# On sequences of homomorphisms into measure algebras and the Efimov problem

Piotr Borodulin–Nadzieja  Instytut Matematyczny, Uniwersytet Wrocławski, pl.
Grunwaldzki 2/4, 50-384 Wrocław, Poland<EMAIL_ADDRESS>and  Damian
Sobota  Kurt Gödel Research Center for Mathematical Logic, Departament of
Mathematics, University of Vienna, Kolingasse 14-16, 1090 Vienna, Austria
<EMAIL_ADDRESS>
###### Abstract.

For given Boolean algebras $\mathbb{A}$ and $\mathbb{B}$ we endow the space
$\mathcal{H}(\mathbb{A},\mathbb{B})$ of all Boolean homomorphisms from
$\mathbb{A}$ to $\mathbb{B}$ with various topologies and study convergence
properties of sequences in $\mathcal{H}(\mathbb{A},\mathbb{B})$. We are in
particular interested in the situation when $\mathbb{B}$ is a measure algebra
as in this case we obtain a natural tool for studying topological convergence
properties of sequences of ultrafilters on $\mathbb{A}$ in random extensions
of the set-theoretical universe. This appears to have strong connections with
Dow and Fremlin’s result stating that there are Efimov spaces in the random
model. We also investigate relations between topologies on
$\mathcal{H}(\mathbb{A},\mathbb{B})$ for a Boolean algebra $\mathbb{B}$
carrying a strictly positive measure and convergence properties of sequences
of measures on $\mathbb{A}$.

###### Key words and phrases:

Efimov problem, measure algebras, weak convergence of measures, algebraic
convergence, random forcing, perfect Hamming codes

###### 2020 Mathematics Subject Classification:

Primary: 28A20, 28A60, 03E40. Secondary: 28E15, 03E75, 28A33

The first author was supported by National Science Center project no.
2018/29/B/ST1/00223. The second author was supported by the Austrian Science
Fund, grant M2500-N35.

##  1\. Introduction

It is a common issue in analysis and topology to compare convergence
properties of sequences in topological spaces, in particular, to study when a
sequence convergent with respect to one topology converges also with respect
to another one. For instance, one can ask what properties a Banach space must
have so that its every weakly convergent sequence is also norm convergent
(studying so called Schur property), or investigate when a given topological
space does not contain non-trivial convergent sequences, i.e. when every
convergent sequence is eventually constant. In this paper we also deal with
this kind of problems—our setting are spaces of homomorphisms between Boolean
algebras, especially measure algebras, endowed with several miscellaneous
topologies.

There are several motivations standing behind our research, but before we
present them, together with our main results, we introduce some basic
notations. For a cardinal $\kappa$ let $\mathbb{M}_{\kappa}$ denote the
measure algebra of the standard product measure $\lambda_{\kappa}$ on the
product space $2^{\kappa}$. If $\kappa=1$, then put
$\mathbbm{2}=\mathbb{M}_{1}$. If $\mathbb{A}$ is a Boolean algebra, then by
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ we denote the family of all
Boolean homomorphisms from $\mathbb{A}$ into $\mathbb{M}_{\kappa}$. The
measure $\lambda_{\kappa}$ induces the so-called Fréchet–Nikodym metric
$d_{\lambda_{\kappa}}$ on $\mathbb{M}_{\kappa}$, and thus
$\mathbb{M}_{\kappa}$ may be treated as a metric topological space. Since
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is simply a subset of the family
of all functions from $\mathbb{A}$ into $\mathbb{M}_{\kappa}$, this further
leads to endowing $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ with two
natural topologies—the pointwise topology and the uniform topology. In this
article we will investigate these structures and their variations as well as
the relations between them.

Our first motivation for this research is the observation that for $\kappa>1$
the space $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ with the pointwise
topology may be treated as a natural generalization of the Stone space of the
Boolean algebra $\mathbb{A}$, or a Stone space of $\mathbb{A}$ of higher
order. Indeed, recall that every ultrafilter on the Boolean algebra
$\mathbb{A}$ induces a homomorphism from $\mathbb{A}$ into the two-element
algebra $\mathbbm{2}$, and vice versa, and thus the Stone space
$St(\mathbb{A})$ of $\mathbb{A}$ may be seen as the space
$\mathcal{H}(\mathbb{A},\mathbbm{2})$, endowed with the pointwise topology
originating from the Fréchet–Nikodym metric $d_{\lambda_{1}}$, induced by the
trivial measure $\lambda_{1}$ on $\mathbbm{2}$. As the literature concerning
convergence properties of sequences in Stone spaces of Boolean algebras, or
more generally in totally disconnected topological spaces, is quite vast, cf.
e.g. [BD19], [Har07], [HS20], we were motivated to conduct similar research in
the setting of the spaces $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ for
general $\kappa$’s and various topologies. Note that similar generalizations
of topological spaces are quite common in analysis, e.g. one can think of
Radon measures of norm $1$ on compact spaces as generalizations of points in
those spaces, or, as it is done in non-commutative topology, of projections in
general C*-algebras as being analogous to characteristic functions of clopen
subsets of locally compact spaces.

For every $\kappa$ the uniform topology on the space
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is of course finer than the
pointwise topology, hence every uniformly convergent sequence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is pointwise metric convergent.
As one can suspect, the converse in general does not hold—cf. Section 6 for
relevant counterexamples (especially interesting is the example described in
Section 6.4, to obtain which we used tools from information theory such as
perfect Hamming codes). In Section 4.3 we introduce and study a strengthening
of pointwise metric convergence—called by us the pointwise algebraic
convergence—that implies under certain conditions put on the Boolean algebra
$\mathbb{A}$ (such as $\sigma$-completeness) also the uniform convergence
(Corollary 8.4). This mode of convergence is related to the well-known notion
of the algebraic convergence in $\sigma$-complete Boolean algebras, studied
e.g. in [BGJ98], [BFH99], [Jec18].

In Section 4.4, using the dual functions to homomorphisms from $\mathbb{A}$
into $\mathbb{M}_{\kappa}$ and Borel subsets of the Stone space $\mathbb{A}$,
we endow the space $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ with yet
another topology, which we call the pointwise Borel metric topology. This
topology is finer than the topology of pointwise convergence and has turned
out to be useful in investigations related to our second motivation which is
the fact that each homomorphism
$\varphi\in\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ naturally induces a
measure on $\mathbb{A}$ of the form $\lambda_{\kappa}\circ\varphi$, as well
as, by the virtue of the celebrated Maharam theorem, for each strictly
positive measure $\mu$ on $\mathbb{A}$ there are a cardinal $\gamma$ and a
(usually not unique) homomorphism
$\psi\in\mathcal{H}(\mathbb{A},\mathbb{M}_{\gamma})$ such that
$\mu=\psi\circ\lambda_{\gamma}$. This correspondence has inspired us to study
relations between sequences $(\varphi_{n})$ of homomorphisms in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ and sequences of measures of the
form $\lambda_{\kappa}\circ\varphi_{n}$. In particular, we show in Corollary
5.2 connections between the uniform convergence, pointwise Borel metric
convergence, and pointwise metric convergence of a sequence $(\varphi_{n})$ in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ and the norm convergence, weak
convergence, and weak* convergence of the sequence
$(\lambda_{\kappa}\circ\varphi_{n})$ of measures on $\mathbb{A}$,
respectively. In Section 5 we prove also that whenever $\mathbb{A}$ has the
Grothendieck property (see the end of Section 3.1 for the definition), then
every pointwise metric convergent sequence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is pointwise Borel metric
convergent—a fact that does not hold for general Boolean algebras $\mathbb{A}$
(cf. the example in Section 6.1).

Our last main motivation lies in the theory of forcing, where actually our
interest in studying the subject of this article started. Imagine that we
force with a Boolean algebra $\mathbb{B}$ over the ground model $V$ and that
we are interested in the behaviour of sequences of ultrafilters in the forcing
extension $V^{\mathbb{B}}$ which are defined on a fixed ground model Boolean
algebra $\mathbb{A}$. This is actually a quite common situation—e.g. reals in
the extension can be seen as ultrafilters on the Cantor algebra $\mathbb{C}$,
i.e. the free countable Boolean algebra. The usual problem is that the access
to objects in forcing extensions is quite _remote_ , through
$\mathbb{B}$-names only. However, the $\mathbb{B}$-names for ultrafilters on
$\mathbb{A}$ naturally correspond to homomorphisms
$\varphi\colon\mathbb{A}\to\mathbb{B}$ (see Section 7). Then, the convergence
of ultrafilters (treated as elements of the Stone space of $\mathbb{A}$ in
$V^{\mathbb{B}}$) appears to be strongly connected to certain convergence
properties of sequences of homomorphisms in the ground model. Namely, we prove
in Proposition 7.3 that for a given sequence $(\dot{\mathcal{U}}_{n})$ of
$\mathbb{M}_{\kappa}$-names for ultrafilters on $\mathbb{A}$ the sequence of
the corresponding homomorphisms $(\varphi_{\dot{\mathcal{U}}_{n}})$ in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is algebraically convergent if
and only if the sequence $((\dot{\mathcal{U}}_{n})_{G})$ of the
interpretations converges in $V[G]$ for every $\mathbb{M}_{\kappa}$-generic
filter $G$. It turns out that this proposition has also variants regarding
uniform convergence of sequences of homomorphisms—Proposition 7.4 asserts that
if $((\dot{\mathcal{U}}_{n})_{G})$ is eventually constant in every generic
extension $V[G]$, then the sequence $(\varphi_{\dot{\mathcal{U}}_{n}})$ is
uniformly convergent. The converse holds partially—in Theorem 7.5 we show that
if a sequence of homomorphisms $(\varphi_{n})$ in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is uniformly convergent to some
$\varphi\in\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$, then for almost all
$n\in\omega$ and the corresponding names $\dot{\mathcal{U}}_{\varphi_{n}}$ and
$\dot{\mathcal{U}}_{\varphi}$ for ultrafilters on $\mathbb{A}$ there is a
condition $p_{n}\in\mathbb{M}_{\kappa}$ forcing that
$\dot{\mathcal{U}}_{\varphi_{n}}=\dot{\mathcal{U}}_{\varphi}$. This latter
theorem is proved with the aid of Theorem 7.6 asserting that if for two
$\mathbb{M}_{\kappa}$-names $\dot{\mathcal{U}}$ and $\dot{\mathcal{V}}$ for
ultrafilters on $\mathbb{A}$ it holds
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}\neq\dot{\mathcal{V}}$, then
there exists a large condition $p\in\mathbb{M}_{\kappa}$, where large means of
measure as close to $1/4$ as one wishes, and an element $A\in\mathbb{A}$ such
that $p\Vdash A\in\dot{\mathcal{U}}\triangle\dot{\mathcal{V}}$—a result
interesting on its own, since a typical argument based on the countable chain
condition of $\mathbb{M}_{\kappa}$ provides us only with a countable antichain
$\\{p_{n}\colon\ n\in\omega\\}$ of conditions witnessing that the two
ultrafilters are different, but with no control over the value of
$\lambda_{\kappa}(p_{n})$ for any $n\in\omega$. The proof of Theorem 7.6 is
also interesting, because it boils down to purely combinatorial Proposition
7.7 having a nice real-life interpretation, see Remark 7.12.

The aforementioned results have connection with the famous long-standing open
question, called the Efimov problem, asking whether there exists an Efimov
space, that is, an infinite compact Hausdorff space which does not contain
copies of $\beta\omega$, the Čech–Stone compactification of $\omega$, nor non-
trivial convergent sequences. No ZFC example of an Efimov space is known,
however several consistent examples have been obtained either under some
additional set-theoretic assumptions, e.g. by Fedorchuk [Fed76] (under
Jensen’s diamond principle), Dow and Pichardo-Mendoza [DPM09] (under the
Continuum Hypothesis), Dow and Shelah [DS13] (under Martin’s axiom), or using
forcing, see e.g. Sobota and Zdomskyy [SZ19] or Dow and Fremlin [DF07].
Results in the latter paper are of particular interest to us as they concern
forcing with $\mathbb{M}_{\kappa}$; namely, Dow and Fremlin proved that if
$\mathbb{A}$ is a ground model Boolean algebra such that its Stone space is an
F-space, which occurs e.g. in the case of $\sigma$-complete Boolean algebras
or the algebra $\mathcal{P}(\omega)/Fin$, then, in $V^{\mathbb{M}_{\kappa}}$,
the Stone space $St(\mathbb{A})$ does not have any non-trivial convergent
sequences (which yields, e.g., that if $V$ is a model of set theory satisfying
the Continuum Hypothesis, then $St(\mathcal{P}(\omega)\cap V)$ is an Efimov
space in any $\mathbb{M}_{\omega_{2}}$-generic extension of $V$). In Section 8
we prove that $\mathbb{M}_{\kappa}$ forces that the Stone space of a given
ground model Boolean algebra $\mathbb{A}$ does not contain any non-trivial
convergent sequences if and only if in $V$ every pointwise algebraically
convergent sequence in $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is
uniformly convergent (Theorem 8.2). This, together with Dow and Fremlin’s
theorem, implies that if the Stone space of an infinite Boolean algebra
$\mathbb{A}$ is an F-space, then every pointwise algebraically convergent
sequence in $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is uniformly
convergent—note that this result does not hold for ’simple’ Boolean algebras
such as the Cantor algebra, see the example in Section 6.5.

The structure of the paper is as follows. In the next section we present basic
notations, terminology and facts used in the paper. In Section 4 we introduce
four types of topologies and modes of convergence of sequences of
homomorphisms, mostly into measure algebras. Section 5 is devoted to study
relations between convergence properties of sequences $(\varphi_{n})$ of
homomorphisms into a Boolean algebra carrying a strictly positive measure
$\mu$ and convergence properties of sequences of measures of the form
$(\mu\circ\varphi_{n})$. In Section 6 we present a series of examples of
sequences of homomorphisms between Boolean algebras being convergent with
respect to one topology but not with respect to another one. In Section 7 we
study relations between various types of convergence of sequences of
homomorphisms into measure algebras and convergence of corresponding
ultrafilters in random generic extensions of the ground model. This study is
continued in Section 8, where we characterize those Boolean algebras whose
Stone spaces contain no non-trivial convergent sequences in random extensions
with the aid of convergence properties of sequences of homomorphisms in the
ground model. The last section provides several open questions and problems.

##  2\. Acknowledgements

We would like to thank Grzegorz Plebanek and Krzysztof Majcher for valuable
discussions concerning homomorphisms, Hamming codes and painting fences. We
also thank Lyubomyr Zdomskyy for reading the first draft of the paper and
sharing with us many relevant comments.

##  3\. Notations and terminology

All compact spaces considered in the paper are assumed to be Hausdorff. If $K$
is a compact space, then by $Clopen(K)$ and $Bor(K)$ we denote the Boolean
algebra of clopen subsets of $K$ and the $\sigma$-field of all Borel subsets
of $K$, respectively. If $\mathbb{A}$ is a Boolean algebra, then by
$St(\mathbb{A})$ we denote its Stone space. Note that $\mathbb{A}$ and
$Clopen(St(\mathbb{A}))$ are isomorphic.

Unless otherwise stated, all measures considered by us are probability
measures. A measure on a Boolean algebra is always meant to be finitely
additive. On the other hand, a measure on a compact space is always a Radon
measure, i.e. it is countably additive, Borel and inner regular with respect
to compact subsets. Note that every measure $\mu$ on a Boolean algebra
$\mathbb{A}$ has a unique extension to a Radon measure $\widehat{\mu}$ on the
Stone space $St(\mathbb{A})$, i.e. $\mu=\widehat{\mu}\restriction
Clopen(St(\mathbb{A}))$, where $\mathbb{A}$ is identified with
$Clopen(St(\mathbb{A}))$. When there should be no confusion, we will usully
drop $\ \widehat{}\ $ and write simply $\mu$ for $\widehat{\mu}$.

