continuous functions on a compact
space $K$ equipped with the supremum norm. $C_{I}(K)$ denotes the subset of
$C(K)$ of functions with the range included in $[0,1]$. For Banach spaces $X$
and $Y$, a bounded linear operator $T:X\rightarrow Y$ is said to be weakly
compact if the closure of $T[B_{X}]$ is compact in the weak topology in $Y$
(here $B_{X}$ stands for the unit ball in $X$). $X\sim Y$ means that $X$ and
$Y$ are isomorphic as Banach spaces. $\mathcal{B}(X)$ denotes the algebra of
all bounded operators on a Banach space $X$ (with the operator norm). An
operator $T:C(K)\rightarrow C(L)$ is multiplicative, if $T(fg)=T(f)T(g)$. We
will use one symbol $\|\cdot\|$ to denote norms in all considered Banach
spaces - this should not lead to misunderstandings.

ZFC stands for Zermelo-Fraenkel set theory with the axiom of choice. CH is the
continuum hypothesis i.e. the sentence $2^{\omega}=\omega_{1}$. Jensen’s
diamond principle ($\diamondsuit$) stands for the following sentence (for
other equivalent formulations see [6]): there is a sequence of sets
$A\subseteq\alpha$ for $\alpha<\omega_{1}$ such that for any subset
$A\subseteq\omega_{1}$ the set $\\{\alpha:A\cap\alpha=A_{\alpha}\\}$ is
stationary in $\omega_{1}$. It is a well-known fact, that $\diamondsuit$
implies CH.

### Radon measures on compact spaces

For a compact space $K$ we will identify the space of bounded linear
functionals on $C(K)$ with the space of Radon measures on $K$ (the
identification is given by the Riesz representation theorem). For every
$\alpha<\omega_{1}$ we have an embedding
$E_{\alpha}:C([0,1]^{\alpha})\rightarrow C([0,1]^{\omega_{1}})$ given by
$E_{\alpha}(f)=f\circ\pi_{\alpha}$, where
$\pi_{\alpha}:[0,1]^{\omega_{1}}\rightarrow[0,1]^{\alpha}$ is the natural
projection. For a Radon measure $\mu$ on $[0,1]^{\omega_{1}}$ we will denote
by $\mu|C([0,1]^{\alpha})$ the restriction of $\mu$ treated as a functional on
$C([0,1]^{\omega_{1}})$ to the subspace $E_{\alpha}[C([0,1]^{\alpha})]$.
Equivalently, $\mu|C([0,1]^{\alpha})$ is a measure on $[0,1]^{\alpha}$ given
by

$\mu|C([0,1]^{\alpha})(A)=\mu(\pi_{\alpha}^{-1}(A)).$

For any measure $\mu$ we denote by $|\mu|$ its variation.

##  3\. Covering dimension

This section is devoted to the basic properties of covering dimension and its
behavior in inverse limits of compact spaces. We start with several basic
definitions. Recall that for a family $\mathcal{A}$ of sets we define its
order as the largest integer $n$ such that $\mathcal{A}$ contains $n+1$ sets
with non-empty intersection. If there is no such $n$, then we say that the
order of $\mathcal{A}$ is $\infty$.

######  Definition 3.1.

[9, Definition 1.6.7] Let $X$ be a topological space. We say that covering
dimension of $X$ (denoted by $\dim X$) is not greater than $n$, if every
finite open cover of $X$ has a finite open refinement of order at most $n$. We
say that $\dim X=n$ if $\dim X\leq n$, but not $\dim X\leq n-1$. If there is
no $n$ such that $\dim X=n$, then we say that $\dim X=\infty$.

######  Definition 3.2.

[9, Definition 1.1.3] Let $X$ be a topological space. A closed set $P\subseteq
X$ is a partition between $A$ and $B$ if there are disjoint open sets
$U\supseteq A,V\supseteq B$ such that $X\backslash P=U\cup V$.

######  Definition 3.3.

[5, p. 16] A family $\\{(A_{i},B_{i}):i=1,2,\dots,n\\}$ of pairs of disjoint
closed subsets of a space $X$ is called essential if for every family
$\\{C_{i}:i=1,2,\dots,n\\}$ such that for each $i\leq n$ the set $C_{i}$ is a
partition between $A_{i}$ and $B_{i}$ we have

$\bigcap_{i=1}^{n}C_{i}\neq\varnothing.$

For the proof of the following theorems see [5, Lemma 3.2, Theorem 3.3].

######  Theorem 3.4.

For a normal space $X$ the following conditions are equivalent:

  1. (1)

a family $\\{(A_{i},B_{i}):i=1,2,\dots,n\\}$ of pairs of disjoint closed sets
is not essential in $X$,

  2. (2)

for each $i=1,2,\dots n$ there are disjoint open sets $U_{i},V_{i}$ such that
$A_{i}\subseteq U_{i},B_{i}\subseteq V_{i}$ and

$\bigcup_{i=1}^{n}(U_{i}\cup V_{i})=X,$

  3. (3)

for each $i=1,2,\dots n$ there are disjoint closed sets $C_{i},D_{i}$ such
that $A_{i}\subseteq C_{i},B_{i}\subseteq D_{i}$ and

$\bigcup_{i=1}^{n}(C_{i}\cup D_{i})=X.$

######  Theorem 3.5.

For a normal space $X$ the following conditions are equivalent:

  1. (1)

$\dim X\geq n$,

  2. (2)

there is an essential family in $X$ consisting of $n$ pairs.

######  Definition 3.6.

Let $\pi:L\rightarrow K$ be a continuous function between compact Hausdorff
spaces. We will say that $\pi$ is essential-preserving if for every family
$\\{(A_{i},B_{i}):i=1,2,\dots,n\\}$ essential in $K$, the family
$\\{(\pi^{-1}(A_{i}),\pi^{-1}(B_{i})):i=1,2,\dots,n\\}$ is essential in $L$.

Note that Theorem 3.5 immediately implies that if $\pi:L\rightarrow K$ is
essential-preserving, then $\dim L\geq\dim K$.

######  Lemma 3.7.

[5, Lemma 16.1] Assume that $K_{\gamma}$ is an inverse limit of a system
$\\{K_{\alpha}:\alpha<\gamma\\}$, where $K_{\alpha}$ are compact Hausdorff
spaces. If $A,B$ are closed disjoint subsets of $K_{\gamma}$ then there is
$\alpha<\gamma$ such that $\pi^{\gamma}_{\alpha}[A],\pi^{\gamma}_{\alpha}[B]$
are disjoint subsets of $K_{\alpha}$, where $\pi^{\gamma}_{\alpha}$ stands for
the canonical projection from $K_{\gamma}$ into $K_{\alpha}$.

######  Theorem 3.8.

Let $\\{K_{\alpha}:\alpha<\gamma\\}$ be an inverse system of compact Hausdorff
spaces with inverse limit $K_{\gamma}$ such that for each limit ordinal
$\beta<\gamma$, $K_{\beta}$ is an inverse limit of
$\\{K_{\alpha}:\alpha<\beta\\}$. Assume that for each $\alpha<\gamma$ the map
$\pi^{\alpha+1}_{\alpha}:K_{\alpha+1}\rightarrow K_{\alpha}$ is surjective and
essential-preserving. Then the canonical projection
$\pi^{\gamma}_{1}:K_{\gamma}\rightarrow K_{1}$ is essential-preserving. In
particular $\dim K_{\gamma}\geq\dim K_{1}$.

###### Proof.

We will prove by induction on $\alpha$ that
$\pi^{\alpha}_{1}:K_{\alpha}\rightarrow K_{1}$ is essential-preserving. For
successor ordinal $\alpha+1$ it is enough to observe that if
$\\{(A_{i},B_{i}):i=1,\dots,n\\}$ is essential in $K_{1}$, then
$\\{((\pi^{\alpha}_{1})^{-1}(A_{i}),(\pi^{\alpha}_{1})^{-1}(B_{i})):i=1,\dots,n\\}$
is essential in $K_{\alpha}$ and hence
$\\{((\pi^{\alpha+1}_{1})^{-1}(A_{i}),(\pi^{\alpha+1}_{1})^{-1}(B_{i})):i=1,\dots,n\\}=\\{((\pi^{\alpha+1}_{\alpha})^{-1}((\pi^{\alpha}_{1})^{-1}(A_{i})),\\\
(\pi^{\alpha+1}_{\alpha})^{-1}((\pi^{\alpha}_{1})^{-1}(B_{i}))):i=1,\dots,n\\}$
is essential in $K_{\alpha+1}$.

Let $\alpha$ be a limit ordinal and that for each $\beta<\alpha$ the map
$\pi^{\beta}_{1}:K_{\beta}\rightarrow K_{1}$ is essential-preserving. Let
$\\{(A_{i},B_{i}):i=1,\dots,n\\}$ be an essential family in $K_{1}$ and assume
that
$\\{(\pi^{\alpha}_{1})^{-1}(A_{i}),(\pi^{\alpha}_{1})^{-1}(B_{i})):i=1,\dots,n\\}$
is not essential in $K_{\alpha}$. Then by Theorem 3.4 for each $i\leq n$ there
are closed disjoint sets
$C_{i}\supseteq(\pi^{\alpha}_{1})^{-1}(A_{i}),D_{i}\supseteq(\pi^{\alpha}_{1})^{-1}(B_{i})$
such that

$\bigcup_{i=1}^{n}(C_{i}\cup D_{i})=K_{\alpha}.$

By Lemma 3.7 for each $i$ there is $\beta_{i}<\alpha$ such that
$\pi^{\alpha}_{\beta_{i}}[C_{i}],\pi^{\alpha}_{\beta_{i}}[D_{i}]$ are disjoint
subsets of $K_{\beta_{i}}$. In particular
$\pi^{\alpha}_{\beta}[C_{i}],\pi^{\alpha}_{\beta}[D_{i}]$ are disjoint closed
subsets of $K_{\beta}$, where $\beta=\max\\{\beta_{i}:i\leq n\\}$. Since
$K_{\alpha}$ is an inverse limit of surjective maps $\pi^{\alpha}_{\beta}$ is
also surjective and so

$\bigcup_{i=1}^{n}(\pi^{\alpha}_{\beta}[C_{i}]\cup\pi^{\alpha}_{\beta}[D_{i}])=K_{\beta}.$

Moreover, $(\pi^{\beta}_{1})^{-1}(A_{i})\subseteq\pi^{\alpha}_{\beta}[C_{i}]$
and $(\pi^{\beta}_{1})^{-1}(B_{i})\subseteq\pi^{\alpha}_{\beta}[D_{i}]$, so
$\\{(\pi^{\beta}_{1})^{-1}(A_{i}),\pi^{\alpha}_{\beta}[C_{i}]:i\leq n\\}$ is
not essential in $K_{\beta}$ which contradicts the inductive assumption. ∎

We will need some basic but important properties of the covering dimension.

######  Theorem 3.9.

[9, Theorem 3.1.3] If $M$ is a closed subspace of a normal space $X$, then
$\dim M\leq\dim X$.

######  Theorem 3.10.

[9, Theorem 3.1.8] Let $n\in\omega\cup\\{\infty\\}$. If a normal space $X$ is
a union of countably many closed subspaces $\\{F_{i}\\}_{i\in\omega}$ with
$\dim F_{i}\leq n$, then $\dim X\leq n$.

######  Theorem 3.11.

[9, Theorem 3.2.13] If $X,Y$ are non-empty compact Hausdorff spaces, then
$\dim(X\times Y)\leq\dim X+\dim Y$.

######  Theorem 3.12.

[9, Theorem 3.4.11] If $K$ is an inverse limit of compact Hausdorff spaces of
dimension at most $n$, then $\dim K\leq n$.

######  Definition 3.13.

[9, p. 170] Let $A$ be a subspace of a space $X$. We define the relative
dimension of $A$ as

$\operatorname{rd}_{X}A=\sup\\{\dim F:F\subseteq A,F\ \text{closed in}\ X\\}.$

######  Lemma 3.14.

Let $n\in\omega\cup\\{\infty\\}$. Assume that a normal space $X$ can be
represented as a union $U\cup F$ where $F$ is finite and
$\operatorname{rd}_{X}U\leq n$. Then $\dim X\leq n$.

###### Proof.

This is a special case of [9, Lemma 3.1.6] (which says that if
$X=\bigcup_{i=0}^{\infty}F_{i}$ and for each $k\in\omega$ the subspace
$\bigcup_{i=0}^{k}F_{i}$ is closed in $X$, and $\operatorname{rd}_{X}F_{k}\leq
n$, then $\dim X\leq n$) where $F_{0}=F,F_{1}=U$ and $F_{n}=\varnothing$ for
$n>1$. ∎

######  Theorem 3.15.

Assume that compact Hausdorff spaces $X$ and $Y$ can be represented as
$X=U\cup F,Y=V\cup E$ where $U,V$ are open, $E,F$ are finite, $U\cap F=V\cap
E=\varnothing$ and $U$ is homeomorphic to $V$. Then $\dim X=\dim Y$.

