# Selective separability properties of Fréchet-Urysohn spaces and their
products

Serhii Bardyla  Universität Wien, Institut für Mathematik, Kurt Gödel Research
Center, Kolingasse 14-16, 1090 Vienna, Austria<EMAIL_ADDRESS>http://www.logic.univie.ac.at/ bardylas55/ ,  Fortunato Maesano  MIFT -
Matematica e Informatica, Scienze Fisiche e Scienze della Terra Department,
University of Messina, Viale F. Stagno d’Alcontres 31, 98166 Messina, Italy.
<EMAIL_ADDRESS>and  Lyubomyr Zdomskyy  Institut für Diskrete Mathematik
und Geometrie, Technische Universität Wien, Wiedner Hauptstrasse 8-10/104,
1040 Wien, Austria<EMAIL_ADDRESS>https://dmg.tuwien.ac.at/zdomskyy/

###### Abstract.

In this paper we study the behaviour of selective separability properties in
the class of Frechét-Urysohn spaces. We present two examples, the first one
given in ZFC proves the existence of a countable Frechét-Urysohn (hence
$R$-separable and selectively separable) space which is not $H$-separable;
assuming $\mathfrak{p}=\mathfrak{c}$, we construct such an example which is
also zero-dimensional and $\alpha_{4}$. Also, motivated by a result of Barman
and Dow stating that the product of two countable Frechét-Urysohn spaces is
$M$-separable under PFA, we show that the MA is not sufficient here. In the
last section we prove that in the Laver model, the product of any two
$H$-separable spaces is $mH$-separable.

###### Key words and phrases:

Frechét-Urysohn space, Martin’s Axiom, separated mad family, $M$-separability,
$H$-separability, $R$-separability, $\alpha_{i}$-space.

###### 2020 Mathematics Subject Classification:

Primary: 03E50, 54D65, 03E35; Secondary: 54D10, 03E17, 03E65.

The first author was supported by the Austrian Science Fund FWF (Grant ESP
399). The second author would like to thank the “National Group for Algebraic
and Geometric Structures, and their Applications” (GNSAGA – INdAM) for
generous support for this research. The third author would like to thank the
Austrian Science Fund FWF (Grants I 3709 and I 5930) for generous support for
this research.

##  1\. Introduction

A space $X$ is said to be _Frechét-Urysohn_ , briefly FU, if for every
$A\subset X$ and $x\in\overline{A}\setminus A$ there exists a sequence
$S\subset A\setminus\\{x\\}$ converging to $x$, i.e.,
$S\in[A\setminus\\{x\\}]^{\omega}$ and $|U\setminus A|<\omega$ for every open
neighbourhood $U$ of $x$. It has been known since 50 years ago that the
product of two FU spaces does not have to be a FU space, neither for countable
nor for compact spaces, see [8, 17, 18, 21] and references therein. However,
it is still interesting which properties products of two FU spaces must (at
least consistently) possess. In this paper we study various combinatorial
versions of the separability of such products. Our motivation comes from the
following result of Barman and Dow:

######  Theorem 1.1.

[3, 3.3] (PFA) The product of finitely many countable FU spaces is
M-separable.

Recall from [20] that a topological space $X$ is said to be _selectively
separable_ (or _M-separable_) if for every sequence $\langle
D_{n}:n\in\omega\rangle$ of dense subsets of $X$, there are finite sets
$F_{n}\subset D_{n}$, $n\in\omega$, such that $\bigcup\\{F_{n}:n\in\omega\\}$
is dense in $X$.

In contrast with Theorem 1.1, Barman and Dow proved in [2] that every
separable FU space is $M$-separable and that under CH, there are two countable
FU spaces whose product is not $M$-separable. In [16, 6.1 and 6.2] this result
has been improved by showing that under CH there are two countable FU
$H$-separable topological groups whose product is not $M$-separable. The main
result of this paper is the following

######  Theorem 1.2.

It is consistent with (MA+$\neg\mathit{CH}$) that there are two countable
regular FU spaces whose product is not $M$-separable.

Among other things Theorem 1.2 shows that Theorem 1.1 cannot be proved using
only the equality between some standard cardinal characteristics of the
continuum, as all these characteristics are also equal to $\mathfrak{c}$ under
MA. The proof of Theorem 1.2 is based on the application of certain kind of
separated mad families, whose existence is consistent with MA by essentially
the same argument as in [9], see Section 4.

The notion of $M$-separability motivated Bella, Bonanzinga and Matveev to
introduce in [4] further variations thereof of this property. In particular, a
topological space $X$ is said to be _H-separable_ (resp., _R-separable_), if
for every sequence $\langle D_{n}:n\in\omega\rangle$ of dense subsets of $X$,
there are finite sets $F_{n}\subset D_{n}$ (resp., points $x_{n}\in D_{n}$),
$n\in\omega$, such that every nonempty open set in $X$ meets all but finitely
many $F_{n}$ (resp., $\\{x_{n}:n\in\omega\\}$ is dense in $X$).

Gruenhage and Sakai in [11] proved that separable FU spaces are $R$-separable
(this improves the result of Barman and Dow asserting that separable FU spaces
are $M$-separable). In Section 2 we prove that the $H$-separability behaves
differently, i.e., in ZFC there exists a countable Hausdorff FU space which is
not $H$-separable, see Theorem 4.5. This gives new examples of a space which
is $R$-separable but not $H$-separable, see [4] and [5] for more examples of
this kind.

In [11], using different terminology, the following weaker versions of
combinatorial density properties were introduced: A topological space $X$ is
said to be _$mP$ -separable,_ where $P\in\\{M,H,R\\}$, if it satisfies the
conclusion of the $P$-separability for any decreasing sequence of dense
subsets of $X$. For instance, $X$ is _mR-separable_ if for every decreasing
sequence $\langle D_{n}:n\in\omega\rangle$ of dense subsets of $X$, there are
points $x_{n}\in D_{n}$ such that $\\{x_{n}:n\in\omega\\}$ is dense in $X$.
The following diagram sums up the implications between the cited selective
properties.

${R\text{-}sep.}$${M\text{-}sep.}$${H\text{-}sep.}$${mR\text{-}sep.}$${mM\text{-}sep.}$${mH\text{-}sep.}$

It is proved in [11] that the $M$-separability and $mM$-separability are in
fact equivalent, which motivated the authors to ask in [11, Question 2.10(1)]
_whether there exists an $mH$-separable space which is not $H$-separable._
Since [11, Lemma 2.7(2)] combined with [11, Corollary 4.2] implies that every
countable FU space is $mH$-separable, the FU non-$H$-separable space that we
build in the proof of Theorem 4.5 is $mH$-separable, and thus our Theorem 4.5
answers the aforementioned question posed in [11] in the affirmative. We do
not know how to get a regular (equivalently, zero-dimensional) example
outright in ZFC.

######  Question 1.3.

  1. (1)

Is there a ZFC example of a regular countable FU (resp. $mH$-separable) space
which is not $H$-separable?

  2. (2)

Is there a (ZFC example of) a [regular] countable $mR$-separable space which
is not $R$-separable?

Given a space $X$ and $x\in X$, we denote by $\Gamma_{x}$ the set of all
$A\in[X\setminus\\{x\\}]^{\omega}$ which converge to $x$. In [1],
Arhangel’skiĭ introduced the following local properties of a space $X$ at some
point $x\in X$:

  * $(\alpha_{1})$

For each $\langle S_{n}:n\in\omega\rangle\in\Gamma^{\omega}_{x}$, there is
$S\in\Gamma_{x}$ such that $S_{n}\subset^{*}S$ for all $n\in\omega$;

  * $(\alpha_{2})$

For each $\langle S_{n}:n\in\omega\rangle\in\Gamma^{\omega}_{x}$, there is
$S\in\Gamma_{x}$ such that $S_{n}\cap S$ is infinite for all $n\in\omega$;

  * $(\alpha_{3})$

For each $\langle S_{n}:n\in\omega\rangle\in\Gamma^{\omega}_{x}$, there is
$S\in\Gamma_{x}$ such that $S_{n}\cap S$ is infinite for infinitely many
$n\in\omega$;

  * $(\alpha_{4})$

For each $\langle S_{n}:n\in\omega\rangle\in\Gamma^{\omega}_{x}$, there is
$S\in\Gamma_{x}$ such that $S_{n}\cap S\neq\emptyset$ for infinitely many
$n\in\omega$.

A space $X$ is an $\alpha_{i}$ space, where $i\in\\{1,2,3,4\\}$, if it is an
$\alpha_{i}$ space at each $x\in X.$ Each of these $\alpha_{i}$-properties
obviously imply the next one. The idea behind the introduction of $\alpha_{i}$
spaces was to find additional properties which together with the FU property
guarantee that the product of two spaces is again FU, see [1, 17] for more
details.

In Section 3 we prove that every separable $\alpha_{2}$ FU space is
$H$-separable, and $\mathfrak{p}=\mathfrak{c}$ yields a countable regular FU
$\alpha_{4}$ which is not $H$-separable. This leaves the following question
open, which actually consists of several subquestions.

######  Question 1.4.

Is it consistent that every (regular) $\alpha_{4}$ [resp. $\alpha_{3}$] FU
space is $H$-separable?

In Section 5 we show that it is consistent that the product of two countable
$H$-separable spaces is $mH$-separable. More precisely, this is the case in
the classical Laver model introduced in [14]. This improved an earlier result
from [19] stating that in the same model, the product of any two countable
$H$-separable spaces is $M$-separable. Consequently, in the Laver model the
product of any two $H$-separable spaces is $mH$-separable provided that it is
hereditarily separable. This result motivates the following

######  Question 1.5.

  1. (1)

Is there a countable (regular) $mH$-separable space in the Laver model which
is not $H$-separable?

  2. (2)

Is the class of $H$-separable ($mH$-separable) spaces consistently closed
under finite products? What about the Laver model?

  3. (3)

Is the product of 3 (resp. finitely many) $H$-separable spaces $mH$-separable
in the Laver model?

Finally, the proof of Theorem 1.1 gives actually the $mR$-separability of the
product, but we do not know whether it can be pushed further.

######  Question 1.6.

Does PFA imply that the product of two countable FU spaces is $mH$\- (resp.
$R$-) separable?

All the undefined notions can be found in [6], [10] and [13].

##  2\. A Hausdorff Fréchet-Urysohn space which is not $H$-separable in ZFC

In the proof of the main result of this section we shall use a fundamental
result of Mathias stating that there is no analytic mad family
$\mathcal{A}\subset[\omega]^{\omega}$, see [15, Corollary 4.7]. Recall that an
infinite $\mathcal{A}\subset[\omega]^{\omega}$ is _mad_ , if any two different
elements of $\mathcal{A}$ have finite intersection, and $\mathcal{A}$ is not
included into to a strictly bigger family with this property.

######  Theorem 2.1.

There exists a countable Hausdorff Frechet-Urysohn space $X$ without isolated
points which is not $H$-separable.

###### Proof.

The underlying set of $X$ will be $\omega$, and let $\tau_{0}$ be a topology
on $\omega$ such that $\langle\omega,\tau_{0}\rangle$ is homeomorphic to the
rationals.

The final topology $\tau=\tau_{\mathfrak{c}}$ turning $\omega$ into a space
with the needed properties will be constructed recursively over ordinals
$\alpha\in\mathfrak{c}$ as an increasing union
$\tau_{\mathfrak{c}}=\bigcup_{\alpha<\mathfrak{c}}\tau_{\alpha}$ of
intermediate topologies.

Let $\langle E_{n}:n\in\omega\rangle$ be a sequence of mutually disjoint dense
subsets of $\langle\omega,\tau_{0}\rangle$. We shall make sure that $E_{n}$’s
remain dense in $\langle\omega,\tau_{\alpha}\rangle$ for all
$\alpha\leq\mathfrak{c}$. Let $\\{\langle
S_{\alpha},x_{\alpha}\rangle:\alpha<\mathfrak{c}\\}$ be an enumeration of
$[\omega]^{\omega}\times\omega$ such that for each $\langle
S,x\rangle\in[\omega]^{\omega}\times\omega$ there are cofinally many ordinals
$\alpha$ such that $\langle S,x\rangle=\langle S_{\alpha},x_{\alpha}\rangle$.
Fix an enumeration $\\{\langle
F^{\alpha}_{n}:n\in\omega\rangle:\alpha<\mathfrak{c}\\}$ of
$\prod_{n\in\omega}[E_{n}]^{<\omega}$. We shall also need a mad family on
$\omega$ of size $\mathfrak{c}$ obtained as follows. Let
$\mathcal{A}\subset[\omega]^{\omega}$ be a compact almost disjoint family of
size $\mathfrak{c}$. It can easily be constructed by, e.g., considering the
family of all branches through $2^{<\omega}$, and then copying it via any
bijection between $\omega$ and $2^{<\omega}$. Let $\mathcal{C}$ be any mad
family extending $\mathcal{A}$. Note that $\mathcal{C}\neq\mathcal{A}$ since
there are no analytic mad families by [15, Corollary 4.7].

Suppose that for some $\alpha<\mathfrak{c}$ and all $\delta\in\alpha$ we have
already constructed a topology $\tau_{\delta}$ on $\omega$, a family
$\mathcal{Y}_{\delta}\subset[\omega]^{\omega}$, and for every
$Y\in\mathcal{Y}_{\delta}$ either $n(Y)\in\omega$ (such $Y$ will be called
_vertical_) or $C(Y)\in\mathcal{C}$ (such $Y$ will be called _horizontal_)
such that the following conditions are satisfied:

  * $(1)$

$\tau_{\beta}\subset\tau_{\delta}$ for all $\beta\leq\delta$;

  * $(2)$

$\mathcal{Y}_{\beta}\subset\mathcal{Y}_{\delta}$ for all $\beta\leq\delta$ and
$\mathcal{Y}_{\delta}$ consists of sequences convergent in
$\langle\omega,\tau_{\delta}\rangle$;

  * $(3)$

For every $\beta<\delta$, if $x_{\beta}$ is a limit point of $S_{\beta}$, then
there exists $Y\in\mathcal{Y}_{\delta}$ converging to $x_{\beta}$ with $|Y\cap
S_{\beta}|=\omega$;

  * $(4)$

Every vertical $Y\in\mathcal{Y}_{\delta}$ is contained in $E_{n(Y)}$, and each
horizontal $Y\in\mathcal{Y}_{\delta}$ is a subset of $\bigcup_{n\in
C(Y)}E_{n}$; Moreover, in the latter case $|Y\cap E_{n}|\leq 1$ for all $n\in
C(Y)$;

  * $(5)$

For every $\beta<\delta$ there exists $U_{\beta}\in\tau_{\delta}$ and
$A_{\beta}\in\mathcal{A}$ such that $U_{\beta}=\omega\setminus\bigcup_{n\in
A_{\beta}}F^{\beta}_{n},$ and
$\\{A_{\beta}:\beta<\delta\\}\cap\\{C(Y):Y\in\mathcal{Y}_{\delta}\\}=\emptyset$;

  * $(6)$

$\tau_{\delta}$ is generated by $\tau_{0}\cup\\{U_{\beta}:\beta<\delta\\}$ as
a subbase;

  * $(7)$

$E_{n}$ is dense in $\langle\omega,\tau_{\delta}\rangle$ for all $n\in\omega$.

Let us note that $(5)$ and $(6)$ imply $(7)$, as well as the fact that for
every $\mathcal{C}^{\prime}\in[\mathcal{C}]^{<\omega}$ and
$K\in[\omega]^{<\omega}$ we have that
$\tau_{\delta}\upharpoonright\bigcup_{n\in
K\cup\bigcup\mathcal{C}^{\prime}}E_{n}$ has a countable base. Moreover, if
$\mathcal{C}^{\prime}\in[\mathcal{C}\setminus\\{A_{\beta}:\beta<\delta\\}]^{<\omega}$,
then

$\tau_{\delta}\upharpoonright\bigcup_{n\in
K\cup\bigcup\mathcal{C}^{\prime}}E_{n}=\tau_{0}\upharpoonright\bigcup_{n\in
K\cup\bigcup\mathcal{C}^{\prime}}E_{n}.$

We shall consider several cases.

I. $\alpha$ is limit.
It is easily checked that the topology $\tau_{\alpha}$ generated by
$\bigcup_{\delta<\alpha}\tau_{\delta}$ as a base, along with
$\mathcal{Y}_{\alpha}=\bigcup_{\delta<\alpha}\mathcal{Y}_{\delta}$ satisfies
$(1)$-$(7)$ for $\delta=\alpha$.

II. $\alpha=\delta+1$ and $x_{\delta}$ is a limit point of $S_{\delta}$ in
$\langle\omega,\tau_{\delta}\rangle$.
If $|Y\cap S_{\delta}|=\omega$ for some $Y\in\mathcal{Y}_{\delta}$ converging
to $x_{\delta}$, then we set $\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta}$, pick
any

$A_{\delta}\in\mathcal{A}\setminus\\{C(Y):Y\in\mathcal{Y}_{\delta}\\},$

let $U_{\delta}$ be such as in item $(6)$, denote by $\tau_{\alpha}$ the
topology generated by $\tau_{\delta}\cup\\{U_{\delta}\\}$ as a subbase, and
note that all the conditions $(1)$-$(7)$ are satisfied. E.g., each
$Y\in\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta}$ is convergent also in
$\langle\omega,\tau_{\alpha}\rangle$ since $U_{\delta}$ almost contains all
such $Y$.