If $\mathbb{A}$ and $\mathbb{B}$ are two Boolean algebras, then
$\mathcal{H}(\mathbb{A},\mathbb{B})$ denotes the family of all homomorphisms
from $\mathbb{A}$ to $\mathbb{B}$. In Section 4 we will endow
$\mathcal{H}(\mathbb{A},\mathbb{B})$ with various topologies and consider
various types of convergences of sequences in
$\mathcal{H}(\mathbb{A},\mathbb{B})$. If
$\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$, then the function
$f_{\varphi}\colon St(\mathbb{B})\to St(\mathbb{A})$ defined as
$f_{\varphi}(x)=\varphi^{-1}[x]$ is continuous (here $x$ is an ultrafilter on
$\mathbb{B}$). On the other hand, if $f\colon St(\mathbb{B})\to
St(\mathbb{A})$ is a continuous function, then the function
$\varphi_{f}\colon\mathbb{A}\to\mathbb{B}$ defined by the formula
$\varphi_{f}(A)=f^{-1}[A]$ is a homomorphism between Boolean algebras
$\mathbb{A}$ and $\mathbb{B}$. It follows that $f=f_{(\varphi_{f})}$ and
$\varphi=\varphi_{(f_{\varphi})}$.

###  3.1. Topologies on spaces of measures.

Let $\mathbb{A}$ be a Boolean algebra and let $K$ be a compact space. The
space of all measures on $K$ is denoted by $P(K)$. By $C(K)$ we denote the
Banach space of continuous real-valued functions on $K$ endowed with the
supremum topology. If $\mu\in P(K)$ and $f\in C(K)$, then $\mu(f)$ is defined
as the integral $\mu(f)=\int_{K}fd\mu$.

We endow $P(K)$ with three topologies: the norm topology, the weak topology
and the weak* topology. The norm topology on $P(K)$ is induced by the
variation metric $d_{var}$ on $P(K)$ defined by the formula

$d_{var}(\mu,\nu)=\sup_{\begin{subarray}{c}A,B\in Bor(K)\\\ A\cap
B=\emptyset\end{subarray}}\big{(}|\mu(A)-\nu(A)|+|\mu(B)-\nu(B)|\big{)}$

for every $\mu,\nu\in P(K)$. Note that if $K$ is totally-disconnected, then in
the above formula for $d_{var}(\mu,\nu)$ we may confine ourselves only to
pairs of disjoint clopen subsets of $K$.

The weak topology on the dual space $C(K)^{*}$ (and hence on $P(K)$) is the
weakest topology which makes all functionals from $C(K)^{*}$ continuous and so
it is induced by the subbase given by sets of the form

$V(\mu,\varphi,\varepsilon)=\big{\\{}\nu\in P(K)\colon\
|\varphi(\mu)-\varphi(\nu)|<\varepsilon\big{\\}},$

where $\mu\in P(K)$, $\varphi\in C(K)^{**}$ and $\varepsilon>0$. Similarly,
the weak* topology on $P(K)$ is defined by sets of the form

$V(\mu,f,\varepsilon)=\big{\\{}\nu\in P(K)\colon\
|\mu(f)-\nu(f)|<\varepsilon\big{\\}},$

where $\mu\in P(K)$, $f\in C(K)$ and $\varepsilon>0$. Recall that $P(K)$ with
the weak* topology is a compact space. Of course, the weak* topology on $P(K)$
is weaker than the weak topology which is on the other hand weaker than the
norm topology.

The following facts are well-known.

######  Proposition 3.1.

Let $K$ be a totally disconnected compact space. Let $\mathcal{V}$ be the
collection of the sets of the form:

$V(\mu,C,\varepsilon)=\big{\\{}\nu\in
P(K)\colon|\nu(C)-\mu(C)|<\varepsilon\big{\\}},$

where $\mu\in P(K)$, $C$ is a clopen subset of $K$ and $\varepsilon>0$. Then
$\mathcal{V}$ is a subbase of the weak* topology on $P(K)$.

######  Proposition 3.2.

[Die84, Theorem 11 in Section VII] Let $K$ be a compact space. Then the
sequence $(\mu_{n})$ of measures on $K$ converges weakly to some $\mu$ if and
only if $\mu_{n}(B)$ converges to $\mu(B)$ for each Borel subset $B\subseteq
K$.

We say that a Boolean algebra $\mathbb{A}$ has the Grothendieck property if
every weak* convergent sequence of (signed) Radon measures on $St(\mathbb{A})$
is weakly convergent. It is well-known that $\sigma$-complete Boolean algebras
have the property, but countable ones do not (or more generally those algebras
whose Stone spaces contain non-trivial convergent sequences).

###  3.2. Measure algebras.

If $\mu$ is a measure on a Boolean algebra $\mathbb{A}$, then we put
$\mathcal{N}_{\mu}=\big{\\{}A\in\mathbb{A}\colon\ \mu(A)=0\big{\\}}$.
Similarly, if $\mu$ is a Radon measure on a compact space $K$, then we denote
$\mathcal{N}_{\mu}=\big{\\{}A\in Bor(K)\colon\ \mu(A)=0\big{\\}}$. For an
infinite cardinal number $\kappa$, by $\mathbb{M}_{\kappa}$ we will denote the
(standard) measure algebra of Maharam type $\kappa$, i.e.
$\mathbb{M}_{\kappa}=\mathrm{Bor}(2^{\kappa})/\mathcal{N}_{\kappa}$, where
$\mathcal{N}_{\kappa}$ is the $\sigma$-ideal of null sets with respect to the
standard product measure $\lambda_{\kappa}$ on the space $2^{\kappa}$, i.e.
$\mathcal{N}_{\kappa}=\mathcal{N}_{\lambda_{\kappa}}$. By $\lambda$ and
$\mathbb{M}$ we mean simply $\lambda_{\omega}$ and $\mathbb{M}_{\omega}$,
respectively.

By $\mathbbm{2}$ we mean the 2-point Boolean algebra $\\{0,1\\}$. Notice that
$\mathbbm{2}=\mathbb{M}_{1}$, i.e. it is a measure algebra (with measure
$\lambda_{1}$).

$\mathbb{C}$ will denote the Cantor algebra, i.e. the free algebra generated
by $\omega$ generators. Note that $\mathbb{C}$ is isomorphic to the algebra of
clopen subsets of the Cantor space $2^{\omega}$ and hence $St(\mathbb{C})$ is
homeomorphic to $2^{\omega}$. We will use this identification frequently.

In general we denote elements of Boolean algebras, including
$\mathbb{M}_{\kappa}$’s, with capital letters. However, speaking about an
element of $\mathbb{M}_{\kappa}$ sometimes we understand it as a condition of
the forcing notion $\mathbb{M}_{\kappa}\setminus\\{0\\}$. In this case, we
will rather use the standard forcing notation: $p$, $q$ and so on.

###  3.3. Metric Boolean algebras.

We will say, after Kolmogorov [Kol95], that a Boolean algebra $\mathbb{B}$ is
_metric_ if it supports a strictly positive measure, i.e. there is a measure
$\mu$ on $\mathbb{B}$ such that $\mu(A)>0$ for every
$A\in\mathbb{B}\setminus\\{0\\}$. Note that a metric algebra need not to be
$\sigma$-complete, so it is not necessarily a measure algebra. The word
_metric_ is explained by the following simple fact.

######  Fact 3.3.

If $\mu$ is a strictly positive measure on a Boolean algebra $\mathbb{B}$,
then the function $d_{\mu}\colon\mathbb{B}\times\mathbb{B}\to\mathbb{R}$
defined for every $A,B\in\mathbb{B}$ by the formula

$d_{\mu}(A,B)=\mu(A\triangle B)$

is a metric on $\mathbb{B}$ (called _Fréchet–Nikodym metric_).

Note that we may define easily a Radon version of the Fréchet–Nikodym metric.
Namely, if $\mu$ is a strictly positive measure on a Boolean algebra
$\mathbb{B}$, then the function $d_{\mu}^{Bor}\colon Bor(St(\mathbb{B}))\times
Bor(St(\mathbb{B}))\to\mathbb{R}$ defined as

$d_{\mu}^{Bor}(A,B)=\widehat{\mu}(A\triangle B),$

where $A,B\in Bor(St(\mathbb{B}))$, is a pseudometric. Call $d_{\mu}^{Bor}$
the Borel Fréchet–Nikodym pseudometric on $(\mathbb{B},\mu)$.

Recall that there is a natural and well-studied notion of convergence in
complete Boolean algebras. Let $\mathbb{A}$ be a complete Boolean algebra and
let $(A_{n})$ be a sequence of its elements. We say that $(A_{n})$
_algebraically converges_ to $A\in\mathbb{A}$ if

$\bigvee_{n}\bigwedge_{m>n}A_{m}=\bigwedge_{n}\bigvee_{m>n}A_{m}=A.$

Such a notion was deeply studied e.g. in [BGJ98], [BFH99], [Jec18]. _The
sequential topology_ on $\mathbb{A}$ is the largest topology with respect to
which all the algebraically convergent sequences converge. If $\mathbb{A}$ is
a measure algebra, then the sequential topology coincides with the topology
introduced by the Fréchet–Nikodym metric (see [BGJ98]).

##  4\. Convergences and topologies in $\mathcal{H}(\mathbb{A},\mathbb{B})$

In the next subsections we will see that for given Boolean algebras
$\mathbb{A}$ and $\mathbb{B}$ we may endow the space
$\mathcal{H}(\mathbb{A},\mathbb{B})$ with several topologies and convergences.

###  4.1. Pointwise metric topology

Let $(\mathbb{B},\mu)$ be a metric algebra. Then, there is a natural topology
on $\mathcal{H}(\mathbb{A},\mathbb{B})$ given by the metric $d_{\mu}$, which
we call the pointwise metric topology. Namely, an element of the subbase is of
the form

$V(\varphi,A,\varepsilon)=\big{\\{}\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})\colon
d_{\mu}(\varphi(A),\psi(A))<\varepsilon\big{\\}}$

for $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$, $A\in\mathbb{A}$ and
$\varepsilon>0$.

Let $(\varphi_{n})$ be a sequence of homomorphisms from $\mathbb{A}$ to
$\mathbb{B}$. It follows that $(\varphi_{n})$ is convergent in the pointwise
metric topology to some $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$ if and
only if for every $A\in\mathbb{A}$ and every $\varepsilon>0$ there is
$N\in\omega$ such that
$d_{\mu}\big{(}\varphi_{n}(A),\varphi(A)\big{)}<\varepsilon$ for every
$n>N$—in this case, we will say that $(\varphi_{n})$ is pointwise metric
convergent to $\varphi$. Note that if $(\varphi_{n})$ is not convergent to
$\varphi$, then there are $A\in\mathbb{A}$, $\varepsilon>0$ and a subsequence
$(\varphi_{n_{k}})$ such that for every $k\in\omega$ we have

$\varepsilon\leq
d_{\mu}\big{(}\varphi_{n_{k}}(A),\varphi(A)\big{)}=\mu\big{(}\varphi_{n_{k}}(A)\setminus\varphi(A)\big{)}+\mu\big{(}\varphi(A)\setminus\varphi_{n_{k}}(A)\big{)},$

so either
$\mu\big{(}\varphi_{n_{k}}(A)\setminus\varphi(A)\big{)}\geq\varepsilon/2$ for
infinitely many $k\in\omega$, or
$\mu\big{(}\varphi(A)\setminus\varphi_{n_{k}}(A)\big{)}\geq\varepsilon/2$ for
infinitely many $k\in\omega$. Since

$\mu\big{(}\varphi(A)\setminus\varphi_{n_{k}}(A)\big{)}=\mu\big{(}\varphi_{n_{k}}(A^{c})\setminus\varphi(A^{c})\big{)},$

it follows that if $(\varphi_{n})$ does not converge pointwise metric to
$\varphi$, then we may always find $A\in\mathbb{A}$, $\varepsilon>0$ and a
subsequence $(\varphi_{n_{k}})$ such that
$\mu\big{(}\varphi_{n_{k}}(A)\setminus\varphi(A)\big{)}\geq\varepsilon$. We
will use this observation regularly.

######  Remark 4.1.

Using the natural bijection between the set of ultrafilters on a Boolean
algebra $\mathbb{A}$ and the set of homomorphisms from
$\mathcal{H}(\mathbb{A},\mathbbm{2})$, we see that the pointwise metric
topology coincides with the Stone topology on the set of ultrafilters on
$\mathbb{A}$, i.e. $St(\mathbb{A})$ and $\mathcal{H}(\mathbb{A},\mathbbm{2})$
are homeomorphic. Thus, since $\mathbbm{2}$ embeds into every (metric) Boolean
algebra $\mathbb{B}$, the space $\mathcal{H}(\mathbb{A},\mathbb{B})$ with the
pointwise metric topology always contains a closed copy of the Stone space of
$\mathbb{A}$, which is immediately yielded by the following trivial fact.

######  Lemma 4.2.

Let $\mathbb{A}$, $\mathbb{B}$ and $\mathbb{C}$ be Boolean algebras. Assume
that $(\mathbb{B},\mu)$ and $(\mathbb{C},\nu)$ are metric and such that
$\mathbb{B}$ is a subalgebra of $\mathbb{C}$ with
$\mu=\nu\restriction\mathbb{B}$. Endow $\mathcal{H}(\mathbb{A},\mathbb{B})$
and $\mathcal{H}(\mathbb{A},\mathbb{C})$ with the pointwise metric topologies.
If for every $A\in\mathbb{C}\setminus\mathbb{B}$ there is $\varepsilon>0$ such
that $d_{\nu}(A,B)<\varepsilon$ for no $B\in\mathbb{B}$, then
$\mathcal{H}(\mathbb{A},\mathbb{B})$ is closed in
$\mathcal{H}(\mathbb{A},\mathbb{C})$.

###  4.2. Uniform topology

Let us again assume that $(\mathbb{B},\mu)$ is a metric algebra. We may define
a metric $d_{hom}$ on the set $\mathcal{H}(\mathbb{A},\mathbb{B})$ in the
following way:

$d_{hom}(\varphi,\psi)=\sup\big{\\{}d_{\mu}(\varphi(A),\psi(A))\colon
A\in\mathbb{A}\big{\\}},$

where $\varphi,\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. According to the
topology induced by $d_{hom}$, which we call the uniform topology, a sequence
$(\varphi_{n})$ converges to $\varphi$ iff it converges uniformly with respect
to the metric $d_{\mu}$, i.e. for each $\varepsilon>0$ there is $N$ such that
for every $n>N$ we have
$d_{\mu}\big{(}\varphi_{n}(A),\varphi(A)\big{)}<\varepsilon$ for each
$A\in\mathbb{A}$. We will say in this case that $(\varphi_{n})$ converges
uniformly to $\varphi$. Of course, if a sequence of homomorphisms converges
uniformly, then it converges pointwise—the converse however does not
necessarily hold, see Sections 6.1, 6.3 and 6.4 for appropriate examples.

######  Remark 4.3.

Contrary to the pointwise metric topology (see Remark 4.1), for every Boolean
algebra $\mathbb{A}$ the uniform topology on
$\mathcal{H}(\mathbb{A},\mathbbm{2})$ is discrete.