###### Proof.

By Theorem 3.9 we have $\operatorname{rd}_{X}U\leq\dim X$ and by Lemma 3.14
$\dim X\leq\operatorname{rd}_{X}U$, so $\dim X=\operatorname{rd}_{X}U$. By the
same argument $\dim Y=\operatorname{rd}_{Y}V$. Since $X,Y$ are compact we have

$\operatorname{rd}_{X}U=\sup\\{\dim F:F\subseteq U,F\ \text{compact}\\}$

and

$\operatorname{rd}_{Y}V=\sup\\{\dim F:F\subseteq V,F\ \text{compact}\\}.$

But $U$ and $V$ are homeomorphic, so every compact subset of $U$ is
homeomorphic to some compact subset of $V$ and vice versa, and hence
$\operatorname{rd}_{X}U=\operatorname{rd}_{Y}V$. This gives $\dim
X=\operatorname{rd}_{X}U=\operatorname{rd}_{Y}V=\dim Y$. ∎

######  Theorem 3.16.

Suppose that $K$ is a metrizable compact space and $\mu$ is a non-zero Radon
measure on $K$. Then there is a compact zero-dimensional subset $Z\subseteq K$
such that $\mu(Z)\neq 0$.

###### Proof.

Let $\\{d_{n}\\}_{n\in\omega}$ be a countable dense subset of $K$. For every
$n\in\omega$ pick a countable local base $\\{U_{i}^{n}\\}_{i\in\omega}$ at
$d_{n}$ such that $\overline{U}_{i+1}\subseteq U_{i}$ for $i\in\omega$. Then
for every $n\in\omega$ there is $k_{n}\in\omega$ such that

$\sum_{i=k_{n}}^{\infty}|\mu|(\partial U_{i}^{n})<\frac{\|\mu\|}{2^{n+1}}.$

In particular we have

$|\mu|(Y)\neq 0$

where

$Y=K\backslash\bigcup_{n=0}^{\infty}\bigcup_{i=k_{n}}^{\infty}\partial
U_{i}^{n}.$

Moreover, $Y$ is zero-dimensional, since $\\{U_{i}^{n}\cap Y:n\in\omega,i\geq
k_{n}\\}=\\{(U_{i}^{n}\backslash\partial U_{i}^{n})\cap Y:n\in\omega,i\geq
k_{n}\\}$ forms a base of $Y$ consisting of clopen sets. By regularity of
$\mu$ there is a compact set $Z\subseteq Y$ with $\mu(Z)\neq 0$ which is zero-
dimensional as a compact subset of zero-dimensional space $Y$. ∎

##  4\. Spaces $C(K)$ with few operators

We will follow the terminology from [23]. We say that a bounded linear
operator $T:C(K)\rightarrow C(K)$ is a weak multiplication, if it is of the
form $T=gI+S$, where $g$ is a continuous function on $K$, $I$ is the identity
operator and $S:C(K)\rightarrow C(K)$ is weakly compact. $T$ is called a weak
multiplier, if $T^{*}=gI+S$ for some bounded Borel map
$g:K\rightarrow\mathbb{R}$ and weakly compact $S:C(K)^{*}\rightarrow
C(K)^{*}$.

######  Definition 4.1.

Let $K$ be a compact Hausdorff space. We say that the Banach space $C(K)$ has
few operators if every bounded linear operator $T:C(K)\rightarrow C(K)$ is a
weak multiplication.

######  Lemma 4.2.

Suppose that $K$ is a c.c.c. compact Hausdorff space and that $C(K)\sim C(L)$
for a compact Hausdorff space $L$. Then $L$ is also c.c.c.

###### Proof.

By [35, Theorem 4.5(a)] a compact space $M$ is c.c.c. if and only if $C(M)$
contains no isomorphic copy of $c_{0}(\omega_{1})$, so in particular given
property is an isomorphism invariant. ∎

######  Lemma 4.3.

Let $K$ be a compact Hausdorff space. If $K$ has a non-trivial convergent
sequence, then $C(K)$ admits a complemented copy of $c_{0}$. In particular, if
$C(K)$ has few operators, then $K$ has no non-trivial convergent sequences.

###### Proof.

The fact that non-trivial convergent sequences give rise to complemented
copies of $c_{0}$ is well-known (cf. [16]). The second part of the lemma
follows from [21, Theorem 13 (3)]. ∎

######  Lemma 4.4.

Assume that $K$ is a separable connected compact Hausdorff space such that
$C(K)$ has few operators and $L$ is a compact Hausdorff space such that
$C(K)\sim C(L)$. Let $J$ be the set of isolated points in $L$ and
$L^{\prime}=L\backslash J$. Then $J$ is a countable set and $L^{\prime}$ has
no isolated points.

###### Proof.

Since $K$ is separable, it is c.c.c., so by Lemma 4.2 $L$ is also c.c.c. In
particular $J$ is countable.

Obviously, if $J$ is finite, then $L^{\prime}$ has no isolated points, so we
may assume that $J$ is infinite. Suppose that $x\in L^{\prime}$ is an isolated
point. Then $L^{\prime}\backslash\\{x\\}$ is a closed subspace of $L$, so
there is an open set $V\subseteq L$ such that $x\in V$ and
$\overline{V}\cap(L^{\prime}\backslash\\{x\\})=\varnothing$.
$\overline{V}\subseteq J\cup\\{x\\}$, so $\overline{V}$ is an infinite
countable compact space with exactly one non-isolated point i.e. it is a
convergent sequence. By Lemma 4.3 $C(L)$ admits a complemented copy of
$c_{0}$, and so $C(K)$ admits a complemented copy of $c_{0}$. However, it is
impossible since by [21, Theorem 13 (a)] $C(K)$ is indecomposable. ∎

######  Definition 4.5.

For a compact space $K$ and a function $f\in C(K)$ we denote by $M_{f}$ the
operator $M_{f}:C(K)\rightarrow C(K)$ given by $M_{f}(g)=fg$.

In the next lemmas we will use the following characterization of weakly
compact operators on Banach spaces of continuous functions from [7, p. 160].

######  Theorem 4.6.

If $K$ is a compact Hausdorff space, then an operator $T$ on $C(K)$ is weakly
compact if and only if for every bounded sequence $(e_{n})_{n\in\omega}$ of
pairwise disjoint functions (i.e. $e_{n}\cdot e_{m}=0$ for $n\neq m$) we have
$\lim_{n\rightarrow\infty}\|T(e_{n})\|=0$.

######  Lemma 4.7.

Let $L$ be a compact Hausdorff space, $J$ the set of isolated points in $L$,
and $L^{\prime}=L\backslash J$. Assume that $f\in C(L)$ is such that
$f|L^{\prime}=0$. Then $M_{f}$ is weakly compact.

###### Proof.

Fix any bounded pairwise disjoint sequence $(e_{n})_{n\in\omega}$ of elements
of $C(L)$. Without loss of generality we may assume that $\|e_{n}\|\leq 1$ for
each $n$. Let $\varepsilon>0$. Since $f$ is continuous and equal to $0$ on
$L^{\prime}$ there is only finitely many points $x$ such that
$|f(x)|\geq\varepsilon$. Hence for $n$ large enough we have
$\|M_{f}(e_{n})\|=\|fe_{n}\|<\varepsilon$, which means that
$\lim_{n\rightarrow\infty}\|M_{f}(e_{n})\|=0$. Now Theorem 4.6 says that
$M_{f}$ is weakly compact. ∎

######  Lemma 4.8.

Assume that $K$ has no isolated points and $f\in C(K)$ is such that $M_{f}$ is
weakly compact. Then $f=0$.

###### Proof.

Assume that $f\neq 0$. Then there is non-empty open set $U\subset K$ such that
$|f(x)|\geq\varepsilon$ for $x\in U$ and some $\varepsilon>0$. Since there are
no isolated points in $K$, $U$ is infinite, so there are pairwise disjoint
open subsets $U_{n}\subseteq U$. Let $e_{n}\in C(K)$ be such that $e_{n}(x)=1$
for some $x\in U_{n},e_{n}(x)=0$ for $x\in K\backslash U_{n}$ and
$\|e_{n}\|=1$. Then for each $n\in\omega$ we have
$\|M_{f}e_{n}\|\geq\varepsilon$, so by Theorem 4.6 $M_{f}$ is not weakly
compact. ∎

######  Lemma 4.9.

Let $f\in C(L)$ for $L$ compact Hausdorff and assume that there is a non-
isolated point $x_{0}\in L$ such that $|f(x_{0})|=\|f\|$. If
$R:C(L)\rightarrow C(L)$ is a weakly compact operator, then
$\|f\|\leq\|M_{f}+R\|$.

###### Proof.

Since $x_{0}$ is non-isolated there are distinct points $x_{n}\in L$ such that
$|f(x_{n})|>\|f\|-1/n$. By passing to a subsequence we may assume that
$\\{x_{n}:n\in\omega\\}$ is a relatively discrete subset of $L$.

Take pairwise disjoint open sets $U_{n}\subseteq\\{x\in
K:|f(x)|>\|f\|-1/n\\},x_{n}\in U_{n}$. For each $n\in\omega$ pick $e_{n}\in
C(L)$ such that $\|e_{n}\|=1$ and $e_{n}|(L\backslash U_{n})=0$. In particular
$(e_{n})_{n\in\omega}$ are pairwise disjoint functions, so by Theorem 4.6
$\lim_{n\rightarrow\infty}\|R(e_{n})\|=0$. Moreover,
$\|M_{f}(e_{n})\|=\|fe_{n}\|\geq\|f\|-1/n$ (from the property of $U_{n}$).
Hence we get that
$\|M_{f}+R\|\geq\|(M_{f}+R)(e_{n})\|=\|M_{f}(e_{n})+R(e_{n})\|\geq\|M_{f}(e_{n})\|-\|R(e_{n})\|\geq\|f\|-1/n-\|R(e_{n})\|$.
By taking limit with $n\rightarrow\infty$ we get $\|M_{f}+R\|\geq\|f\|$. ∎

######  Remark 4.10.

If $K$ and $L$ are compact Hausdorff spaces, and $T:C(K)\rightarrow C(L)$ is
an isomorphism of Banach spaces, then $T$ induces an isomorphism of the Banach
algebras $\Phi_{T}:\mathcal{B}(C(L))\rightarrow\mathcal{B}(C(K))$ given by

$\Phi_{T}(U)=T^{-1}UT.$

If $R\in\mathcal{B}(C(L))$ is a weakly compact operator, then $\Phi_{T}(R)$ is
also weakly compact as a composition of a weakly compact operator with bounded
operators. Similarly, if $S\in\mathcal{B}(C(K))$ is weakly compact, then
$\Phi_{T}^{-1}(S)$ is weakly compact.

For the rest of this section we will assume that $K$ and $L$ are compact
Hausdorff spaces, $L^{\prime}$ is the set of non-isolated points of $L$,
$C(K)$ has few operators and $T:C(K)\rightarrow C(L)$ is an isomorphism of
Banach spaces.

######  Definition 4.11.

Let $\Phi_{T}$ be such as in Remark 4.10. We define an operator
$\Psi_{T}:C(L^{\prime})\rightarrow C(K)$ by putting for each $f^{\prime}\in
C(L^{\prime})$

$\Psi_{T}(f^{\prime})=g,$

for $g\in C(K)$ satisfying $\Phi_{T}(M_{f})=M_{g}+R,$ where $R$ is weakly
compact and $f\in C(L)$ is such that $f^{\prime}=f|L^{\prime}$.

In other words, $\Psi_{T}$ is defined in the way such that the following
diagram commutes:

${C(L)}$${\mathcal{B}(C(L))}$${\mathcal{B}(C(K))}$${\mathcal{B}(C(K))/\mathcal{WC}(C(K))}$${C(L^{\prime})}$${C(K)}$$\scriptstyle{\mathcal{R}}$$\scriptstyle{M}$$\scriptstyle{\Phi_{T}}$$\scriptstyle{\pi}$$\scriptstyle{\mathcal{I}}$$\scriptstyle{\Psi_{T}}$

Here $\mathcal{R}$ stands for the restriction operator (i.e.
$\mathcal{R}(f)=f|L^{\prime}$), $M(f)=M_{f}$, $\pi$ is the natural surjection
onto the quotient algebra $\mathcal{B}(C(K))/\mathcal{WC}(C(K))$, where
$\mathcal{WC}(C(K))$ is the closed ideal in $\mathcal{B}(C(K))$ consisting of
weakly compact operators and
$\mathcal{I}:\mathcal{B}(C(K))/\mathcal{WC}(C(K))\rightarrow C(K)$ is the
isometry given by $\mathcal{I}([M_{g}])=g$.

######  Lemma 4.12.

Suppose that $K$ has no isolated points. Then the induced operator
$\Psi_{T}:C(L^{\prime})\rightarrow C(K)$ from Definition 4.11 is a well-
defined bounded linear and multiplicative operator.