So consider the case $|Y\cap S_{\delta}|<\omega$ for all
$Y\in\mathcal{Y}_{\delta}$ converging to $x_{\delta}$. If there exists
$n\in\omega$ such that $x_{\delta}\in\overline{S_{\delta}\cap
E_{n}}^{\tau_{\delta}}$, let $Y_{\delta}\in[S_{\delta}\cap E_{n}]^{\omega}$ be
a sequence converging to $x_{\delta}$ (remember that all the $\tau_{\beta}$’s
restricted to $E_{n}$ are the same, and thus turn $E_{n}$ into a copy of the
rationals), and in this case we set $n(Y_{\delta})=n$. Let us assume now that
there is no such $n$. We claim that there exists a sequence
$Y\in[S_{\delta}]^{\omega}$ converging to $x_{\delta}$ in
$\langle\omega,\tau_{\delta}\rangle$. Let $\mathcal{C}^{\prime}$ be the family
of all those $C\in\mathcal{C}$ such that there exists a sequence
$Y_{C}\in[S_{\delta}]^{\omega}$ convergent to $x_{\delta}$ in
$\langle\omega,\tau_{0}\rangle$ such that $Y_{C}\subset\bigcup_{n\in C}E_{n}$
and $|Y_{C}\cap E_{n}|\leq 1$ for all $n\in C$. Two cases are possible.

a) $\mathcal{C}^{\prime}$ is finite.
It follows that $x_{\delta}$ is not in the closure of
$S_{\delta}^{\prime}=S_{\delta}\setminus\bigcup\\{E_{n}:n\in\cup\mathcal{C}^{\prime}\\}$
with respect to $\tau_{0}$. Indeed, otherwise there exists a sequence
$Y\in[S_{\delta}^{\prime}]^{\omega}$ converging to $x_{\delta}$ in
$\langle\omega,\tau_{0}\rangle$, and since it cannot have infinite
intersection with any $E_{n}$, we conclude that there exists
$L\in[\omega]^{\omega}$ with $|Y\cap E_{n}|\neq\emptyset$ for all $n\in L$.
Then $L\cap\cup\mathcal{C}^{\prime}=\emptyset$, but on the other hand any
$C\in\mathcal{C}$ with $|C\cap L|=\omega$ must be in $C^{\prime}$, a
contradiction. Consequently, $x_{\delta}$ is also not in the closure of
$S_{\delta}^{\prime}$ with respect to $\tau_{\delta}$, and hence it must be in
the closure of
$S_{\delta}^{\prime\prime}=S_{\delta}\cap\bigcup\\{E_{n}:n\in\cup\mathcal{A}^{\prime}\\}$
with respect to $\tau_{\delta}$. However, $\tau_{\delta}\upharpoonright
S_{\delta}^{\prime\prime}$ has countable base, so there exists a sequence
$Y_{\delta}\in[S_{\delta}^{\prime\prime}]^{\omega}$ convergent to $x_{\delta}$
with respect to $\tau_{\delta}$. By shrinking $Y_{\delta}$, if necessary, we
may assume find $C(Y_{\delta})\in\mathcal{C}^{\prime}$ satisfying $(4)$ along
with $Y_{\delta}$. It remains to set
$\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta}\cup\\{Y_{\delta}\\}$, pick any
$A_{\delta}\in\mathcal{A}\setminus\\{C(Y):Y\in\mathcal{Y}_{\alpha}\\}$, and
let $U_{\delta}$ and $\tau_{\alpha}$ be such as required in $(5)$, $(6)$ for
$\alpha$.

b) $\mathcal{C}^{\prime}\supset\\{C_{k}:k\in\omega\\}$, where $C_{k}\neq
C_{m}$ for $k\neq m$.
For every $k\in\omega$ let $Y_{C_{k}}$ be witnessing for
$C_{k}\in\mathcal{C}^{\prime}$. By shrinking $Y_{C_{k}}$’s, if necessary, we
may assume that $|\bigcup_{k\in\omega}Y_{C_{k}}\cap E_{n}|\leq 1$ for all
$n\in\omega$. Let $Y\subset\bigcup_{k\in\omega}Y_{C_{k}}$ be such that
$Y_{C_{k}}\subset^{*}Y$ for all $k\in\omega$ and $Y$ converges to $x_{\delta}$
with respect to $\tau_{0}$. Such $Y$ obviously exists since $\tau_{0}$ has a
countable base. Set $L=\\{n\in\omega:Y\cap E_{n}\neq\emptyset\\}$ and note
that $L$ cannot be covered by finitely many elements of $\mathcal{C}$ because
$|L\cap C_{k}|=\omega$ for all $k\in\omega$. Thus

$\\{A\cap L\>:\>A\in\mathcal{A},|A\cap L|=\omega\\}$

cannot be a mad family of infinite subsets of $L$ since it is analytic, and
hence there exists $L^{\prime}\in[L]^{\omega}$ which is almost disjoint from
all $A\in\mathcal{A}$. Let $C\in\mathcal{C}$ be such that $|C\cap
L^{\prime}|=\omega$, denote by $Y_{\delta}$ the sequence $Y\cap\bigcup_{n\in
C\cap L^{\prime}}E_{n}$, set $C(Y_{\delta})=C$, and
$Y_{\alpha}=\mathcal{Y}_{\delta}\cup\\{Y_{\delta}\\}$. Finally, pick any
$A_{\delta}\in A\setminus\\{C(Y):Y\in\mathcal{Y}_{\alpha}\\}$ and let
$U_{\delta}$ and $\tau_{\alpha}$ be such as required in $(5)$, $(6)$ for
$\alpha$. Since $A_{\beta}\cap C(Y)$ is finite for all
$Y\in\mathcal{Y}_{\alpha}$ and $\beta<\alpha$, we conclude that
$Y\subset^{*}U_{\beta}$ for all $Y\in\mathcal{Y}_{\alpha}$ and $\beta<\alpha$,
and hence all $Y\in\mathcal{Y}_{\alpha}$ remain convergent with respect to
$\tau_{\alpha}$.

III. $\alpha=\delta+1$ and $x_{\delta}$ is not a limit point of $S_{\delta}$
in $\langle\omega,\tau_{\delta}\rangle$.
In this case we set $x_{\delta}^{\prime}=0$ (any $i\in\omega$ instead of $0$
would work), $S_{\delta}^{\prime}=\omega\setminus\\{0\\}$, and repeat what we
have done in item $II$, with $\langle
S_{\delta}^{\prime},x_{\delta}^{\prime}\rangle$ instead of $\langle
S_{\delta},x_{\delta}\rangle$. Note that $x_{\delta}^{\prime}$ must be in the
closure of $S_{\delta}^{\prime}$ in $\langle\omega,\tau_{\delta}\rangle$ since
this topological space has no isolated points, because it has disjoint dense
subsets by $(7)$.

This completes our recursive construction of the objects mentioned in
$(1)$-$(7)$ so that these conditions are satisfied. The space
$\langle\omega,\tau_{\mathfrak{c}}\rangle$ is as required: $(2),(3)$ imply the
Frechet-Urysohn property, and $(5)$ entails the failure of the
$H$-separability. ∎

##  3\. A Fréchet-Urysohn $\alpha_{4}$ space which is not $H$-separable under
$\mathfrak{p}=\mathfrak{c}$

The main result of this section may be thought of as a version of Theorem 2.1,
namely we shall get a zero-dimensional $\alpha_{4}$ example like in Theorem
2.1, but at the cost of an additional set-theoretic assumption.

The next easy statement can be considered as a folklore and we shall refer to
it as “there are no $(\omega,<\mathfrak{b})$-gaps”. A very similar argument
can be found in [12, page 578].

######  Lemma 3.1.

Suppose that $\mathcal{A}\subset[\omega]^{\omega}$,
$|\mathcal{A}|<\mathfrak{b}$,
$\\{B_{i}:i\in\omega\\}\subset[\omega]^{\omega}$, and $|A\cap B_{i}|<\omega$
for all $A\in\mathcal{A}$ and $i\in\omega$. Then there exists $X\subset\omega$
such that $A\subset^{*}\omega\setminus X$ and $B_{i}\subset^{*}X$ for any
$A\in\mathcal{A}$ and $i\in\omega$.

###### Proof.

For each $A\in\mathcal{A}$ find an increasing function
$f_{A}\in\omega^{\omega}$ with $A\cap B_{i}\subset f_{A}(i)$ for all
$i\in\omega$, and let $f\in\omega^{\omega}$ be such that $f_{A}\leq^{*}f$ for
all $A\in\mathcal{A}$. Then $X=\bigcup_{i\in\omega}(B_{i}\setminus f(i))$ is
as required. ∎

######  Theorem 3.2.

($\mathfrak{p}=\mathfrak{c}$) There exists a countable zero-dimensional
$\alpha_{4}$ Frechet-Urysohn space $X$ without isolated points which is not
$H$-separable.

###### Proof.

Let $\mathcal{G}$ be a $P_{\mathfrak{c}}$-point, i.e., an ultrafilter such
that for any $\mathcal{G}^{\prime}\in[\mathcal{G}]^{<\mathfrak{c}}$ there
exists $G\in\mathcal{G}$ such that $G\subset^{*}G^{\prime}$ for all
$G^{\prime}\in\mathcal{G}^{\prime}$. It is well-known and easy to check that
$\mathfrak{p}=\mathfrak{c}$ implies the existence of such an ultrafilter. The
underlying set of $X$ will be $\omega$, and let $\tau_{0}$ be a topology on
$\omega$ such that $\langle\omega,\tau_{0}\rangle$ is homeomorphic to the
rationals.

The final topology $\tau=\tau_{\mathfrak{c}}$ turning $\omega$ into a space
with the needed properties will be constructed recursively over ordinals
$\alpha\in\mathfrak{c}$ as an increasing union
$\tau_{\mathfrak{c}}=\bigcup_{\alpha<\mathfrak{c}}\tau_{\alpha}$ of
intermediate zero-dimensional topologies.

Let $\langle E_{n}:n\in\omega\rangle$ be a sequence of mutually disjoint dense
subsets of $\langle\omega,\tau_{0}\rangle$. We shall make sure that $E_{n}$’s
remain dense in $\langle\omega,\tau_{\alpha}\rangle$ for all
$\alpha\leq\mathfrak{c}$. Let $\\{\langle
S_{\alpha},\mathcal{A}_{\alpha},x_{\alpha}\rangle:\alpha<\mathfrak{c}\\}$ be
an enumeration of
$[\omega]^{\omega}\times[[\omega]^{\omega}]^{\omega}\times\omega$ such that
for each $\langle
S,\mathcal{A},x\rangle\in[\omega]^{\omega}\times[[\omega]^{\omega}]^{\omega}\times\omega$
there are cofinally many ordinals $\alpha$ such that $\langle
S,\mathcal{A},x\rangle=\langle
S_{\alpha},\mathcal{A}_{\alpha},x_{\alpha}\rangle$. Fix also an enumeration
$\\{\langle F^{\alpha}_{n}:n\in\omega\rangle:\alpha<\mathfrak{c}\\}$ of
$\prod_{n\in\omega}[E_{n}]^{<\omega}$.

Suppose that for some $\alpha<\mathfrak{c}$ and all $\delta\in\alpha$ we have
already constructed a zero-dimensional topology $\tau_{\delta}$ on $\omega$,
an almost disjoint family $\mathcal{Y}_{\delta}\subset[\omega]^{\omega}$, and
for every $Y\in Y_{\delta}$ and element
$I_{Y}\in\mathcal{G}^{*}=\mathcal{P}(\omega)\setminus\mathcal{G}$ such that
the following conditions are satisfied:

  * $(1)$

The weight of $\tau_{\delta}$ is $<\mathfrak{c}$;

  * $(2)$

For every $n,x\in\omega$ there exists $Y\in\mathcal{Y}_{0}$ such that
$Y\in[E_{n}]^{\omega}$ and $Y$ converges to $x$;

  * $(3)$

$\mathcal{Y}_{\beta}\subset\mathcal{Y}_{\delta}$ for all
$\beta\leq\delta<\alpha$ and $\mathcal{Y}_{\delta}$ consists of sequences
convergent in $\langle\omega,\tau_{\delta}\rangle$;

  * $(4)$

For every $\beta<\delta$, if $x_{\beta}$ is a limit point of $S_{\beta}$ and
every $A\in\mathcal{A}_{\beta}$ is a sequence converging to $x_{\beta}$ in
$\langle\omega,\tau_{\delta}\rangle$, then

    * –

there exists $Y\in\mathcal{Y}_{\delta}$ converging to $x_{\beta}$ with $|Y\cap
S_{\beta}|=\omega$;

    * –

there exists $Y\in\mathcal{Y}_{\delta}$ such that
$Y\cap(A\setminus\\{x_{\beta}\\})\neq\emptyset$ for infinitely many
$A\in\mathcal{A}_{\beta}$;

  * $(5)$

Every $Y\in\mathcal{Y}_{\delta}$ is contained in $\bigcup_{n\in I_{Y}}E_{n}$;
Moreover, either $|Y\cap E_{n}|=1$ for all $n\in I_{Y}$, or $|I_{Y}|=1$;

  * $(6)$

For every $\beta<\delta$ there exists $U_{\beta}\in\tau_{\delta}$ and
$G_{\beta}\in\mathcal{G}$ such that $U_{\beta}\cap\bigcup_{n\in
G_{\beta}}F^{\beta}_{n}=\emptyset;$

  * $(7)$

$E_{n}$ is dense in $\langle\omega,\tau_{\delta}\rangle$ for all $n\in\omega$.

We shall consider several cases.

I. $\alpha$ is limit.
It is easily checked that the topology $\tau_{\alpha}$ generated by
$\bigcup_{\delta<\alpha}\tau_{\delta}$ as a base, along with
$\mathcal{Y}_{\alpha}=\bigcup_{\delta<\alpha}\mathcal{Y}_{\delta}$ satisfies
$(1)$-$(7)$ for $\delta=\alpha$.

II. $\alpha=\delta+1$, $x_{\delta}$ is a limit point of $S_{\delta}$ in
$\langle\omega,\tau_{\delta}\rangle$, and $\mathcal{A}_{\delta}$ consists of
mutually disjoint sequences convergent to $x_{\delta}$.
If $|Y\cap S_{\delta}|=\omega$ for some $Y\in\mathcal{Y}_{\delta}$ converging
to $x_{\delta}$, then we denote by $Y_{0,\delta}$ one of these $Y$’s.
Similarly, if
$\big{\\{}A\in\mathcal{A}_{\beta}\>:\>Y\cap(A\setminus\\{x_{\beta}\\})\neq\emptyset\big{\\}}$
is infinite for some $Y\in\mathcal{Y}_{\delta}$, we denote by $Y_{1,\delta}$
one of these $Y$’s.

So consider the case $|Y\cap S_{\delta}|<\omega$ for all
$Y\in\mathcal{Y}_{\delta}$ converging to $x_{\delta}$. Since the weight of
$\langle\omega,\tau_{\delta}\rangle$ is $<\mathfrak{c}=\mathfrak{p}$, there
exists a sequence $Y_{0,\delta}\in[S_{\delta}]^{\omega}$ convergent to
$x_{\delta}$. Passing to an infinite subset of $Y_{0,\delta}$, if necessary,
we may assume that either there exists $n\in\omega$ such that
$Y_{0,\delta}\subset E_{n}$, in which case we set $I_{Y_{0,\delta}}=\\{n\\}$,
or there exists an infinite $I_{Y_{0,\delta}}\in\mathcal{G}^{*}$ such that
$Y_{0,\delta}\subset\bigcup_{n\in I_{Y_{0,\delta}}}E_{n}$ and
$|Y_{0,\delta}\cap E_{n}|=1$ for all $n\in I_{Y_{0,\delta}}$. Note that
$Y_{0,\delta}$ is almost disjoint from any $Y\in\mathcal{Y}_{\delta}$.

Now suppose that
$\\{A\in\mathcal{A}_{\beta}\>:\>Y\cap(A\setminus\\{x_{\beta}\\})\neq\emptyset\\}$
is finite for all $Y\in\mathcal{A}$. Replacing each $A\in\mathcal{A}_{\delta}$
with an infinite subset thereof, we may assume that for each
$A\in\mathcal{A}_{\delta}$, either there exists $n(A)\in\omega$ such that
$A\subset E_{n(A)}$ (such $A$ will be called _vertical_), or $|A\cap
E_{n}|\leq 1$ for all $n\in\omega$ (such $A$ will be called _horizontal_).
Since every space of character $<\mathfrak{b}$ is $\alpha_{1}$, there exists a
sequence $Y\subset\cup\mathcal{A}_{\delta}$ convergent to $x_{\delta}$ such
that $|Y\cap A|=1$ for all $A\in\mathcal{A}_{\delta}$. The fact that no
element of $\mathcal{Y}_{\delta}$ converging to $x_{\delta}$ intersects
infinitely many $A\in\mathcal{A}_{\delta}$ yields that $Y$ is almost disjoint
from all elements of $\mathcal{Y}_{\delta}$. Note also that
$\mathcal{A}_{\delta}$ is a disjoint family, and hence each infinite subset of
$Y$ intersects infinitely many elements of $\mathcal{A}_{\delta}$.