###  4.3. Pointwise algebraic convergence

This kind of convergence can be introduced in
$\mathcal{H}(\mathbb{A},\mathbb{B})$ if $\mathbb{B}$ is $\sigma$-complete (but
not necessarily metric). We say that a sequence of homomorphisms
$(\varphi_{n})$ from $\mathbb{A}$ to $\mathbb{B}$ converges pointwise
algebraically to $\varphi$ if for every $A\in\mathbb{A}$ we have

$\bigvee_{n}\bigwedge_{m>n}\varphi_{m}(A)=\bigwedge_{n}\bigvee_{m>n}\varphi_{m}(A)=\varphi(A).$

In fact, by the de Morgan laws, $(\varphi_{n})$ converges pointwise
algebraically to $\varphi$ if and only if any of the following two equivalent
conditions hold:

  * •

for every $A\in\mathbb{A}$ we have
$\bigvee_{n}\bigwedge_{m>n}\varphi_{m}(A)=\varphi(A)$;

  * •

for every $A\in\mathbb{A}$ we have
$\bigwedge_{n}\bigvee_{m>n}\varphi_{m}(A)=\varphi(A)$.

If we assume that $\mathbb{B}$ is metric (with a strictly positive measure
$\mu$), then if $(\varphi_{n})$ converges pointwise algebraically to
$\varphi$, then for each $A\in\mathbb{A}$ we have

$\lim_{n\to\infty}\mu\big{(}\bigwedge_{m>n}\varphi_{m}(A)\big{)}=\lim_{n\to\infty}\mu\big{(}\bigvee_{m>n}\varphi_{m}(A)\big{)}=\mu(\varphi(A)).$

The following fact binds the pointwise algebraic convergence with the
pointwise metric convergence. Recall that for every $\kappa$ the measure
$\lambda_{\kappa}$ is continuous, i.e. for every decreasing sequence $(A_{n})$
in $\mathbb{M}_{\kappa}$ we have
$\lim_{n\to\infty}\lambda_{\kappa}(A_{n})=\lambda_{\kappa}\big{(}\bigwedge_{n\in\omega}A_{n}\big{)}$.

######  Proposition 4.4.

Let $\mathbb{A}$ be a Boolean algebra and $\kappa$ a cardinal number. Let
$(\varphi_{n})$ be a sequence of homomorphisms from $\mathbb{A}$ to
$\mathbb{M}_{\kappa}$. If $(\varphi_{n})$ is pointwise algebrically convergent
to some $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$, then
$\varphi_{n}$ converges pointwise metric to $\varphi$.

###### Proof.

Let $A\in\mathbb{A}$. We will show that
$\lim_{n\to\infty}\lambda_{\kappa}\big{(}\varphi_{n}(A)\triangle\varphi(A)\big{)}=0$.
For the sake of contradiction assume that there is $\varepsilon>0$ and a
subsequence $(\varphi_{n_{k}})$ such that
$\lambda_{\kappa}(\varphi_{n_{k}}(A)\setminus\varphi(A))\geq\varepsilon$ for
every $k\in\omega$. Then, by the continuity of $\lambda_{\kappa}$,

$\lambda_{\kappa}\Big{(}\bigwedge_{k\in\omega}\bigvee_{l>k}\big{(}\varphi_{n_{l}}(A)\setminus\varphi(A)\big{)}\Big{)}\geq\varepsilon.$

However, since
$\bigwedge_{k\in\omega}\bigvee_{l>k}\varphi_{n_{l}}(A)=\varphi(A)$, we get
that
$\bigwedge_{k\in\omega}\bigvee_{l>k}\big{(}\varphi_{n_{l}}(A)\setminus\varphi(A)\big{)}=0$,
so

$0=\lambda_{\kappa}\Big{(}\bigwedge_{k\in\omega}\bigvee_{l>k}\big{(}\varphi_{n_{l}}(A)\setminus\varphi(A)\big{)}\Big{)}\geq\varepsilon>0,$

a contradiction. ∎

As we will see in Sections 6.2 and 6.3, the converse to Proposition 4.4 does
not hold in general. However, for $\mathbb{B}=\mathbbm{2}$ we have the
following corollary.

######  Corollary 4.5.

Let $\mathbb{A}$ be a Boolean algebra. Then, in
$\mathcal{H}(\mathbb{A},\mathbbm{2})$ the pointwise algebraic convergence
coincides with the pointwise metric convergence.

###### Proof.

Let $(\varphi_{n})$ be a sequence in $\mathcal{H}(\mathbb{A},\mathbbm{2})$. If
$(\varphi_{n})$ is pointwise algebraically convergent, then by Proposition 4.4
it is also pointwise metric convergence.

Conversely, if $(\varphi_{n})$ is pointwise metric convergent to
$\varphi\in\mathcal{H}(\mathbb{A},\mathbbm{2})$, then for every
$A\in\mathbb{A}$ there is $N\in\omega$ such that $\varphi_{n}(A)=\varphi(A)$
for every $n>N$. It follows that $\bigwedge_{n>N}\varphi_{n}(A)=\varphi(A)$,
so in particular
$\bigvee_{N\in\omega}\bigwedge_{n>N}\varphi_{n}(A)=\varphi(A)$. Hence,
$(\varphi_{n})$ converges pointwise algebraically to $\varphi$. ∎

###  4.4. Pointwise Borel metric topology and uniform Borel topology

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ a metric Boolean
algebra. To introduce yet another topology on
$\mathcal{H}(\mathbb{A},\mathbb{B})$, called by us pointwise Borel metric
topology, we need to appeal to dual continuous functions and the pseudometric
$d_{\mu}^{Bor}$. Namely, we define the subbase of the topology to be given by
the sets of the form

$V^{Bor}(\varphi,A,\varepsilon)=\big{\\{}\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})\colon
d_{\mu}^{Bor}\big{(}f_{\varphi}^{-1}[A],f_{\psi}^{-1}[A]\big{)}<\varepsilon\big{\\}}$

for $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$, $A\in Bor(St(\mathbb{A}))$
and $\varepsilon>0$.

Let $(\varphi_{n})$ be a sequence of homomorphisms from $\mathbb{A}$ to
$\mathbb{B}$. For each $n\in\omega$ let $f_{n}=f_{\varphi_{n}}$ and let
$f=f_{\varphi}$. We say that $(\varphi_{n})$ converges pointwise Borel metric
to $\varphi$ if $\lim_{n\to\infty}\widehat{\mu}\big{(}f_{n}^{-1}[B]\triangle
f^{-1}[B]\big{)}=0$ for every $B\in Bor(St(\mathbb{A}))$. Obviously, if a
sequence of homomorphisms converges pointwise Borel metric, then it converges
pointwise metric, however the converse may not hold—we provide relevant
counterexamples in Sections 6.1 and 6.3 as well as some positive results in
Section 5 (see, in particular, Corollary 5.6).

Similarly to the uniform topology on $\mathcal{H}(\mathbb{A},\mathbb{B})$, we
may define the uniform Borel topology. Namely, we define a Borel version
$d_{hom}^{Bor}$ of the metric $d_{hom}$:

$d_{hom}^{Bor}(\varphi,\psi)=\sup\big{\\{}d_{\mu}^{Bor}\big{(}f_{\varphi}^{-1}[A],f_{\psi}^{-1}[A]\big{)}\colon
A\in Bor(St(\mathbb{A}))\big{\\}},$

where $\varphi,\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. Note that, since
$\mu$ is strictly positive on $\mathbb{A}$, $d_{hom}^{Bor}$ is a metric, even
though $d_{\mu}^{Bor}$ is only a pseudometric. As before, we will say that a
sequence $(\varphi_{n})$ converges uniformly Borel to $\varphi$ if for every
$\varepsilon>0$ there is $N\in\omega$ such that
$d_{hom}^{Bor}(\varphi_{n},\varphi)<\varepsilon$ for every $n>N$, i.e.
$\widehat{\mu}\big{(}f_{n}^{-1}[B]\triangle f^{-1}[B]\big{)}<\varepsilon$ for
every $n>N$ and every Borel subset $B\subseteq St(\mathbb{A})$. Of course, if
a sequence converges uniformly Borel, then it converges pointwise Borel metric
and uniformly. It appears that, conversely, the uniform convergence easily
implies the Borel uniform convergence, because, in fact, the uniform topology
and uniform Borel topology coincide.

######  Lemma 4.6.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ a metric Boolan
algebra. Then, for the metrics $d_{hom}$ and $d_{hom}^{Bor}$ on
$\mathcal{H}(\mathbb{A},\mathbb{B})$ we have $d_{hom}=d_{hom}^{Bor}$.

###### Proof.

Fix $\varphi,\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. Since every clopen
set in $St(\mathbb{A})$ is Borel, we have $d_{hom}(\varphi,\psi)\leq
d_{hom}^{Bor}(\varphi,\psi)$. Let $\varepsilon>0$—we will show that
$d_{hom}^{Bor}(\varphi,\psi)\leq d_{hom}(\varphi,\psi)+\varepsilon$, which
will prove that $d_{hom}^{Bor}(\varphi,\psi)\leq d_{hom}(\varphi,\psi)$. By
the regularity of $\widehat{\mu}\circ f_{\varphi}^{-1}$ and
$\widehat{\mu}\circ f_{\psi}^{-1}$, for every $B\in Bor(St(\mathbb{A}))$ there
is $A_{B}\in\mathbb{A}$ such that $B\subseteq A_{B}$,
$\widehat{\mu}\big{(}f_{\varphi}^{-1}[A_{B}\setminus B]\big{)}<\varepsilon/2$
and $\widehat{\mu}\big{(}f_{\psi}^{-1}[A_{B}\setminus
B]\big{)}<\varepsilon/2$. Thus, for every Borel $B\in Bor(St(\mathbb{A}))$ we
have:

$\widehat{\mu}\big{(}f_{\varphi}^{-1}[B]\triangle f_{\psi}^{-1}[B]\big{)}\leq$
$\widehat{\mu}\big{(}f_{\varphi}^{-1}[B]\triangle
f_{\varphi}^{-1}\big{[}A_{B}\big{]}\big{)}+\widehat{\mu}\big{(}f_{\varphi}^{-1}\big{[}A_{B}\big{]}\triangle
f_{\psi}^{-1}\big{[}A_{B}\big{]}\big{)}+\widehat{\mu}\big{(}f_{\psi}^{-1}\big{[}A_{B}\big{]}\triangle
f_{\psi}^{-1}[B]\big{)}<$
$\varepsilon/2+d_{\mu}\big{(}\varphi\big{(}A_{B}\big{)},\psi\big{(}A_{B}\big{)}\big{)}+\varepsilon/2\leq
d_{hom}(\varphi,\psi)+\varepsilon,$

so $d_{hom}^{Bor}(\varphi,\psi)\leq d_{hom}(\varphi,\psi)+\varepsilon$. ∎

######  Corollary 4.7.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ a metric Boolan
algebra. Then:

  1. (1)

the uniform topology and the uniform Borel topology coincide;

  2. (2)

a sequence of homomorphisms from $\mathbb{A}$ to $\mathbb{B}$ converges
uniformly if and only if it converges uniformly Borel.

##  5\. Homomorphisms and measures

By the Maharam theorem for every measure $\mu$ on a Boolean algebra
$\mathbb{A}$ there are a cardinal number $\kappa$ and an injective
homomorphism $\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ such that
$\mu=\lambda_{\kappa}\circ\varphi$. On the other hand, if
$\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ is a homomorphism, then it is
plain to check that the function
$\lambda_{\kappa}\circ\varphi\colon\mathbb{A}\to\mathbb{R}$ is a (finitely
additive) measure. It follows that there is a natural mapping from the family
of homomorphisms $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ into the family
of measures on $\mathbb{A}$. This observation allows us to use notions from
the theory of measures on Boolean algebras to study
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$.

######  Proposition 5.1.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ a metric algebra.
Consider the space $\mathcal{H}(\mathbb{A},\mathbb{B})$ with the topology
$\tau$ and the space $P(St(\mathbb{A}))$ with topology $\tau^{\prime}$, both
given below. Then, the function
$F_{\mu}\colon\mathcal{H}(\mathbb{A},\mathbb{B})\to P(St(\mathbb{A}))$, given
by $F_{\mu}(\varphi)=\widehat{\mu}\circ f_{\varphi}^{-1}$, is continuous in
each of the following cases:

  1. (1)

$\tau$ is the uniform topology and $\tau^{\prime}$ is the norm topology;

  2. (2)

$\tau$ is the pointwise metric topology and $\tau^{\prime}$ is the weak*
topology.

###### Proof.

We will use below the fact that $|\mu(C)-\mu(D)|\leq\mu(C\triangle D)$ for
every $C,D\in\mathbb{B}$.

We first prove (1). Using the above inequality, for every $A\in\mathbb{A}$ we
have

$\big{|}\mu(\varphi(A))-\mu(\psi(A))\big{|}\leq\mu\big{(}\varphi(A)\triangle\psi(A)\big{)},$

and thus, by the definition of the variation metric on $P(St(\mathbb{A}))$, it
holds:

$d_{var}\big{(}F_{\mu}(\varphi),F_{\mu}(\psi)\big{)}=\sup_{\begin{subarray}{c}A,B\in\mathbb{A}\\\
A\wedge
B=0_{\mathbb{A}}\end{subarray}}\Big{(}\big{|}F_{\mu}(\varphi)(A)-F_{\mu}(\psi)(A)\big{|}+\big{|}F_{\mu}(\varphi)(B)-F_{\mu}(\psi)(B)\big{|}\Big{)}\leq$
$2\sup_{A\in\mathbb{A}}\big{|}F_{\mu}(\varphi)(A)-F_{\mu}(\psi)(A)\big{|}\leq
2\sup_{A\in\mathbb{A}}d_{\mu}(\varphi(A),\psi(A))=2d_{hom}(\varphi,\psi),$

which actually shows that $F_{\mu}$ is $2$-Lipschitz.

To prove (2) let $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$ and fix an
element $V$ of the subbase of the weak* topology on $P(St(\mathbb{A}))$ of the
form
$V=V(F_{\mu}(\varphi),A,\varepsilon)=\big{\\{}\nu\colon\big{|}F_{\mu}(\varphi)(A)-\nu(A)\big{|}<\varepsilon\big{\\}}$,
where $A\in\mathbb{A}$ and $\varepsilon>0$. Put:

$U=\big{\\{}\psi\in\mathcal{H}(\mathbb{A},\mathbb{B})\colon
d_{\mu}(\varphi(A),\psi(A))<\varepsilon\big{\\}}.$

Then, $U$ is an open set in $\mathcal{H}(\mathbb{A},\mathbb{B})$. For every
$\psi\in U$ we have:

$\big{|}F_{\mu}(\varphi)(A)-F_{\mu}(\psi)(A)\big{|}\leq
d_{\mu}\big{(}\varphi(A),\psi(A)\big{)}<\varepsilon,$

so $\varphi\in F_{\mu}[U]\subseteq V$, which proves that $F_{\mu}$ is
continuous.

∎

Consequently, the equivalence relation on $\mathcal{H}(\mathbb{A},\mathbb{B})$
defined by the formula

$\varphi\sim\psi\iff F_{\mu}(\varphi)=F_{\mu}(\psi),$

has closed equivalence classes, provided that
$\mathcal{H}(\mathbb{A},\mathbb{B})$ is endowed with any of the topologies
$\tau$ mentioned in the proposition.

######  Corollary 5.2.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ a metric algebra.
Let $\varphi_{n}\in\mathcal{H}(\mathbb{A},\mathbb{B})$, $n\in\omega$, and
$\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. The following hold:

  1. (1)

if $(\varphi_{n})$ converges uniformly to $\varphi$, then
$\big{(}\widehat{\mu}\circ f_{\varphi_{n}}^{-1}\big{)}$ converges to
$\widehat{\mu}\circ f_{\varphi}^{-1}$ in norm;

  2. (2)

if $(\varphi_{n})$ converges pointwise Borel metric to $\varphi$, then
$\big{(}\widehat{\mu}\circ f_{\varphi_{n}}^{-1}\big{)}$ converges weakly to
$\widehat{\mu}\circ f_{\varphi}^{-1}$;

  3. (3)

if $(\varphi_{n})$ converges pointwise metric to $\varphi$, then
$\big{(}\widehat{\mu}\circ f_{\varphi_{n}}^{-1}\big{)}$ converges weakly* to
$\widehat{\mu}\circ f_{\varphi}^{-1}$.