###### Proof.

Take any $f^{\prime}\in C(L^{\prime})$ and let $f_{1},f_{2}\in C(L)$ and
$g_{1},g_{2}\in C(K)$ be such that
$f_{1}|L^{\prime}=f_{2}|L^{\prime}=f^{\prime}$ and

$\Phi_{T}(M_{f_{i}})=M_{g_{i}}+R_{i}\ \text{for}\ i=1,2,$

where $R_{1},R_{2}$ are weakly compact. Then $(f_{1}-f_{2})|L^{\prime}=0$, so
by Lemma 4.7 $M_{f_{1}}-M_{f_{2}}=M_{f_{1}-f_{2}}$ is weakly compact. This
implies that

$\displaystyle
M_{g_{1}-g_{2}}=M_{g_{1}}-M_{g_{2}}=R_{1}-\Phi_{T}(M_{f_{1}})-R_{2}+\Phi_{T}(M_{f_{2}})=$
$\displaystyle=R_{1}-R_{2}-\Phi_{T}(M_{f_{1}}-M_{f_{2}})$

is weakly compact since $\Phi_{T}(M_{f_{1}}-M_{f_{2}})$ is weakly compact (cf.
Remark 4.10). Since $K$ has no isolated points, Lemma 4.8 implies that
$g_{1}-g_{2}=0$, so $\Psi_{T}$ is well-defined.

For the linearity and multiplicativeness fix
$f^{\prime}_{1}=f_{1}|L^{\prime},f^{\prime}_{2}=f_{2}|L^{\prime}\in
C(L),a,b\in\mathbb{R}$ and put
$\Psi_{T}(f^{\prime}_{1})=g_{1},\Psi_{T}(f^{\prime}_{2})=g_{2}$. We have

$\displaystyle\Phi_{T}(M_{af_{1}+bf_{2}})=\Phi_{T}(aM_{f_{1}}+bM_{f_{2}})=a\Phi_{T}(M_{f_{1}})+b\Phi_{T}(M_{f_{2}})=$
$\displaystyle=M_{ag_{1}}+aR_{1}+M_{bg_{2}}+bR_{2}=M_{ag_{1}+bg_{2}}+aR_{1}+bR_{2}$

and

$\displaystyle\Phi_{T}(M_{f_{1}f_{2}})=\Phi_{T}(M_{f_{1}}M_{f_{2}})=\Phi_{T}(M_{f_{1}})\Phi_{T}(M_{f_{2}})=$
$\displaystyle=(M_{g_{1}}+R_{1})(M_{g_{2}}+R_{2})=M_{g_{1}g_{2}}+R_{1}M_{g_{2}}+M_{g_{1}}R_{2}+R_{1}R_{2}.$

But $aR_{1}+bR_{2}$ and $R_{1}M_{g_{2}}+M_{g_{1}}R_{2}+R_{1}R_{2}$ are weakly
compact as the sums of weakly compact operators composed with bounded
operators. Hence $\Psi_{T}(af^{\prime}_{1}+bf^{\prime}_{2})=ag_{1}+bg_{2}$ and
$\Psi_{T}(f^{\prime}_{1}f^{\prime}_{2})=g_{1}g_{2}$.

Now we will show that $\Psi_{T}$ is bounded. Pick any $f^{\prime}\in
C(L^{\prime})$. By the Tietze theorem $f^{\prime}$ has an extension $f\in
C(L)$ satisfying $\|f\|=\|f^{\prime}\|$. From Lemma 4.9 we get that if
$\Phi_{T}(M_{f})=M_{g}+R$, then
$\|g\|\leq\|M_{g}+R\|\leq\|\Phi_{T}\|\|M_{f}\|=\|\Phi_{T}\|\|f\|=\|\Phi_{T}\|\|f^{\prime}\|$,
so $\|\Psi_{T}\|\leq\|\Phi_{T}\|$. ∎

######  Lemma 4.13.

Suppose that $K$ is separable and connected. Then there is $c>0$ such that for
every $f^{\prime}\in C(L^{\prime})$ we have $\|\Psi_{T}(f^{\prime})\|\geq
c\|f^{\prime}\|$ i.e. $\Psi_{T}$ is an isomorphism onto its range. In
particular $\Psi_{T}$ has closed range.

###### Proof.

Assume that $\Psi_{T}(f^{\prime})=g$. Let $f\in C(L)$ be an extension of
$f^{\prime}$ such that $\|f\|=\|f^{\prime}\|$. We have
$\Phi_{T}(M_{f})=M_{g}+R$ for some weakly compact $R$, so
$\Phi_{T}^{-1}(M_{g})=M_{f}-\Phi_{T}^{-1}(R)$. $\Phi_{T}^{-1}(R)$ is weakly
compact by Remark 4.10, so from Lemma 4.9 we get

$\displaystyle\|f\|\leq\|M_{f}-\Phi_{T}^{-1}(R)\|=\|\Phi_{T}^{-1}\circ\Phi_{T}(M_{f}-\Phi_{T}^{-1}(R))\|=\|\Phi_{T}^{-1}(M_{g}+R-R)\|=$
$\displaystyle=\|\Phi_{T}^{-1}(M_{g})\|\leq\|\Phi_{T}^{-1}\|\|M_{g}\|=\|\Phi_{T}^{-1}\|\|g\|.$

Hence it is enough to take $c=\frac{1}{\|\Phi_{T}^{-1}\|}$. ∎

######  Proposition 4.14.

Suppose that $K$ is separable and connected.
Let $S:C(K)\rightarrow C(K)$ be given by $S(f)=\Psi_{T}(T(f)|L^{\prime})$.
Then

$\ker(S)=T^{-1}(\\{g\in C(L):g|L^{\prime}=0\\})$

and it is a separable subspace of $C(K)$.

${C(K)}$${C(L)}$${C(L^{\prime})}$${C(K)}$$\scriptstyle{T}$$\scriptstyle{S}$$\scriptstyle{\mathcal{R}}$$\scriptstyle{\Psi_{T}}$

###### Proof.

By Lemma 4.4 the set $J$ of isolated points in $L$ is countable, so we may
write $J=\\{x_{n}:n\in\omega\\}$. Let $\chi_{\\{x_{n}\\}}$ be the
characteristic function of $\\{x_{n}\\}$. Observe that
$\overline{\operatorname{span}\\{\chi_{\\{x_{n}\\}}:n\in\omega\\}}=\\{g\in
C(L):g|L^{\prime}=0\\}$ is a separable subspace of $C(L)$, so it is enough to
show that $\ker(S)=T^{-1}(\\{g\in C(L):g|L^{\prime}=0\\})$, since $T$ is an
isomorphism.

Assume that $S(f)=0$. Then $\Psi_{T}(T(f)|L^{\prime})=0$, so
$\Phi_{T}(M_{T(f)})=M_{0}+R=R$ is weakly compact and hence
$M_{T(f)}=T\Phi_{T}(M_{T(f)})T^{-1}$ is also weakly compact as a composition
of a weakly compact operator with bounded operators. From Theorem 4.6
$\lim_{n\rightarrow\infty}\|T(f)e_{n}\|=0$ for every bounded disjoint sequence
$(e_{n})_{n\in\omega}$. This implies that
$\lim_{n\rightarrow\infty}\|(T(f)|L^{\prime})e_{n}\|=0$ for every bounded
disjoint sequence $(e_{n})_{n\in\omega}$. By applying Theorem 4.6 once again
we get that $M_{T(f)|L^{\prime}}$ is weakly compact as an operator on
$C(L^{\prime})$. Since $L^{\prime}$ has no isolated points (cf. Lemma 4.4) we
get that $T(f)|L^{\prime}=0$ by Lemma 4.8 i.e. $f\in T^{-1}(\\{g\in
C(L):g|L^{\prime}=0\\})$, so $\ker(S)\subseteq T^{-1}(\\{g\in
C(L):g|L^{\prime}=0\\})$.

If $g\in C(L)$ is such that $g|L^{\prime}=0$, then by Lemma 4.7 $M_{g}$ is
weakly compact, so $S(T^{-1}(g))=\Psi(g|L^{\prime})=\Psi(0)=0$ and hence
$T^{-1}(g)\in\ker(S)$. ∎

######  Proposition 4.15.

Suppose that $K$ is separable and connected. Let
$S=\Psi_{T}(T(f)|L^{\prime})$. Write $S$ as a sum $S=M_{e}+W$ with $W$ weakly
compact. Then $M_{e}$ is an isomorphism of $C(K)$.

###### Proof.

It is enough to prove that $e(x)\neq 0$ for every $x\in K$. Indeed, if it is
the case, then $M_{g}$ is the inverse of $M_{e}$ for $g=\frac{1}{e}$.

Assume that $e(z)=0$ for some $z\in K$ and aim for a contradiction. Then using
the technique from the proof of Lemma 4.9 we construct pairwise disjoint non-
empty open subsets $U_{n}\subseteq K$ such that $\|e|U_{n}\|\leq\frac{1}{n}$
for each $n\in\omega$. Let $V_{n}$ be non-empty open sets such that
$\overline{V}_{n}\subseteq U_{n}$.

By Lemma 4.3 $K$ has no convergent sequences and hence for every $n\in\omega$
the space $\overline{V}_{n}$ is non-metrizable as an infinite (because $V_{n}$
has no isolated points) compact set without convergent sequences. We get that
points in $\overline{V}_{n}$ cannot be separated by countable family of
continuous functions (otherwise, if $(f_{n})_{n\in\omega}$ separated points of
$\overline{V}_{n}$,
$(f_{1},f_{2},\dots):\overline{V}_{n}\rightarrow\mathbb{R}^{n}$ would be a
homeomorphism onto a compact subspace of metrizable space), so since $\ker(S)$
is separable, there are points $x_{n},y_{n}\in\overline{V}_{n}\subseteq U_{n}$
such that $d(x_{n})=d(y_{n})$ for all $d\in\ker(S)$. Let $f_{n}\in C(K)$ be
such that $\|f_{n}\|=1,f_{n}(x_{n})=1,f_{n}(y_{n})=0$ and $f_{n}|(K\backslash
U_{n})=0$. Then for all $d\in\ker(S)$

$\displaystyle\|f_{n}-d\|\geq\max\\{|f_{n}(x_{n})-d(x_{n})|,|f_{n}(y_{n})-d(y_{n})|\\}=$
$\displaystyle=\max\\{|1-d(x_{n})|,|d(x_{n})|\\}\geq 1/2.$

Since $f_{n}|(K\backslash U_{n})=0$ and $\|e|U_{n}\|\leq\frac{1}{n}$ we have
$\|ef_{n}\|\leq\frac{1}{n}$, so $\lim_{n\rightarrow\infty}\|ef_{n}\|=0$.

$\Psi_{T}$ has closed range (Lemma 4.13) and $T,\mathcal{R}$ are surjective,
so $S$ has also closed range. By the first isomorphism theorem (see e.g. [10,
Corollary 2.26]) $S[C(K)]$ is isomorphic to $C(K)/\ker(S)$, so since the
distance of $f_{n}$ from $\ker(S)$ is greater than $1/2$ for all $n\in\omega$,
there is $c>0$ such that $\|S(f_{n})\|>c$ for all $n\in\omega$.

${C(K)}$${{S[C(K)]}}$${C(K)/\ker(S)}$$\scriptstyle{S}$$\scriptstyle{\sim}$

But on the other hand we have

$\|S(f_{n})\|=\|ef_{n}+W(f_{n})\|\leq\|ef_{n}\|+\|W(f_{n})\|\rightarrow 0$

when $n\rightarrow\infty$, since we have
$\lim_{n\rightarrow\infty}\|ef_{n}\|=0$ and
$\lim_{n\rightarrow\infty}\|W(f_{n})\|=0$ (because $W$ is weakly compact and
$(f_{n})$ are bounded and pairwise disjoint), so we get a contradiction. ∎

Recall that an operator $R:X\rightarrow Y$ is called strictly singular, if for
every infinite-dimensional subspace $X^{\prime}\subseteq X$ the restriction
$R|X^{\prime}$ is not isomorphism. We cite the result from [28].

######  Theorem 4.16.

Let $X$ be a compact Hausdorff space. A bounded operator $R:C(X)\rightarrow
C(X)$ is weakly compact if and only if it is strictly singular.

If we apply the above theorem to [24, Proposition 2.c.10] we get the
following.

######  Theorem 4.17.

Let $E:C(X)\rightarrow C(X)$ be an operator with closed range for which
$\dim\ker(E)<\infty$ and $\dim(C(X)/E(C(X)))<\infty$. Let $R:C(X)\rightarrow
C(X)$ be weakly compact. Then $E+R$ also has closed range and
$\dim((C(X))/(E+R)(C(X))<\infty$.

######  Corollary 4.18.

Suppose that $K$ is separable and connected. Let
$S=\Psi_{T}(T(f)|L^{\prime})$.