If there are infinitely many horizontal sequences, then $|Y\cap
E_{n}|\neq\emptyset$ for infinitely many $n$, and hence by shrinking $Y$ to
some infinite $Y_{1,\delta}$ we may assume that $|Y_{1,\delta}\cap E_{n}|\leq
1$ for all $n$ and $|Y_{1,\delta}\cap E_{n}|=1$ if and only if $n\in I$ for
some $I\in\mathcal{G}^{*}$. In this case we set $I_{Y_{1,\delta}}=I$.

Now suppose that all but finitely many $A\in\mathcal{A}_{\delta}$ are
vertical. If $\\{n(A):A\in\mathcal{A}_{\delta},A$ is vertical$\\}$ is
infinite, then as before by shrinking $Y$ to some infinite $Y_{1,\delta}$ we
may assume that $|Y_{1,\delta}\cap E_{n}|\leq 1$ for all $n$ and
$|Y_{1,\delta}\cap E_{n}|=1$ if and only if $n\in I$ for some
$I\in\mathcal{G}^{*}$. Also in this case we set $I_{Y_{1,\delta}}=I$.

In the remaining case there exists $n\in\omega$ with
$\\{A\in\mathcal{A}_{\delta}:n(A)=n\\}$ infinite. Then we set
$Y_{1,\delta}=Y\cap E_{n}$, $I_{Y_{1,\delta}}=\\{n\\}$, and note that
$Y_{1,\delta}$ is infinite.

Shrinking $Y_{0,\delta}$ and $Y_{1,\delta}$, if necessary, we may assume that
they are disjoint. Finally, set
$\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta+1}=\mathcal{Y}_{\delta}\cup\\{Y_{0,\delta},Y_{1,\delta}\\}$
and note that $(3),(4),$ and $(5)$ are satisfied for $\delta+1$ instead of
$\delta$.

Next, we pass to the construction of $U_{\delta},G_{\delta}$ satisfying $(6)$.
This will require that if $x\in U_{\delta}$ and $Y\in\mathcal{Y}_{\delta}$
converges to $x$, then $Y\subset^{*}U_{\delta}$. Let
$G_{\delta}\in\mathcal{G}$ be such that $|G_{\delta}\cap I_{Y}|<\omega$ for
all $Y\in\mathcal{Y}_{\alpha}$. Then $F_{\delta}:=\bigcup_{n\in
G_{\delta}}F^{\delta}_{n}$ is almost disjoint from any
$Y\in\mathcal{Y}_{\alpha}$. We need the following auxiliary

######  Claim 3.2.1.

For every $n\in\omega$ there exists $C_{n}\subset\omega$ such that

  * $(8)$

$Y\subset^{*}C_{n}$ for any $Y\in\mathcal{Y}_{\alpha}$ converging to $n$ in
$\langle\omega,\tau_{\delta}\rangle$;

  * $(9)$

$C_{n}\cap C_{m}=\emptyset$ for all $n\neq m$;

  * $(10)$

$C_{n}\cap F_{\delta}=\emptyset$ for all $n\neq m$.

###### Proof.

For every $n\in\omega$ let us denote by $\mathcal{Y}_{\alpha,n}$ the family of
all $Y\in\mathcal{Y}_{\alpha}$ converging to $n$, and fix a family
$\\{O^{n}_{k}:k\in\omega\setminus\\{n\\}\\}\subset\tau_{0}$ such that
$O^{n}_{k}$ is a clopen neighbourhood of $k$ not containing $n$. Then $|Y\cap
O^{n}_{k}|<\omega$ for all $Y\in\mathcal{Y}_{\alpha,n}$ and $k\neq n$, and
hence Lemma 3.1 implies that there exists $C^{0}_{n}\subset\omega$ such that
$Y\subset^{*}C^{0}_{n}$ and $|C^{0}_{n}\cap O^{n}_{k}|<\omega$ for all $Y,k$
as above. Let us note that $|C^{0}_{n}\cap Y|<\omega$ for all
$Y\in\mathcal{Y}_{\alpha}\setminus\mathcal{Y}_{\alpha,n}$ because
$Y\subset^{*}O^{n}_{k}$ for $k\neq n$ being the limit point of $Y$. Thus
letting $C_{0}=C^{0}_{0}\setminus F_{\delta}$ and
$C_{n}=C^{0}_{n}\setminus(\bigcup_{n^{\prime}<n}C_{n^{\prime}}\cup
F_{\delta})$ we get that $\\{C_{n}:n\in\omega\\}$ is a disjoint family
satisfying $(8)-(10)$. ∎

The construction of $V_{\delta}:=\omega\setminus U_{\delta}$ will be done
recursively over $k\in\omega$, namely it will be constructed as an increasing
union $\bigcup_{k\in\omega}V^{\delta}_{k}$. Moreover, we shall make sure that
adding its complement to $\tau_{\delta}$ can also be done without any harm.
More precisely, set $V^{\delta}_{-1}=\emptyset$, $V^{\delta}_{0}=F_{\delta}$
and assuming that $V^{\delta}_{k}$ is constructed we let

$V^{\delta}_{k+1}=V^{\delta}_{k}\cup\bigcup\\{C_{n}\setminus\\{\min
C_{n}\\}:n\in V^{\delta}_{k}\setminus V^{\delta}_{k-1}\\}.$

In the sequel we call a subset $B$ of $\omega$ _saturated_ if $n\in B$ implies
$C_{n}\subset^{*}B$. It follows that $V_{\delta}$ defined above is saturated.
We claim that $U_{\delta}:=\omega\setminus V_{\delta}$ is saturated as well.
Indeed, otherwise there exists $n\in U_{\delta}$ such that $C_{n}\cap
V_{\delta}\neq\emptyset$. Let $k\in\omega$ be the minimal such that $C_{n}\cap
V^{\delta}_{k+1}\neq\emptyset$ (note that $C_{n}\cap V^{\delta}_{0}=C_{n}\cap
F_{\delta}=\emptyset$ by $(9)$). Then there exists $m\in
V^{\delta}_{k}\setminus V^{\delta}_{k-1}$ with $C_{n}\cap C_{m}\neq\emptyset$,
which is impossible since $n\neq m$ (because $n\in U_{\delta}$ and $m\in
V_{\delta}$).

Finally, $U_{\delta}\neq\emptyset$ since $\min C_{n}\in U_{\delta}$ for all
$n\in V_{\delta}.$

Let $\tau_{\alpha}$ be the topology generated by
$\tau_{\delta}\cup\\{U_{\delta},V_{\delta}\\}$ as a base. This way we get a
$0$-dimensional topology satisfying $(6)$. Since both $U_{\alpha},V_{\alpha}$
are saturated, it is easy to see that all $Y\in\mathcal{Y}_{\alpha}$ are
convergent also in $\langle\omega,\tau_{\alpha}\rangle$. Indeed, suppose that
$y$ is the limit of $Y\in\mathcal{Y}_{\alpha}$ and $y\in U\cap U_{\delta}$,
where $U\in\tau_{\delta}$. Then $Y\subset^{*}U$ (because $Y$ converges in
$\langle\omega,\tau_{\delta}\rangle$) and
$Y\subset^{*}C_{y}\subset^{*}U_{\delta}$, and therefore $Y\subset^{*}U\cap
U_{\delta}$. The same argument works also for $V_{\delta}$ instead of
$U_{\delta}$. Since $U\in\tau_{\delta}$ was arbitrary, we conclude that $Y$
converges to $y$ also in $\langle\omega,\tau_{\alpha}\rangle$.

To see that also $(7)$ holds, let us fix $n\in\omega$ and $y\in U$, where
$U\in\tau_{\alpha}$. Let
$Y\in\mathcal{Y}_{0}\cap[E_{n}]^{\omega}\subset\mathcal{Y}_{\alpha}$ be
convergent to $y$. Then $Y\subset^{*}U$, and hence $Y\subset^{*}U\cap E_{n}$,
which yields $U\cap E_{n}\neq\emptyset$.

This completes our construction of the objects mentioned in $(1)$-$(7)$ for
all $\delta<\mathfrak{c}$, so that these conditions are satisfied. Condition
$(4)$ implies that $X=\langle\omega,\tau_{\mathfrak{c}}\rangle$ is Frechet-
Urysohn and $\alpha_{4}$. Indeed, if $\mathcal{A}$ is a countable family of
mutually disjoint sequences convergent to some $x\in\omega$ in
$X=\langle\omega,\tau_{\mathfrak{c}}\rangle$, then there exists
$\delta<\mathfrak{c}$ such that
$\langle\omega\setminus\\{x\\},\mathcal{A},x\rangle=\langle
S_{\delta},\mathcal{A}_{\delta},x_{\delta}\rangle$. Then
$Y_{1,\delta}\in\mathcal{Y}_{\delta+1}$ is convergent to $x_{\alpha}$ in
$\langle\omega,\tau_{\mathfrak{c}}\rangle$, and it intersects infinitely many
elements of $\mathcal{A}_{\delta}=\mathcal{A}$. Similarly one can check the
Frechet-Urysohn property. Finally, $(6)$ implies the failure of the
$H$-separability, which completes the proof. ∎

The assumption of being $\alpha_{4}$ cannot be much improved in Theorem 3.2 as
the following fact shows.

######  Proposition 3.3.

Every separable Frechet-Urysohn $\alpha_{2}$ space $X$ is $H$-separable.

###### Proof.

Let $\langle E_{n}:n\in\omega\rangle$ be a sequence of dense subsets of $X$,
and $D$ be a countable dense subset of $X$. Let $\\{d_{i}:i\in\omega\\}$ be an
enumeration of all non-isolated elements of $D$, and
$\\{d^{\prime}_{i}:i\in\omega\\}=D\setminus\\{d_{i}:i\in\omega\\}$. For every
$i,n\in\omega$ let us fix an injective sequence $S^{i}_{n}$ of elements of
$E_{n}\setminus\\{d_{i}\\}$ convergent to $d_{i}$. Applying $\alpha_{2}$ we
can find a sequence $\langle s^{i}_{n}:n\in\omega\rangle$ convergent to
$d_{i}$ such that $s^{i}_{n}\in S^{i}_{n}\subset E_{n}$. Let us note that
$\\{d^{\prime}_{i}:i\in\omega\\}$ consists of isolated points of $X$ and hence
it is a subset of $E_{n}$ for all $n\in\omega$. It follows that the sequence

$\big{\langle}F_{n}:=\\{d^{\prime}_{i}:i\leq n\\}\cup\\{s^{i}_{n}:i\leq
n\\}\>:\>n\in\omega\big{\rangle}$

is a witness for the $H$-separability of $X$. Indeed, given any open $U\subset
X$, either there exists $i\in\omega$ such that $d^{\prime}_{i}\in U$, and thus
$U\cap F_{n}\neq\emptyset$ for all $n\geq i$; or there exists $i\in\omega$
with $d_{i}\in U$, and then $U\cap F_{n}\neq\emptyset$ as soon as
$s^{i}_{n}\in U$ and $n\geq i$, and $s^{i}_{n}\in U$ for all but finitely many
$n$ since $\langle s^{i}_{n}:n\in\omega\rangle$ convergent to $d_{i}$. ∎

##  4\. Products of Fréchet-Urysohn spaces and $M$-separability

In [3], the authors proved the next important result.

######  Theorem 4.1.

(PFA) The product of finitely many countable FU spaces is $M$-separable.

It is worth noting that the proof given by the authors provides a stronger
result: their argument proves that the product of finitely many countable FU
space under PFA is in fact $mR$-separable.

The aim of this section is to show that MA is not sufficent in Theorem 4.1.
Namely we shall define a new set-theoretic principle which is consistent with
MA, and, when combined with it, it implies the negation of the conclusion of
Theorem 4.1.

Let $\tau_{0}$ be a topology on $\omega$ turning it into a space homeomorphic
to the rationals $\mathbb{Q}$. In what follows $(\ast_{\mathbb{Q}})$ stands
for the following statement:

> MA holds and there exists a mad family $\mathcal{A}$ on $\omega$ such that
>
>   * •
>
> every disjoint pair
> $\mathcal{A}^{\prime},\mathcal{A}^{\prime\prime}\in[\mathcal{A}]^{<\mathfrak{c}}$
> is _separated_ , i.e. there exists $S\subset\omega$ such that
> $A^{\prime}\subset^{*}S$ and $A^{\prime\prime}\subset^{*}\omega\setminus S$
> for any $A^{\prime}\in\mathcal{A}^{\prime}$ and
> $A^{\prime\prime}\in\mathcal{A}^{\prime\prime}$;
>
>   * •
>
> Every $A\in\mathcal{A}$ is either closed discrete or a convergent sequence
> in $\langle\omega,\tau_{0}\rangle$.
>
>

The formally weaker statement $(\ast)$ obtained from $(\ast_{\mathbb{Q}})$ by
dropping the second item is known to be consistent. More precisely, the
following result was proved by Dow and Shelah.

######  Theorem 4.2.

[9] It is consistent with MA and $\mathfrak{c}=\omega_{2}$ that there is a
maximal almost disjoint family such that any disjoint pair of its
$\leq\omega_{1}$-sized subsets are separated.

To prove the ”whole statement”, namely that $(\ast_{\mathbb{Q}})$ is
consistent, first we remind the definitions of standard posets that will be
involved. Every filter $\mathcal{F}$ gives rise to a natural forcing notion
$\mathbb{M}_{\mathcal{F}}$ introducing a generic subset
$X\in[\omega]^{\omega}$ such that $X\subset^{*}F$ for all $F\in\mathcal{F}$ as
follows: $\mathbb{M}_{\mathcal{F}}$ consists of pairs $\langle s,F\rangle$
such that $s\in[\omega]^{<\omega}$, $F\in\mathcal{F}$, and $\max s<\min F$. A
condition $\langle s,F\rangle$ is stronger than $\langle t,G\rangle$ if
$F\subset G$, $s$ is an end-extension of $t$, and $s\setminus t\subset G$.
$\mathbb{M}_{\mathcal{F}}$ is usually called _Mathias forcing associated with_
$\mathcal{F}$.

Every infinite almost disjoint family $\mathcal{A}\subset[\omega]^{\omega}$
generates the dual filter $\mathcal{F}(\mathcal{A})$ consisting of sets $F$
such that there exists a finite $\mathcal{A}^{\prime}\subset\mathcal{A}$ with
the property $\omega\setminus\cup\mathcal{A}^{\prime}\subset^{*}F$. We shall
be interested in Mathias forcing associated with filters which are usually
larger than $\mathcal{F}(\mathcal{A})$ for certain almost disjoint families
$\mathcal{A}$ defined as follows. Recall that $\tau_{0}$ is a topology on
$\omega$ such that $\langle\omega,\tau_{0}\rangle$ is homeomorphic to the
rationals in $[0,1]$, and let $\theta:\omega\to\mathbb{Q}\cap[0,1]$ be the
corresponding homeomorphism. In what follows we shall refer to the topology
$\tau_{0}$ whenever we speak about topological properties of subsets of
$\omega$, e.g., convergence, being closed discrete, etc.

Let us fix an almost disjoint family $\mathcal{A}\subset[\omega]^{\omega}$
consisting of convergent sequences as well as closed discrete subsets. Given
any $n\in\omega$, denote by $\mathcal{F}_{n}(\mathcal{A})$ the filter
generated by $\mathcal{F}(\mathcal{A})\cup\\{O\in\tau_{0}:n\in O\\}$ and note
that if $G$ is $\mathbb{M}_{\mathcal{F}_{n}(A)}$-generic, then
$X_{G}:=\bigcup_{\langle s,F\rangle\in G}s$ is a sequence convergent to $n$,
and $|X_{G}\cap Y|=\omega$ for any sequence $Y\in V$ convergent to $n$.

Fix $r\in[0,1]\setminus\mathbb{Q}$ and let $\mathcal{F}_{r}(\mathcal{A})$ be
the filter generated by

$\mathcal{F}(\mathcal{A})\cup\\{\theta^{-1}[O\cap\mathbb{Q}]\>:\>r\in O\mbox{
is an open subset of }\mathbb{R}\\}.$

By genericity, if $G$ is $\mathbb{M}_{\mathcal{F}_{r}(A)}$-generic, then
$X_{G}:=\bigcup_{\langle s,F\rangle\in G}s\subset\omega$ is closed and
discrete because $\theta[X_{G}]$ converges to an irrational number
$r\in[0,1]$, and $|X_{G}\cap Y|=\omega$ for any closed discrete subset $Y$ of
$\omega$ such that $r$ is a limit point of $\theta[Y]$.