###### Proof.

(1) and (3) are direct consequences of Proposition 5.1. (2) follows from
Proposition 3.2. ∎

######  Remark 5.3.

Note here that the converse to Corollary 5.2 may not hold even in the simplest
case when all homomorphisms $\varphi_{n}$’s are the same. Indeed, let
$(\mathbb{B},\mu)$ be a metric algebra and let each $\varphi_{n}$ be the
identity homomorphism on $\mathbb{B}$. If
$\varphi\in\mathcal{H}(\mathbb{B},\mathbb{B})$ is a homomorphism such that
$\mu(\varphi(A))=\mu(A)$ and $\varphi(A)\neq A$ for some $A\in\mathbb{B}$,
then the sequence $(\varphi_{n})$ does not converge to $\varphi$ in any of the
considered topologies, even though $\mu(\varphi_{n}(B))=\mu(\varphi(B))$ for
every $n\in\omega$ and $B\in\mathbb{B}$.

We now prove that if a Boolean algebra $\mathbb{A}$ has the Grothendieck
property, then every pointwise metric convergent sequence of homomorphisms
from $\mathbb{A}$ into a metric algebra is also pointwise Borel metric
convergent. Recall that a sequence $(\mu_{k})$ of Radon measures on a compact
Hausdorff space $K$ is uniformly countably additive (also, uniformly
exhaustive) if for every descending sequence $(E_{n})$ of Borel subsets of $K$
such that $\bigcap_{n}E_{n}=\emptyset$ and every $\varepsilon>0$ there is
$N\in\omega$ such that $\big{|}\mu_{k}(E_{n})\big{|}<\varepsilon$ for every
$n\geq N$ and $k\in\omega$. Equivalently, there is $N\in\omega$ such that
$\big{|}\mu_{k}(E_{m})-\mu_{k}(E_{n})\big{|}<\varepsilon$ for every $n,m\geq
N$ and $k\in\omega$ (see [Die84, Chapter VII, Theorem 10] for other equivalent
definitions). We also say that $(\mu_{n})$ is weakly convergent to a Radon
measure $\mu$ on $K$ if $\lim_{n\to\infty}\mu_{n}(B)=\mu(B)$ for every Borel
subset $B$ of $K$. The Nikodym Convergence Theorem ([Die84, Chapter VII, page
90]) asserts that every weakly convergent sequence of Radon measures is
uniformly countably additive.

######  Proposition 5.4.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ be a metric
Boolean algebra. Let $(\varphi_{n})$ be a sequence in
$\mathcal{H}(\mathbb{A},\mathbb{B})$ pointwise metric convergent to a
homomorphism $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. If the sequence
$\big{(}\widehat{\mu}\circ f_{\varphi_{n}}^{-1}\big{)}$ is uniformly countably
additive, then $(\varphi_{n})$ converges pointwise Borel metric to $\varphi$.

###### Proof.

Write simply $f_{n}$ for $f_{\varphi_{n}}$ and $f$ for $f_{\varphi}$. Fix
$\varepsilon>0$ and a Borel subset $B$ of $St(\mathbb{A})$. Let
$\delta=\varepsilon/2$. We will show first that
$\widehat{\mu}\big{(}f^{-1}[B]\setminus f_{k}^{-1}[B]\big{)}<\delta$ for
almost all $k\in\omega$.

Since each measure $\widehat{\mu}\circ f_{k}^{-1}$ is (outer) regular on
$St(\mathbb{A})$, for each $k\in\omega$ we can find $A_{k}\in\mathbb{A}$ such
that $B\subseteq A_{k}$ and

$\widehat{\mu}\big{(}f_{k}^{-1}[A_{k}\setminus B]\big{)}<\delta/4.$

By taking intersections, we may assume that $A_{k+1}\leq A_{k}$ for every
$k\in\omega$. By the uniform countable additiveness of
$\big{(}\widehat{\mu}\circ f_{n}^{-1}\big{)}$, there is $k_{0}\in\omega$ such
that

$\widehat{\mu}\big{(}f_{k}^{-1}[A_{l}\setminus A_{k}]\big{)}<\delta/4$

for every $k>l>k_{0}$. Since obviously the measure $\widehat{\mu}\circ f^{-1}$
is also countably additive, there is $k_{1}\geq k_{0}$ such that

$\widehat{\mu}\big{(}f^{-1}[A_{l}\setminus A_{k}]\big{)}<\delta/4$

for every $k>l>k_{1}$. Now, fix $l=k_{1}+1$. By the pointwise metric
convergence of $(\varphi_{k})$ to $\varphi$, there is $k_{2}\geq l$ such that

$\widehat{\mu}\big{(}f^{-1}[A_{l}]\triangle f_{k}^{-1}[A_{l}]\big{)}<\delta/4$

for every $k>k_{2}$. Using (3), (4) and (2), for every $k>k_{2}$ we have:

$\widehat{\mu}\big{(}f^{-1}[A_{k}]\triangle f_{k}^{-1}[A_{k}]\big{)}\leq$
$\widehat{\mu}\big{(}f^{-1}[A_{k}]\triangle
f^{-1}[A_{l}]\big{)}+\widehat{\mu}\big{(}f^{-1}[A_{l}]\triangle
f_{k}^{-1}[A_{l}]\big{)}+\widehat{\mu}\big{(}f_{k}^{-1}[A_{l}]\triangle
f_{k}^{-1}[A_{k}]\big{)}<\frac{3\delta}{4}.$

(5) and (1) yield that for every $k>k_{2}$ it holds:

$\widehat{\mu}\big{(}f^{-1}[B]\setminus
f_{k}^{-1}[B]\big{)}\leq\widehat{\mu}\big{(}f^{-1}[A_{k}]\setminus
f_{k}^{-1}[B]\big{)}\leq\widehat{\mu}\big{(}f^{-1}[A_{k}]\triangle
f_{k}^{-1}[B]\big{)}\leq$ $\widehat{\mu}\big{(}f^{-1}[A_{k}]\triangle
f_{k}^{-1}[A_{k}]\big{)}+\widehat{\mu}\big{(}f_{k}^{-1}[A_{k}]\triangle
f_{k}^{-1}[B]\big{)}\leq\frac{3\delta}{4}+\frac{\delta}{4}=\delta,$

so

$\widehat{\mu}\big{(}f^{-1}[B]\setminus f_{k}^{-1}[B]\big{)}<\varepsilon/2.$

Changing $B$ to $B^{c}$ and doing exactly the same computations that led us to
(6), we get that there is $k_{3}\geq k_{2}$ such that

$\widehat{\mu}\big{(}f_{k}^{-1}[B]\setminus
f^{-1}[B]\big{)}=\widehat{\mu}\big{(}f^{-1}[B^{c}]\setminus
f_{k}^{-1}[B^{c}]\big{)}<\varepsilon/2$

for every $k>k_{3}$. Thus, by (6) and (7), for every $k>k_{3}$ we get:

$\widehat{\mu}\big{(}f^{-1}[B]\triangle f_{k}^{-1}[B]\big{)}<\varepsilon.$

This proves that $(\varphi_{n})$ converges pointwise Borel metric to
$\varphi$. ∎

######  Corollary 5.5.

Let $\mathbb{A}$ be a Boolean algebra and $(\mathbb{B},\mu)$ be a metric
Boolean algebra. Let $(\varphi_{n})$ be a sequence in
$\mathcal{H}(\mathbb{A},\mathbb{B})$ pointwise metric convergent to a
homomorphism $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. If $\mathbb{A}$
has the Grothendieck property, then $(\varphi_{n})$ converges pointwise Borel
metric to $\varphi$.

###### Proof.

Assume that $\mathbb{A}$ has the Grothendieck property. Let $(f_{k})$ and $f$
be the continuous functions dual to $(\varphi_{k})$ and $\varphi$,
respectively. Since
$\lim_{k\to\infty}\widehat{\mu}\big{(}f_{k}^{-1}[A]\big{)}=\widehat{\mu}\big{(}f^{-1}[A]\big{)}$
for every $A\in\mathbb{A}$, the sequence $\big{(}\widehat{\mu}\circ
f_{k}^{-1}\big{)}$ is weakly* convergent to $\widehat{\mu}\circ f^{-1}$ (as it
is uniformly bounded). The Grothendieck property of $\mathbb{A}$ implies that
$\big{(}\widehat{\mu}\circ f_{k}^{-1}\big{)}$ is weakly convergent to
$\widehat{\mu}\circ f^{-1}$. By the Nikodym Convergence Theorem, the sequence
is uniformly countably additive, so the conclusion follows by Proposition 5.4.
∎

Notice here that the measures $\widehat{\mu}\circ f_{k}^{-1}$ considered in
the proof of Corollary 5.5 are non-negative, i.e. $\widehat{\mu}\circ
f_{k}^{-1}(B)\geq 0$ for every $B\in Bor(St(\mathbb{A}))$. It follows that in
the proof we do not need the full Grothendieck property of the algebra
$\mathbb{A}$, but only the property for non-negative measures, i.e. if a
sequence of non-negative measures on $\mathbb{A}$ is weakly* convergent, then
it is weakly convergent. Such a variant of the Grothendieck property was
introduced in Koszmider and Shelah [KS13], where it was called the positive
Grothendieck property.

######  Corollary 5.6.

Let $\mathbb{A}$ be a Boolean algebra with the positive Grothendieck property
and $(\mathbb{B},\mu)$ be a metric Boolean algebra. Let $(\varphi_{n})$ be a
sequence in $\mathcal{H}(\mathbb{A},\mathbb{B})$ pointwise metric convergent
to a homomorphism $\varphi\in\mathcal{H}(\mathbb{A},\mathbb{B})$. Then,
$(\varphi_{n})$ converges pointwise Borel metric to $\varphi$.

The following weaker form of the $\sigma$-completeness was also introduced in
[KS13]: a Boolean algebra $\mathbb{A}$ is said to have the Weak Subsequential
Separation Property (the WSSP) if for every sequence $(A_{n})$ of pairwise
disjoint elements of $\mathbb{A}$ there is an element $A\in\mathbb{A}$ such
that both of the sets $\big{\\{}n\in\omega\colon A_{n}\leq A\big{\\}}$ and
$\big{\\{}n\in\omega\colon A_{n}\wedge A=0\big{\\}}$ are infinite. [KS13,
Proposition 2.4] asserts that every Boolean algebra with the WSSP has the
positive Grothendieck property. Note that there exists a Boolean algebra with
the WSSP (and hence with the positive Grothendieck property), but without the
Grothendieck property, see [KS13, Proposition 2.5].

Recall that by the Eberlein–Šmulian theorem and the Dieudonné–Grothendieck
characterization of weakly compact sets of Radon measures on compact spaces
(see [Die84, Chapter VII, Theorem 14]), a weakly* convergent sequence
$(\mu_{n})$ of measures on a given compact space $K$ is weakly convergent if
and only if there is no antichain $(U_{k})$ of open subsets of $K$,
$\varepsilon>0$ and subsequence $(\mu_{n_{k}})$ such that
$|\mu_{n_{k}}(U_{k})|\geq\varepsilon$ for every $k\in\omega$.

######  Proposition 5.7.

Let $\mathbb{A}$ be a Boolean algebra with the positive Grothendieck property,
$(\mathbb{B},\mu)$ a metric Boolean algebra, and
$\varphi,\varphi_{n}\in\mathcal{H}(\mathbb{A},\mathbb{B})$, $n\in\omega$. Let
$(A_{n})$ be an antichain in $\mathbb{A}$ such that for some $\varepsilon>0$
and every $n\in\omega$ we have
$d_{\mu}\big{(}\varphi_{n}(A_{n}),\varphi(A_{n})\big{)}\geq\varepsilon$. Then,
$(\varphi_{n})$ is not pointwise metric convergent to $\varphi$.

###### Proof.

For the sake of contradiction assume that $(\varphi_{n})$ is pointwise metric
convergent to $\varphi$. Then, by Corollary 5.2.(3), the sequence
$\big{(}\widehat{\mu}\circ f^{-1}_{\varphi_{n}}\big{)}$ is weakly* convergent
to $\widehat{\mu}\circ f^{-1}_{\varphi}$. Since $(A_{n})$ is an antichain in
$\mathbb{A}$, the sequence $(\varphi(A_{n}))$ is an antichain in $\mathbb{B}$
and thus $\lim_{n\to\infty}\mu(\varphi(A_{n}))=0$. From the assumption that
$d_{\mu}\big{(}\varphi_{n}(A_{n}),\varphi(A_{n})\big{)}\geq\varepsilon$ for
every $n\in\omega$ it follows that there is $N\in\omega$ such that
$\mu(\varphi_{n}(A_{n}))\geq\varepsilon/2$ for every $n>N$, so
$\big{(}\widehat{\mu}\circ
f_{\varphi_{n}}^{-1}\big{)}(A_{n})\geq\varepsilon/2$ for every $n>N$. By the
aforementioned characterisation of weakly convergent sequences of measures on
compact spaces, the sequence $\big{(}\widehat{\mu}\circ
f^{-1}_{\varphi_{n}}\big{)}$ is not weakly convergent to $\widehat{\mu}\circ
f_{\varphi}^{-1}$, and hence, by the positive Grothendieck property of
$\mathbb{A}$, not weakly* convergent, which is a contradiction. ∎

##  6\. Examples of sequences of homomorphisms

In this section we will present certain examples distinguishing different
types of convergence. First, we summarize in the following self-explanatory
diagram all the implications which we have already proved or which we are
going to prove.

uniformpointwise Borel metricpointwise metricpointwise
algebraic$\scriptstyle{1}$$\scriptstyle{2}$ the positive Grothendieck
property$\scriptstyle{4}$ Seever’s interpolation
property$\scriptstyle{5}$$\scriptstyle{3}$

.

Of those, (2) is trivial, (1) follows from Corollary 4.7 and (3) is proved in
Proposition 4.4. No other implication holds in general. (4) holds for Boolean
algebras with the positive Grothendieck property and hence, in particular, for
Boolean algebras with the WSSP (Corollary 5.6). (5) is true for Boolean
algebras with the interpolation property (or, the property (I)) introduced by
Seever (see Section 8). Corollary 4.5 implies that (3) may be reversed for
$\mathbb{B}=\mathbbm{2}$, but we skipped that in the diagram, since it is
trivial.

Note that it is easy to obtain a sequence which satisfies simultaneously all
of the above types of convergence (e.g. a trivial sequence). Below we present
examples of sequences of homomorphisms satisfying various other combinations
of types of convergence and witnessing that the above diagram is complete.

In what follows we will often abuse the notation identifying equivalence
classes (particularly elements of $\mathbb{M}$) with their representatives.