Then the range of $S$ is finite-codimensional in $C(K)$. In particular the
range of $\Psi_{T}$ is finite-codimensional in $C(K)$.

###### Proof.

Since $M_{e}$ is an isomorphism (by Proposition 4.15) and $W$ is weakly
compact we may apply Theorem 4.17 to $S=M_{e}+W$. ∎

Since $\Psi_{T}:C(L^{\prime})\rightarrow C(K)$ is a bounded linear
multiplicative operator (Lemma 4.12), there is $\varphi:K\rightarrow
L^{\prime}$ such that $\Psi_{T}(f)=f\circ\varphi$ for $f\in C(L^{\prime})$
(see e.g. [37, Theorem 7.7.1]). From Lemma 4.13 and Corollary 4.18 we get that
$\Psi_{T}$ is an embedding with finite-codimensional range, so the induced map
$\varphi$ is surjective and has only finitely many fibers containing more than
one element and each of these fibers is finite. In particular $K=U\cup F$
where $F$ is a finite set and $\varphi|U$ is a homeomorphism and we get the
following theorem.

######  Theorem 4.19.

Suppose that $K$ is a separable connected compact Hausdorff space such that
$C(K)$ has few operators and $L$ is a compact Hausdorff space such that
$C(K)\sim C(L)$.

Then $K$ and $L$ are homeomorphic modulo finite set i.e. there are open
subsets $U\subseteq K,V\subseteq L$ and finite sets $E\subseteq K,F\subseteq
L$ such that $U,V$ are homeomorphic and $K=U\cup E,L=V\cup F$.

######  Corollary 4.20.

If $\dim(K)=n$ and $K$ is a compact, separable and connected Hausdorff space
such that $C(K)$ has few operators, then for each compact Hausdorff space $L$
such that $C(K)\sim C(L)$ we have $\dim(L)=n$.

###### Proof.

Use Theorem 4.19 and Theorem 3.15. ∎

##  5\. Extensions of compact spaces

In this section we consider the notion of strong extension from [19]. We
describe the methods of controlling the dimension in constructions of compact
spaces using strong extensions. We prove that strong extensions cannot lower
the dimension of initial space and we show how to construct extensions that
cannot rise the dimension.

######  Definition 5.1.

Let $K$ be a compact Hausdorff space and $(f_{n})_{n\in\omega}$ be a sequence
of pairwise disjoint continuous functions $f_{n}:K\rightarrow[0,1]$. Define

$D((f_{n})_{n\in\omega})=\bigcup\\{U:U\ \text{is open and}\
\\{n:\operatorname{supp}(f_{n})\cap U\neq\varnothing\\}\ \text{is finite}\\}.$

We say that $L\subseteq K\times[0,1]$ is the extension of $K$ by
$(f_{n})_{n\in\omega}$ if and only if $L$ is the closure of the graph of
$(\sum_{n\in\omega}f_{n})|D((f_{n})_{n\in\omega})$. We say that this is a
strong extension, if the graph of $\sum_{n\in\omega}f_{n}$ is a subset of $L$.

######  Lemma 5.2.

[19, Lemma 4.1] If $(f_{n})_{n\in\omega}$ are pairwise disjoint continuous
functions on $K$ with values in $[0,1]$, then $\sum_{n\in\omega}f_{n}$ is
well-defined and continuous in the dense open set $D((f_{n})_{n\in\omega})$.

######  Lemma 5.3.

[19, Lemma 4.4] Strong extension of a connected compact Hausdorff space is
connected.

Note that there are known examples of extensions of connected compact spaces
which are not connected (see [2]), so the assumption that considered
extensions are strong is necessary.

######  Lemma 5.4.

Let $K$ be a separable compact Hausdorff space with countable dense set
$Q=\\{q_{n}:\in\omega\\}$ and let $L$ be an extension of $K$ with the natural
projection $\pi:L\rightarrow K$. Assume that
$Q^{\prime}=\\{q^{\prime}_{n}:n\in\omega\\}$ is a subset of $L$ such that
$\pi(q^{\prime}_{n})=q_{n}$ for every $n\in\omega$. Then $Q^{\prime}$ is a
dense subset of $L$.

###### Proof.

Let $(f_{n})_{n\in\omega}$ be a sequence of pairwise disjoint continuous
functions such that $L$ is the extension of $K$ by $(f_{n})_{n\in\omega}$. By
[19, Lemma 4.3 a)] $\pi^{-1}(D((f_{n})_{n\in\omega})$ is dense in $L$.
Moreover, $\pi|\pi^{-1}(D((f_{n})_{n\in\omega}))$ is homeomorphism as a
projection of graph of continuous function onto its domain. Since $Q$ is dense
in $K$ and $D((f_{n})_{n\in\omega})$ is open, $Q\cap D((f_{n})_{n\in\omega})$
is dense in $D((f_{n})_{n\in\omega})$. Hence we get that $\pi^{-1}(Q\cap
D((f_{n})_{n\in\omega}))$ is dense in $L$. But if $q_{n}\in
D((f_{n})_{n\in\omega})$, then $\pi^{-1}(q_{n})=\\{q^{\prime}_{n}\\}$, so
$Q^{\prime}\supseteq\pi^{-1}(Q\cap D((f_{n})_{n\in\omega})$ is also dense in
$L$. ∎

The following lemma is a special case of [19, Lemma 4.5].

######  Lemma 5.5.

Suppose that $K$ is a compact metric space and that for every $n\in\omega$
$X^{n}_{1},X^{n}_{2}$ are disjoint relatively discrete subsets of $K$ such
that $\overline{X^{n}_{1}}\cap\overline{X^{n}_{2}}\neq\varnothing$. Let
$(f_{n})_{n\in\omega}$ be a pairwise disjoint sequence of continuous functions
from $K$ into $[0,1]$. For an infinite subset $B\subseteq\omega$ denote by
$K(B)$ the extension of $K$ by $(f_{n})_{n\in B}$. For $i=0,1$ and
$n\in\omega$ put

$X^{n}_{i}(B)=\\{(x,t):x\in X^{n}_{i},t=\sum_{k\in B}f_{k}(x)\\}.$

Then there is an infinite $N\subseteq\omega$ such that for every $B\subseteq
N$:

  1. (1)

$K(B)$ is a strong extension of $K$ by $(f_{n})_{n\in B}$,

  2. (2)

$\overline{X^{n}_{1}(B)}\cap\overline{X^{n}_{2}(B)}\neq\varnothing$ for every
$n\in\omega$, where the closures are taken in $K(B)$.

######  Proposition 5.6.

If $L$ is a strong extension of a compact Hausdorff space $K$ with the natural
projection $\pi:L\rightarrow K$, then $\pi$ is essential-preserving.

###### Proof.

Let $(f_{k})_{k\in\omega}$ be such that $L$ is a strong extension of $K$ by
$(f_{k})_{k\in\omega}$.
Let $\\{(A_{i},B_{i}):i=1,2,\dots,n\\}$ be an essential family in $K$ and
assume that $\\{(\pi^{-1}(A_{i}),\pi^{-1}(B_{i})):i=1,2,\dots,n\\}$ is not
essential in $L$. By Theorem 3.4 there are closed sets
$C_{i}\supseteq\pi^{-1}(A_{i}),D_{i}\supseteq\pi^{-1}(B_{i})$ such that
$C_{i}\cap D_{i}=\varnothing$ for each $i\leq n$ and

$\bigcup_{i=1}^{n}(C_{i}\cup D_{i})=L.$

Since $C_{i},D_{i}$ are compact, there are sets $U_{i},V_{i}$ open in
$K\times[0,1]$ such that $C_{i}\subseteq U_{i}$, $D_{i}\subseteq V_{i}$ and
$U_{i}\cap V_{i}=\varnothing$ for every $i\leq n$.

For each $k\in\omega$ denote by $L_{k}$ the graph of $\sum_{i\leq k}f_{i}$ and
let $\pi_{k}:L_{k}\rightarrow K$ be the projection onto $K$.

######  Claim.

For every $k\in\omega$ we have

$L_{k}\backslash\bigcup_{i=1}^{n}(U_{i}\cup V_{i})\neq\varnothing.$

###### Proof of the claim.

Assume that there is $N$ such that

$L_{N}\subseteq\bigcup_{i=1}^{n}(U_{i}\cup V_{i}).$

Then for every $k\geq N$

(1) $\displaystyle L_{k}\backslash
L_{N}=\operatorname{graph}({\sum_{i=N+1}^{k}f_{i}|\operatorname{supp}(\sum_{i=N+1}^{k}f_{i})})\subseteq
L\subseteq\bigcup_{i=1}^{n}(U_{i}\cup V_{i})$

(the first equality holds, because the supports of $f_{i}$’s are pairwise
disjoint), so we have

$L_{k}\subseteq\bigcup_{i=1}^{n}(U_{i}\cup V_{i}).$

Put $A_{i}^{k}=\pi_{k}^{-1}(A_{i}),B_{i}^{k}=\pi_{k}^{-1}(B_{i})$ and observe
that the family $\\{(A_{i}^{k},B_{i}^{k}):i\leq n\\}$ is essential in $L_{k}$
since $\pi_{k}$ is a homeomorphism. Hence there is $i\leq n$ such that
$A_{i}^{k}\nsubseteq U_{i}$ or $B_{i}^{k}\nsubseteq V_{i}$. Indeed, otherwise
$U_{i}\cap L_{k},V_{i}\cap L_{k}$ would be disjoint open subsets of $L_{k}$
with

$\bigcup_{i=1}^{n}((U_{i}\cap L_{k})\cup(V_{i}\cap L_{k}))=L_{k},$

which contradicts the fact that $\\{(A_{i}^{k},B_{i}^{k}):i\leq n\\}$ is
essential (cf. Theorem 3.4). Without loss of generality there are infinitely
many $k$ such that $A_{1}^{k}\backslash U_{1}\neq\varnothing$. For every
$k\in\omega$ we have

$A_{1}^{k+1}\backslash
A_{1}^{k}=\pi_{k+1}^{-1}(A_{1})\backslash\pi_{k}^{-1}(A_{1})=\operatorname{graph}(f_{k+1}|(A_{1}\cap\operatorname{supp}(f_{k+1}))\subseteq\pi^{-1}(A_{1})\subseteq
U_{1}.$

In particular $(A_{1}^{k}\backslash U_{1})_{k\in\omega}$ form a decreasing
sequence of non-empty compact sets. Hence

$A=\bigcap_{k=1}^{\infty}A_{1}^{k}\backslash U_{1}\neq\varnothing.$

We have $A\subseteq L$ since if $(x,t)\in A$, then $f_{k}(x)=0$ for all $k$,
so $\sum_{k\in\omega}f_{k}(x)=0$ and hence $(x,t)=(x,0)$ is an element of the
graph of $\sum_{k\in\omega}f_{k}$ which is a subset of $L$. Moreover
$A\subseteq A_{1}\times[0,1]$, so $A\subseteq(A_{1}\times[0,1])\cap
L=\pi^{-1}(A_{1})$ which contradicts the assumption that
$\pi^{-1}(A_{1})\subseteq U_{1}$ and completes the proof of the claim. ∎

To finish the proof of the proposition put

$F_{k}=L_{k}\backslash\bigcup_{i=1}^{n}(U_{i}\cup V_{i})$

and observe that $(F_{k})_{k\in\omega}$ is a decreasing sequence of non-empty
compact sets (by (1) from the claim), so as in the case of the set $A$ from
the claim we get that

$F=\bigcap_{k=1}^{\infty}F_{k}$

is a non-empty subset of the graph of $\sum_{k\in\omega}f_{k}$, so $F\subseteq
L$ (because the extension is strong), which is a contradiction, since $F$ is
disjoint from $\bigcup_{i\leq n}(U_{i}\cup V_{i})\supseteq L$. ∎

######  Lemma 5.7.

Suppose that $K$ is a compact metric space with $0<\dim(K)\leq n$ and
$f_{k}:K\rightarrow[0,1]$ are pairwise disjoint continuous functions such that
the set

$Z=K\backslash D((f_{k})_{k\in\omega})$

is zero-dimensional. Assume that $L$ is a strong extension of $K$ by
$(f_{k})_{k\in\omega}$. Then $\dim L\leq n$.