Recall from [9] that an almost disjoint family
$\mathcal{A}\subset[\omega]^{\omega}$ is called _special_ , if there exists
$c:[\mathcal{A}]^{<\omega}\to\omega$ and a linear ordering $<$ of
$\mathcal{A}$ such that for each $n\in\omega$ and sequences

$\langle B_{0},B_{1},\ldots,B_{n-1}\rangle,\langle
C_{0},C_{1},\ldots,C_{n-1}\rangle\in\mathcal{A}^{n}$

increasing with respect to $<$, if

$c\langle B_{0},B_{1},\ldots,B_{n-1}\rangle=c\langle
C_{0},C_{1},\ldots,C_{n-1}\rangle=k,$

then for all $i\neq j$, $i,j\in n$ we have $B_{i}\cap C_{j}\subset k$. There
is a natural poset introduced in [9, Def. 2.4] and denoted by
$\mathbb{Q}_{\mathcal{A},<},$ which for an almost disjoint family
$\mathcal{A}$ and a linear order $<$ thereof, introduces a function $c$ having
the above property: A condition in this poset is a finite function
$d:\mathcal{P}(\mathcal{A}_{d})\to\omega$, where
$\mathcal{A}_{d}\in[\mathcal{A}]^{<\omega}$, such that for each $n\in\omega$
and $<$-increasing sequences

$\langle B_{0},B_{1},\ldots,B_{n-1}\rangle,\langle
C_{0},C_{1},\ldots,C_{n-1}\rangle\in\mathcal{A}_{d}^{n},$

if $d\langle B_{0},B_{1},\ldots,B_{n-1}\rangle=d\langle
C_{0},C_{1},\ldots,C_{n-1}\rangle=k,$ then for all $i\neq j$, $i,j\in n$ we
have $B_{i}\cap C_{j}\subset k$. It is clear that $\mathcal{A}$ is special in
$V^{\mathbb{Q}_{\mathcal{A},<}}$. However, this poset may collapse cardinals:
In ZFC one can construct an almost disjoint family $\mathcal{A}$ of size
$\omega_{1}$ which cannot be made special by any forcing which preserves
$\omega_{1}$, see [9, p. 108] and references therein.

An almost disjoint family $\mathcal{A}\subset[\omega]^{\omega}$ is called
_$\omega_{1}$ -special_, if any
$\mathcal{A}^{\prime}\in[\mathcal{A}]^{\omega_{1}}$ is special. The next fact
is [9, Prop. 1.5]. It has been inspired by earlier results of Silver, as it is
stated in [9].

######  Lemma 4.3.

If $\mathit{MA}_{\omega_{1}}$ holds and
$\mathcal{A}\in[[\omega]^{\omega}]^{\omega_{1}}$ is $\omega_{1}$-special, then
$\mathcal{A}$ is separated, i.e., for any
$\mathcal{A}^{\prime}\subset\mathcal{A}$ there exists $X\subset\omega$ such
that $A\subset^{*}X$ for all $A\in\mathcal{A}^{\prime}$ and $|A\cap X|<\omega$
for all $A\in\mathcal{A}\setminus\mathcal{A}^{\prime}$.

We are in a position now to present the sketch of the proof of the following

######  Theorem 4.4.

$(*_{\mathbb{Q}})$ is consistent.

###### Proof.

We shall present only the main ideas of the proof, the remaining details are
exactly the same as in the proof of [9, Theorem 2.6]. Let us denote by
$\Lambda$ the set of all limit ordinals below $\omega_{2}$. Assuming GCH in
$V$, we shall construct a finitely supported iteration
$\langle\mathbb{P}_{\alpha},\dot{\mathbb{Q}}_{\beta}:\alpha\leq\omega_{2},\beta<\omega_{2}\rangle$
of ccc posets of size $\omega_{1}$, along with a sequence
$\langle\dot{\mathcal{A}}_{\beta}:\beta<\omega_{2},\beta\in\Lambda\rangle$,
where
$\dot{\mathcal{A}_{\beta}}=\\{\dot{X}_{\xi}:\xi\in(\beta+1)\cap\Lambda\\}$ is
a $\mathbb{P}_{\beta+1}$-name for an almost disjoint family, as follows:

  * •

If $\beta$ is a successor ordinal, then $\dot{\mathbb{Q}}_{\beta}$ is
($\mathbb{P}_{\beta}$-name for) a ccc poset of size $\omega_{1}$;

  * •

For $\beta\in\Lambda$ the poset $\dot{\mathbb{Q}}_{\beta}$ is a
$\mathbb{P}_{\beta}$-name for the product
$\dot{\mathbb{Q}}^{0}_{\beta}\times\dot{\mathbb{Q}}^{1}_{\beta}$, where

    * $(a)$

Letting $\dot{\mathcal{A}}^{-}_{\beta}$ be a $\mathbb{P}_{\beta}$-name for
$\bigcup_{\xi\in\beta\cap\Lambda}\dot{\mathcal{A}}_{\xi}=\\{\dot{X}_{\xi}:\xi\in\beta\cap\Lambda\\}$,
$\dot{\mathbb{Q}}^{0}_{\beta}$ is either
$\mathbb{M}_{\mathcal{F}_{\dot{r}_{\beta}}(\dot{\mathcal{A}}^{-}_{\beta})}$
for some $\mathbb{P}_{\beta}$-name $\dot{r}_{\beta}$ for an irrational number
in $[0,1]$, or
$\mathbb{M}_{\mathcal{F}_{n_{\beta}}(\dot{\mathcal{A}}^{-}_{\beta})}$ for some
$n_{\beta}\in\omega$, decided by a bookkeeping function as described below;
$\dot{\mathbb{Q}}^{0}_{\beta}$ produces a generic subset $\dot{X}_{\beta}$
almost disjoint from all elements of $\dot{\mathcal{A}}^{-}_{\beta}$, and we
set
$\dot{\mathcal{A}}_{\beta}=\dot{\mathcal{A}}^{-}_{\beta}\cup\\{\dot{X}_{\beta}\\}$;

    * $(b)$

$\dot{\mathbb{Q}}^{1}_{\beta}=\dot{\mathbb{Q}}_{\dot{\mathcal{A}}^{-}_{\beta},<}$,
where $<$ is the linear order (actually wellorder) of
$\dot{\mathcal{A}}^{-}_{\beta}$ generated by the indices of its elements,
i.e., $\dot{X}_{\xi}<\dot{X}_{\zeta}$ iff $\xi<\zeta,$
$\xi,\zeta\in\beta\cap\Lambda$.

Almost literally following the argument in the proof of [9, Th. 2.6], we can
show that $\mathbb{P}_{\omega_{2}}$ is ccc, the only non-trivial ingredient
being that $\dot{\mathbb{Q}}^{1}_{\beta}$ is forced to be ccc by
$1_{\mathbb{P}_{\beta}}$. This, roughly speaking, happens because elements of
$\dot{\mathcal{A}}^{-}_{\beta}$ are added as generic reals for corresponding
Mathias posets. Now a standard choice of a suitable bookkeeping function
“delivering” the names for posets $\dot{\mathbb{Q}}_{\beta}$ at successor
steps $\beta$, and $n_{\beta}$ or $\dot{r}_{\beta}$ for limit $\beta$,
guarantees that $(*_{\mathbb{Q}})$ holds in $V^{\mathbb{P}_{\omega_{2}}}.$
More precisely, MA is guaranteed at successor stages, and
$\mathcal{A}=\bigcup_{\beta\in\Lambda}\mathcal{A}_{\beta}$ is a mad family
consisting of convergent sequences (added at stages $\beta\in\Lambda$ for
which the bookkeeping function gives $n_{\beta}\in\omega$) and closed discrete
sets (added at stages $\beta\in\Lambda$ for which the bookkeeping function
gives $\dot{r}_{\beta}$); The presence of $\dot{\mathbb{Q}}^{1}_{\beta}$’s for
$\beta\in\Lambda$ implies that any
$\mathcal{A}^{\prime}\in[\mathcal{A}]^{\omega_{1}}$ is special, which together
with Lemma 4.3 implies that any such $\mathcal{A}^{\prime}$ is separated, and
thus completes the proof of the theorem. ∎

Remark. Even though the main part of our proof of Theorem 4.4, namely that
$\dot{\mathbb{Q}}^{1}_{\beta}$ is forced to be ccc by
$1_{\mathbb{P}_{\beta}}$, can be done in exactly the same way as the
corresponding step in the proof of [9, Th. 2.6], the overall scheme of the
proof of Theorem 4.4 is very different from that of [9, Th. 2.1]. More
precisely, the proof of the latter gives a so-called _tight_ mad family, i.e.,
a mad family $\mathcal{A}$ such that for every countable
$\mathcal{X}\subset\mathcal{I}(\mathcal{A})^{+}$ (recall that
$X\in\mathcal{I}(\mathcal{A})^{+}$ iff $X$ has infinite intersection with
infinitely many elements of $\mathcal{A}$) there exists $A\in\mathcal{A}$ such
that $|A\cap X|=\omega$ for all $X\in\mathcal{X}$. On the other hand, no mad
family $\mathcal{A}$ of subsets of $\mathbb{Q}$ consisting of convergent
sequences and closed discrete sets can be tight. Indeed, it is an easy
exercise to prove that for every $0<\delta<\epsilon$ there exists a closed
discrete $D\subset\mathbb{Q}$ such that $D\in\mathcal{I}(\mathcal{A})^{+}$ and
$\delta<|x|<\epsilon$ for all $x\in D$. Now let $\langle
D_{n}:n\in\omega\rangle$ be a sequence of closed discrete subsets of
$\mathbb{Q}$ which lie in $\mathcal{I}(\mathcal{A})^{+}$ and such that
$\frac{3}{5\cdot 2^{n}}<|x|<\frac{4}{5\cdot 2^{n}}$ for all $x\in D_{n}$. Then
it is clear that there is neither a closed discrete subset of $\mathbb{Q}$ nor
a convergent sequence there which has infinite intersection with each (even
infinitely many) of $D_{n}$’s. $\Box$

The overall scheme of the proof of the next theorem is patterned after that of
[2, Theorem 2.24]. However, there are also essential differences because our
construction lasts $>\omega_{1}$-many steps and hence we have to make sure
that we do not face a kind of Hausdorff gaps consisting of convergent
sequences in our construction, which is done with the help of a mad family
witnessing $(\ast_{\mathbb{Q}})$.

######  Theorem 4.5.

$(\ast_{\mathbb{Q}})$ There exist two countable Frechet-Urysohn spaces
$X_{0},X_{1}$ without isolated points whose product is not $M$-separable.

###### Proof.

Let the underlying sets of $X_{0},X_{1}$ be $\omega$ and $\tau_{0}=\sigma_{0}$
be a topology on $\omega$ such that $\langle\omega,\tau_{0}\rangle$ is
homeomorphic to the rationals. Let $\mathcal{A}$ be a mad family on $\omega$
whose existence is guaranteed by $(\ast_{\mathbb{Q}})$.

The final topologies $\tau=\tau_{\mathfrak{c}}$ and
$\sigma=\sigma_{\mathfrak{c}}$ turning $\omega$ into Frechet-Urysohn-spaces
with non-$M$-separable product will be constructed recursively over ordinals
$\alpha\in\mathfrak{c}$ as increasing unions
$\tau_{\mathfrak{c}}=\bigcup_{\alpha<\mathfrak{c}}\tau_{\alpha}$ and
$\sigma_{\mathfrak{c}}=\bigcup_{\alpha<\mathfrak{c}}\sigma_{\alpha}$ of
intermediate zero-dimensional topologies. Let $\langle
E_{n}:n\in\omega\rangle$ be a sequence of mutually disjoint dense subsets of
$\langle\omega\times\omega,\tau_{0}\otimes\sigma_{0}\rangle$ such that
$E:=\bigcup_{n\in\omega}E_{n}$ is (the graph of) a permutation of $\omega$. We
shall make sure that $E_{n}$’s remain dense in
$\langle\omega\times\omega,\tau_{\alpha}\otimes\sigma_{\alpha}\rangle$ for all
$\alpha\leq\mathfrak{c}$. Let $\\{\langle
S_{\alpha},x_{\alpha}\rangle:\alpha<\mathfrak{c}\\}$ be an enumeration of
$[\omega]^{\omega}\times\omega$ such that for each $\langle
S,x\rangle\in[\omega]^{\omega}\times\omega$ there are cofinally many even as
well as odd111An ordinal is _even_ (resp. _odd_) if it can be written in the
form $\omega\cdot\alpha+i$ for some even (resp. odd) $i\in\omega$.  ordinals
$\alpha$ such that $\langle S,x\rangle=\langle S_{\alpha},x_{\alpha}\rangle$.
Fix also an enumeration $\\{\langle
F^{\alpha}_{n}:n\in\omega\rangle:\alpha<\mathfrak{c}\\}$ of
$\prod_{n\in\omega}[E_{n}]^{<\omega}$. For the convenience of those readers
familiar with the proof of [2, Theorem 2.24], for a subset $A$ of $\omega$ we
denote $E[A]$ and $E^{-1}[A]$ (here we view $E$ as a map from $\omega$ to
$\omega$) by $E(A,1)$ and $E(A,0)$, respectively.

Suppose that for some $\alpha<\mathfrak{c}$ and all $\delta\in\alpha$ we have
already constructed topologies $\tau_{\delta},\sigma_{\delta}$ on $\omega$,
almost disjoint families
$\mathcal{Y}_{\delta},\mathcal{Z}_{\delta}\subset[\omega]^{\omega}$, such that
the following conditions are satisfied:

  * $(1)$

The weight of $\tau_{\delta},\sigma_{\delta}$ is $<\mathfrak{c}$;

  * $(2)$

    * $(a)$

$\mathcal{Y}_{\beta}\subset\mathcal{Y}_{\delta}$ for all $\beta\leq\delta$ and
$\mathcal{Y}_{\delta}$ consists of sequences convergent in
$\langle\omega,\tau_{\delta}\rangle$;

    * $(b)$

$\mathcal{Z}_{\beta}\subset\mathcal{Z}_{\delta}$ for all $\beta\leq\delta$ and
$\mathcal{Z}_{\delta}$ consists of sequences convergent in
$\langle\omega,\sigma_{\delta}\rangle$;

  * $(3)$

    * $(a)$

For every $\beta<\delta$, if $\beta$ is even and $x_{\beta}$ is a limit point
of $S_{\beta}$ in $\langle\omega,\tau_{\delta}\rangle$, then there exists
$Y\in\mathcal{Y}_{\delta}$ converging to $x_{\beta}$ in
$\langle\omega,\tau_{\delta}\rangle$ with $|Y\cap S_{\beta}|=\omega$;

    * $(b)$

For every $\beta<\delta$, if $\beta$ is odd and $x_{\beta}$ is a limit point
of $S_{\beta}$ in $\langle\omega,\sigma_{\delta}\rangle$, then there exists
$Z\in\mathcal{Z}_{\delta}$ converging to $x_{\beta}$ in
$\langle\omega,\sigma_{\delta}\rangle$ and such that $|Z\cap
S_{\beta}|=\omega$;

  * $(4)$

    * $(a)$

There are injective maps
$\phi^{\delta}_{0},\phi^{\delta}_{1}:\mathcal{Y}_{\delta}\to\mathcal{A}$ such
that $Y\subset\phi^{\delta}_{0}(Y)$ and $E(Y,1)\subset\phi^{\delta}_{1}(Y)$
for all $Y\in\mathcal{Y}_{\delta}$;

    * $(b)$

There are injective maps
$\psi^{\delta}_{1},\psi^{\delta}_{0}:\mathcal{Z}_{\delta}\to\mathcal{A}$ such
that $Z\subset\psi^{\delta}_{1}(Y)$ and $E(Z,0)\subset\psi^{\delta}_{0}(Z)$
for all $Z\in\mathcal{Z}_{\delta}$;

    * $(c)$

$\psi^{\delta}_{1}[\mathcal{Z}_{\delta}]\cap\phi^{\delta}_{1}[\mathcal{Y}_{\delta}]=\emptyset$
and
$\psi^{\delta}_{0}[\mathcal{Z}_{\delta}]\cap\phi^{\delta}_{0}[\mathcal{Y}_{\delta}]=\emptyset$;

  * $(5)$

    * $(a)$

$E(\phi^{\delta}_{0}(Y),1)$ and $\phi^{\delta}_{1}(Y)$ are closed and discrete
in $\langle\omega,\sigma_{\delta}\rangle$ for all $Y\in\mathcal{Y}_{\delta}$;

    * $(b)$

$E(\psi^{\delta}_{1}(Z),0)$ and $\psi^{\delta}_{0}(Z)$ are closed and discrete
in $\langle\omega,\tau_{\delta}\rangle$ for all $Z\in\mathcal{Z}_{\delta}$;

  * $(6)$

For every $\beta<\delta$ there are $U_{\beta}\in\tau_{\delta}$ and
$V_{\beta}\in\sigma_{\delta}$ such that222In particular,
$U_{\beta},V_{\beta},I_{\beta}$ depend only on $\beta$ and not on $\delta$.

$(U_{\beta}\times V_{\beta})\cap\bigcup_{n\in\omega}F^{\beta}_{n}=\emptyset.$

  * $(7)$

$E_{n}$ is dense in
$\langle\omega\times\omega,\tau_{\delta}\otimes\sigma_{\delta}\rangle$ for all
$n\in\omega$.

We shall consider several cases.