###  6.1. A pointwise algebraically convergent sequence which is not pointwise
Borel metric convergent

To see that the pointwise algebraic convergence does not imply the pointwise
Borel metric convergence, consider the Cantor algebra $\mathbb{C}$ and a non-
trivial sequence $(x_{n})$ of points in the Cantor space
$St(\mathbb{C})=2^{\omega}$ convergent to some $x\in 2^{\omega}$. For each
$n\in\omega$ define $\varphi_{n}\in\mathcal{H}(\mathbb{C},\mathbbm{2})$ by the
condition: $\varphi_{n}(A)=1$ iff $x_{n}\in A$, where $A\in\mathbb{C}$.
Similarly, define $\varphi\in\mathcal{H}(\mathbb{C},\mathbbm{2})$ by the
condition: $\varphi(A)=1$ iff $x\in A$, where $A\in\mathbb{C}$. Since
$\lim_{n\to\infty}x_{n}=x$ in $St(\mathbb{C})$, $(\varphi_{n})$ is pointwise
algebraically convergent. On the other hand, $(\varphi_{n})$ does not converge
pointwise Borel metric to $\varphi$, since for the Borel set $\\{x\\}$ and
every $n\in\omega$ we have:

$\widehat{\lambda}_{1}\big{(}f_{\varphi_{n}}^{-1}\big{[}\\{x\\}\big{]}\triangle
f_{\varphi}^{-1}\big{[}\\{x\\}\big{]}\big{)}=\lambda_{1}(0\triangle 1)=1.$

Of course, in the above example instead of $\mathbb{C}$ we may use any Boolean
algebra such that its Stone space contains a non-trivial convergent sequence.

###  6.2. A uniformly convergent sequence which is not pointwise algebraically
convergent

Here we will present an example of a sequence of homomorphisms in
$\mathcal{H}(\mathbb{C},\mathbb{M})$ which converges uniformly but not
pointwise algebraically.

Let $\\{s_{n}\colon n\in\omega\\}$ be any enumeration of $2^{<\omega}$. For
$x\in 2^{\omega}$ define $\overline{x}\in 2^{\omega}$ by

$\overline{x}(k)=\begin{cases}1-x(k),\mbox{ if }k=0,\\\ x(k),\mbox{
otherwise,}\end{cases}$

so $\overline{x}$ is just $x$ but with the $0$-th bit flipped. For $s\in
2^{<\omega}$ and $x\in 2^{\omega}$ we will say that $x$ _agrees with_ $s$, if
$x(k+1)=s(k)$ for each $k<|s|$ (so, in other words, $s$ is a segment of $x$
starting with the $1$-st bit).

Now for $n\in\omega$ define $g_{n}\colon 2^{\omega}\to 2^{\omega}$ by

$g_{n}(x)=\begin{cases}\overline{x},\mbox{ if }x\mbox{ agrees with }s_{n},\\\
x,\mbox{ otherwise.}\end{cases}$

It is easy to check that $\psi_{n}\colon\mathbb{C}\to\mathbb{M}$ defined by
$\psi_{n}(A)=g_{n}[A]=g_{n}^{-1}[A]$ is a homomorphism. Let $\psi$ be the
natural embedding of $\mathbb{C}$ into $\mathbb{M}$, i.e. $\psi(A)=A$ for
every $A\in\mathbb{C}$. (Note that here we treat $\mathbb{C}$ as the algebra
of clopen subsets of $2^{\omega}$ and writing $g_{n}[A]$ we mean the
appropriate equivalence class, that is, an element of $\mathbb{M}$.)

######  Proposition 6.1.

$(\psi_{n})$ converges uniformly to $\psi$ but it does not converge pointwise
algebraically.

###### Proof.

For each $s\in 2^{<\omega}$ let $A_{s}=\\{x\in 2^{\omega}\colon x\mbox{ agrees
with }s\\}$. Clearly, $\lambda(A_{s})=1/2^{|s|}$ and so
$\lim_{n\to\infty}\lambda(A_{s_{n}})=0$. But $\psi_{n}$ coincides with $\psi$
on $2^{\omega}\setminus A_{s_{n}}$ and so $(\psi_{n})$ converges uniformly to
$\psi$.

Let $A=\\{x\in 2^{\omega}\colon x(0)=0\\}$. Fix $x\in 2^{\omega}$ and
$n\in\omega$. If $x\in A$, then there is $m>n$ such that $x$ agrees with
$s_{m}$ and so $x\not\in g_{m}[A]$. Hence,
$A\cap\bigcap_{m>n}g_{m}[A]=\emptyset$. If $x\not\in A$, then there is $m>n$
such that $x$ does not agree with $s_{m}$ and so $x\not\in g_{m}[A]$. Hence,
$A^{c}\cap\bigcap_{m>n}g_{m}[A]=\emptyset$, and so
$\bigcap_{m>n}g_{m}[A]=\emptyset$. Since $n$ was arbitrary, we get that
$\bigcup_{n}\bigcap_{m>n}g_{m}[A]=\emptyset$ and hence
$\bigvee_{n}\bigwedge_{m>n}\psi_{n}(A)=0\neq A=\psi(A)$, which implies that
$(\varphi_{n})$ does not converge pointwise algebraically to $\psi$. ∎

Note that since the pointwise algebraic convergence implies the pointwise
metric convergence, it is not possible that $(\psi_{n})$ converges pointwise
algebraically to a homomorphism other than $\psi$.

###  6.3. A pointwise metric convergent sequence which is neither pointwise
Borel metric convergent nor pointwise algebraically convergent

Combining methods used in Sections 6.1 and 6.2, we may easily obtain an
example of a sequence of homomorphisms in $\mathcal{H}(\mathbb{C},\mathbb{M})$
which is only pointwise metric convergent.

For each $n\in\omega$, let $\varphi_{n}\colon\mathbb{C}\to\mathbbm{2}$ be
defined as in Section 6.1 and let $\psi_{n}\colon\mathbb{C}\to\mathbb{M}$ be
defined as in Section 6.2. Define
$\rho_{n}\colon\mathbb{C}\to\mathbbm{2}\times\mathbb{M}$ by the formula
$\rho_{n}(A)=(\varphi_{n}(A),\psi_{n}(A))$ for every $A\in\mathbb{C}$.

By the same arguments as in Sections 6.1 and 6.2, the sequence $(\rho_{n})$ is
pointwise metric convergent to the homomorphism $\rho=(\varphi,\psi)$, yet,
again by the same arguments, it is not pointwise Borel metric convergent nor
pointwise algebraically convergent. Of course, $\mathbbm{2}\times\mathbb{M}$
can be embedded into $\mathbb{M}$, so $(\rho_{n})$ can be treated as a
sequence in $\mathcal{H}(\mathbb{C},\mathbb{M})$.

###  6.4. A pointwise Borel metric convergent sequence which is neither
uniformly convergent nor pointwise algebraically convergent

For every $n\in\omega$ define $flip_{n}\colon 2^{\omega}\to 2^{\omega}$ by
$flip_{n}(x)=x+\chi_{\\{n\\}}$, where $+$ is pointwise modulo $2$ (so, in
short, $flip_{n}$ flips the $n$-th coordinate of $x$). In the analogous way we
define $flip_{n}$ also on $2^{<\omega}\setminus 2^{<n}$ (of course,
$flip_{n}(s)$ makes sense only if $n<|s|$). Let
$\varphi_{n}\colon\mathbb{M}\to\mathbb{M}$ be the homomorphism induced by
$flip_{n}$, i.e. the homomorphism defined by $\varphi_{n}(B)=flip_{n}[B]$.
Then $(\varphi_{n})$ is convergent to the identity on $\mathbb{M}$ in the
pointwise metric topology. Indeed, first notice that $(\varphi_{n}(A))$
converges trivially to $A$ if $A$ is a clopen subset of $2^{\omega}$. Also,
for every $\varepsilon>0$ and every measurable set $B$ there is a clopen set
$A$ such that $d_{\lambda}(B,A)<\varepsilon$. Since $\varphi_{n}$ is an
isometry for each $n$, we get that $\varphi_{n}(B)$ converges with respect to
the pointwise metric topology for every Borel subset $B$ of $2^{\omega}$.

As $\mathbb{M}$ has the Grothendieck property, $(\varphi_{n})$ is
automatically pointwise Borel metric convergent. On the other hand,
$d_{\lambda}\big{(}\varphi_{n}(X_{n}),X_{n}\big{)}=1$ for each $n$, where
$X_{n}=\\{x\in 2^{\omega}\colon x(n)=0\\}$, so $(\varphi_{n})$ does not
converge uniformly.

The rest of this section will be devoted to the proof that $(\varphi_{n})$
does not converge pointwise algebraically. The crucial fact is the following.

######  Proposition 6.2.

There is a closed subset $B$ of $2^{\omega}$ such that $\lambda(B)>1/2$ and
for each $x\in B$ we have $flip_{n}(x)\notin B$ for infinitely many $n$’s.

Perhaps quite unexpectedly, to prove it we will use several notions from
coding theory. For $n\in\omega$ the function
$\rho_{n}\colon\\{0,1\\}^{n}\to\omega$ defined by

$\rho_{n}(x,y)=\big{|}\\{i\leq n\colon x(i)\neq y(i)\\}\big{|}$

is clearly a metric (called _the Hamming metric_). We will call a subset
$A\subseteq\\{0,1\\}^{n}$ _a code_ if for every $x\in\\{0,1\\}^{n}$ there is
$y\in A$ such that $\rho_{n}(x,y)\leq 1$. In coding theory this property is
one of the conditions defining so-called 1-error correcting codes.

In what follows we will use the fact that there is a perfect sphere packing in
$\\{0,1\\}^{n}$ for each $n$ of the form $n=2^{m}-1$, $m\in\omega$, i.e.
$\\{0,1\\}^{n}$ is a disjoint union of the closed balls of radius $1$ with
respect to the Hamming metric (see [Ham50]). The set of centers of those balls
is a code, which we will call _a perfect code_ (in the coding theory
terminology this is the same as a perfect 1-error-correcting Hamming code).
There is a rich literature concerning Hamming codes and it is quite easy to
learn _how_ to find perfect codes. It seems, however, that it is not so easy
to find a proof _why_ those algorithms work, so we sketch here the argument,
using the algorithm for creating perfect codes presented e.g. in [AK92,
Section 2.5].

######  Proposition 6.3.

If $n=2^{m}-1$, for $m\in\mathbb{N}$, then $\\{0,1\\}^{n}$ is a disjoint union
of the closed balls of radius $1$ with respect to the Hamming metric.

###### Proof.

Let $n=2^{m}-1$. We endow the set $\\{0,1\\}^{n}$ with the algebraic structure
and see it as $\mathbb{F}^{n}_{2}$ (i.e. $n$-dimensional vector space over
$\mathbb{F}_{2}$, the field consisting of two elements). Let $H$ be a matrix
created by putting all the non-zero elements of $\mathbb{F}^{m}_{2}$ as its
columns. Then $H$ is an $(n\times m)$-matrix. Define

$C=\big{\\{}v\in\mathbb{F}^{n}_{2}\colon Hv=0_{\mathbb{F}^{n}_{2}}\big{\\}}.$

In other words, $v\in C$ if and only if $\sum_{l<n}H(k,l)\cdot v(l)=0$ for
each $k<m$.

We claim that $C$ is a perfect code, i.e. the collection
$\mathcal{K}=\\{K_{1}(c)\colon c\in C\\}$, where $K_{r}(x)$ is a closed ball
of radius $r$ centered at $x$, forms a partition of $\mathbb{F}^{n}_{2}$.

First, notice that if $v\in C$ and $1\leq\rho_{n}(v,w)\leq 2$ for some
$w\in\mathbb{F}_{2}^{n}$, then $w\notin C$. Indeed, suppose first that there
is exactly one $i<n$ such that $v(i)\neq w(i)$. There is $k<m$ such that
$H(k,i)=1$ and then $\sum_{l<n}H(k,l)\cdot w(l)=1$ and so $w\notin C$. If
there are exactly two indices $i<j<n$ such that $v(i)\neq w(i)$ and $v(j)\neq
w(j)$, then for $k<m$ such that $H(k,i)\neq H(k,j)$ we again have
$\sum_{l<n}H(k,l)\cdot w(l)=1$, and $w\notin C$.

So, by the above, the family $\mathcal{K}$ is pairwise disjoint. Furthermore,
we may arrange columns of $H$ as we wish, so we may suppose that the last $m$
columns of $H$ consist of vectors with only one $1$. In this way it is easy to
see that every vector in $\mathbb{F}_{2}^{n-m}$ has exactly one extension to
the vector in $\mathbb{F}_{2}^{n}$ which belongs to $C$ and so $|C|=2^{n-m}$.
As every element of $\mathcal{K}$ consists of $n+1$ points and every point is
covered by at most one element of $\mathcal{K}$, we see that $\mathcal{K}$
covers $(n+1)\cdot 2^{n-m}=2^{n}$ points, i.e. the whole space
$\mathbb{F}_{2}^{n}$. ∎

Let $c_{n}$ be the smallest size of a code in $\\{0,1\\}^{n}$ and let
$a_{n}=c_{n}/2^{n}$ (so $a_{n}$ measures how big portion of whole
$\\{0,1\\}^{n}$ we have to take to obtain a code). The following fact is a
consequence of the existence of perfect codes.

######  Proposition 6.4.

$\lim_{n\to\infty}a_{n}=0$.

###### Proof.

First, notice that $(a_{n})$ is non-increasing. Indeed, if $A$ is a code in
$\\{0,1\\}^{n}$, then the family $A^{\prime}=\\{x^{\frown}0\colon x\in
A\\}\cup\\{x^{\frown}1\colon x\in A\\}$ is a code in $\\{0,1\\}^{n+1}$ and
$|A^{\prime}|=2|A|$.

To show that $(a_{n})$ can be arbitrarily close to $0$, fix $m\in\omega$ and
let $n=2^{m}-1$. By Proposition 6.3, there is a perfect code $C$ in
$\\{0,1\\}^{n}$, and from the proof we know that $|C|=2^{n-m}=2^{n}/(n+1)$ and
so $c_{n}\leq 2^{n}/(n+1)$ and hence $a_{n}\leq 1/(n+1)$. As $(a_{n})$ is non-
increasing, $\lim_{n\to\infty}a_{n}=0$. ∎

###### Proof of Proposition 6.2.

We are going to construct inductively an increasing sequence $(d_{n})$ of
natural numbers and a sequence $(C_{n})$ of sets such that for every
$n\in\omega$ the following conditions hold:

  1. (1)

$C_{n}\subseteq\\{0,1\\}^{d_{n}}$,

  2. (2)

$|C_{n}|<2^{d_{n}-1}$,

  3. (3)

for every $c\in C_{n}$ and $t\in\\{0,1\\}^{d_{n+1}-d_{n}}$ we have
$c^{\frown}t\in C_{n+1}$,

  4. (4)

For each $s\in\\{0,1\\}^{d_{n+1}}\setminus C_{n+1}$ there is $d_{n}\leq
m<d_{n+1}$ such that $flip_{m}(s)\in C_{n+1}$.

Let $d_{0}=3$ and $C_{0}$ be a perfect code in $\\{0,1\\}^{3}$ (e.g. $C_{0}$
may consist of two constant sequences). Suppose now that we have constructed
$d_{n}$ and $C_{n}$. Let

$\alpha=\frac{2^{d_{n}-1}-|C_{n}|}{2^{d_{n}}-|C_{n}|}.$

By (2) $\alpha>0$ and so, using Proposition 6.4, we may find $k$ so that
$a_{k}<\alpha$. Let $C$ be a code in $\\{0,1\\}^{k}$ such that
$|C|=a_{k}2^{k}$. Define $C_{n+1}$ in the following way:

$C_{n+1}=\big{\\{}t^{\frown}x\colon t\in
C_{n},x\in\\{0,1\\}^{k}\big{\\}}\cup\big{\\{}t^{\frown}c\colon
t\in\\{0,1\\}^{d_{n}}\setminus C_{n},c\in C\big{\\}}.$

Thus, $C_{n+1}$ consists of two parts: all the possible extensions of elements
of $C_{n}$ to $\\{0,1\\}^{d_{n}+k}$ and all the extensions of the rest of
$\\{0,1\\}^{d_{n}}$ by elements of $C$. Put $d_{n+1}=d_{n}+k$. We claim that
$C_{n+1}$ and $d_{n+1}$ satisfy all the desired properties. Indeed, (1) and
(3) are clear, and concerning (2) we have:

$|C_{n+1}|=2^{k}|C_{n}|+(2^{d_{n}}-|C_{n}|)\cdot a_{k}2^{k},$

so

$\frac{|C_{n+1}|}{2^{d_{n+1}}}=\frac{|C_{n+1}|}{2^{d_{n}+k}}=\frac{|C_{n}|}{2^{d_{n}}}+\frac{(2^{d_{n}}-|C_{n}|)\cdot
a_{k}}{2^{d_{n}}}<\\\
<\frac{|C_{n}|}{2^{d_{n}}}+\frac{(2^{d_{n}}-|C_{n}|)}{2^{d_{n}}}\cdot\frac{(2^{d_{n}-1}-|C_{n}|)}{(2^{d_{n}}-|C_{n}|)}=\frac{2^{d_{n}-1}}{2^{d_{n}}}=\frac{1}{2},$

hence $|C_{n+1}|<2^{d_{n+1}-1}$.