###### Proof.

Let $\pi$ be the natural projection from $L$ onto $K$.
$\pi^{-1}(D((f_{k})_{k\in\omega}))$ is an open subset of a metric space, so it
is a union of countably many closed sets, each of dimension at most $n$ since
$\pi^{-1}(D((f_{k})_{k\in\omega}))$ is homeomorphic to
$D((f_{k})_{k\in\omega})$ (cf. Theorem 3.9). The set $\pi^{-1}(Z)$ is included
in $Z\times[0,1]$ so $\dim\pi^{-1}(Z)\leq 1\leq n$ by Theorem 3.11. Hence
$L=\pi^{-1}(D((f_{k})_{k\in\omega}))\cup\pi^{-1}(Z)$ is a countable union of
closed sets of dimension at most $n$. Now Theorem 3.10 gives the inequality
$\dim L\leq n$. ∎

######  Corollary 5.8.

Let $\gamma$ be an ordinal number. Suppose that
$\\{K_{\alpha}:\alpha<\gamma\\}$ is an inverse system of compact Hausdorff
spaces such that:

  * •

for every $\alpha$ the map $\pi^{\alpha+1}_{\alpha}:K_{\alpha+1}\rightarrow
K_{\alpha}$ is a strong extension by pairwise disjoint continuous functions
$(f_{n}^{\alpha})_{n\in\omega}$ and the set $Z_{\alpha}=K_{\alpha}\backslash
D((f_{n}^{\alpha})_{n\in\omega})$ is zero-dimensional,

  * •

if $\alpha$ is a limit ordinal, then $K_{\alpha}$ is the inverse limit of
$\\{K_{\beta}:\beta<\alpha\\}$.

Denote by $K_{\gamma}$ the inverse limit of $\\{K_{\alpha}:\alpha<\gamma\\}$.
Then $\dim K_{\gamma}=\dim K_{1}$.

###### Proof.

The inequality $\dim K_{\gamma}\geq\dim K_{1}$ follows from Proposition 5.6
and Theorem 3.8. The inequality $\dim K_{\gamma}\leq\dim K_{1}$ follows from
Lemma 5.7 and Theorem 3.12. ∎

##  6\. The main construction

######  Theorem 6.1.

[23, Lemma 2.4] Suppose that $K$ is a compact Hausdorff space. If a bounded
linear operator $T:C(K)\rightarrow C(K)$ is not a weak multiplier, then there
are $\delta>0$, a pairwise disjoint sequence $(g_{n})_{n\in\omega}\subseteq
C_{I}(K)$ and pairwise disjoint open sets $(V_{n})_{n\in\omega}$ such that

$\operatorname{supp}(g_{n})\cap V_{m}=\varnothing$

for all $n,m\in\omega$ and

$|T(g_{n})|V_{n}|>\delta$

for all $n\in\omega$.

In particular, if $x_{n}\in V_{n}$ and $\mu_{n}=T^{*}(\delta_{x_{n}})$ for
$n\in\omega$, then $|\int g_{n}d\mu_{n}|=|T(g_{n})(q_{l_{n}})|>\delta$, and so
$|\mu_{n}|(\operatorname{supp}(g_{n}))\geq|\int g_{n}d\mu_{n}|>\delta$.

The idea behind the construction is as follows. We will construct a compact
space $K$ as the inverse limit of spaces $K_{\alpha}\subseteq[0,1]^{\alpha}$
(so the final space is a subset of $[0,1]^{\mathfrak{c}}$). For each bounded
sequence $(\mu_{n})_{n\in\omega}$ of Radon measures on $[0,1]^{\mathfrak{c}}$
and a sequence of pairwise disjoint open sets $(V_{n})_{n\in\omega}$ we want
to use a strong extension in such a way that in the final space there will be
no sequence $(g_{n})_{n\in\omega}$ for which the properties from Theorem 6.1
are satisfied. However, we need to consider $2^{\mathfrak{c}}$ sequences of
Radon measures on $[0,1]^{\mathfrak{c}}$, while there are only $\mathfrak{c}$
steps in the construction. In order to handle this we will use $\diamondsuit$
(cf. Lemma 6.4).

######  Proposition 6.2.

Let $K$ be a compact metrizable space and $(\mu_{n})_{n\in\omega}$ be a
bounded sequence of Radon measures on $K$. Assume that $(U_{n})_{n\in\omega}$
is a sequence of pairwise disjoint open sets and $\delta>0$ is such that
$|\mu_{n}|(U_{n})>\delta$ for $n\in\omega$. Then there is an infinite set
$N\subseteq\omega$, continuous pairwise disjoint functions
$f_{n}:K\rightarrow[0,1]$ and $\varepsilon>0$ such that

  1. (1)

$\operatorname{supp}(f_{n})\subseteq U_{n}$ for $n\in N$,

  2. (2)

$|\int f_{n}d\mu_{n}|>\varepsilon$ for $n\in N$,

  3. (3)

$\sum\\{|\int f_{m}d\mu_{n}|:n\neq m,m\in N\\}<\varepsilon/3$ for $n\in N$,

  4. (4)

$K\backslash D((f_{n})_{n\in N})$ is zero-dimensional.

###### Proof.

Since $\mu_{n}$’s are Radon measures there is $\delta^{\prime}>0$ and open
sets $U^{\prime}_{n}\subseteq U_{n}$ such that
$|\mu_{n}(U^{\prime}_{n})|>\delta^{\prime}$ for $n\in\omega$. Without loss of
generality we may assume that $U^{\prime}_{n}=U_{n}$ and
$\delta^{\prime}=\delta$.

Put $\nu_{n}=\mu_{n}|U_{n}$ for $n\in\omega$. Let $N^{\prime}$ be such that
the sequence $(\nu_{n})_{n\in N^{\prime}}$ has the weak* limit $\nu$. Since
$|\int 1d\nu_{n}|>\delta$ for every $n$, we have $|\int 1d\nu|\geq\delta$, so
$\nu$ is a non-zero measure. By Theorem 3.16 there is a compact zero-
dimensional subset $Z\subseteq K$ and $\varepsilon>0$ such that
$|\nu(Z)|>2\varepsilon$. Since $Z$ is a closed subset of a metrizable space
and $\nu$ is a regular measure, there is a decreasing sequence of open sets
$(G_{n})_{n\in N^{\prime}}$ such that $Z=\bigcap G_{n}$ and
$|\nu(G_{n})|>2\varepsilon$ for all $n\in N^{\prime}$.

Note that if $f\in C_{I}(K)$ is such that $\operatorname{supp}(f)\subseteq
G_{n}$ and $|\int fd\nu|>2\varepsilon$, then for big enough $l\in N^{\prime}$
we have $|\int fd\nu_{l}|>2\varepsilon$ and so
$|\nu_{l}|(G_{n})=|\nu_{l}|(G_{n}\cap U_{l})>2\varepsilon$. Hence for each
$l\in N^{\prime}$ we may pick $f_{l}\in C_{I}(K)$ such that
$\operatorname{supp}f_{l}\subseteq G_{n}\cap U_{l}$ and $|\int
f_{l}d\nu_{l}|=|\int f_{l}d\mu_{l}|>\varepsilon$.

For each $n\in N^{\prime}$ let $l_{n}\in N^{\prime}$ be such that
$\operatorname{supp}f_{l_{n}}\subseteq G_{n}\cap U_{l_{n}}$, $|\int
f_{l_{n}}d\mu_{l_{n}}|>\varepsilon$ and $(l_{n})_{n\in N^{\prime}}$ is an
increasing sequence. Let $N^{\prime\prime}=\\{l_{n}:n\in N^{\prime}\\}$. For
every $M\subseteq N^{\prime\prime}$ denote $Z_{M}=K\backslash
D((f_{l_{n}})_{n\in M})$. If $x\in K\backslash Z$, then there is an open
neighbourhood $V\ni x$ such that for big enough $n\in M$ we have $V\cap
G_{n}=\varnothing$ and so $V\cap\operatorname{supp}(f_{l_{n}})=\varnothing$.
Hence $V\subseteq D((f_{l_{n}})_{n\in M})$, which gives $x\notin Z_{M}$. This
implies that $Z_{M}\subseteq Z$, so in particular $Z_{M}$ is zero-dimensional
and condition (4) is satisfied for any choice of $M\subseteq
N^{\prime\prime}$. Now we use Rosenthal’s lemma (see [8, p. 82] or [38]) to
obtain an infinite $N\subseteq N^{\prime\prime}$ such that the 3rd condition
is also satisfied. ∎

We will need the following strengthening of [19, Lemma 6.2].

######  Lemma 6.3.

Let $K$ be a compact, connected metrizable space with a countable dense set
$Q=\\{q_{n}:n\in\omega\\}$. Let $U,V$ be open subsets of $K$ such that
$\overline{U}\cap\overline{V}\neq\varnothing$. Then there is a sequence
$(f_{n})_{n\in\omega}$ of pairwise disjoint functions $f_{n}\in C_{I}(K)$ and
infinite sets $A_{0},A_{1},S_{0},S_{1}\subseteq\omega$ such that:

  1. (1)

the sets $\\{q_{n}:n\in S_{0}\\}\subseteq U,\\{q_{n}:n\in S_{1}\\}\subseteq V$
are relatively discrete,

  2. (2)

$A_{i}\subseteq S_{i}$ and $|S_{i}\backslash A_{i}|=\omega$ for $i=0,1$,

  3. (3)

for every infinite $B\subseteq\omega$ in the extension $K(B)$ of $K$ by
$(f_{n})_{n\in B}$ there are disjoint closed sets $F_{0},F_{1}\subseteq K(B)$
and distinct $x_{0},x_{1}\in K(B)$ such that for $i=0,1$

$x_{i}\in\overline{\pi^{-1}(U)\cap\\{q^{B}_{n}:n\in
A_{i}\\}}\cap\overline{\pi^{-1}(V)\cap\\{q^{B}_{n}:n\in S_{i}\backslash
A_{i}\\}}$

and

$\\{q^{B}_{n}:n\in A_{i}\\}\subseteq F_{i},$

where $q^{B}_{j}=(q_{j},t)$ and $t=\sum_{n\in B}f_{n}(q_{j})$,

  4. (4)

$|K\backslash D(f_{n})_{n\in B}|=1$ (in particular $K\backslash D(f_{n})_{n\in
B}$ is zero-dimensional).

###### Proof.

Fix any compatible metric $d$ on $K$. Pick any
$x\in\overline{U}\cap\overline{V}$. Since $K$ is connected, $x$ is not an
isolated point. For $n\in\omega$ put
$U^{\prime}_{n}=U\cap\mathcal{B}(x,1/n),V^{\prime}_{n}=V\cap\mathcal{B}(x,1/n)$
(where $\mathcal{B}(x,\varepsilon)$ is the open ball with $x$ as the center
and radius $\varepsilon$ with respect to $d$) and let $U_{n}\subseteq
U^{\prime}_{n},V_{n}\subseteq V^{\prime}_{n}$ be non-empty open sets such that
the members of the family $\\{U_{n},V_{n}:n\in\omega\\}$ are pairwise
disjoint. Take continuous functions $f_{n}\in C(K)$ and $k_{n},l_{n}\in\omega$
such that:

  * •

$q_{k_{n}}\in U_{n},q_{l_{n}}\in V_{n}$,

  * •

$f_{n}(q_{k_{2n}})=f_{n}(q_{l_{2n}})=1$,

  * •

$\operatorname{supp}(f_{n})\subseteq U_{2n}\cup V_{2n}$.

Let $B\subseteq\omega$ be infinite. For (2) and (3) it is enough to take
$S_{0}=\\{{k_{2n+1}},{l_{2n+1}}:n\in\omega\\},A_{0}=\\{{k_{2n+1}}:n\in\omega\\},S_{1}=\\{{k_{2n}},{l_{2n}}:n\in\omega\\},A_{1}=\\{{k_{2n}}:n\in\omega\\}$,
$x_{0}=(x,0),x_{1}=(x,1)$ and
$F_{0}=K(B)\cap(K\times[0,1/3]),F_{1}=K(B)\cap(K\times[2/3,1])$. (1) is
satisfied since $U_{n},V_{m}$ are pairwise disjoint for $n,m\in\omega$.

(4) follows from the fact that $x$ is the only point all of whose
neighborhoods intersect all but finitely many $U_{n}$’s and $V_{n}$’s, so we
have $K\backslash D(f_{n})_{n\in B}=\\{x\\}$.

∎

######  Lemma 6.4.

Assume $\diamondsuit$. Then there is a sequence
$(M^{\alpha},\mathcal{U}^{\alpha},L^{\alpha})_{\alpha<\omega_{1}}$ such that:

  * •

$M^{\alpha}=(\mu^{\alpha}_{n})_{n\in\omega}$ is a bounded sequence of Radon
measures on $[0,1]^{\alpha}$,

  * •

$\mathcal{U}^{\alpha}=(U_{n,m}^{\alpha})_{n,m\in\omega}$ is a sequence of
basic open sets in $[0,1]^{\alpha}$,

  * •

$L^{\alpha}=(l_{n}^{\alpha})_{n\in\omega}$ is a sequence of distinct natural
numbers,

and for every:

  * •

bounded sequence $(\mu_{n})_{n\in\omega}$ of Radon measures on
$[0,1]^{\omega_{1}}$,

  * •

sequence $(U_{n,m})_{n,m\in\omega}$ of basic open sets in
$[0,1]^{\omega_{1}}$,

  * •

increasing sequence $l_{n}$ of natural numbers

there is a stationary set $S\subseteq\omega_{1}$ such that for $\beta\in S$ we
have

  * •

$\mu_{n}|C([0,1]^{\beta})=\mu_{n}^{\beta}$,

  * •

$\pi_{\beta}[U_{n,m}]=U_{n,m}^{\beta}$,

  * •

$l_{n}=l_{n}^{\beta}$,

where $\pi_{\beta}$ denotes the natural projection from $[0,1]^{\omega_{1}}$
onto $[0,1]^{\beta}$.