I. $\alpha$ is limit.
It is easily checked that topologies $\tau_{\alpha}$ and $\sigma_{\alpha}$
generated by $\bigcup_{\delta<\alpha}\tau_{\delta}$ and
$\bigcup_{\delta<\alpha}\sigma_{\delta}$ as a base, respectively, along with
$\mathcal{Y}_{\alpha}=\bigcup_{\delta<\alpha}\mathcal{Y}_{\delta}$ and
$\mathcal{Z}_{\alpha}=\bigcup_{\delta<\alpha}\mathcal{Z}_{\delta}$ satisfy
$(1)$-$(7)$ for $\delta=\alpha$.

II. $\alpha=\delta+1$, $\delta$ is even, $x_{\delta}$ is a limit point of
$S_{\delta}$ in $\langle\omega,\tau_{\delta}\rangle$, and $|Y\cap
S_{\delta}|<\omega$ for all $Y\in\mathcal{Y}_{\delta}$ converging to
$x_{\delta}$.
Since the weight of $\langle\omega,\tau_{\delta}\rangle$ is
$<\mathfrak{c}=\mathfrak{p}$, there exists a sequence
$T\in[S_{\delta}]^{\omega}$ convergent to $x_{\delta}$. Let
$A^{\delta}_{0}\in\mathcal{A}$ be such that $|A^{\delta}_{0}\cap T|=\omega$.

Analogously, since $|\psi^{\delta}_{0}(Z)\cap T|<\omega$ for all
$Z\in\mathcal{Z}_{\delta}$ (because $\psi^{\delta}_{0}(Z)$ is closed discrete
in $\langle\omega,\tau_{\delta}\rangle$), we have $\psi^{\delta}_{0}(Z)\neq
A^{\delta}_{0}$ for all $Z\in\mathcal{Z}_{\delta}$. For the same reason
$|E(\psi^{\delta}_{1}(Z),0)\cap T|<\omega$ for all $Z\in\mathcal{Z}_{\delta}$.
Since $E$ is a permutation of $\omega$, $\mathcal{A}_{1\downarrow
0}:=\\{E(A,0):A\in\mathcal{A}\\}$ is a mad family, and hence there exists
$A^{\delta}_{1}\in\mathcal{A}$ such that $|E(A^{\delta}_{1},0)\cap
A^{\delta}_{0}\cap T|=\omega$. It follows from the above that
$E(A^{\delta}_{1},0)\neq E(\psi^{\delta}_{1}(Z),0)$ for all
$Z\in\mathcal{Z}_{\delta}$, and hence also
$A^{\delta}_{1}\neq\psi^{\delta}_{1}(Z)$.

Finally, set $Y_{\delta}=T\cap A^{\delta}_{0}\cap E(A^{\delta}_{1},0)$,
$\mathcal{Y}_{\alpha}:=\mathcal{Y}_{\delta}\cup\\{Y_{\delta}\\}$,
$\mathcal{Z}_{\alpha}=\mathcal{Z}_{\delta}$,
$\phi^{\alpha}_{0}(Y_{\delta})=A^{\delta}_{0}$,
$\phi^{\alpha}_{1}(Y_{\delta})=A^{\delta}_{1}$,
$\phi^{\alpha}_{0}\upharpoonright\mathcal{Y}_{\delta}=\phi^{\delta}_{0}$, and
$\phi^{\alpha}_{1}\upharpoonright\mathcal{Y}_{\delta}=\phi^{\delta}_{1}$. It
follows from the construction that item $(4)$ is satisfied. To proceed further
we need the following direct consequence of [2, Lemma 2.23].

######  Claim 4.5.1.

Suppose that $X$ is a countable crowded space of weight less than
$\mathfrak{p}$, $\mathcal{T}\subset\mathcal{P}(X)$ is a family of almost
disjoint converging sequences in $X$, $|\mathcal{T}|<\mathfrak{p}$, and
$\mathcal{R}\subset\mathcal{P}(X)$ is a countable family of subsets of $X$
such that each $R\in\mathcal{R}$ has dense complement and is almost disjoint
from each member of $\mathcal{T}$, with all but possibly one elements of
$\mathcal{R}$ being discrete. Then there is an expansion of the topology of
$X$ to a larger crowded one, obtained by adding countably many sets along with
their complements, in which each $R\in\mathcal{R}$ is a closed nowhere dense
set, and each member of $\mathcal{T}$ is again a converging sequence.

For every $\langle x,y\rangle\in\omega^{2}$ and $n\in\omega$ find an injective
sequence

$\big{\langle}\langle s^{\langle x,y\rangle,n}_{i},t^{\langle
x,y\rangle,n}_{i}\rangle:i\in\omega\big{\rangle}\in\big{(}E_{n}\cap\big{(}(\omega\setminus
A^{\delta}_{0})\times(\omega\setminus A^{\delta}_{1})\big{)}^{\omega}$

convergent to $\langle x,y\rangle$. This is possible because all elements of
$\mathcal{A}$ are nowhere dense in $\tau_{0}=\sigma_{0}$, and hence also in
all further topologies we construct. It follows that $T^{x,y}:=\\{t^{\langle
x,y\rangle,n}_{i}:i\in\omega\\}$ is almost disjoint from each
$Z\in\mathcal{Z}_{\alpha}$. Indeed, $E(T^{\langle
x,y\rangle,n},0)=\\{s^{\langle x,y\rangle,n}_{i}:i\in\omega\\}$ is a
convergent sequence in $\langle\omega,\tau_{\delta}\rangle$, whereas $E(Z,0)$
is closed discrete in this space for all $Z\in\mathcal{Z}_{\delta}$. Also,
$E(T^{\langle x,y\rangle,n},0)\cap A^{\delta}_{0}=\emptyset$ by the choice of
$T^{\langle x,y\rangle,n}$, and hence $T^{\langle x,y\rangle,n}\cap
E(A^{\delta}_{0},1)=\emptyset$ for all $x,y$ as above. In particular, since
$y\in\omega$ is arbitrary, it shows that $\omega\setminus E(A^{\delta}_{0},1)$
is dense333This is of course easy and follows directly from the fact that
$A^{\delta}_{0}$ is nowhere dense in $\langle\omega,\tau_{\delta}\rangle$. in
$\langle\omega,\sigma_{\delta}\rangle$. Also, $T^{\langle x,y\rangle,n}$ is
almost disjoint from $A^{\delta}_{1}$ for all $x,y$ by construction. Applying
Lemma 4.5.1 to $X=\langle\omega,\sigma_{\delta}\rangle$,
$\mathcal{T}=\mathcal{Z}_{\delta}\cup\\{T^{\langle
x,y\rangle,n}:x,y,n\in\omega\\},$
$\mathcal{R}=\\{E(A^{\delta}_{0},1),A^{\delta}_{1}\\}$, we get a
$0$-dimensional topology $\sigma^{-}_{\alpha}\supset\sigma_{\delta}$ on
$\omega$ in which elements of $\mathcal{T}$ are converging, and those of
$\mathcal{R}$ are closed discrete. By the choice of $\mathcal{T}$ we have that
$E_{n}$ is dense in
$\langle\omega\times\omega,\tau_{\delta}\otimes\sigma^{-}_{\alpha}\rangle$ for
all $n$, because there is a sequence in $E_{n}$ converging to each $\langle
x,y\rangle$. Thus condition $(5)$ will be satisfied because the topology
$\sigma_{\alpha}$ we plan to construct is going to be crowded and stronger
than $\sigma^{-}_{\alpha}$. For convenience we set
$\tau^{-}_{\alpha}=\tau_{\delta}$.

Next, we pass to the construction of $U_{\delta},V_{\delta}$ satisfying $(6)$.
This condition requires that if $x\in U_{\delta}$ and
$Y\in\mathcal{Y}_{\delta}$ converges to $x$, then $Y\subset^{*}U_{\delta}$. To
achieve this we shall use the following auxiliary

######  Claim 4.5.2.

There are families $\\{C_{n}:n\in\omega\\}$ and $\\{D_{n}:n\in\omega\\}$ such
that

  * $(8)$

$Y\subset^{*}C_{n}$ for any $Y\in\mathcal{Y}_{\alpha}$ converging to $n$ in
$\langle\omega,\tau^{-}_{\alpha}\rangle$;

  * $(9)$

$Z\subset^{*}D_{n}$ for any $Z\in\mathcal{Z}_{\alpha}$ converging to $n$ in
$\langle\omega,\sigma^{-}_{\alpha}\rangle$;

  * $(10)$

$C_{n}\cap C_{m}=\emptyset$ and $D_{n}\cap D_{m}=\emptyset$ for all $n\neq m$;

  * $(11)$

$E(C_{n},1)\cap D_{m}=\emptyset$ (or equivalently, $E(D_{m},0)\cap
C_{n}=\emptyset$) for any $n,m\in\omega$.

###### Proof.

For every $n\in\omega$ let us denote by $\mathcal{Y}_{\alpha,n}$ the family of
all $Y\in\mathcal{Y}_{\alpha}$ converging to $n$, and fix a family
$\\{O^{n}_{k}:k\in\omega\setminus\\{n\\}\\}\subset\tau_{0}$ such that
$O^{n}_{k}$ is a clopen neighbourhood of $k$ not containing $n$. Then $|Y\cap
O^{n}_{k}|<\omega$ for all $Y\in\mathcal{Y}_{\alpha,n}$ and $k\neq n$, and
hence Lemma 3.1 implies that there exists $C^{0}_{n}\subset\omega$ such that
$Y\subset^{*}C^{0}_{n}$ and $|C^{0}_{n}\cap O^{n}_{k}|<\omega$ for all $Y,k$
as above. Let us note that $|C^{0}_{n}\cap Y|<\omega$ for all
$Y\in\mathcal{Y}_{\alpha}\setminus\mathcal{Y}_{\alpha,n}$ because
$Y\subset^{*}O^{n}_{k}$ for $k$ being the limit point of $Y$. Thus letting
$C^{1}_{0}=C^{0}_{0}$ and
$C^{1}_{n}=C^{0}_{n}\setminus\bigcup_{n^{\prime}<n}C^{0}_{n^{\prime}}$ we get
that $\\{C^{1}_{n}:n\in\omega\\}$ is a disjoint family such that
$Y\subset^{*}C^{1}_{n}$ for all $Y\in\mathcal{Y}_{\alpha,n}$. Similarly we can
get a disjoint family $\\{D^{1}_{n}:n\in\omega\\}$ such that
$Z\subset^{*}D^{1}_{n}$ for all $Z\in\mathcal{Z}_{\alpha,n}$, where
$\mathcal{Z}_{\alpha,n}=\\{Z\in\mathcal{Z}_{\alpha}:Z$ converges to $n\\}$.
Finally, by item $(4)$ and the property of $\mathcal{A}$ stated in
$(\ast_{\mathbb{Q}})$ we can find $C,D\subset\omega$ such that
$\phi^{\alpha}_{0}(Y)\subset^{*}C$ and $\phi^{\alpha}_{1}(Y)\subset^{*}D$ for
all $Y\in\mathcal{Y}_{\alpha}$, and $|\psi^{\alpha}_{1}(Z)\cap D|<\omega$ and
$|\psi^{\alpha}_{0}(Z)\cap C|<\omega$ for all $Z\in\mathcal{Z}_{\alpha}$.

Set $C_{n}=C^{1}_{n}\cap C\cap E(D,0)$ and
$D_{n}=D^{1}_{n}\cap(\omega\setminus D)\cap(\omega\setminus E(C,1))$. We claim
that the families $\\{C_{n}:n\in\omega\\}$ and $\\{D_{n}:n\in\omega\\}$ are as
required. Indeed, for $(8)$ let us note that for all
$Y\in\mathcal{Y}_{\alpha,n}$ we have $Y\subset^{*}C^{1}_{n}$,
$Y\subset\phi^{\delta}_{0}(Y)\subset^{*}C$, and
$E(Y,1)\subset\phi^{\alpha}_{1}(Y)\subset^{*}D$, the latter implying
$Y\subset^{*}E(D,0)$, and hence $Y\subset^{*}C_{n}$. $(9)$ is analogous: for
all $Z\in\mathcal{Z}_{\alpha,n}$ we have $Z\subset^{*}D^{1}_{n}$,
$Z\subset\psi^{\delta}_{1}(Z)\subset^{*}\omega\setminus D$, and
$E(Z,0)\subset\psi^{\alpha}_{0}(Z)\subset^{*}\omega\setminus C$, the latter
implying $Z\subset^{*}\omega\setminus E(C,1)$, and hence $Z\subset^{*}D_{n}$.
Condition $(10)$ follows from $C_{n}\subset C^{1}_{n}$ and $D_{n}\subset
D^{1}_{n}$. And finally,

$E(C_{n},1)\cap D_{m}\subset E\big{(}E(D,0),1\big{)}\cap(\omega\setminus
D)=D\cap(\omega\setminus D)=\emptyset$

for all $n,m\in\omega$, which yields $(11).$ ∎

For every $\langle x,y\rangle\in\omega^{2}$ and $n\in\omega$ we fix an
injective sequence

$\big{\langle}\langle s^{\langle x,y\rangle,n}_{k},t^{\langle
x,y\rangle,n}_{k}\rangle:k\in\omega\big{\rangle}\in(E_{n}\setminus
F^{\delta}_{n})^{\omega}$

converging to $\langle x,y\rangle$ in
$\langle\omega\times\omega,\tau^{-}_{\alpha}\otimes\sigma^{-}_{\alpha}\rangle$.
We shall in addition assume that these sequences are mutually disjoint, i.e.,
$s^{\langle x,y\rangle,n}_{k}=s^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}_{k^{\prime}}$ (resp. $t^{\langle
x,y\rangle,n}_{k}=t^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}_{k^{\prime}}$) iff $x=x^{\prime}$,
$y=y^{\prime}$, $n=n^{\prime}$, and $k=k^{\prime}$. Set $S^{\langle
x,y\rangle,n}=\\{s^{\langle x,y\rangle,n}_{k}:k\in\omega\\}$ and $T^{\langle
x,y\rangle,n}=\\{t^{\langle x,y\rangle,n}_{k}:k\in\omega\\}$ for all $\langle
x,y\rangle$ and $n$ as above.

Replacing the $D_{m}$’s and $C_{m}$’s with smaller sets, if necessary, we may
additionally assume that

$|S^{\langle x,y\rangle,n}\cap C_{m}|<\omega\mbox{ \ and\ \ }|T^{\langle
x,y\rangle,n}\cap D_{m}|<\omega$

for all $\langle x,y\rangle$, $n$, and $m$. Indeed, since $E(Y,1)$ is closed
and discrete in $\langle\omega,\sigma^{-}_{\alpha}\rangle$ for every
$Y\in\mathcal{Y}_{\alpha}$ and $E(S^{\langle x,y\rangle,n},1)=T^{\langle
x,y\rangle,n}$ is a convergent sequence in this topology, we conclude that
$|S^{\langle x,y\rangle,n}\cap Y|<\omega$ for all $\langle x,y\rangle,n$ as
above. Since there are no $(\omega,<\mathfrak{b})$ gaps, there exists
$S\subset\omega$ such that $S^{\langle x,y\rangle,n}\subset^{*}S$ and $|Y\cap
S|<\omega$ for all $\langle x,y\rangle,n$. Thus replacing $C_{m}$ with
$C_{m}\setminus S$ for all $m\in\omega$, if necessary, we may assume that
$|S^{\langle x,y\rangle,n}\cap C_{m}|<\omega$ for all $\langle x,y\rangle,n$,
in addition to all properties of $C_{m}$ stated above. Analogously with
$|T^{\langle x,y\rangle,n}\cap D_{m}|<\omega$ for all $\langle x,y\rangle,n$.