Finally, to check (4) let $s\in\\{0,1\\}^{d_{n+1}}\setminus C_{n+1}$ and
$t=s\restriction d_{n}$. If $t\in C_{n}$, then $flip_{m}(s)\in C_{n+1}$ for
every $d_{n}\leq m<d_{n+1}$. If $t\not\in C_{n}$, then, as $C$ is a code,
there is $c\in C$ such that $\rho_{d_{n+1}}(t^{\frown}c,s)\leq 1$, which means
that there is $d_{n}\leq m<d_{n+1}$ such that $flip_{m}(s)=t^{\frown}c$ and so
$flip_{m}(s)\in C_{n+1}$.

Let $B_{n}=2^{\omega}\setminus\bigcup_{c\in C_{n}}[c]$. By (3), the sequence
$(B_{n})$ is $\subseteq$-decreasing. Let $B=\bigcap_{n}B_{n}$. Because of (2)
we have that $\lambda(B_{n})>1/2$ for each $n$ and so $\lambda(B)\geq 1/2$.

Let $x\in 2^{\omega}$. If $x\notin B$, then $c=x\restriction d_{n}\in C_{n}$
for some $n$. Then $flip_{m}(x)\in[c]$ for every $m>d_{n}$ and so
$flip_{m}(x)\notin B$ for every $m>d_{n}$. Now, assume that $x\in B$, so
$x\restriction d_{n}\not\in C_{n}$ for any $n\in\omega$. Fix $N\in\omega$.
Find $n$ such that $d_{n}>N$ and let $s=x\restriction d_{n+1}$, so that
$s\in\\{0,1\\}^{d_{n}+1}\setminus C_{n+1}$. By (4), there is $m\geq d_{n}$
such that $flip_{m}(s)\in C_{n+1}$. This means that $flip_{m}(x)\notin
B_{n+1}$ and so $flip_{m}(x)\notin B$. Since $N$ was arbitrary, we are done. ∎

######  Corollary 6.5.

$(\varphi_{n})$ does not converge pointwise algebraically.

###### Proof.

Let $B$ be a set given by Proposition 6.2. For each $m$ we have

$\bigcap_{n>m}flip_{n}[B]=\emptyset.$

Indeed, if $x\in\bigcap_{n>m}flip_{n}[B]$, then for every $n>m$ there is
$y_{n}\in B$ such that $x=flip_{n}(y_{n})$. But this means that
$flip_{n}(x)=y_{n}\in B$ for each $n>m$, which is a contradiction with the
property of $B$ promised by Proposition 6.2.

It follows that $\bigvee_{m}\bigwedge_{n>m}\varphi_{n}(B)=0$, but $B\neq 0$
and so $(\varphi_{n})$ does not converge pointwise algebraically to the
identity $\psi$. ∎

######  Remark 6.6.

Notice that in fact in Proposition 6.2 we can ask for $B$ of arbitrarily large
measure (smaller than $1$)—it is enough to adjust the definition of $\alpha$
in the proof.

######  Remark 6.7.

Using the Vitali equivalence relation, it is easy to construct a _non-
measurable_ set $A$ such that $\lambda^{*}(A)=1$ with even stronger property:
for each $x\in A$ and every $n\in\omega$ we have $flip_{n}(x)\notin A$. Using
this set is probably the easiest method of showing the existence of a infinite
game with perfect information which is not determined (see e.g. [KN14]). It
seems however that to obtain a Borel set, we have to be much more careful and
we do not know any way to do this without using Hamming codes.

###  6.5. A sequence which is pointwise Borel metric convergent and pointwise
algebraically convergent but not uniformly convergent

The above example may be adapted to obtain a sequence which is pointwise Borel
metric convergent and pointwise algebraically convergent but not uniformly
convergent.

Let $(\varphi_{n})$ be the sequence defined in Section 6.4. Define
$\psi_{n}\colon\mathbb{C}\to\mathbb{M}$ by
$\psi_{n}=\varphi_{n}\restriction\mathbb{C}$. Here, we treat $\mathbb{C}$ as a
subalgebra of $\mathbb{M}$, identifying clopen subsets of $2^{\omega}$ with
its equivalence relations (i.e. the elements of $\mathbb{M}$). By the same
argument which showed that $(\varphi_{n})$ was pointwise metric convergent to
the identity on $\mathbb{M}$, $(\psi_{n})$ is pointwise Borel metric
convergent to the identity on $\mathbb{C}$. Since the sets $X_{n}$’s from
Section 6.4 are clopen in $2^{\omega}$, $(\psi_{n})$ is not uniformly
convergent. It is however pointwise algebraically convergent, since
$flip_{n}[C]$ is eventually constant for $C$ being a clopen subset of
$2^{\omega}$.

In Section 8 we prove however that if a Boolean algebra $\mathbb{A}$ is
somewhat more complicated than $\mathbb{C}$, e.g. $\mathbb{A}$ is
$\sigma$-complete, then there is no similar example in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$.

##  7\. Homomorphisms and forcing names for ultrafilters

Fix, once and for all, an infinite cardinal $\kappa$ and suppose that we force
with the notion $\mathbb{M}_{\kappa}$ over a model $V$ of set theory. If
$\mathbb{A}$ is a Boolean algebra in $V$ and $G$ is an
$\mathbb{M}_{\kappa}$-generic filter over $V$, then we can consider
$\mathbb{A}$ as an element of $V[G]$, however, there might be new ultrafilters
on $\mathbb{A}$ in $V[G]$. It appears nevertheless that to each
$\mathbb{M}_{\kappa}$-name $\dot{\mathcal{U}}$ for an ultrafilter, i.e. to
such $\dot{\mathcal{U}}$ that
$\Vdash_{\mathbb{M}_{\kappa}}``\dot{\mathcal{U}}$ is an ultrafilter on
$\mathbb{A}$”, we can assign a homomorphism from $\mathbb{A}$ to
$\mathbb{M}_{\kappa}$ in $V$ in a quite natural way. Namely, if
$\dot{\mathcal{U}}$ is such an $\mathbb{M}_{\kappa}$-name, then the function
$\varphi_{\dot{\mathcal{U}}}\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ defined fo
every $A\in\mathbb{A}$ (recall that $\mathbb{A}\in V$) by the formula

$\varphi_{\dot{\mathcal{U}}}(A)=\llbracket A\in\dot{\mathcal{U}}\rrbracket$

is a Boolean homomorphism belonging to $V$. (We skip writting the symbol
$\check{}$ over canonical names for elements of $V$.)

Now fix a homomorphism $\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ in $V$
and define an $\mathbb{M}_{\kappa}$-name $\tau_{\varphi}$ for a family of
subsets $\mathbb{A}$ by

$\tau_{\varphi}=\big{\\{}\langle A,\varphi(A)\rangle\colon
A\in\mathbb{A}\big{\\}}.$

Then, $\Vdash_{\mathbb{M}_{\kappa}}$“$\tau_{\varphi}$ is an ultrafilter on
$\mathbb{A}$”; in fact, we actually get that
$\Vdash_{\mathbb{M}_{\kappa}}\tau_{\varphi}=\varphi^{-1}[\Gamma]$, where
$\Gamma$ is the canonical $\mathbb{M}_{\kappa}$-name for an
$\mathbb{M}_{\kappa}$-generic filter over $V$.

######  Lemma 7.1.

For every $\mathbb{M}_{\kappa}$-name $\dot{\mathcal{U}}$ for an ultrafilter on
a Boolean algebra $\mathbb{A}$ and for every homomorphism
$\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ it holds
$\varphi_{(\tau_{\varphi})}=\varphi$ and
$\Vdash_{\mathbb{M}_{\kappa}}\tau_{(\varphi_{\dot{\mathcal{U}}})}=\dot{\mathcal{U}}$.

In this way we obtain a bijective correspondence between elements of
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ and ultrafilters on $\mathbb{A}$
in the forcing extension by $\mathbb{M}_{\kappa}$. One of the advantages of
such an approach is that we may use properties of measures associated to
homomorphisms (as in Section 4) to study $\mathbb{M}_{\kappa}$-names for
ultrafilters on _old_ Boolean algebras (in particular, to study ultrafilters
on the Cantor algebra, i.e. reals).

######  Example 7.2.

Suppose that $\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ is such that the
measure $\lambda_{\kappa}\circ\varphi$ is a point mass measure
$\delta_{\mathcal{V}}$ for some $\mathcal{V}\in St(\mathbb{A})\cap V$. Then,
$\Vdash_{\mathbb{M}_{\kappa}}\tau_{\varphi}=\mathcal{V}$.

A study of the interplay between the properties of $\mathbb{M}_{\kappa}$-names
for ultrafilters and measures is a subject of forthcoming paper [BNC].

In what follows $(\varphi_{n})$ and $\varphi$ will denote (a sequence of)
elements of $\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ associated—as
described above—to (a sequence of) $\mathbb{M}_{\kappa}$-names
$(\dot{\mathcal{U}}_{n})$ and $\dot{\mathcal{U}}$ for ultrafilters on a fixed
Boolean algebra $\mathbb{A}\in V$, respectively. Note that by Lemma 7.1, it
has no meaning whether we associate homomorphisms in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ to $\mathbb{M}_{\kappa}$-names,
or $\mathbb{M}_{\kappa}$-names to homomorphisms in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$.

Recall here that for every sequence $(\phi_{n})$ of formulae the following
equalities hold:

$\llbracket\exists^{\infty}n\colon\ \phi_{n}\rrbracket=\llbracket\forall
n\exists m>n\
\phi_{n}\rrbracket=\bigwedge_{n}\bigvee_{m>n}\llbracket\phi_{n}\rrbracket$

and

$\llbracket\forall^{\infty}n\colon\ \phi_{n}\rrbracket=\llbracket\exists
n\forall m>n\
\phi_{n}\rrbracket=\bigvee_{n}\bigwedge_{m>n}\llbracket\phi_{n}\rrbracket;$

we will use them frequently.

###  7.1. Pointwise algebraic convergence

There is a strong connection between the pointwise algebraic convergence of
sequences of homomorphisms from a Boolean algebra $\mathbb{A}$ into
$\mathbb{M}_{\kappa}$ in the ground model $V$ and the convergence of the
sequences of corresponding ultrafilters in the Stone space $St(\mathbb{A})$ in
the $\mathbb{M}_{\kappa}$-generic extension $V[G]$.

######  Proposition 7.3.

The following conditions are equivalent:

  * •

$\Vdash_{\mathbb{M}_{\kappa}}``(\dot{\mathcal{U}}_{n})$ converges to
$\dot{\mathcal{U}}$”,

  * •

$(\varphi_{n})$ converges to $\varphi$ pointwise algebraically.

###### Proof.

Suppose that it is not true that
$\Vdash_{\mathbb{M}_{\kappa}}``(\dot{\mathcal{U}}_{n})$ converges to
$\dot{\mathcal{U}}$”. It means that there is a non-zero
$p\in\mathbb{M}_{\kappa}$ which forces the opposite, i.e. that there is
$A\in\mathbb{A}$ such that $p\Vdash\exists^{\infty}n\
A\in\dot{\mathcal{U}}\setminus\dot{\mathcal{U}}_{n}$. It follows that

$p\leq\bigwedge_{m}\bigvee_{n>m}\llbracket
A\in\dot{\mathcal{U}}\setminus\dot{\mathcal{U}}_{n}\rrbracket,$

and hence

$\bigwedge_{m}\bigvee_{n>m}\big{(}\varphi(A)\setminus\varphi_{n}(A)\big{)}\neq
0.$

Thus, $(\varphi_{n}(A))$ does not converge to $\varphi(A)$ and hence
$(\varphi_{n})$ does not converge pointwise algebraically to $\varphi$.

If $(\varphi_{n}(A))$ does not converge algebraically to $\varphi(A)$ for some
$A\in\mathbb{A}$, then

$p=\varphi(A)\setminus\big{(}\bigwedge_{m}\bigvee_{n>m}\varphi_{n}(A)\big{)}\neq
0\quad\text{or}\quad
q=\big{(}\bigwedge_{m}\bigvee_{n>m}\varphi_{n}(A)\big{)}\setminus\varphi(A)\neq
0.$

Assume first that the former case holds, i.e. $p\neq 0$. We have:

$p=\varphi(A)\wedge\big{(}\bigvee_{m}\bigwedge_{n>m}\varphi_{n}(A^{c})\big{)},$

so $p\Vdash A\in\dot{\mathcal{U}}$ and $p\Vdash\exists^{\infty}n\
A^{c}\in\dot{\mathcal{U}}_{n}$. It follows that
$p\Vdash``(\dot{\mathcal{U}}_{n})$ does not converge to $\dot{\mathcal{U}}$”.

We proceed similarly in the latter case, i.e. when $q\neq\emptyset$.

∎

###  7.2. Uniform convergence

In this section we will show that the notion of uniform convergence of
homomorphisms is connected to trivial convergence of ultrafilters in the
random extension.

######  Proposition 7.4.

If $\Vdash_{\mathbb{M}_{\kappa}}\forall^{\infty}n\in\omega\
\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$, then $(\varphi_{n})$ converges
uniformly to $\varphi$.

###### Proof.

Assume that $(\varphi_{n})$ does not converge uniformly to $\varphi$. Then,
without loss of generality, there is $\varepsilon>0$ such that for each
$n\in\omega$ there is $A_{n}\in\mathbb{A}$ such that
$\lambda\big{(}\varphi(A_{n})\setminus\varphi_{n}(A_{n})\big{)}>\varepsilon$.
For every $n\in\omega$ put $p_{n}=\varphi(A_{n})\setminus\varphi_{n}(A_{n})$.
Since $\lambda\big{(}\bigvee_{n>m}p_{n}\big{)}>\varepsilon$ for every
$m\in\omega$, for $p=\bigwedge_{m}\bigvee_{n>m}p_{n}$ we have
$\lambda(p)\geq\varepsilon$ and thus $p\neq 0$.

Let $G$ be an $\mathbb{M}_{\kappa}$-generic filter over $V$ containing $p$. We
work in $V$. There is $q\in G$ below $p$ such that
$q\Vdash_{\mathbb{M}_{\kappa}}\forall^{\infty}n\in\omega\
\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$, and hence there is $r\in G$ below
$q$ and $m\in\omega$ such that for every $n>m$ we have
$r\Vdash\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$. It follows immediately that
$p_{n}\perp r$ for every $n>m$ and thus $p\perp r$, a contradiction, since
$r\leq p$. ∎

It appears that a fact which is in a sense converse to Proposition 7.4 also
holds and constitutes actually the main result of this section (Theorem 7.5).
Note here that in the proof of Proposition 7.4, as well as of Proposition 7.3,
we actually did not use that the Boolean algebra $\mathbb{M}_{\kappa}$ carries
a strictly positive measure and thus those results remain in fact true for any
$\sigma$-complete Boolean algebra $\mathbb{B}$ in place of
$\mathbb{M}_{\kappa}$ (and hence, of course, for any reasonable notion of
forcing).