###### Proof.

Firstly we will show that there is a sequence
$(M_{0}^{\alpha})_{\alpha<\omega_{1}}$ such that
$M_{0}^{\alpha}=((\nu^{\alpha}_{n})_{n\in\omega})$ is a bounded sequence of
Radon measures on $[0,1]^{\alpha}$ and for every bounded sequence
$(\nu_{n})_{n\in\omega}$ of Radon measures on $[0,1]^{\omega_{1}}$ there is a
stationary set $S\subseteq\omega_{1}$ such that for $\beta\in S$ we have
$\nu_{n}|C([0,1]^{\beta})=\nu_{n}^{\beta}$.

We will use the identification of Radon measures on $[0,1]^{\omega_{1}}$ with
bounded functionals on $C([0,1]^{\omega_{1}})$ described in Section 2. For a
finite set $F\in[\omega_{1}]^{<\omega}$ denote by $w_{F}$ the product
$\prod_{\alpha\in F}w_{\alpha}$, where $w_{\alpha}\in
C([0,1]^{\omega_{1}}),w_{\alpha}(x)=x(\alpha)$. Observe that finite linear
combinations of $w_{F}$’s form a subalgebra of $C([0,1]^{\omega_{1}})$. If
$x,y\in[0,1]^{\omega_{1}}$ are distinct points with $x(\alpha)\neq y(\alpha)$,
then $w_{\alpha}(x)\neq w_{\alpha}(y)$, so by the Stone-Weierstrass theorem
this subalgebra is dense in $C([0,1]^{\omega_{1}})$. Hence if $\nu$ is a Radon
measure on $[0,1]^{\omega_{1}}$ then it is determined by the values of
$\nu(w_{F})$ for $F\in[\omega_{1}]^{<\omega}$ (note also that in the same way
if $\beta<\omega_{1}$, then $\nu|C([0,1]^{\beta})$ is determined by the values
of $\nu(w_{F})$ for $F\in[\beta]^{<\omega}$). So we can represent each Radon
measure $\nu$ on $[0,1]^{\omega_{1}}$ by the function

$\varphi_{\nu}:[\omega_{1}]^{<\omega}\rightarrow\mathbb{R},\
\varphi(F)=\nu(w_{F})$

(and then $\nu|C([0,1]^{\beta})$ is represented by
$\varphi_{\nu}|[\beta]^{<\omega}$), and each countable sequence
$M=(\nu_{n})_{n\in\omega}$ we can represent by the function

$\varphi_{M}:[\omega_{1}]^{<\omega}\times\omega\rightarrow\mathbb{R},\
\varphi_{M}(F,n)=\nu_{n}(w_{F}).$

Let $\Phi_{1}:\omega_{1}\rightarrow[\omega_{1}]^{<\omega}\times\omega$ be a
bijection such that for each limit ordinal $\gamma\in Lim\cap\omega_{1}$ the
restriction $\Phi_{1}|\gamma$ is bijection onto
$[\gamma]^{<\omega}\times\omega$ (to see that such a bijection exists it is
enough to note that for every $\gamma\in Lim\cap\omega_{1}$ there is a
bijection
$\phi_{\gamma}:[\gamma,\gamma+\omega)\rightarrow([\gamma+\omega]^{<\omega}\times\omega)\backslash([\gamma]^{<\omega}\times\omega)$
and take $\Phi_{1}|[\gamma,\gamma+\omega)=\phi_{\gamma}$). We need to fix one
more bijection $\Phi_{2}:\mathbb{R}\rightarrow\omega_{1}$ ($\diamondsuit$
implies CH, so such a bijection exists). Put

$\psi_{M}=\Phi_{2}\circ\varphi_{M}\circ\Phi_{1},\
\psi_{M}:\omega_{1}\rightarrow\omega_{1}.$

Since $\Phi_{1}|\gamma$ is a bijection onto $[\gamma]^{<\omega}\times\omega$
for all limit $\gamma$ we may treat $\psi_{M}|\gamma$ as a representation of
the sequence of measures $(\nu_{n}|C([0,1]^{\gamma}))_{n\in\omega}$.

We will use the following characterization of $\diamondsuit$ (cf. [6, Theorem
2.7]):
There exists a sequence
$(f_{\alpha})_{\alpha<\omega_{1}},f_{\alpha}:\alpha\rightarrow\alpha$ such
that for for each $f:\omega_{1}\rightarrow\omega_{1}$ the set
$\\{\alpha:f|\alpha=f_{\alpha}\\}$ is stationary.

For $\alpha\in\omega_{1}$ let $M_{0}^{\alpha}$ be a sequence of Radon measures
on $[0,1]^{\alpha}$ represented by $f_{\alpha}$, if $f_{\alpha}$ is a
representation for some such sequence (otherwise we pick $M_{0}^{\alpha}$ in
any way). Let $M$ be a bounded sequence of of measures on $[0,1]^{\omega_{1}}$
and let $S=\\{\alpha:\psi_{M}|\alpha=f_{\alpha}\\}$. Since for limit
$\gamma<\omega_{1}$ the function $\psi_{M}|\gamma$ is a representation of some
sequence of measures we get that for $\alpha\in Lim\cap S$ the function
$\psi_{M}|\alpha$ is the representation of a sequence $M_{0}^{\alpha}$.
Moreover the set $S\cap Lim$ is a stationary subset of $\omega_{1}$, so the
first part of the proof is complete.

To show the existence of sequence
$(M^{\alpha},\mathcal{U}^{\alpha},L^{\alpha})_{\alpha<\omega_{1}}$ required in
the Lemma, we need to observe that each triple $(M,\mathcal{U},L)$ may be
represented as a bounded countable sequence of Radon measures on
$[0,1]^{\omega_{1}}$. Indeed, any basic open set $U\in\mathcal{U}$ may be
treated as a measure $\lambda_{U}$ on $[0,1]^{\omega_{1}}$, given by
$\lambda_{U}(A)=\lambda(A\cap U)$, where $\lambda$ is a product measure of
$\omega_{1}$ Lebesgue measures on $[0,1]$ (note that if $U,V$ are different
basic open sets, then some of their sections differ on a non-trivial interval,
so we have $\lambda_{U}\neq\lambda_{V}$) and $L$ may be represented as
$\delta_{x_{L}}$ where $x_{L}=(y_{l},0,0,\dots)$ and $y_{l}=g(L)$ for some
fixed bijection $g$ between the set of sequences of natural numbers and
$[0,1]$. ∎

######  Proposition 6.5.

Assume $\diamondsuit$. Then for every $k>0,k\in\omega\cup\\{\infty\\}$ there
is a compact Hausdorff space $K$ satisfying the following properties:

  1. (1)

$\dim K=k$,

  2. (2)

$K$ is separable with a countable dense set $Q=\\{q_{n}:n\in\omega\\}$,

  3. (3)

$K$ is connected,

  4. (4)

for every:

     * •

sequence $(U_{n})_{n\in\omega}$ of pairwise disjoint open sets which are
countable unions of basic open sets (basic open set in $K$ is a set of the
form $W\cap K$, where $W$ is a basic open set in $[0,1]^{\omega_{1}}$),

     * •

relatively discrete sequence $(q_{l_{n}}:n\in\omega)\subseteq Q$ with
$q_{l_{n}}\notin U_{m}$ for $n,m\in\omega$,

     * •

bounded sequence $(\mu_{n})_{n\in\omega}$ of Radon measures on $K$ such that
$|\mu_{n}|(U_{n})>\delta$ for some $\delta>0$,

there is $\varepsilon>0$, continuous functions $(f_{n})_{n\in\omega}\subseteq
C_{I}(K)$ and infinite sets $B\subseteq N\subseteq\omega$ such that:

    1. (a)

$(f_{n})$ is a sequence of pairwise disjoint functions with
$\operatorname{supp}(f_{n})\subseteq U_{n}$ for $n\in\omega$,

    2. (b)

$|\int f_{n}d\mu_{n}|>\varepsilon$ for $n\in B$,

    3. (c)

$\sum\\{|\int f_{m}d\mu_{n}|:m\in B\backslash\\{n\\}\\}<\varepsilon/3$ for
$n\in N$,

    4. (d)

$\\{f_{n}:n\in B\\}$ has supremum in the lattice $C(K)$,

    5. (e)

$\overline{\\{q_{l_{n}}:n\in B\\}}\cap\overline{\\{q_{l_{n}}:n\in N\backslash
B\\}}\neq\varnothing,$

  5. (5)

whenever $U,V$ are open subsets of $K$ such that
$\overline{U}\cap\overline{V}\neq\varnothing$, then
$\overline{U}\cap\overline{V}$ contains at least two points.

We will start with the description of the construction. Then we will prove
that the constructed space satisfies the required conditions.

######  Construction 6.6.

Assume $\diamondsuit$. We will construct by induction on $\alpha<\omega_{1}$
an inverse system $(K_{\alpha})_{\alpha<\omega_{1}}$ with the limit $K$, where
$K_{\alpha}\subseteq[0,1]^{\alpha}$ and countable dense sets
$Q_{\alpha}=\\{q_{n}|\alpha:n\in\omega\\}\subseteq K_{\alpha}$.

We start with $K_{k}=[0,1]^{k}$ (or $K_{\omega}=[0,1]^{\omega}$ in the case
$k=\infty$) and we pick $Q_{k}$ to be any countable dense subset of $K_{k}$.
If $\alpha$ is a limit ordinal then we take as $K_{\alpha}$ the inverse limit
of $(K_{\beta})_{\beta<\alpha}$.

Denote by $\operatorname{Even}$ and $\operatorname{Odd}$ the sets consisting
of even and odd (respectively) countable ordinals greater than $k$. Let
$(M^{\alpha},\mathcal{U}^{\alpha},L^{\alpha})_{\alpha<\omega_{1}}$ be as in
Lemma 6.4 and fix an enumeration
$(U_{\alpha},V_{\alpha})_{\alpha\in\operatorname{Odd}}$ of pairs of open
subsets of $[0,1]^{\omega_{1}}$ which are countable unions of basic open sets,
and require that each such a pair occurs in the sequence uncountably many
times (such an enumeration exists since by CH there is
$\omega_{1}^{\omega}=\omega_{1}$ open sets, which are countable unions of
basic open sets in $[0,1]^{\omega_{1}}$).

Firstly we describe the construction of $K_{\alpha+1}$ where $\alpha$ is an
even ordinal. We assume that $K_{\alpha}$ is already constructed and for each
$\beta<\alpha$ the following are satisfied:

  1. (1)

if $\beta\in\operatorname{Even}$ then we have infinite sets
$b_{\beta}^{*}\subseteq a_{\beta}^{*}\subseteq\omega$ such that
$\\{q_{n}|\alpha:n\in a_{\beta}^{*}\\}$ is relatively discrete and

$\overline{\\{q_{n}|\alpha:n\in
b_{\beta}^{*}\\}}\cap\overline{\\{q_{n}|\alpha:n\in a_{\beta}^{*}\backslash
b_{\beta}^{*}\\}}\neq\varnothing.$

  2. (2)

if $\beta\in\operatorname{Odd}$ then we have infinite sets
$b_{\beta}^{i}\subseteq a_{\beta}^{i}\subseteq\omega$ for $i=0,1$ such that
the set $\\{q_{n}|\alpha:n\in a_{\beta}^{i}\\}$ is relatively discrete and

$\overline{\\{q_{n}|\alpha:n\in
b_{\beta}^{i}\\}}\cap\overline{\\{q_{n}|\alpha:n\in a_{\beta}^{i}\backslash
b_{\beta}^{i}\\}}\neq\varnothing$

for $i=0,1$.

Put $U_{n}^{\alpha}=\bigcup_{m\in\omega}U_{n,m}^{\alpha}$. We will say that
even step $\alpha$ is non-trivial if

  * •

there is $\delta>0$ such that $|\mu_{n}^{\alpha}|(U_{n}^{\alpha}\cap
K_{\alpha})>\delta$ for each $n\in\omega$,

  * •

$(U_{n}^{\alpha}\cap K_{\alpha})_{n\in\omega}$ are pairwise disjoint,

  * •

$\\{q_{l_{n}^{\alpha}}:n\in\omega\\}$ is relatively discrete in $K_{\alpha}$,

  * •

$\\{q_{l_{n}^{\alpha}}:n\in\omega\\}\cap U_{m}^{\alpha}=\varnothing$ for
$m\in\omega$.

Otherwise we call this step trivial and we put
$K_{\alpha+1}=K_{\alpha}\times\\{0\\}$ and
$q_{n}|\alpha+1=q_{n}|\alpha^{\frown}0$.