Finally, we pass to the construction of $U_{\delta}$ and $V_{\delta}$. This
will be done recursively over $k\in\omega$, namely they will be constructed as
increasing unions $\bigcup_{k\in\omega}U^{\delta,1}_{k}$ and
$\bigcup_{k\in\omega}V^{\delta,1}_{k}$, respectively. Moreover, we shall make
sure that adding their complements to the corresponding topologies can also be
done without any harm, by constructing these as increasing unions
$\bigcup_{k\in\omega}U^{\delta,0}_{k}$ and
$\bigcup_{k\in\omega}V^{\delta,0}_{k}$, respectively. We construct these
objects along with

  * •

non-decreasing sequences $\langle H^{j}_{k}:k\in\omega\rangle$ and $\langle
G^{j}_{k}:k\in\omega\rangle$ of finite subsets of $\omega$, where $j\in 2$;
and

  * •

for each $x\in\bigcup_{k\in\omega}H^{1}_{k}\cup\bigcup_{k\in\omega}H^{0}_{k}$,
$y\in\bigcup_{k\in\omega}G^{1}_{k}\cup\bigcup_{k\in\omega}G^{0}_{k}$ and
$n\in\omega$ cofinal subsets $\tilde{C}_{x}$, $\tilde{D}_{y}$,
$\tilde{S}^{\langle x,y\rangle,n}$ and $\tilde{T}^{\langle x,y\rangle,n}$ of
$C_{x}$, $D_{y}$, $S^{\langle x,y\rangle,n}$ and $T^{\langle x,y\rangle,n}$,
respectively,

such that $H^{1}_{0}=\\{x^{1}_{0}\\}$ and $G^{1}_{0}=\\{y^{1}_{0}\\}$ for some
$\langle x^{1}_{0},y^{1}_{0}\rangle\not\in\bigcup_{n\in\omega}F^{\delta}_{n}$,
$H^{0}_{0}=G^{0}_{0}=\emptyset$, and the following conditions are satisfied
for all $k\in\omega$ and $i\in 2$:

  * $(i)$

$H^{1}_{k}\cap H^{0}_{k}=G^{1}_{k}\cap G^{0}_{k}=\emptyset$;

  * $(ii)$

$U^{\delta,i}_{k}=\big{[}H^{i}_{k}\cup\bigcup\\{\tilde{C}_{x}:x\in
H^{i}_{k}\\}\cup\\\ \cup\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k,\langle x,y\rangle\in H^{i}_{k}\times(G^{1}_{k}\cup
G^{0}_{k})\\}\big{]}\cup\Delta^{i}_{k}$,
$V^{\delta,i}_{k}=\big{[}G^{i}_{k}\cup\bigcup\\{\tilde{D}_{y}:y\in
G^{i}_{k}\\}\cup\\\ \cup\bigcup\\{\tilde{T}^{\langle x,y\rangle,n}:n\leq
k,\langle x,y\rangle\in(H^{1}_{k}\cup H^{0}_{k})\times
G^{i}_{k}\\}\big{]}\cup\Sigma^{i}_{k}$;

  * $(iii)$

$\Delta^{1}_{k}=\emptyset$ and $\Delta^{0}_{k}=\\{k\\}$ if $k\not\in
U^{\delta,1}_{k}$, otherwise $\Delta^{0}_{k}=\emptyset$;
$\Sigma^{1}_{k}=\emptyset$ and $\Sigma^{0}_{k}=\\{k\\}$ if $k\not\in
V^{\delta,1}_{k}$, otherwise $\Sigma^{0}_{k}=\emptyset$;

  * $(iv)$

$U^{\delta,1}_{k}\cap U^{\delta,0}_{k}=V^{\delta,1}_{k}\cap
V^{\delta,0}_{k}=\emptyset$;

  * $(v)$

If $\langle x,y\rangle\in(H^{1}_{k}\cup H^{0}_{k})\times(G^{1}_{k}\cup
G^{0}_{k})$, then $\tilde{T}^{\langle x,y\rangle,n}=E(\tilde{S}^{\langle
x,y\rangle,n},1)$ for all $n\leq k$;

  * $(vi)$

$(U^{\delta,1}_{k}\times
V^{\delta,1}_{k})\cap\bigcup_{n\in\omega}F^{\delta}_{n}=\emptyset$;

  * $(vii)$

$H^{i}_{k+1}=H^{i}_{k}\cup\\{\min(U^{\delta,i}_{k}\setminus H^{i}_{k})\\}$
provided that $H^{i}_{k}\neq\emptyset$, and $H^{i}_{k+1}=H^{i}_{k}=\emptyset$
otherwise; $G^{i}_{k+1}=G^{i}_{k}\cup\\{\min(V^{\delta,i}_{k}\setminus
G^{i}_{k})\\}$ provided that $G^{i}_{k}\neq\emptyset$, and
$G^{i}_{k+1}=G^{i}_{k}=\emptyset$ otherwise.

To start off the inductive construction, set
$\tilde{C}_{x^{1}_{0}}=C_{x^{1}_{0}}\setminus E(\\{y^{1}_{0}\\},0),$
$\tilde{D}_{y^{1}_{0}}=D_{y^{1}_{0}}\setminus E(\\{x^{1}_{0}\\},1),$

$\hat{T}^{\langle x^{1}_{0},y^{1}_{0}\rangle,0}=T^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\setminus\big{(}E(C_{x^{1}_{0}},1)\cup
E(\\{x^{1}_{0}\\},1)\big{)},$ $\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}=S^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\setminus\big{(}E(D_{y^{1}_{0}},0)\cup
E(\\{y^{1}_{0}\\},0)\big{)}$

and then

$\tilde{T}^{\langle x^{1}_{0},y^{1}_{0}\rangle,0}=\hat{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\cap E(\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},1)$

and

$\tilde{S}^{\langle x^{1}_{0},y^{1}_{0}\rangle,0}=\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\cap E(\hat{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},0).$

The equation

$\displaystyle E(\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},1)=E\big{(}\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\cap E(\hat{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},0),1\big{)}=$ $\displaystyle E(\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},1)\cap E\big{(}E(\hat{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},0),1\big{)}=$ $\displaystyle=E(\hat{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0},1)\cap\hat{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}=\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}$

proves $(v)$ for $k=0$. To check $(vi)$ note that

$\displaystyle U^{0,1}_{0}\times
V^{0,1}_{0}=\big{(}\\{x^{1}_{0}\\}\cup\tilde{C}_{x^{1}_{0}}\cup\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\big{)}\times\big{(}\\{y^{1}_{0}\\}\cup\tilde{D}_{y^{1}_{0}}\cup\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\big{)}=$ $\displaystyle\\{\langle
x^{1}_{0},y^{1}_{0}\rangle\\}\cup(\\{x^{1}_{0}\\}\times\tilde{D}_{y^{1}_{0}})\cup(\\{x^{1}_{0}\\}\times\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0})\cup$
$\displaystyle\cup(\tilde{C}_{x^{1}_{0}}\times\\{y^{1}_{0}\\})\cup(\tilde{C}_{x^{1}_{0}}\times\tilde{D}_{y^{1}_{0}})\cup(\tilde{C}_{x^{1}_{0}}\times\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0})\cup$ $\displaystyle\cup(\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\times\\{y^{1}_{0}\\})\cup(\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\times\tilde{D}_{y^{1}_{0}})\cup(\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\times\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}),$

i.e., $U^{0,1}_{0}\times V^{0,1}_{0}$ is a union of 9 sets, of which 7 ones
“in the middle” have empty intersection with $\bigcup_{n\in\omega}E_{n}$
because $A\times B\cap\bigcup_{n\in\omega}E_{n}=\emptyset$ provided that
$A\cap E(B,0)=\emptyset$, which is the case for these 7 products by the choice
or definition of the corresponding sets. Also, $\langle
x^{1}_{0},y^{1}_{0}\rangle\not\in\bigcup_{n\in\omega}F^{\delta}_{n}$ by the
choice, whereas

$\tilde{S}^{\langle x^{1}_{0},y^{1}_{0}\rangle,0}\times\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\cap\bigcup_{n\in\omega}E_{n}=\\{\langle
s^{\langle x^{1}_{0},y^{1}_{0}\rangle,0}_{k},t^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}_{k}\rangle:k\in\omega\\}\subset E_{0}\setminus
F^{\delta}_{0},$

and hence $\tilde{S}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}\times\tilde{T}^{\langle
x^{1}_{0},y^{1}_{0}\rangle,0}$ also has empty intersection with
$\bigcup_{n\in\omega}F^{\delta}_{n}$. This completes the case $k=0$ since all
other items are easily checked.

Suppose now that items $(i)-(vii)$ hold for $k$ and note that $(vii)$ gives
the unique way to define $H^{i}_{k+1}$ and $G^{i}_{k+1}$ for $i\in 2$, and
$(iv)$ for $k$ yields $(i)$ for $k+1$. Given $i\in 2$ and $x\in
H^{i}_{k+1}\setminus H^{i}_{k}$ and $y\in G^{i}_{k+1}\setminus G^{i}_{k}$, set

$\displaystyle\tilde{C}_{x}=C_{x}\setminus\Big{(}E(G^{1}_{k+1},0)\cup\bigcup\big{\\{}E(T^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},0)\>:$ $\displaystyle\langle
x^{\prime},y^{\prime}\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times(G^{1}_{k+1}\cup G^{0}_{k+1}),n^{\prime}\leq k\big{\\}}\cup
H^{0}_{k+1}\Big{)}=$
$\displaystyle=C_{x}\setminus\Big{(}E(G^{1}_{k+1},0)\cup\bigcup\big{\\{}S^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\>:$ $\displaystyle\langle
x^{\prime},y^{\prime}\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times(G^{1}_{k+1}\cup G^{0}_{k+1}),n^{\prime}\leq k\big{\\}}\cup
H^{0}_{k+1}\Big{)}$

and

$\displaystyle\tilde{D}_{y}=D_{y}\setminus\Big{(}E(H^{1}_{k+1},1)\cup\bigcup\big{\\{}E(S^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},1)\>:$ $\displaystyle\langle
x^{\prime},y^{\prime}\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times(G^{1}_{k+1}\cup G^{0}_{k+1}),n^{\prime}\leq k\big{\\}}\cup
G^{0}_{k+1}\Big{)}=$
$\displaystyle=D_{y}\setminus\Big{(}E(H^{1}_{k+1},1)\cup\bigcup\big{\\{}T^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\>:$ $\displaystyle\langle
x^{\prime},y^{\prime}\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times(G^{1}_{k+1}\cup G^{0}_{k+1}),n^{\prime}\leq k\big{\\}}\cup
G^{0}_{k+1}\Big{)}.$

For every

$\displaystyle\langle\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}\rangle\in\big{(}(H^{1}_{k+1}\cup
H^{0}_{k+1})\times(G^{1}_{k+1}\cup G^{0}_{k+1})\times(k+2)\big{)}\setminus$
$\displaystyle\setminus\big{(}(H^{1}_{k}\cup H^{0}_{k})\times(G^{1}_{k}\cup
G^{0}_{k})\times(k+1)\big{)}$

we let

$\displaystyle\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}=T^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\setminus\big{[}E\big{(}\bigcup\big{\\{}C_{x}:x\in
H^{1}_{k+1}\big{\\}}\cup H^{1}_{k+1},1\big{)}\cup G^{0}_{k+1}\cup$
$\displaystyle\cup\bigcup\big{\\{}D_{y}:y\in G^{1}_{k+1}\cup
G^{0}_{k+1}\big{\\}}\big{]},$ $\displaystyle\hat{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}=S^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\setminus\big{[}E\big{(}\bigcup\big{\\{}D_{y}:y\in
G^{1}_{k+1}\big{\\}}\cup G^{1}_{k+1},0\big{)}\cup H^{0}_{k+1}\cup$
$\displaystyle\cup\bigcup\big{\\{}C_{x}:x\in H^{1}_{k+1}\cup
H^{0}_{k+1}\big{\\}}\big{]},$

and finally

$\displaystyle\tilde{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}:=\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\cap E(\hat{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},1),$ $\displaystyle\tilde{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}:=\hat{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\cap E(\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},0).$

The equality

$\displaystyle E(\tilde{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},1)=E\big{(}\hat{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\cap E(\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},0),1\big{)}=$ $\displaystyle
E(\hat{S}^{\langle x^{\prime},y^{\prime}\rangle,n^{\prime}},1)\cap
E\big{(}E(\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},0),1\big{)}=$
$\displaystyle=E(\hat{S}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}},1)\cap\hat{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}=\tilde{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}$

for all $\langle\langle x^{\prime},y^{\prime}\rangle,n^{\prime}\rangle$ as
above yields $(v)$ for $k+1$. Let $U^{\delta,i}_{k+1}$ and
$V^{\delta,i}_{k+1}$ be defined according to $(ii)$ and $(iii)$ for $k+1$. To
check $(vi)$ for $(k+1)$, let us consider the product

$\begin{array}[]{ll}&U^{\delta,1}_{k+1}\times V^{\delta,1}_{k+1}=\\\
(\alpha)&=\big{(}H^{1}_{k+1}\times
G^{1}_{k+1}\big{)}\cup\big{(}H^{1}_{k+1}\times\bigcup\\{\tilde{D}_{y}:y\in
G^{1}_{k+1}\\}\big{)}\bigcup\\\
(\beta)&\bigcup\big{(}H^{1}_{k+1}\times\bigcup\\{\tilde{T}^{\langle
x,y\rangle,n}:n\leq k+1,\langle x,y\rangle\in\\\ &\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times G^{1}_{k+1}\\}\big{)}\bigcup\\\
(\gamma)&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in H^{1}_{k+1}\\}\times
G^{1}_{k+1}\big{)}\bigcup\\\
(\zeta)&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in
H^{1}_{k+1}\\}\times\bigcup\\{\tilde{D}_{y}:y\in
G^{1}_{k+1}\\}\big{)}\bigcup\\\
(\varepsilon)&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in
H^{1}_{k+1}\\}\times\\\ &\times\bigcup\\{\tilde{T}^{\langle
x,y\rangle,n}:n\leq k+1,\langle x,y\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times G^{1}_{k+1}\\}\big{)}\bigcup\\\
(\eta)&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in\\\ &\in H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\times G^{1}_{k+1}\big{)}\bigcup\\\
(\theta)&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\times\\\ &\times\bigcup\\{\tilde{D}_{y}:y\in
G^{1}_{k+1}\\}\big{)}\bigcup\\\
(\lambda)&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\times\\\ &\times\bigcup\\{\tilde{T}^{\langle
x,y\rangle,n}:n\leq k+1,\langle x,y\rangle\in(H^{1}_{k+1}\cup
H^{0}_{k+1})\times G^{1}_{k+1}\\}\big{)}.\end{array}$

To show that this product is disjoint from
$F=\bigcup_{n\in\omega}F^{\delta}_{n}$ we shall analyze sets appearing in the
above long formula one by one. $H^{1}_{k+1}\times G^{1}_{k+1}$ is disjoint
from $F$ by our inductive assumption, namely $(vi)$ for $k$, since
$H^{1}_{k+1}\subset U^{\delta,1}_{k}$ and $G^{1}_{k+1}\subset
V^{\delta,1}_{k}$. The second product in $(\alpha)$ is disjoint from $F$
because we have explicitly made $E(H^{1}_{k+1},1)$ disjoint from
$\tilde{D}_{y}$ if $y\in G^{1}_{k+1}\setminus G^{1}_{k}$, and hence
$H^{1}_{k+1}\times\tilde{D}_{y}$ is disjoint even from $E$ in this case. And
if $y\in G^{1}_{k}$ then we can use $(vi)$ for $k$ since $H^{1}_{k+1}\subset
U^{\delta,1}_{k}$ and $\tilde{D}_{y}\subset V^{\delta,1}_{k}$. Analogously we
can show that also the product in item $(\gamma)$ is disjoint from $F$.

The product in $(\beta)$ is disjoint from $F$ because we have explicitly made
$E(H^{1}_{k+1},1)$ disjoint from $\tilde{T}^{\langle x,y\rangle,n}$ if
$\langle\langle x,y\rangle,n\rangle$ lies in the difference $N:=$

$\big{(}((H^{1}_{k+1}\cup H^{0}_{k+1})\times
G^{1}_{k+1})\times(k+2)\big{)}\setminus\big{(}((H^{1}_{k}\cup H^{0}_{k})\times
G^{1}_{k})\times(k+1)\big{)},$

and if

$\langle\langle x,y\rangle,n\rangle\in((H^{1}_{k}\cup H^{0}_{k})\times
G^{1}_{k})\times(k+1)$

then we can use $(vi)$ for $k$ since $H^{1}_{k+1}\subset U^{\delta,1}_{k}$ and
$\tilde{T}^{\langle x,y\rangle,n}\subset V^{\delta,1}_{k}$. Analogously we can
show that also the product in item $(\eta)$ is disjoint from $F$.

The product in $(\zeta)$ is disjoint even from $E$ since $E(C_{x},1)\cap
D_{y}=\emptyset$ for any $x,y\in\omega$ by Claim 4.5.2$(11)$.

The product in item $(\varepsilon)$ is disjoint from $F$ because we have
explicitly made $E(\tilde{C}_{x},1)$ disjoint from $\tilde{T}^{\langle
x^{\prime},y^{\prime}\rangle,n}$ if $x\in H^{1}_{k+1}\setminus H^{1}_{k}$ or
$\langle\langle x^{\prime},y^{\prime}\rangle,n\rangle\in N$, and if both $x\in
H^{1}_{k}$ and

$\langle\langle x^{\prime},y^{\prime}\rangle,n\rangle\in(H^{1}_{k}\cup
H^{0}_{k})\times(G^{1}_{k}\cup G^{0}_{k})\times(k+1)$

then we can use $(vi)$ for $k$ since $\tilde{C}_{x}\subset U^{\delta,1}_{k}$
and $\tilde{T}^{\langle x^{\prime},y^{\prime}\rangle,n}\subset
V^{\delta,1}_{k}$. Analogously we can show that also the product in item
$(\theta)$ is disjoint from $F$.

Finally, consider the product in $(\lambda)$. Given $\langle\langle
x,y\rangle,n\rangle$ and $\langle\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}\rangle$ in

$((H^{1}_{k+1}\cup H^{0}_{k+1})\times G^{1}_{k+1})\times(k+2),$

two cases are possible. If $\langle\langle x,y\rangle,n\rangle=\langle\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}\rangle$, then $\tilde{T}^{\langle
x,y\rangle,n}=E(\tilde{S}^{\langle x,y\rangle,n},1)$, and hence

$\displaystyle(\tilde{S}^{\langle x,y\rangle,n}\times\tilde{T}^{\langle
x,y\rangle,n})\bigcap E\subset\\{\langle s^{\langle
x,y\rangle,n}_{k},t^{\langle x,y\rangle,n}_{k}\rangle:k\in\omega\\}\subset
E_{n}\setminus F^{\delta}_{n},$

i.e., $\tilde{S}^{\langle x,y\rangle,n}\times\tilde{T}^{\langle x,y\rangle,n}$
is disjoint from $F$. If $\langle\langle x,y\rangle,n\rangle\neq\langle\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}\rangle$, then

$E(\tilde{S}^{\langle x,y\rangle,n},1)=\tilde{T}^{\langle x,y\rangle,n}\subset
T^{\langle x,y\rangle,n},$

and hence

$E(\tilde{S}^{\langle x,y\rangle,n},1)\cap\tilde{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}}\subset T^{\langle x,y\rangle,n}\cap
T^{\langle x^{\prime},y^{\prime}\rangle,n^{\prime}}=\emptyset,$

which implies that $(\tilde{S}^{\langle x,y\rangle,n}\cap\tilde{T}^{\langle
x^{\prime},y^{\prime}\rangle,n^{\prime}})\cap E=\emptyset$. This completes our
proof of $(vi)$ for $(k+1)$.