######  Theorem 7.5.

If $(\varphi_{n})$ converges uniformly to $\varphi$, then for almost all $n$
there is $p_{n}\in\mathbb{M}_{\kappa}$ such that
$p_{n}\Vdash\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$.

We will prove this theorem in a series of lemmas and propositions. One of
them, Theorem 7.6, is interesting on its own sake.

######  Theorem 7.6.

Let $\mathbb{A}$ be an atomless Boolean algebra in $V$. If $\dot{\mathcal{U}}$
and $\dot{\mathcal{V}}$ are $\mathbb{M}_{\kappa}$-names for ultrafilters on
$\mathbb{A}$ such that
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}\neq\dot{\mathcal{V}}$, then for
every $\varepsilon>0$ there is $p\in\mathbb{M}_{\kappa}$ and $C\in\mathbb{A}$
such that $\lambda(p)>1/4-\varepsilon$ and $p\Vdash
C\in\dot{\mathcal{U}}\triangle\dot{\mathcal{V}}$.

Theorem 7.6 requires a brief comment. Assume that $\mathbb{A}$ is a Boolean
algebra and $\dot{\mathcal{U}}$ and $\dot{\mathcal{V}}$ are
$\mathbb{M}_{\kappa}$-names for ultrafilters on $\mathbb{A}$ such that
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}\neq\dot{\mathcal{V}}$. Since
$\mathbb{M}_{\kappa}$ is a ccc forcing, there is a maximal antichain $(p_{n})$
in $\mathbb{M}_{\kappa}$ and a sequence $(A_{n})$ of elements of $\mathbb{A}$
such that $p_{n}\Vdash A_{n}\in\dot{\mathcal{U}}\setminus\dot{\mathcal{V}}$.
The problem with this general statement is that _a priori_ we do not have any
control on the measures of $p_{n}$’s—it might as lief happen that all of them
are very small in terms of the measure. Theorem 7.6 shows however that we can
always find one “large” condition in $\mathbb{M}_{\kappa}$ which distinguishes
the ultrafilters.

The proof of Theorem 7.6 requires some auxiliary lemmas.

######  Proposition 7.7.

Assume that $(X,\Sigma,\mu)$ is a measure space with a non-negative (not
necessarily probability) measure $\mu$, $K$ is a compact space, and $f,g\colon
X\to K$ are simple measurable functions. If $f\neq g$ $\mu$-almost everywhere,
then there exists $L\in\Sigma$ such that $\mu(L)\geq\mu(X)/4$ and $f[L]\cap
g[L]=\emptyset$.

###### Proof.

Let $\mathcal{P}$ be a finite partition of $X$ into measurable pieces such
that for every $A\in\mathcal{P}$ the functions $f$ and $g$ are constant on $A$
and $\mu(A)>0$. Define the functions
$f_{\mathcal{P}},g_{\mathcal{P}}\colon\mathcal{P}\to K$ as follows:

$f_{\mathcal{P}}(A)=x\quad\text{if and only if}\quad f[A]=\\{x\\},$

and similarly

$g_{\mathcal{P}}(A)=x\quad\text{if and only if}\quad g[A]=\\{x\\}.$

For every $A\in\mathcal{P}$ we have $f_{\mathcal{P}}(A)\neq
g_{\mathcal{P}}(A)$ (since $\mu(A)>0$). Put $H=f[X]\cap g[X]$. Define an
auxiliary measure $\nu$ on $\mathcal{P}(H\times H)$ as follows:

$\nu\big{(}\\{(x,y)\\}\big{)}=\sum\big{\\{}\mu(A)\colon\
f_{\mathcal{P}}(A)=x,\ g_{\mathcal{P}}(A)=y,A\in\mathcal{P}\big{\\}}$

for every $(x,y)\in H\times H$. It follows that
$\nu\big{(}\\{(x,x)\\}\big{)}=0$ for every $x\in H$.

For every $N\subseteq H$ let

$N_{0}=(N\times N)\cup(N^{c}\times N^{c})\quad\text{and}\quad N_{1}=(N\times
N^{c})\cup(N^{c}\times N).$

It follows that for every $(x,y)\in H\times H$ with $x\neq y$ we have:

$\big{|}\\{N\subseteq H\colon\ (x,y)\in N_{0}\\}\big{|}=\big{|}\\{N\subseteq
H\colon\ (x,y)\in N_{1}\\}\big{|},$

and so

$\sum_{N\in\mathcal{P}(H)}\nu(N_{0})=\sum_{N\in\mathcal{P}(H)}\nu(N_{1}).$

There exists $N\subseteq H$ such that $\nu(N_{1})\geq\nu(N_{0})$. Since
$N_{0}\cup N_{1}=H\times H$, $\nu(N_{1})\geq\nu(H\times H)/2$ and hence either
$\nu(N\times N^{c})\geq\nu(H\times H)/4$ or $\nu(N^{c}\times N)\geq\nu(H\times
H)/4$. Without loss of generality we may assume that $\nu(N\times
N^{c})\geq\nu(H\times H)/4$ as this assumption makes no difference for the
following arguments.

For the set $L^{\prime}\subseteq X$ defined as follows:

$L^{\prime}=\bigcup\big{\\{}A\in\mathcal{P}\colon\ f_{\mathcal{P}}(A)\in N,\
g_{\mathcal{P}}(A)\in N^{c}\big{\\}},$

we have:

$\mu(L^{\prime})=\sum\big{\\{}\mu(A)\colon\ f_{\mathcal{P}}(A)\in N,\
g_{\mathcal{P}}(A)\in N^{c},\ A\in\mathcal{P}\big{\\}}=\nu(N\times N^{c})\geq$
$\geq\nu(H\times H)/4=\frac{1}{4}\sum\big{\\{}\mu(A)\colon
f_{\mathcal{P}}(A)\in H,\ g_{\mathcal{P}}(A)\in H,\ A\in\mathcal{P}\big{\\}}.$

Note that for every $x\in L^{\prime}$ we have $f(x)\in N$ and $g(x)\in N^{c}$,
so neither $f(x)\in g[L^{\prime}]$, nor $g(x)\in f[L^{\prime}]$. Finally, for
the set $L\subseteq X$ given by the formula:

$L=L^{\prime}\cup\bigcup\big{\\{}A\in\mathcal{P}\colon\
f_{\mathcal{P}}(A)\not\in H\text{ or }g_{\mathcal{P}}(A)\not\in H\big{\\}},$

we also have that $f[L]\cap g[L]=\emptyset$ as well as it holds:

$\mu(L)=\mu(L^{\prime})+\sum\big{\\{}\mu(A)\colon\ f_{\mathcal{P}}(A)\not\in
H\text{ or }g_{\mathcal{P}}(A)\not\in H\big{\\}}\geq$
$\geq\frac{1}{4}\sum\big{\\{}\mu(A)\colon f_{\mathcal{P}}(A)\in H,\
g_{\mathcal{P}}(A)\in H\big{\\}}+\frac{1}{4}\sum\big{\\{}\mu(A)\colon\
f_{\mathcal{P}}(A)\not\in H\text{ or }g_{\mathcal{P}}(A)\not\in H\big{\\}}=$
$=\frac{1}{4}\sum\big{\\{}\mu(A)\colon\ A\in\mathcal{P}\big{\\}}=\mu(X)/4.$

∎

Note that in the above proposition we may in fact assume that $K$ is finite.

######  Lemma 7.8.

Assume that $(X,\Sigma,\mu)$ is a probability space, $K$ is a closed subset of
$2^{\omega}$, and $f,g\colon X\to K$ are measurable and $f\neq g$ $\mu$-almost
everywhere. Then, for every $\varepsilon>0$ there is $L\in\Sigma$ such that
$\mu(L)>1/4-\varepsilon$ and $f[L]\cap g[L]=\emptyset$.

###### Proof.

For every $s\in 2^{<\omega}$ let $s_{0}\in 2^{\omega}$ be the leftmost branch
with root $s$ (think of $s_{0}$ as of a fixed representative of the clopen
$\\{x\in 2^{\omega}\colon s\subseteq x\\}$). For $n\in\omega$ define the
simple measurable functions $f_{n},g_{n}\colon X\to K$ by putting
$f_{n}(x)=s_{0}$ and $g_{n}(x)=t_{0}$, where $s=f(x)\restriction n$ and
$t=g(x)\restriction n$.

For every $n\in\omega$ put $A_{n}=\big{\\{}x\in X\colon
f_{n}(x)=g_{n}(x)\big{\\}}$ and note that $A_{n+1}\subseteq A_{n}$. It follows
that $\mu\big{(}A_{n_{0}}\big{)}<\varepsilon$ for some $n_{0}\in\omega$.
Indeed, if $\mu(A_{n})\geq\varepsilon$ for every $n\in\omega$, then
$\mu(\bigcap_{n}A_{n})\geq\varepsilon$, and since $f(x)=g(x)$ for every
$x\in\bigcap_{n}A_{n}$, we get a contradiction with the assumption that $f\neq
g$ $\mu$-almost everywhere.

Let $X_{0}=X\setminus A_{n_{0}}$, so $\mu(X_{0})>1-\varepsilon$ and for every
$x\in X_{0}$ we have $f_{n}(x)\neq g_{n}(x)$. By Lemma 7.7 there is a
measurable subset $L\subseteq X_{0}$ such that $\mu(L)\geq
1/4\mu(X_{0})>1/4-\varepsilon$ and $f_{n}[L]\cap g_{n}[L]=\emptyset$. It
follows that for every $x,x^{\prime}\in L$ we have $f_{n}(x)\neq
g_{n}(x^{\prime})$ and hence $f(x)\neq g(x^{\prime})$, so $L$ is as desired. ∎

######  Proposition 7.9.

Let $\mathbb{A}$ be a countable Boolean algebra. Assume that
$\varphi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ and
$\psi\colon\mathbb{A}\to\mathbb{M}_{\kappa}$ are such homomorphisms that
$\bigvee\big{\\{}\varphi(A)\triangle\psi(A)\colon A\in\mathbb{A}\big{\\}}=1$.
Then, for each $\varepsilon>0$ there is $C\in\mathbb{A}$ such that
$\lambda_{\kappa}\big{(}\varphi(C)\triangle\psi(C)\big{)}>1/4-\varepsilon$.

###### Proof.

Let $X$ and $Y$ be the Stone spaces of $\mathbb{M}_{\kappa}$ and $\mathbb{A}$,
respectively. In what follows, we will treat $\lambda_{\kappa}$ as a Radon
measure on $X$. The functions $f_{\varphi},f_{\psi}\colon X\to Y$ are
$\lambda_{\kappa}$-measurable. We claim that $f_{\varphi}\neq f_{\psi}$
$\lambda_{\kappa}$-almost everywhere. If not, then there is a Borel subset
$B\subseteq X$ of positive measure such that $f_{\varphi}\restriction
B=f_{\psi}\restriction B$. By the regularity of $\lambda_{\kappa}$, there is a
compact subset $K$ of $X$ such that $K\subseteq B$ and
$\lambda_{\kappa}(K)>0$. Recall that $\lambda_{\kappa}$ is a normal measure,
i.e. for every nowhere dense subset $Z$ of $X$ we have
$\lambda_{\kappa}(Z)=0$, so $K$ cannot be nowhere dense and thus there is
$U\in\mathbb{M}_{\kappa}$ such that $U\subseteq K$ and
$\lambda_{\kappa}(U)>0$. But for every $A\in\mathbb{A}$ we have
$(\varphi(A)\setminus\psi(A))\cap U=0$ (since otherwise for each
$x\in(\varphi(A)\setminus\psi(A))\cap U$ we would have $f_{\varphi}(x)\in A$
and $f_{\psi}(x)\notin A$ and so $f_{\varphi}(x)\neq f_{\psi}(x)$ despite the
fact that $x\in B$) and $(\psi(A)\setminus\varphi(A))\cap U=0$ (by a similar
argument). This is a contradiction with the assumption that
$\bigvee\big{\\{}\varphi(A)\triangle\psi(A)\colon A\in\mathbb{A}\big{\\}}=1$.

Now fix $\varepsilon>0$ and use Lemma 7.8 to find a measurable subset
$L\subseteq X$ such that $\lambda_{\kappa}(L)>1/4-\varepsilon$ and
$f_{\varphi}[L]\cap f_{\psi}[L]=\emptyset$ (notice that $Y$ may be treated as
a closed subset of $2^{\omega}$). Using the regularity of $\lambda_{\kappa}$,
we may find a compact subset $M$ of $X$ such that $M\subseteq L$ and
$\lambda_{\kappa}(M)>1/4-\varepsilon$. By the normality of $\lambda_{\kappa}$,
the boundary of $M$ has measure $0$, so there is $V\in\mathbb{M}_{\kappa}$
such that $V\subseteq M$ and $\lambda_{\kappa}(V)>1/4-\varepsilon$. Let $C$ be
a clopen in $Y$ separating $f_{\varphi}[V]$ and $f_{\psi}[V]$. Then,
$V\subseteq\varphi(C)$ but $V\cap\psi(C)=\emptyset$, and hence
$\lambda(\varphi(C)\setminus\psi(C))>1/4-\varepsilon$. ∎

###### Proof of Theorem 7.6.

Since $\mathbb{M}_{\kappa}$ is a ccc poset, there are a sequence $(A_{n})$ in
$\mathbb{A}$ and a maximal antichain $(p_{n})$ in $\mathbb{M}_{\kappa}$ such
that
$p_{n}\leq\varphi_{\dot{\mathcal{U}}}(A_{n})\triangle\varphi_{\dot{\mathcal{V}}}(A_{n})$
for every $n\in\omega$. It follows that

$\bigvee\big{\\{}\varphi_{\dot{\mathcal{U}}}(A_{n})\triangle\varphi_{\dot{\mathcal{V}}}(A_{n})\colon\
n\in\omega\big{\\}}=1.$

Let $\mathbb{B}$ be the subalgebra of $\mathbb{A}$ generated by the set
$\\{A_{n}\colon\ n\in\omega\\}$. By Proposition 7.9, there is $C$ in
$\mathbb{B}$ (and hence in $\mathbb{A}$) such that
$\lambda_{\kappa}\big{(}\varphi_{\dot{\mathcal{U}}}(C)\triangle\varphi_{\dot{\mathcal{V}}}(C)\big{)}>1/4-\varepsilon$.
Put $p=\varphi_{\dot{\mathcal{U}}}(C)\triangle\varphi_{\dot{\mathcal{V}}}(C)$
to finish the proof. ∎

Finally, we are in the position to prove the main theorem of this section.

###### Proof of Theorem 7.5.

Since $(\varphi_{n})$ converges uniformly to $\varphi$, there is $m\in\omega$
such that for each $n>m$ and $A\in\mathbb{A}$ we have
$\lambda_{\kappa}\big{(}\varphi_{n}(A)\triangle\varphi(A)\big{)}<1/5$. We get
that there is $m\in\omega$ such that for each $n>m$ there are no
$p\in\mathbb{M}_{\kappa}$ and no $A\in\mathbb{A}$ such that
$p\leq\varphi_{n}(A)\triangle\varphi(A)$ and $\lambda_{\kappa}(p)\geq 1/5$,
which, by Theorem 7.6, means that for every $n>m$ we have a non-zero
$p_{n}\in\mathbb{M}_{\kappa}$ such that
$p_{n}\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$.

∎

Surprisingly, in general we cannot hope to find anything better than the
constant $1/4$ in Proposition 7.7 and in Theorem 7.6.