Assume that we are in a non-trivial case. Apply proposition 6.2 for
$U_{n}=U_{n}^{\alpha}\cap K_{\alpha},\mu_{n}=\mu_{n}^{\alpha}$ to obtain
$(f_{n}^{\alpha})_{n\in\omega}\subseteq C_{I}(K_{\alpha})$, infinite
$N\subseteq\omega$ and $\varepsilon>0$ such that

  * •

$\operatorname{supp}(f_{n}^{\alpha})\subseteq U_{n}^{\alpha}\cap K_{\alpha}$
for $n\in N$,

  * •

$|\int f_{n}^{\alpha}d\mu_{n}^{\alpha}|>\varepsilon$ for $n\in N$,

  * •

$\sum\\{|\int f_{m}^{\alpha}d\mu_{n}|:n\neq m,m\in N\\}<\varepsilon/3$ for
$n\in N$,

  * •

$K_{\alpha}\backslash D((f_{n}^{\alpha})_{n\in N})$ is zero-dimensional.

By Lemma 5.5, without loss of generality (by passing to an infinite subset of
$N$) we may assume that for all infinite $B\subseteq N$ the extension
$K_{\alpha}(B)$ of $K_{\alpha}$ by $(f_{n}^{\alpha})_{n\in B}$ is strong and
for each $\beta<\alpha$ and $i\in\\{*,0,1\\}$ we have

$\overline{\\{q_{n}^{B}|\alpha+1:n\in
b_{\beta}^{i}\\}}\cap\overline{\\{q_{n}^{B}|\alpha+1:n\in
a_{\beta}^{i}\backslash b_{\beta}^{i}\\}}\neq\varnothing,$

where

$q_{l}^{B}|\alpha+1=q_{l}|\alpha^{\frown}t,t=\sum_{n\in
B}f_{n}^{\alpha}(q_{l}|\alpha),$

and the closures are taken in $K_{\alpha}(B)$.

Let $a_{\alpha}^{*}=\\{l_{n}^{\alpha}:n\in N\\}$. Then

($*$) $\displaystyle N=\\{n\in\omega:l_{n}^{\alpha}\in a_{\alpha}^{*}\\}.$

We will show that there is infinite $b_{\alpha}^{*}\subseteq a_{\alpha}^{*}$
such that

$\overline{\\{q_{n}|\alpha:n\in
b_{\alpha}^{*}\\}}\cap\overline{\\{q_{n}|\alpha:n\in a_{\alpha}^{*}\backslash
b_{\alpha}^{*}\\}}\neq\varnothing.$

Suppose otherwise. Then since $K_{\alpha}$ is a compact metrizable space, for
each $X\subseteq a_{\alpha}^{*}$ there are disjoint open sets $U_{X},V_{X}$
such that

$\overline{\\{q_{n}|\alpha:n\in X\\}}\subseteq
U_{X},\overline{\\{q_{n}|\alpha:n\in a_{\alpha}^{*}\backslash X\\}}\subseteq
V_{X},$

and $U_{X},V_{X}$ are finite unions of members of some fixed countable base in
$K_{\alpha}$. There are uncountably many choices of $X$ and only countably
many pairs of such open sets in $K_{\alpha}$, so for some $X\neq Y$ we have
$\\{U_{X},V_{X}\\}=\\{U_{Y},V_{Y}\\}$ which is a contradiction.

Let $b_{\alpha}^{*}$ be such that

$\overline{\\{q_{n}|\alpha:n\in
b_{\alpha}^{*}\\}}\cap\overline{\\{q_{n}|\alpha:n\in a_{\alpha}^{*}\backslash
b_{\alpha}^{*}\\}}\neq\varnothing$

and define

($**$) $\displaystyle B=\\{n\in N:l_{n}^{\alpha}\in b_{\alpha}^{*}\\}.$

To finish the construction at this step we put
$K_{\alpha+1}=K_{\alpha}(B),q_{n}|\alpha+1=q_{n}^{B}|\alpha+1$ and observe
that (1) is satisfied for $a_{\alpha}^{*},b_{\alpha}^{*}$, because if

$x\in\overline{\\{q_{n}|\alpha:n\in
b_{\alpha}^{*}\\}}\cap\overline{\\{q_{n}|\alpha:n\in a_{\alpha}^{*}\backslash
b_{\alpha}^{*}\\}},$

then

$(x,0)\in\overline{\\{q_{n}|\alpha+1:n\in
b_{\alpha}^{*}\\}}\cap\overline{\\{q_{n}|\alpha+1:n\in
a_{\alpha}^{*}\backslash b_{\alpha}^{*}\\}},$

since $f_{n}^{\alpha}(q_{k}|\alpha)=0$ for all $n\in B$ and $k\in a_{\alpha}$.

At step $\alpha\in\operatorname{Odd}$ we assume that we are given
$a_{\beta}^{i},b_{\beta}^{i}$ satisfying (1) and (2) from the even step for
all $\beta<\alpha$ (where $i=*$ if $\beta\in\operatorname{Odd}$ and
$i\in\\{0,1\\}$ if $\beta\in\operatorname{Even}$). We call this step non-
trivial, if the closures of $\pi_{\alpha}[U_{\alpha}]$ and
$\pi_{\alpha}[V_{\alpha}]$ have non-empty intersection. If the case is non-
trivial we use Lemma 6.3 (note that Lemma 5.3 implies that $K_{\alpha}$ is
connected) to find appropriate $(f_{n})_{n\in\omega}\subseteq
C_{I}(K_{\alpha}),A_{i}$ and $S_{i}$ for $i=0,1$. In the same way as in the
even step we find $B\subseteq\omega$ such that $K_{\alpha}(B)$ is a strong
extension of $K_{\alpha}$ and the conditions (1) and (2) are preserved in
$K_{\alpha}(B)$ for $\beta<\alpha$. To finish this step we define
$K_{\alpha+1}=K_{\alpha}(B),a_{\alpha}^{i}=S_{i},b_{\alpha}^{i}=A_{i}$ and
$q_{n}|\alpha+1=q_{n}^{B}|\alpha+1$. Lemma 6.3 guarantees that the condition
(2) holds at the step $\alpha+1$.

In both cases the density of $Q_{\alpha+1}=\\{q_{n}|\alpha+1:n\in\omega\\}$ in
$K_{\alpha+1}$ follows from Lemma 5.4.

###### Proof of Proposition 6.5.

We will show that the space constructed above satisfies the required
conditions. (1) follows from Corollary 5.8 and the fact that $[0,1]^{k}$ is a
$k$-dimensional space. $Q$ is a countable dense set in $K$, since each
$Q_{\alpha}$ is dense in $K_{\alpha}$ for $\alpha<\omega_{1}$. Connectedness
follows from inductive argument using Lemma 5.3.

Let $U_{n},l_{n},\mu_{n}$ be as in (4). Let
$U_{n}=\bigcup_{m\in\omega}U_{n,m}\cap K$ where $U_{n,m}$ are basic open sets
in $[0,1]^{\omega_{1}}$. Every $U_{n,m}$ is determined by finitely many
coordinates, so there is $\gamma<\omega_{1}$ such that
$\pi_{\gamma}^{-1}(\pi_{\gamma}[U_{n,m}])=U_{n,m}$ for $n\in\omega$, where
$\pi_{\gamma}$ is the natural projection from $[0,1]^{\omega_{1}}$ onto
$[0,1]^{\gamma}$ (so $U_{n,m}$ are determined by first $\gamma$ coordinates).
By Lemma 6.4 there is $\alpha>\gamma,\alpha\in\operatorname{Even}$ such that
for $n\in\omega$

  * •

$\mu_{n}|C(K_{\alpha})=\mu_{n}^{\alpha}$,

  * •

$\pi_{\alpha}[U_{n,m}]=U_{n,m}^{\alpha}$,

  * •

$l_{n}=l_{n}^{\alpha}$.

Let $(f_{n}^{\alpha})_{n\in B}$ be such that in the $\alpha$-th step of
construction. Since $(f_{n}^{\alpha})_{n\in B}$ satisfy conditions of
Proposition 6.2, functions $f_{n}=f_{n}^{\alpha}\circ\pi_{\alpha}$ satisfy
conditions (a-c). (d) follows from [19, Lemma 4.6] and the fact that
$K_{\alpha+1}$ is a strong extension of $K_{\alpha}$ by $(f_{n})_{n\in B}$. By
construction we have

$\overline{\\{q_{n}:n\in b_{\alpha}^{*}\\}}\cap\overline{\\{q_{n}:n\in
a_{\alpha}^{*}\backslash b_{\alpha}^{*}\\}}\neq\varnothing$

and by $(*)$ and $(**)$

$\\{q_{n}:n\in b_{\beta}^{*}\\}=\\{q_{l_{n}}:n\in B\\},$ $\\{q_{n}:n\in
a_{\alpha}^{*}\backslash b_{\alpha}^{*}\\}=\\{q_{l_{n}}:n\in N\backslash
B\\},$

which gives (e).

Now we will prove (5). Fix open sets $U,V\subseteq K$ such that
$\overline{U}\cap\overline{V}\neq\varnothing$. As $K$ is separable it is
c.c.c. so there are open $U^{\prime}\subseteq U,V^{\prime}\subseteq V$ which
are countable unions of basic open sets such that
$\overline{U^{\prime}}=\overline{U}$ and $\overline{V^{\prime}}=\overline{V}$
(namely it is enough to take as $U^{\prime}$ the union of a maximal antichain
of open subsets in $U$, and similarly for $V^{\prime}$). Without loss of
generality we may assume that $U^{\prime}=U$ and $V^{\prime}=V$. Since $U,V$
are countable unions of basic open sets, there is $\gamma<\omega_{1}$ such
that $U,V$ are determined by coordinates less than $\gamma$. Let
$\alpha>\gamma,\alpha\in\operatorname{Odd}$ be such that $U=U_{\alpha}\cap
K,V=V_{\alpha}\cap K$. Then
$\overline{\pi_{\alpha}[U]}\cap\overline{\pi_{\alpha}[V]}$ is nonempty so
$\alpha$-th step in construction is nontrivial. By construction we have for
$i=0,1$

$\overline{\\{q_{n}|\beta:n\in
b_{\alpha}^{i}\\}}\cap\overline{\\{q_{n}|\beta:n\in a_{\alpha}^{i}\backslash
b_{\alpha}^{i}\\}}\neq\varnothing$

for all $\beta>\alpha$, so there are $x_{i}\in\overline{U}\cap\overline{V}$
such that

$x_{i}\in\overline{\\{q_{n}:n\in b_{\alpha}^{i}\\}}\cap\overline{\\{q_{n}:n\in
a_{\alpha}^{i}\backslash b_{\alpha}^{i}\\}}.$

To finish the proof we need only to notice that $x_{0}\neq x_{1}$, but this
follows form the fact that $a_{\alpha}^{i},b_{\alpha}^{i}$ were chosen to
satisfy Lemma 6.3(3). ∎

######  Lemma 6.7.

Suppose that $(U_{n})_{n\in\omega}$ is a sequence of pairwise disjoint open
subsets of a compact Hausdorff space $K$. Let $M,N\subset\omega$ be infinite
sets such that $M\cap N$ is finite. Assume that $(f_{m})_{m\in
M},(g_{n})_{n\in N}\subseteq C_{I}(K)$ are such that
$\operatorname{supp}(f_{m})\subseteq U_{m},\operatorname{supp}(g_{n})\subseteq
U_{n}$ for $m\in M,n\in N$ and the suprema $f_{\sup}=\sup\\{f_{m}:m\in
M\\},g_{\sup}=\sup\\{g_{n}:n\in N\\}$ exist in $C_{I}(K)$. Denote

$f=f_{\sup}-\sum_{m\in M}f_{m},\ g=g_{\sup}-\sum_{n\in N}g_{n}.$

Then $f,g$ are Borel functions with disjoint supports.

###### Proof.

$f$ and $g$ are Borel functions since they are pointwise sums of countably
many continuous functions. Put $D=M\cap N$ and note that since $D$ is finite
the function $\sum_{m\in D}f_{m}$ is continuous. We will show that

($+$) $\displaystyle\sup\\{f_{m}:m\in M\backslash D\\}=\sup\\{f_{m}:m\in
M\\}-\sum_{m\in D}f_{m}.$

Let $x\in K$. If $x\in\operatorname{supp}(f_{n})$ for some $n\in M\backslash
D$, then $\sum_{m\in D}f_{m}(x)=0$, so

$(\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m})(x)=\sup\\{f_{m}:m\in M\\}(x)\geq
f_{n}(x)$

for every $n\in M\backslash D$. If $x\notin\operatorname{supp}(f_{n})$ for
every $n\in M\backslash D$, then since $f_{n}$’s have disjoint supports we get
that

$(\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m})(x)\geq 0=f_{n}(x)$

for $n\in M\backslash D$. Hence

$\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m}\geq f_{n}$

for $n\in M\backslash D$ in the lattice $C(K)$. Let $h\in C(K)$ be such that

$\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m}\geq h\geq f_{n}$

for $n\in M\backslash D$. Since $f_{n}$’s have disjoint supports we have

$\sup\\{f_{m}:m\in M\\}\geq h+\sum_{m\in D}f_{m}\geq\sum_{m\in M}f_{m}.$

But

$\sup\\{f_{m}:m\in M\\}(x)=\sum_{m\in M}f_{m}(x)$

for $x\in D((f_{n})_{n\in M})$, so

$\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m}=h,$

because $\sup\\{f_{m}:m\in M\\}-\sum_{m\in D}f_{m}$ and $h$ are continuous
functions equal on the set $D((f_{n})_{n\in M})$, which is dense in $K$ (cf.
Lemma 5.2). This completes the proof of the equality $(+)$.