Finally, it remains to check $(iv)$ for $k+1$. We can write
$U^{\delta,1}_{k+1}\cap U^{\delta,0}_{k+1}$ as follows:

$\begin{array}[]{ll}&U^{\delta,1}_{k+1}\cap U^{\delta,0}_{k+1}=\\\
(\alpha_{\cap})&=\big{(}H^{1}_{k+1}\cap
H^{0}_{k+1}\big{)}\bigcup\big{(}H^{1}_{k+1}\cap\bigcup\\{\tilde{C}_{x}:x\in
H^{0}_{k+1}\\}\big{)}\bigcup\\\
(\beta_{\cap})&\bigcup\big{(}H^{1}_{k+1}\cap\bigcup\\{\tilde{S}^{\langle
x,y\rangle,n}:n\leq k+1,\langle x,y\rangle\in\\\ &\in
H^{0}_{k+1}\times(G^{1}_{k+1}\cup G^{0}_{k+1})\\}\big{)}\bigcup\\\
(\gamma_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in H^{1}_{k+1}\\}\cap
H^{0}_{k+1}\big{)}\bigcup\\\
(\zeta_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in
H^{1}_{k+1}\\}\cap\bigcup\\{\tilde{C}_{x}:x\in H^{0}_{k+1}\\}\big{)}\bigcup\\\
(\varepsilon_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{C}_{x}:x\in
H^{1}_{k+1}\\}\cap\\\ &\cap\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in H^{0}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\big{)}\bigcup\\\
(\eta_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in\\\ &\in H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\cap H^{0}_{k+1}\big{)}\bigcup\\\
(\theta_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq
k+1,\langle x,y\rangle\in H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\cap\\\ &\cap\bigcup\\{\tilde{C}_{x}:x\in
H^{0}_{k+1}\\}\big{)}\bigcup\\\
(\lambda_{\cap})&\bigcup\big{(}\bigcup\\{\tilde{S}^{\langle
x,y\rangle,n}:n\leq k+1,\langle x,y\rangle\in
H^{1}_{k+1}\times(G^{1}_{k+1}\cup G^{0}_{k+1})\\}\cap\\\
&\cap\bigcup\\{\tilde{S}^{\langle x,y\rangle,n}:n\leq k+1,\langle
x,y\rangle\in H^{0}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\\}\big{)}.\end{array}$

Similarly (but easier) to the proof of $(vi)$ we can check that all the
intersections in the formula displayed above are empty. For instance, let us
consider the intersection in item $(\theta_{\cap})$. Fix any

$\langle\langle x,y\rangle,n\rangle\in\big{(}H^{1}_{k+1}\times(G^{1}_{k+1}\cup
G^{0}_{k+1})\big{)}\times(k+2)$

and $x^{\prime}\in H^{0}_{k+1}$. If $x^{\prime}\not\in H^{0}_{k}$, then we
have explicitly subtracted $S^{\langle x,y\rangle,n}$ from $C_{x^{\prime}}$ in
the definition of $\tilde{C}_{x^{\prime}}$, so
$\tilde{C}_{x^{\prime}}\cap\tilde{S}^{\langle x,y\rangle,n}=\emptyset$. If
$\langle\langle x,y\rangle,n\rangle\in N$ (see the proof of $(vi)$ for the
definition thereof), then we have explicitly subtracted $C_{x^{\prime}}$ from
$S^{\langle x,y\rangle,n}$ in the definition of $\tilde{S}^{\langle
x,y\rangle,n}$, so again $\tilde{C}_{x^{\prime}}\cap\tilde{S}^{\langle
x,y\rangle,n}=\emptyset$. Otherwise $\tilde{S}^{\langle x,y\rangle,n}\subset
U^{\delta,1}_{k}$ and $\tilde{C}_{x^{\prime}}\subset U^{\delta,0}_{k}$ and
therefore we can use $(iv)$ for $k$ in order to get
$\tilde{C}_{x^{\prime}}\cap\tilde{S}^{\langle x,y\rangle,n}=\emptyset$.

Thus the inductive construction of all the objects needed for defining
$U_{\delta}$ and $V_{\delta}$, so that conditions $(i)-(vii)$ are satisfied,
is finished.

As planned, we set $U_{\delta}=\bigcup_{k\in\omega}U^{\delta,1}_{k}$,
$V_{\delta}=\bigcup_{k\in\omega}V^{\delta,1}_{k}$, and note that
$\omega\setminus U_{\delta}=\bigcup_{k\in\omega}U^{\delta,0}_{k}$ and
$\omega\setminus V_{\delta}=\bigcup_{k\in\omega}V^{\delta,0}_{k}$ by $(iii)$
and $(iv)$. Let $\tau_{\alpha}$ and $\sigma_{\alpha}$ be the topologies
generated by $\tau^{-}_{\alpha}\cup\\{U_{\delta},\omega\setminus
U_{\delta}\\}$ and $\sigma^{-}_{\alpha}\cup\\{V_{\delta},\omega\setminus
V_{\delta}\\}$ as a base, respectively. This way we get $0$-dimensional
topologies and $(vi)$ implies that $(6)$ is satisfied. Moreover, $(vii)$ gives
that $U_{\delta}=\bigcup_{k\in\omega}H^{1}_{k}$, $\omega\setminus
U_{\delta}=\bigcup_{k\in\omega}H^{0}_{k}$,
$V_{\delta}=\bigcup_{k\in\omega}G^{1}_{k}$, and $\omega\setminus
V_{\delta}=\bigcup_{k\in\omega}G^{0}_{k}$. Thus by $(ii)$ for every $x\in
U_{\delta}$ the set $C_{x}$ is almost contained in $U_{\delta}$, and hence
also $Y\subset^{*}U_{\delta}$ for all $Y\in\mathcal{Y}_{\alpha}$ converging to
$x$. The same holds for $\omega\setminus U_{\delta}.$ It follows that all
$Y\in\mathcal{Y}_{\alpha}$ remain convergent sequences in
$\langle\omega,\tau_{\alpha}\rangle$. Analogously, all
$Z\in\mathcal{Z}_{\alpha}$ remain convergent sequences in
$\langle\omega,\sigma_{\alpha}\rangle$, and hence $(3)$ holds for $\alpha$.
Finally, whenever $n\in\omega$, $\ell\in 2^{2}$, and $\langle x,y\rangle\in
U^{(\ell(0))}_{\delta}\times V^{(\ell(1))}_{\delta}$, then444For
$I\subset\omega$ we set $I^{(1)}=I$ and $I^{(0)}=\omega\setminus I$., then
$\\{\langle s^{\langle x,y\rangle,n}_{k},s^{\langle
x,y\rangle,n}_{k}\rangle:k\in\omega\\}\subset^{*}U^{(\ell(0))}_{\delta}\times
V^{(\ell(1))}_{\delta}$. Indeed, by $(vii)$ there exists $k\in\omega$ such
that $x\in U^{\delta,\ell(0)}_{k}$ and $y\in V^{\delta,\ell(1)}_{k}$, and
therefore $(ii)$ implies

$\tilde{S}^{\langle x,y\rangle,n}\subset^{*}U^{\delta,\ell(0)}_{k}\subset
U^{\ell(0)}_{\delta}\mbox{\ \ and\ \ }\tilde{T}^{\langle
x,y\rangle,n}\subset^{*}V^{\delta,\ell(1)}_{k}\subset V^{\ell(1)}_{\delta}.$

As a result $E_{n}$ remains dense in
$\langle\omega\times\omega,\tau_{\alpha}\otimes\sigma_{\alpha}\rangle$ for all
$n$, because there is an injective sequence in $E_{n}$ converging to each
$\langle x,y\rangle$. This also proves that there are no isolated points in
$\langle\omega\times\omega,\tau_{\alpha}\otimes\sigma_{\alpha}\rangle$. This
completes the verification of $(1)-(7)$ for $\delta=\alpha$.

III. $\alpha=\delta+1$, $\delta$ is odd, $x_{\delta}$ is a limit point of
$S_{\delta}$ in $\langle\omega,\sigma_{\delta}\rangle$, and $|Z\cap
S_{\delta}|<\omega$ for all $Z\in\mathcal{Z}_{\delta}$ converging to
$x_{\delta}$.
In this case we simply repeat our argument from Case II, with the roles of
$\mathcal{Z}_{\delta}$ and $\mathcal{Y}_{\delta}$ interchanged, so that again
$(1)$-$(7)$ are satisfied for $\delta=\alpha$.

IV. $\alpha=\delta+1$, $\delta$ is even, and either $x_{\delta}$ is _not_ a
limit point of $S_{\delta}$ in $\langle\omega,\tau_{\delta}\rangle$, or it is
and there exists $Y\in\mathcal{Y}_{\delta}$ converging to $x_{\delta}$ in
$\langle\omega,\tau_{\delta}\rangle$ with $|Y\cap S_{\delta}|=\omega$. Then we
set $\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta}$,
$\mathcal{Z}_{\alpha}=\mathcal{Z}_{\delta}$,
$\tau^{-}_{\alpha}=\tau_{\delta},$ $\sigma^{-}_{\alpha}=\sigma_{\delta}$, and
repeat the argument from Case II starting from Claim 4.5.2.

V. $\alpha=\delta+1$, $\delta$ is odd, and either $x_{\delta}$ is _not_ a
limit point of $S_{\delta}$ in $\langle\omega,\sigma_{\delta}\rangle$, or it
is and there exists $Z\in\mathcal{Z}_{\delta}$ converging to $x_{\delta}$ in
$\langle\omega,\sigma_{\delta}\rangle$ with $|Z\cap S_{\delta}|=\omega$. Then
we set $\mathcal{Y}_{\alpha}=\mathcal{Y}_{\delta}$,
$\mathcal{Z}_{\alpha}=\mathcal{Z}_{\delta}$,
$\tau^{-}_{\alpha}=\tau_{\delta},$ $\sigma^{-}_{\alpha}=\sigma_{\delta}$, and
repeat the argument from Case III starting from Claim 4.5.2.

All in all, this completes our construction of the objects mentioned in
$(1)$-$(7)$ for all $\delta<\mathfrak{c}$, so that these conditions are
satisfied. Conditions $(3)$ and $(2)$ imply that
$X=\langle\omega,\tau_{\mathfrak{c}}\rangle$ and
$Y=\langle\omega,\sigma_{\mathfrak{c}}\rangle$ are Frechet-Urysohn, and
$X\times Y$ is not $M$-separable by $(6)$. Moreover, both spaces are zero-
dimensional since we started from zero-dimensional topologies and always
enlarged them by adding new sets along with their complements. This completes
our proof of Theorem 4.5. ∎

Finally, Theorem 1.2 is a direct consequence of Theorems 4.4 and 4.5.

##  5\. Products of $H$-separable spaces in the Laver model

We will need to define the following new class of spaces.

######  Definition 5.1.

A topological space $\langle X,\tau\rangle$ is called _bounded box-separable_
(briefly, _b.b.-separable_) if for every function $R$ assigning to each
countable family $\mathcal{U}$ of non-empty open subsets of $X$ a sequence
$R(\mathcal{U})=\langle F_{n}:n\in\omega\rangle$ of finite non-empty subsets
of $X$ such that $\\{n:F_{n}\subset U\\}$ is infinite for every
$U\in\mathcal{U}$, there exists
$\mathbb{U}\subset[\tau\setminus\\{\emptyset\\}]^{\omega}$ of size
$|\mathbb{U}|=\omega_{1}$ and a sequence $\langle
l_{i}:i\in\omega\rangle\in\omega^{\omega}$ such that for all
$U\in\tau\setminus\\{\emptyset\\}$ there exists $\mathcal{U}\in\mathbb{U}$
such that for all but finite $i\in\omega$ there is $n\in[l_{i},l_{i+1})$ such
that $R(\mathcal{U})(n)\subseteq U$.

The proof of the following statement is close to that of [19, Lemma 2.2], the
only difference being a more careful analysis of sets of the form
$\\{n:R(\mathcal{U})(n)\subset U\\}$.

######  Proposition 5.2.

In the Laver model every countable $H$-separable space is b.b.-separable

###### Proof.

We work in $V[G_{\omega_{2}}]$, where $G_{\omega_{2}}$ is
$\mathbb{P}_{\omega_{2}}$-generic and $\mathbb{P}_{\omega_{2}}$ is the
iteration of length $\omega_{2}$ with countable supports of the Laver forcing,
see [14] for details. Let us fix an $H$-separable space of the form
$\langle\omega,\tau\rangle$ and a function $R$ such as in the definition of
b.b-separability. By a standard argument (see, e.g., the proof of [7, Lemma
5.10]) there exists an $\omega_{1}$-club $C\subseteq\omega_{2}$ such that for
every $\alpha\in C$ the following conditions hold:

  * (i)

$\tau\cap V[G_{\alpha}]\in V[G_{\alpha}]$ and for every sequence $\langle
D_{n}:n\in\omega\rangle\in V[G_{\alpha}]$ of dense subsets of
$\langle\omega,\tau\rangle$ there exists a sequence $\langle
K_{n}:n\in\omega\rangle\in V[G_{\alpha}]$ such that
$K_{n}\in[D_{n}]^{<\omega}$ and for every $U\in\tau\setminus\\{\emptyset\\}$
the intersection $U\cap K_{n}$ is non-empty for all but finitely many
$n\in\omega$;

  * (ii)

$R(\mathcal{U})\in V[G_{\alpha}]$ for any
$\mathcal{U}\in[\tau\setminus\\{\emptyset\\}]^{\omega}\cap V[G_{\alpha}]$; and

  * (iii)

For every $A\in\mathcal{P}(\omega)\cap V[G_{\alpha}]$ the interior $Int(A)$
also belongs to $V[G_{\alpha}]$.

By [14, Lemma 11] there is no loss of generality in assuming that $0\in C$.
Set $\mathbb{U}:=[\tau\setminus\\{\emptyset\\}]^{\omega}\cap V$. It suffices
to prove the following auxiliary

######  Lemma 5.3.

For every $A\in\tau\setminus\\{\emptyset\\}$ there are
$\mathcal{U}\in\mathbb{U},J\in[\omega]^{\omega}\cap V[G_{1}]$ such that for
every consecutive $j,j^{\prime}\in J$ there exists $n\in[j,j^{\prime})$ with
the property $R(\mathcal{U})(n)\subset A$.

###### Proof.

We shall prove an equivalent statement:

> For every $A\in\tau\setminus\\{\emptyset\\}$ there are
> $\mathcal{U}\in\mathbb{U},J\in[\omega]^{\omega}\cap V[G_{1}]$ such that for
> every $m\in\omega$ there exists $n\in[m,J(m))$ with the property
> $R(\mathcal{U})(n)\subset A$, where $J(m)$ is the $m$-th element of $J$ with
> respect to its increasing enumeration.

Suppose that there exists $A\in\tau\setminus\\{\emptyset\\}$ for which the
lemma is false. Let $\dot{A}$ be a $\mathbb{P}_{\omega_{2}}$-name for $A$ and
$p\in\mathbb{P}_{\omega_{2}}$ a condition forcing the negation of the
statement quoted above. Applying [14, Lemma 14] to the sequence
$\langle\dot{a}_{i}:i\in\omega\rangle$ such that $\dot{a}_{i}=\dot{A}$ for all
$i\in\omega$, we get a condition $p^{\prime}\leq p$ such that
$p^{\prime}(0)\leq^{0}p(0)$, and a finite set
$\mathcal{U}_{s}\subset\mathcal{P}(\omega)$ for every $s\in p^{\prime}(0)$
with $p^{\prime}(0)\langle 0\rangle\leq s$, such that for each $n\in\omega$,
$s\in p^{\prime}(0)$ with $p^{\prime}(0)\langle 0\rangle\leq s$, and for all
but finitely many immediate successors $t$ of $s$ in $p^{\prime}(0)$ we have

$p^{\prime}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime}\restriction[1,\omega_{2})\Vdash\exists
U\in\mathcal{U}_{s}(\dot{A}\cap n=U\cap n)$.