######  Lemma 7.10.

For each non-zero even $n\in\omega$ put $X_{n}=\\{1,2,\ldots,n\cdot(n-1)\\}$
and let $\mu_{n}$ be such that $\mu_{n}(\\{x\\})=1/|X_{n}|$ for each $x\in
X_{n}$. Let $f_{n},g_{n}\colon X_{n}\to\omega$ be such functions that for
every distinct $a,b\in\\{1,\ldots,n\\}$ there is $x\in X_{n}$ for which we
have $f_{n}(x)=a$ and $g_{n}(x)=b$. Then, for each subset $L$ of $X_{n}$ such
that $f_{n}[L]\cap g_{n}[L]=\emptyset$ we have:

$\mu_{n}(L)\leq\frac{1}{4}+\frac{1}{16n-4}.$

###### Proof.

Let $L\subseteq X_{n}$ be a set such that $f_{n}[L]\cap g_{n}[L]=\emptyset$.
If $|f_{n}[L]|=k$ for some $1\leq k\leq n$, then $|g_{n}[L]|\leq n-k$. Since
each pair $(a,b)\in\\{1,\ldots,n\\}^{2}$, $a\neq b$, occurs exactly once in
the set $f_{n}[X_{n}]\times g_{n}[X_{n}]$, $L$ may have cardinality at most
$k\cdot(n-k)$. The only maximum of the function
$\rho_{n}\colon\mathbb{R}\to\mathbb{R}$ given by the formula
$\rho_{n}(x)=x\cdot(n-x)$ occurs at $x=n/2$ and is equal to $n^{2}/4$. It
follows that the maximal possible cardinality of $L$ is also $n^{2}/4$ and
thus:

$\mu_{n}(L)\leq\frac{n^{2}}{4}\cdot\frac{1}{n(n-1)}=\frac{1}{4-\frac{1}{n}}=\frac{1}{4}+\frac{1}{16n-4}.$

∎

######  Corollary 7.11.

Let $\mathbb{A}$ be an infinite Boolean algebra. Then, for every
$\varepsilon>0$ there are $\mathbb{M}_{\kappa}$-names $\dot{\mathcal{U}}$ and
$\dot{\mathcal{V}}$ for ultrafilters on $\mathbb{A}$ having the following
properties:

  * •

$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}\neq\dot{\mathcal{V}}$, and

  * •

for every $p\in\mathbb{M}_{\kappa}$ for which there is $A\in\mathbb{A}$ such
that $p\Vdash A\in\dot{\mathcal{U}}\setminus\dot{\mathcal{V}}$ we have
$\lambda_{\kappa}(p)\leq 1/4+\varepsilon$.

###### Proof.

Let $\varepsilon>0$ and fix $n$ such that $\frac{1}{16n-4}<\varepsilon$. Let
$(\mathcal{U}_{i})_{i<n}$ be sequence of distinct ultrafilters on $\mathbb{A}$
(in $V$).

Let $X_{n}$, $f_{n}$, $g_{n}$ be as in Lemma 7.10. Fix a maximal antichain
$\\{P_{k}\colon k\in X_{n}\\}$ in $\mathbb{M}_{\kappa}$ consisting of sets of
measure $1/|X_{n}|$. For $i<n$ denote $F_{i}=\bigvee\\{P_{k}\colon f(k)=i\\}$
and $G_{i}=\bigvee\\{P_{k}\colon g(k)=i\\}$. Notice that both $\\{F_{i}\colon
i<n\\}$ and $\\{G_{i}\colon i<n\\}$ form maximal antichains in
$\mathbb{M}_{\kappa}$ consisting of sets of measure $1/n$.

Now, let $\dot{\mathcal{U}}$ be an $\mathbb{M}_{\kappa}$-name for an
ultrafilter on $\mathbb{A}$ such that
$\llbracket\dot{\mathcal{U}}=\mathcal{U}_{i}\rrbracket=F_{i}$ for every $i<n$.
In other words, $\mathbb{M}_{\kappa}$ forces that $\dot{\mathcal{U}}$ is one
of the $\mathcal{U}_{i}$’s and the elements of the antichain $\\{F_{i}\colon
i<n\\}$ decides which of them. Similarly, choose an $\mathbb{M}_{\kappa}$-name
$\dot{\mathcal{V}}$ in such a way that
$\llbracket\dot{\mathcal{V}}=\mathcal{U}_{i}\rrbracket=G_{i}$. Then, for every
$k$

$P_{k}\Vdash\dot{\mathcal{U}}=\mathcal{U}_{f(k)}\mbox{ and
}\dot{\mathcal{V}}=\mathcal{U}_{g(k)},$

and since $f(k)\neq g(k)$ we have
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mathcal{U}}\neq\dot{\mathcal{V}}$.

Let $p\in\mathbb{M}_{\kappa}$ be such that $\lambda(p)\geq 1/4+\varepsilon$
and let $L=\big{\\{}k\in X_{n}\colon P_{k}\wedge p\neq 0\big{\\}}$. Then,
$|L|\geq|X_{n}|(\frac{1}{4}+\varepsilon)$ and so, in terms of Lemma 7.10,

$\mu_{n}(L)\geq\frac{1}{4}+\varepsilon>\frac{1}{4}+\frac{1}{16n-4}.$

Thus, by Lemma 7.10 we have $f[L]\cap g[L]\neq\emptyset$ and so there are
$k,l\in L$ such that $f(k)=g(l)=i$ for some $i<n$. Then $P_{k}\land p$ and
$P_{l}\land p$ are non-zero and

$P_{k}\land p\leq F_{i}\Vdash\dot{\mathcal{U}}=\mathcal{U}_{i}$

but also

$P_{l}\land p\leq G_{i}\Vdash\dot{\mathcal{V}}=\mathcal{U}_{i}.$

Hence, there is no $A\in\mathbb{A}$ such that $p\Vdash
A\in\dot{\mathcal{U}}\setminus\dot{\mathcal{V}}$. ∎

######  Remark 7.12.

The core argument of Theorem 7.6 is contained in Proposition 7.7, which is a
statement in finite combinatorics. It can be formulated in a popular way as
follows. Adam and Eve are painting a picket fence between their properties.
The fence consists of $n$ many rails. Each rail has two sides—Adam’s and
Eve’s—and those sides have to be painted in such a way that Adam’s side has
different colour than Eve’s. Proposition 7.7 says that no matter how many
colours they use, there is always a set $B$ of at least $n/4$ many rails with
the following property: the set of Adam’s colours used in $B$ is disjoint with
the set of Eve’s colours used in $B$. (The translation of the above into the
terms of Proposition 7.7 is as follows: $X$ is the set of rails, $K$ is the
set of colors, $\mu$ is the counting measure, and the functions $f$ and $g$
assign colors to Adam’s and Eve’s sides of the fence.)

##  8\. The Efimov problem

We now use results from the previous section to obtain a characterization of
those Boolean algebras whose Stone spaces do not contain any non-trivial
convergent sequences in random extensions.

Recall that a Boolean algebra $\mathbb{A}$ has Seever’s interpolation property
(or, property (I)) if for every sequences $(A_{n})$ and $(B_{n})$ in
$\mathbb{A}$ such that $A_{n}\leq B_{m}$ for every $n,m\in\omega$ there is
$C\in\mathbb{A}$ such that $A_{n}\leq C\leq B_{m}$ for every $n,m\in\omega$
(see [See68]). It is immediate that a Boolean algebra $\mathbb{A}$ has the
interpolation property if and only if its Stone space $St(\mathbb{A})$ is an
F-space, so e.g. $\mathcal{P}(\omega)/Fin$ has the interpolation property as
well as all $\sigma$-complete Boolean algebras do. The result of Dow and
Fremlin mentioned in Introduction yields thus the following corollary.

######  Theorem 8.1.

[DF07, Corollary 2.3] Let $\mathbb{A}$ be a Boolean algebra in $V$. If
$\mathbb{A}$ has the interpolation property in $V$, then in
$V^{\mathbb{M}_{\kappa}}$ the space $\mathrm{St}(\mathbb{A})$ does not have
non-trivial convergent sequences.

Theorem 8.1 together with results obtained above allows us to bring down the
question about convergence of ultrafilters in the random extensions to the
question about convergence of homomorphisms. Note that if the Stone space
$St(\mathbb{A})$ of a Boolean algebra $\mathbb{A}$ contains a non-trivial
convergent sequence in $V$, then $St(\mathbb{A})$ contains such a sequence in
$V^{\mathbb{M}_{\kappa}}$, too. The following theorem implies thus e.g. that
every countable Boolean algebra admits a sequence of homomorphisms into
$\mathbb{M}_{\kappa}$ which is pointwise algebraically convergent but not
uniformly convergent (since $St(\mathbb{A})$ is metrizable; cf. the examples
in Sections 6.1 and 6.5).

######  Theorem 8.2.

Let $\mathbb{A}$ be a Boolean algebra in $V$. The following are equivalent:

  1. (1)

$\Vdash_{\mathbb{M}_{\kappa}}$ $St(\mathbb{A})$ does not have non-trivial
convergent sequences;

  2. (2)

every pointwise algebraically convergent sequence of homomorphisms from
$\mathbb{A}$ into $\mathbb{M}_{\kappa}$ converges uniformly.

###### Proof.

(1)$\implies$(2). Suppose that $(\varphi_{n})$ in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ converges pointwise
algebraically to some homomorphism $\varphi$. Then, by Proposition 7.3,
$\Vdash_{\mathbb{M}_{\kappa}}$“$\tau_{\varphi_{n}}$ converges to
$\tau_{\varphi}$”. By the assumption,
$\Vdash_{\mathbb{M}_{\kappa}}\forall^{\infty}n\in\omega\
\tau_{\varphi_{n}}=\tau_{\varphi}$, and so, by Proposition 7.4,
$(\varphi_{n})$ converges uniformly to $\varphi$.

(2)$\implies$(1). Suppose now that there exist $p\in\mathbb{M}_{\kappa}$, and
$\mathbb{M}_{\kappa}$-names $(\dot{\mathcal{U}}_{n})$, $\dot{\mathcal{U}}$ for
ultrafilters on $\mathbb{A}$ such that
$p\Vdash_{\mathbb{M}_{\kappa}}``(\dot{\mathcal{U}}_{n})$ converges to
$\dot{\mathcal{U}}$ and $\forall n\
\dot{\mathcal{U}}_{n}\neq\dot{\mathcal{U}}$”. Since $\mathbb{M}_{\kappa}$ is
isomorphic to the restricted forcing
$p\wedge\mathbb{M}_{\kappa}=\big{\\{}q\in\mathbb{M}_{\kappa}\colon q\leq
p\big{\\}}$ and hence $\mathbb{M}_{\kappa}$-generic extensions of $V$ are the
same as $(p\wedge\mathbb{M}_{\kappa})$-generic extensions, we may assume that
$p=1$.

By Proposition 7.3, the sequence $(\varphi_{\dot{\mathcal{U}}_{n}})$ converges
pointwise algebraically to $\varphi_{\dot{\mathcal{U}}}$ and hence, by the
assumption, uniformly. Theorem 7.5 implies that there is $n$ and a non-zero
$p\in\mathbb{M}_{\kappa}$ such that
$p\Vdash\dot{\mathcal{U}}_{n}=\dot{\mathcal{U}}$, a contradiction. It follows
that there are no non-trivial convergent sequences in $St(\mathbb{A})\cap
V^{\mathbb{M}_{\kappa}}$ . ∎

Since a compact space containing a copy of $\beta\omega$ must necessarily have
weight at least $2^{\omega}$ and the forcing $\mathbb{M}_{\kappa}$ preserves
cardinals, we immediately get the following corollary.

######  Corollary 8.3.

Assume that $\mathbb{A}\in V$ is a Boolean algebra of size $<\kappa$ and such
that in $V$ every pointwise algebraically convergent sequence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is uniformly convergent. Then,
in $V^{\mathbb{M}_{\kappa}}$, the Stone space of $\mathbb{A}$ is an Efimov
space.

The theorem of Dow and Fremlin yields the following corollary.

######  Corollary 8.4.

If $\mathbb{A}$ has the interpolation property, then every pointwise
algebraically convergent sequence of homomorphisms in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ converges uniformly. In
particular, this holds if $\mathbb{A}$ is $\sigma$-complete or
$\mathbb{A}=\mathcal{P}(\omega)/Fin$.

Recall that in Section 6.5 we presented an example of a pointwise algebraic
convergent sequence of homomorphisms from the Cantor algebra $\mathbb{C}$ to
$\mathbb{M}$ which is not uniformly convergent—this example, as contradicting
the conclusion of Corollary 8.4, shows that a Boolean algebra $\mathbb{A}$ in
the corollary cannot be to simple.

##  9\. Open questions and problems

Fix two Boolean algebras $\mathbb{A}$ and $\mathbb{B}$ and an infinite
cardinal number $\kappa$. In this final section we briefly list several open
questions and problems concerning the space
$\mathcal{H}(\mathbb{A},\mathbb{B})$.

######  Question 9.1.

What topological properties does the space
$\mathcal{H}(\mathbb{A},\mathbb{B})$ have when endowed with the topologies
defined in Section 4? In particular, what properties does the space
$\mathcal{H}(\mathbb{M}_{\kappa},\mathbb{M}_{\kappa})$ have?

The following two questions seem natural in the context of Proposition 5.1 and
Corollary 5.2.

######  Question 9.2.

Is the function $F_{\mu}$ from Proposition 5.1 continuous if the spaces
$\mathcal{H}(\mathbb{A},\mathbb{B})$ and $P(St(\mathbb{A}))$ are endowed with
the pointwise Borel metric topology and weak topology, respectively?

######  Problem 9.3.

Let $\mathbb{A}$ be a Boolean algebra and $\kappa$ a cardinal. Is there a
natural convergence on the space $P(St(\mathbb{M}_{\kappa}))$ which
corresponds to the pointwise algebraic convergence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ in the sense of Corollary 5.2?

The following question concerns possible extensions of the diagram from the
beginning of Section 6.

######  Question 9.4.

What algebraic or structural properties of the Boolean algebras $\mathbb{A}$
and $\mathbb{B}$ imply that a sequence in $\mathcal{H}(\mathbb{A},\mathbb{B})$
convergent in one topology is also convergent in some other one?

In Section 6 we have presented several examples of sequences of homomorphisms
witnessing that the convergence in one topology may not be sufficient to imply
the convergence in some other one. We believe that some of those examples may
be generalized to work in more generic situations.

######  Problem 9.5.

Provide a general scheme for constructing a definable sequence of
homomorphisms in $\mathcal{H}(\mathbb{A},\mathbb{B})$ such that it is
convergent in one topology but not in the other one, or a general scheme for
constructing a Borel set or a sequence of Borel sets in $St(\mathbb{A})$
witnessing that a given sequence of homomorphisms in
$\mathcal{H}(\mathbb{A},\mathbb{B})$ is not convergent in some topology.

By the Dow–Fremlin theorem and Theorem 8.2 it follows that if $\mathbb{A}$ has
the interpolation property, then every pointwise algebraic sequence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ is uniformly convergent
(Corollary 8.4). Since the interpolation property implies the Grothendieck
property, we pose the following question being a particular case of Question
9.4.

######  Question 9.6.

Assume that $\mathbb{A}$ has the (positive) Grothendieck property. Is every
pointwise algebraic convergent sequence in
$\mathcal{H}(\mathbb{A},\mathbb{M}_{\kappa})$ also uniformly convergent?

Note that, again by Theorem 8.2, obtaining an algebraic or topological proof
for an affirmative answer to Question 9.6 would yield an alternative proof to
the theorem of Dow and Fremlin.

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