From $(+)$ we get that

$\sup\\{f_{m}:m\in M\backslash D\\}-\sum_{m\in M\backslash
D}f_{m}=\sup\\{f_{m}:m\in M\\}-\sum_{m\in M}f_{m}=f.$

In particular in the definition of $f$ we may replace $M$ with $M\backslash D$
and assume that $M\cap N=\varnothing$.

We will show that in this case we have
$\operatorname{supp}(f_{\sup})\cap\operatorname{supp}(g_{\sup})=\varnothing$,
which will finish the proof since
$\operatorname{supp}(f)\subseteq\operatorname{supp}(f_{\sup})$ and
$\operatorname{supp}(g)\subseteq\operatorname{supp}(g_{\sup})$ (the inclusions
hold because $f\leq f_{\sup},g\leq g_{\sup}$ and $f,g$ are non-negative).
Firstly we observe that for each $n\in N$ we have
$\operatorname{supp}(f_{\sup})\cap\operatorname{supp}(g_{n})=\varnothing$.
Indeed, if it is not the case, then there is $x\in U_{n}$ such that
$f_{\sup}(x)>0$. Then by the Tietze extension theorem we may find $h\in
C_{I}(K)$ such that $h(x)=0$ and $h|K\backslash U_{n}=f_{\sup}|K\backslash
U_{n}$. But then $f_{\sup}>h\geq f_{m}$ for every $m\in M$, which is a
contradiction with the fact that $f_{\sup}$ is the supremum of $f_{m}$’s. Now,
in the same way we show that if
$\operatorname{supp}(f_{\sup})\cap\operatorname{supp}(g_{\sup})\neq\varnothing$,
then there is $h^{\prime}$ such that $g_{\sup}>h^{\prime}>g_{n}$ for $n\in N$.
∎

######  Theorem 6.8.

Assume $\diamondsuit$. For each $k>0$ there is a compact Hausdorff, separable,
connected space $K$ such that $C(K)$ has few operators and $\dim K=k$.

###### Proof.

We will show that if $K$ is the space with properties from Proposition 6.5,
then $C(K)$ has few operators. $K$ satisfies Proposition 6.5(5), so by [19,
Theorem 2.7, Lemma 2.8] it is enough to show that all operators on $C(K)$ are
weak multipliers.

Assume that there is a bounded linear operator $T:C(K)\rightarrow C(K)$, which
is not a weak multiplier. By Theorem 6.1 there is a pairwise disjoint sequence
$(g_{n})_{n\in\omega}\subseteq C_{I}(K)$ and pairwise disjoint open sets
$(V_{n})_{n\in\omega}$ such that $g_{n}|V_{m}=0$ for $n,m\in\omega$ and
$|T(g_{n})|V_{n}|>\delta$ for some $\delta>0$. For $n\in\omega$ let
$U_{n}=\operatorname{supp}(g_{n})$. Let $g^{\prime}_{n}\in
C([0,1]^{\omega_{1}})$ be an extension of $g_{n}$ and
$U_{n}^{\prime}=\operatorname{supp}(g_{n}^{\prime})$. By Mibu’s theorem (see
[25]) for every $n\in\omega$ there is $\alpha_{n}<\omega_{1}$ such that
whenever $x,y\in[0,1]^{\omega_{1}},x|\alpha_{n}=y|\alpha_{n}$, we have
$g^{\prime}_{n}(x)=g^{\prime}_{n}(y)$. Hence $U^{\prime}_{n}$ is an open set
of the form $W_{n}\times[0,1]^{\omega_{1}\backslash\alpha_{n}}$, where $W_{n}$
is an open set in $[0,1]^{\alpha_{n}}$. Since $\alpha_{n}$ is countable,
$W_{n}$ is a union of countably many basic open set in $[0,1]^{\alpha_{n}}$.
Thus for every $n\in\omega$ the set $U^{\prime}_{n}$ is a union of countably
many basic open sets in $[0,1]^{\omega_{1}}$ and $U_{n}=U^{\prime}_{n}\cap K$
is a union of countably many basic open sets in $K$.

Let $(l_{n})_{n\in\omega}$ for $n\in\omega$ be such that $q_{l_{n}}\in V_{n}$
(so in particular $\\{q_{l_{n}}:n\in\omega\\}$ is relatively discrete in $K$)
and define $\mu_{n}=T^{*}(\delta_{q_{l_{n}}})$. Then $|\int
g_{n}d\mu_{n}|=|T(g_{n})(q_{l_{n}})|>\delta$. Since
$\operatorname{supp}(g_{n})\subseteq U_{n}$ and $\|g_{n}\|\leq 1$ we get that
$|\mu_{n}|(U_{n})\geq|\int g_{n}d\mu_{n}|>\delta$.

By Proposition 6.5 for every infinite subset $A\subseteq\omega$ there are
infinite sets $B_{A}\subseteq N_{A}\subseteq A$, continuous functions
$(f_{n,A})_{n\in A}\subseteq C_{I}(K)$ and $\varepsilon_{A}$ such that

  1. (a)

$(f_{n,A})_{n\in A}$ is a sequence of pairwise disjoint functions with
$\operatorname{supp}(f_{n,A})\subseteq U_{n}$ for $n\in A$,

  2. (b)

$|\int f_{n,A}d\mu_{n}|>\varepsilon_{A}$ for $n\in B_{A}$,

  3. (c)

$\sum\\{|\int f_{m,A}d\mu_{n}|:n\neq m,m\in B_{A}\\}<\varepsilon_{A}/3$ for
$n\in N_{A}$,

  4. (d)

$\\{f_{n,A}:n\in B_{A}\\}$ has its supremum in the lattice $C(K)$,

  5. (e)

$\overline{\\{q_{l_{n}}:n\in B_{A}\\}}\cap\overline{\\{q_{l_{n}}:n\in
N_{A}\backslash B_{A}\\}}\neq\varnothing.$

Put $f_{A}=\sup\\{f_{n,A}:n\in B_{A}\\}-\sum_{m\in B_{A}}f_{m,A}$. We will
show that there is an infinite set $M\subseteq\omega$ such that

($++$) $\displaystyle\int f_{M}d\mu_{n}=0.$

Suppose this is not the case. Let $\\{M_{\xi}:\xi<\omega_{1}\\}$ be a family
of infinite subsets of $\omega$ such that for $\xi\neq\xi^{\prime}$ the set
$M_{\xi}\cap M_{\xi^{\prime}}$ is finite. Assume ($++$) does not hold for
every $M_{\xi}$. Then there is $n\in\omega$ such that

$\int f_{M_{\xi}}d\mu_{n}\neq 0$

for uncountably many $\xi$’s. By Lemma 6.7 $f_{M_{\xi}},f_{M_{\xi^{\prime}}}$
have disjoint supports for $\xi\neq\xi^{\prime}$, so in particular there is an
uncountable family of non-null (with respect to $\mu_{n}$) Borel sets in $K$,
which is a contradiction.

Put $f_{n}=f_{n,M},\varepsilon=\varepsilon_{M}$, $B=B_{M}$ and $N=N_{M}$. Let
$f=\sup\\{f_{n}:n\in B\\}$. By (b), (c), ($++$) and the definition of
$\mu_{n}$ we get that for $n\in B$

$\displaystyle|T(f)(q_{l_{n}})|=|\int fd\mu_{n}|=|\int
f_{n}d\mu_{n}+\int\sum_{m\in B\backslash\\{n\\}}f_{m}|\geq$
$\displaystyle|\int f_{n}d\mu_{n}|-|\int\sum_{m\in
B\backslash\\{n\\}}f_{m}|\geq\varepsilon-\varepsilon/3=2\varepsilon/3.$

For $n\in N\backslash B$ (c) gives

$|T(f)(q_{l_{n}})|=|\int\sum_{m\in B}f_{m}d\mu_{n}|<\varepsilon/3.$

As $T(f)$ is a continuous function on $K$ we obtain that

$\overline{\\{q_{l_{n}}:n\in B\\}}\cap\overline{\\{q_{l_{n}}:n\in N\backslash
B\\}}=\varnothing,$

which contradicts (e). ∎

######  Theorem 6.9.

Assume $\diamondsuit$. Then for every $k\in\omega\cup\\{\infty\\}$ there is a
compact Hausdorff space $K$ such that $\dim(K)=k$ and whenever $C(K)\sim
C(L)$, $\dim(L)=k$.

###### Proof.

For $k=0$ every finite space $K$ works. If $k>0$, then the space from Theorem
6.8 satisfies the required property by Corollary 4.20. ∎

##  7\. Remarks and questions

The first natural question concerning our results is whether Theorem 6.9 is
true without any additional assumption.

######  Question 7.1.

Let $k\in\omega\backslash\\{0\\}$. Is there (in ZFC) a compact Hausdorff space
$K$ such that $\dim(K)=k$ and whenever $C(K)\sim C(L)$, $\dim(L)=k$?

In the light of Theorem 4.19 to show that the Question 7.1 has positive answer
it would be enough to prove that the following question has positive answer.

######  Question 7.2.

Let $k\in\omega\backslash\\{0\\}$. Is there (in ZFC) a compact, separable,
connected Hausdorff space $K$ such that $\dim K=k$ and $C(K)$ has few
operators?

The original construction of a Banach space $C(K)$ where all the operators are
weak multipliers was carried out in ZFC ([19]). In this construction we set
all sequences of pairwise disjoint continuous functions on
$[0,1]^{\mathfrak{c}}$ into a sequence of length $\mathfrak{c}$, and the
choice of the strong extension at $\alpha$-th step depends on the $\alpha$-th
sequence of functions. Later, in order to prove that $K$ satisfies the
required conditions, we look at any sequence $(\mu_{n})_{n\in\omega}$ of Radon
measures on $K$ and show that we can find sequences of continuous functions
satisfying properties (a-e) from Proposition 6.5. However, in this approach we
may obtain an infinite-dimensional space, since used strong extensions may
increase the dimension. One can try to proceed in a similar way, by applying
only those extensions that preserve the dimension. The problem is that we do
not know, whether the extension by the sequence of functions given at some
step changes the dimension, since it depends on the earlier steps (i.e. it
depends on the bookkeeping of sequences of continuous functions on
$[0,1]^{\mathfrak{c}}$). Consequently, there may be a sequence of measures on
the final space, for which every suitable sequence of functions appears at a
step, in which using the extension would increase the dimension.

Although the main reason to use the diamond principle is the guessing of
measures in Lemma 6.4, we also needed the continuum hypothesis to ensure that
all intermediate spaces from our construction are metrizable. At that point we
used the fact that for every non-zero Radon measure on metrizable compact
space there is a zero-dimensional $G_{\delta}$ compact subset of non-zero
measure (Theorem 3.16). In the light of this theorem the following problem
seems to be interesting.

######  Problem 7.3.

Describe the class of compact Hausdorff spaces $K$ such that for every non-
zero Radon measure $\mu$ on $K$ there is a zero-dimensional compact subset
$L\subseteq K$ such that $\mu(L)\neq 0$.

Assume that $K$ is such that $C(K)$ has few operators. Then by [36,
Proposition 4.8] there is a space $L$ such that $C(K)\sim C(L)$, but $C(L)$
does not have few operators. However, by Theorem 4.19 the topology of $L$ is
very close to $K$, at least if we assume that $K$ is separable and connected.

######  Question 7.4.

Suppose that $K$ is a compact Hausdorff space such that every operator
$T:C(K)\rightarrow C(K)$ is a weak multiplier and $C(L)\sim C(K)$ for some
compact Hausdorff space. Is it true that $K$ and $L$ are homeomorphic modulo
finitely many points in the sense of Theorem 4.19?

One may also ask, what properties $K$ should have to satisfy Theorem 6.9.
There are known examples of “nice” spaces $K$ such that if $C(K)\sim C(L)$,
then $L$ is not zero-dimensional. For instance Avilés and Koszmider showed
that there is such a space which is quasi Radon-Nikodým ([1]) and Plebanek
gave a consistent example of such a space which is a Corson copmact ([34]).

## Acknowledgements

The author would like to thank his PhD supervisor Professor Piotr Koszmider
for introducing to the topic, constant help and many valuable suggestions.

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