######  Claim 5.3.1.

There exists $p^{\prime\prime}\leq p^{\prime}$ s.t. for every $s\in
p^{\prime\prime}(0),p^{\prime\prime}(0)\langle 0\rangle\leq s$, $n\in\omega$
and every immediate successor $t$ of $s$ in $p^{\prime\prime}(0)$ we have

$p^{\prime\prime}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})\Vdash\exists
U\in\mathcal{U}_{s}(Int(U)\neq\emptyset\land\dot{A}\cap n=U\cap n).$

###### Proof.

Suppose by contradiction that

  * $(**)$

For every $p^{\prime\prime}\leq p^{\prime}$ it is possible to find $s\in
p^{\prime\prime}(0)$ with $p^{\prime\prime}(0)\langle 0\rangle\leq s$,
$n\in\omega$, infinitely many immediate successors $t$ of $s$ in
$p^{\prime\prime}(0)$, for which there exists $r=r(t)\leq
p^{\prime\prime}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})$ forcing
$\forall U\in\mathcal{U}_{s}\>(U\cap n=\dot{A}\cap n\Rightarrow
Int(U)=\emptyset)$.

Let us fix $p^{\prime\prime}\leq p^{\prime}$ and let $s$ be as in $(**)$. Note
that we can replace $n$ with any other bigger number and $(**)$ will be still
satisfied, so we may assume that $U\cap n\neq U^{\prime}\cap n$ for any
distinct $U,U^{\prime}\in\mathcal{U}_{s}$. Let
$p^{(3)}\in\mathbb{P}_{\omega_{2}}$ be the condition obtained by strengthening
$p^{\prime\prime}(0)$ by leaving only those infinitely many immediate
successors of $s$ like in $(**)$, and
$p^{(3)}(\alpha)=p^{\prime\prime}(\alpha)$ for all $\alpha>0$. Next, removing
yet another finite collection of immediate successors of $s$ in $p^{(3)}(0)$
and keeping the other coordinates the same, we may assume that for every
immediate successor $t$ of $s$ in $p^{(3)}(0)$ we have

$p^{(3)}(0)_{t}^{\,\frown}\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})\Vdash\exists
U\in\mathcal{U}_{s}\>(U\cap n=\dot{A}\cap n),$

and hence by strengthening $r(t),$ if necessary, we may additionally assume
that for some $U(t)\in\mathcal{U}_{s}$ with $Int(U(t))=\emptyset$ we have
$r(t)\Vdash\dot{A}\cap n=U(t)\cap n$. By the choice of $n$ we get that for
each $t$ such an $U(t)$ is unique. Furthermore, strengthening $p^{(3)}$ again
in the way described above and using the finiteness of $\mathcal{U}_{s}$, we
may assume that there exists $U_{s}\in\mathcal{U}_{s}$ with $U(t)=U_{s}$ for
all $t$ as above. Summarizing all the modifications of $p^{\prime\prime}$
mentioned above, we get

  * $(**)^{\prime}$

For every $p^{\prime\prime}\leq p^{\prime}$ it is possible to find
$p^{(3)}\leq p^{\prime\prime}$, $s\in p^{(3)}(0)$ with $p^{(3)}(0)\langle
0\rangle\leq s$, $n\in\omega$ with $U\cap n\neq U^{\prime}\cap n$ for any
distinct $U,U^{\prime}\in\mathcal{U}_{s}$, $U_{s}\in\mathcal{U}_{s}$ with
$Int(U_{s})=\emptyset$, and for every immediate successors $t$ of $s$ in
$p^{(3)}(0)$ a condition $r(t)\leq
p^{(3)}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})$ forcing $U_{s}\cap
n=\dot{A}\cap n$.

Let $\langle D_{k}:k\in\omega\rangle\in V$ be a sequence of dense subsets of
$\langle\omega,\tau\rangle$ such that for every
$U\in\bigcup\\{\mathcal{U}_{s}:s\in p^{\prime}(0),p^{\prime}(0)\langle
0\rangle\leq s\\}$ with $Int(U)=\emptyset$ there are infinitely many
$k\in\omega$ such that $D_{k}=\omega\setminus U$. Let $\langle
F_{k}:k\in\omega\rangle\in V$ be a witness of the $H$-separability of $X$ for
$\langle D_{k}:k\in\omega\rangle$. So it is possible to choose
$k_{0}\in\omega$ and $p^{\prime\prime}\leq p^{\prime}$ such that
$p^{\prime\prime}\Vdash(\forall k\geq k_{0})(\dot{A}\cap F_{k}\neq\emptyset)$.

Fix $p^{(3)}$, $s,$ $n$, $U_{s}$, and $p(t)$’s such as in $(**)^{\prime}$. Let
$k_{1}\geq k_{0}$ be such that $D_{k_{1}}=\omega\setminus U_{s}$, and thus
$F_{k_{1}}\cap U_{s}=\emptyset$. Choose now $n_{1}>n,\max F_{k_{1}}$ and an
immediate successor $t$ of $s$ in $p^{(3)}(0)$ with

$p^{(3)}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})\Vdash\exists
U\in\mathcal{U}_{s}(\dot{A}\cap n_{1}=U\cap n_{1})$.

Thus $r(t)$ also forces the above property (being stronger than
$p^{(3)}(0)_{t}^{\,\smallfrown}\mkern-2.0mu\hskip
2.84526ptp^{\prime\prime}\restriction[1,\omega_{2})$) as well as $U_{s}\cap
n=\dot{A}\cap n$. Since all the differences between elements of
$\mathcal{U}_{s}$ show up below $n$, we conclude that $r(t)$ forces $U_{s}\cap
n_{1}=\dot{A}\cap n_{1}$. Since

$F_{k_{1}}\subset D_{k_{1}}\cap n_{1}=(\omega\setminus U_{s})\cap
n_{1}=n_{1}\setminus U_{s},$

we conclude that $r(t)$ forces $F_{k_{1}}\cap\dot{A}=\emptyset$. On the other
hand $k_{1}\geq k_{0}$, $r(t)\leq p^{\prime\prime}$, and the latter forces
$\dot{A}\cap F_{k}\neq\emptyset$ for all $k\geq k_{0}$, a contradiction. ∎

We can now assume that every $U\in\mathcal{U}_{s}$ with $s\in p^{\prime}(0)$
and $p^{\prime}(0)\langle 0\rangle\leq s$ has nonempty interior. Let
$\mathcal{U}=\\{Int(U):U\in\bigcup\\{\mathcal{U}_{s}:s\in
p^{\prime\prime}(0),p^{\prime\prime}(0)\langle 0\rangle\leq s\\}$ (or any
countable family containing it if this family of interiors is finite) and
$R(\mathcal{U})(n)=\langle F_{n}:n\in\omega\rangle$. Replace $p^{\prime}(0)$
with a tree $T$ in the Laver forcing $\mathcal{L}$ such that $T\langle
0\rangle=p^{\prime}(0)\langle 0\rangle$ and for all $s\in T$ with $s\geq
T\langle 0\rangle$ there exists $N_{s}\in\omega$ so that for every immediate
successor $t$ of $s$ in $T$ the following two properties hold:

  * (a)

$\forall U\in\mathcal{U}_{s}$ $\exists n=n(s,U)>|s|\>(F_{n}\subset Int(U)\cap
N_{s})$

  * (b)

$T_{t}\hskip 1.42262pt^{\,\smallfrown}\mkern-2.0mu\hskip
1.42262ptp^{\prime}\restriction[1,\omega_{2})\Vdash\exists
U\in\mathcal{U}_{s}\>(\dot{A}\cap N_{s}=U\cap N_{s})$

For any $s\in T$ set

$j_{s}=\max\\{n(s,U):U\in\mathcal{U}_{s}\\}+1$

Let $G_{1}$ be $\mathcal{L}$-generic over $V$ with $T\in G_{1}$ and
$\ell\in\omega^{\uparrow\omega}$ be the Laver real generated by $G_{1}$.
Finally put

$J=\\{j_{\ell\restriction m}:m\geq|T\langle 0\rangle|\\}$

and note that $J\in V[G_{1}]$.

######  Claim 5.3.2.

$T^{\,\smallfrown}\mkern-2.0mu\hskip
1.42262ptp^{\prime}\restriction[1,\omega_{2})\Vdash\forall m\geq|T\langle
0\rangle|\>\exists n\in[m,j_{\dot{\ell}\restriction
m})\>(F_{n}\subset\dot{A})$

###### Proof.

Suppose the statement is false and pick $r\leq
T^{\,\smallfrown}\mkern-2.0mup^{\prime}\restriction[1,\omega_{2})$ and
$m\geq|T\langle 0\rangle|$ such that

$r\Vdash\forall n\in[m,j_{\dot{\ell}\restriction
m})(F_{n}\not\subset\dot{A}).$

We can write $r=R^{\,\smallfrown}\mkern-2.0mu\hskip
1.42262ptr\restriction[1,\omega_{2})$ and without loss of generality assume
$|R\langle 0\rangle|\geq m+1$. Setting $\\{s\\}=R\cap\omega^{m}$ and
$\\{t\\}=R\cap\omega^{m+1}$ we obtain two elements of $T$ which contradict the
disjunction of $(a)$ and $(b)$ above. ∎

It follows that

$T^{\,\smallfrown}\mkern-2.0mu\hskip
1.42262ptp^{\prime}\restriction[1,\omega_{2})\leq p^{\prime}\leq p$

and $T^{\,\smallfrown}\mkern-2.0mu\hskip
1.42262ptp^{\prime}\restriction[1,\omega_{2})$ forces for $\dot{A}$ the quoted
statement after the formulation of Lemma 5.3, contradicting our assumption
that $p$ forces for $\dot{A}$ the negation of that statement. This
contradiction completes the proof of Lemma 5.3 and thus also of Proposition
5.2. ∎

∎

######  Lemma 5.4.

Suppose $\mathfrak{b}>\omega_{1}$, $X$ is a b.b.-separable space and $Y$ is an
$H$-separable space. Then $X\times Y$ is $mH$-separable, provided it is
separable.

###### Proof.

Let $\langle D_{n}:n\in\omega\rangle$ be a decreasing sequence of countable
dense subsets of $X\times Y$. Fix a countable family $\mathcal{U}$ of open
non-empty subsets of $X$ and a partition $\\{\Omega_{U}:U\in\mathcal{U}\\}$ of
$\omega$ into infinite pieces. For every $U\in\mathcal{U}$ and
$n\in\Omega_{U}$ set

$D_{n}^{\mathcal{U}}=\\{y\in Y:\exists x\in U(\langle x,y\rangle\in
D_{n})\\}.$

Note that every $D_{n}^{\mathcal{U}}$ defined in this way is dense in $Y$.
Since $Y$ is $H$-separable there exists a sequence of sets $\langle
L^{\mathcal{U}}_{n}:n\in\omega\rangle$ such that
$L^{\mathcal{U}}_{n}\in[D^{\mathcal{U}}_{n}]^{<\omega}$ for all $n\in\omega$
and every open subset of $Y$ intersects all but finite
$L^{\mathcal{U}}_{n}$’s. For every $U\in\mathcal{U}$ and $n\in\Omega_{U}$ find
a set $K^{\mathcal{U}}_{n}\in[U]^{<\omega}$ such that for every $y\in
L^{\mathcal{U}}_{n}$ there exists $x\in K^{\mathcal{U}}_{n}$ such that
$\langle x,y\rangle\in D_{n}$, and set $R(\mathcal{U})=\langle
K^{\mathcal{U}}_{n}:n\in\omega\rangle$. Note that $K^{\mathcal{U}}_{n}\subset
U$ for all $n\in\Omega_{U}$, so $R$ is as in the definition of
b.b.-separability. Therefore there exist a family $\mathbb{U}$ of countable
collections of open non-empty subsets of $X$ of cardinality $\omega_{1}$ and a
sequence $\langle l_{i}:i\in\omega\rangle\in\omega^{\omega}$ that witness the
b.b.-separability. By our hypothesis $|\mathbb{U}|<\mathfrak{b}$, so we can
select a sequence $\langle F_{n}:n\in\omega\rangle$ such that
$F_{n}\in[D_{n}]^{<\omega}$ and $K^{\mathcal{U}}_{n}\times
L^{\mathcal{U}}_{n}\subset F_{n}$ for all $\mathcal{U}\in\mathbb{U}$ and all
but finitely many $n\in\omega$. We claim that the sequence

$\langle
F^{\prime}_{i}:=\bigcup\limits_{n\in[l_{i},l_{i+1})}F_{n}\>:\>i\in\omega\rangle$

witnesses the $mH$-separability of $X\times Y$. First of all, note that
$F^{\prime}_{i}\subset D_{l_{i}}\subset D_{i}$ by the monotonicity of the
sequence $\langle D_{n}:n\in\omega\rangle$. Now fix an open non-empty subset
$U\times V\subseteq X\times Y$ and find $\mathcal{U}\in\mathbb{U}$ and
$i_{0}\in\omega$ such that for all $i\geq i_{0}$ there exists
$n\in[l_{i},l_{i+1})$ with $R(\mathcal{U})(n)=K^{\mathcal{U}}_{n}\subset U$.
Suppose that $i\geq i_{0}$, and both of the conditions
$K^{\mathcal{U}}_{n}\times L^{\mathcal{U}}_{n}\subset F_{n}$ and
$L^{\mathcal{U}}_{n}\cap V\neq\emptyset$ hold true for all $n\geq l_{i}$.
Given $n\in[l_{i},l_{i+1})$ with $R(\mathcal{U})(n)=K^{\mathcal{U}}_{n}\subset
U$, it follows from the above that $F_{n}\cap(U\times V)\neq\emptyset$, which
combined with $F_{n}\subset F^{\prime}_{i}$ proves the $mH$-separability of
$X\times Y$. ∎

Proposition 5.2, Lemma 5.4, and the fact that $\mathfrak{b}>\omega_{1}$ is
true in the Laver model all together imply the following

######  Theorem 5.5.

In the Laver model, the product of two $H$-separable spaces is $mH$-separable
provided that it is separable. In particular, in this model the product of two
countable $H$-separable spaces is $mH$-separable.

## References

  * [1] A. Arhangel’skiĭ, Frequency spectrum of a topological space and the classification of spaces, Doklady Akademii Nauk SSSR, 206 (1972), pp. 265–268.
  * [2] D. Barman and A. Dow, Selective separability and SS+, Topology Proceedings, 37 (2011), pp. 181–204.
  * [3] , Proper forcing axiom and selective separability, Topology and its Applications, 159 (2012), pp. 806–813.
  * [4] A. Bella, M. Bonanzinga, and M. Matveev, Variations of selective separability, Topology and its Applications, 156 (2009), pp. 1241–1252.  Special Issue Dedicated To Jan ”Honza” Pelant.
  * [5] A. Bella, M. Bonanzinga, M. Matveev, and V. V. Tkachuk, Selective separability: general facts and behavior in countable spaces, Topological Proceedings, 32 (2008), pp. 15–30.
  * [6] A. Blass, Combinatorial cardinal characteristics of the continuum, in _Handbook of Set Theory_ , M. Foreman and A. Kanamori, eds., Springer Netherlands, (2010), pp. 395–491.
  * [7] A. Blass and S. Shelah, There may be simple $P_{\aleph_{1}}$ and $P_{\aleph_{2}}$-points and the rudin-keisler ordering may be downward directed, Annals of Pure and Applied Logic, 33 (1987), pp. 213–243.
  * [8] C. Costantini, On a problem of Nogura about the product of Fréchet-Urysohn $\langle\alpha_{4}\rangle$-spaces, Commentationes Mathematicae Universitatis Carolinae, 40 (1999), pp. 537–549.
  * [9] A. Dow and S. Shelah, Martin’s axiom and separated mad families, Rendiconti del Circolo Matematico di Palermo (2), 61 (2012), pp. 107–115.
  * [10] R. Engelking, General Topology, Heldermann Verlag, 1989.
  * [11] G. Gruenhage and M. Sakai, Selective separability and its variations, Topology and its Applications, 158 (2011), pp. 1352–1359.  Special Issue: Analysis, Topology and Applications 2010 (ATA 2010).
  * [12] T. Jech, Set theory: The third millennium edition,revised and expanded, Springer Berlin, Heidelberg, 2003.
  * [13] K. Kunen, Set Theory: An Introduction to Independence Proofs, Elsevier, Amsterdam, 1980.
  * [14] R. Laver, On the consistency of Borel’s conjecture, Acta Mathematica, 137 (1976), pp. 151–169.
  * [15] A. Mathias, Happy families, Annals of Mathematical Logic, 12 (1977), pp. 59–111.
  * [16] A. W. Miller, B. Tsaban, and L. Zdomskyy, Selective covering properties of product spaces, II: $\gamma$ spaces, Trans. Amer. Math. Soc., 368 (2016), pp. 2865–2889.
  * [17] T. Nogura, The product of $\langle\alpha_{i}\rangle$-spaces, Topology and its Applications, 21 (1985), pp. 251–259.
  * [18] R. C. Olson, Bi-quotient maps, countably bi-sequential spaces, and related topics, General Topology and Appl., 4 (1974), pp. 1–28.
  * [19] D. Repovs and L. Zdomskyy, Products of H-separable spaces in the Laver model, Topology and its Applications, 239 (2018), pp. 115–119.
  * [20] M. Scheepers, Combinatorics of open covers VI: selectors for sequences of dense sets, Quaestiones Mathematicae, 22 (1999), pp. 109–130.
  * [21] P. Simon, A compact Frèchet space whose square is not Frèchet, Commentationes Mathematicae Universitatis Carolinae, 21 (1980), pp. 749–753.