# Products of Luzin-type sets with combinatorial properties

Piotr Szewczak  Piotr Szewczak, Institute of Mathematics, Faculty of
Mathematics and Natural Science College of Sciences, Cardinal Stefan Wyszyński
University in Warsaw, Wóycickiego 1$/$3, 01–938 Warsaw, Poland
<EMAIL_ADDRESS>http://piotrszewczak.pl and  Grzegorz Wiśniewski  Grzegorz
Wiśniewski, Institute of Mathematics, Faculty of Mathematics and Natural
Science College of Sciences, Cardinal Stefan Wyszyński University in Warsaw,
Wóycickiego 1$/$3, 01–938 Warsaw, Poland<EMAIL_ADDRESS>
###### Abstract.

We construct Luzin-type subsets of the real line in all finite powers
Rothberger, with a non-Menger product. To this end, we use a purely
combinatorial approach which allows to weaken assumptions used earlier to
construct sets with analogous properties. Our assumptions hold, e.g., in the
Random model, where already known category theoretic methods fail.

###### Key words and phrases:

Luzin set, Rothberger property, Menger property, Scheepers property, product
space, scales.

###### 2010 Mathematics Subject Classification:

Primary: 54D20; Secondary: 03E17.

##  1\. Introduction

###  1.1. Combinatorial covering properties

By _space_ we mean a topological space. A space $X$ is _Rothberger_ if, for
every sequence $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$
of open covers of the space $X$, there are sets
$U_{1}\in\mathcal{U}_{1},U_{2}\in\mathcal{U}_{2},\dotsc$ such that the family
$\\{\,U_{n}:n\in\mathbb{N}\,\\}$ covers $X$. In the latter case, Rothberger’s
property can be viewed as a topological version of _strong measure zero_. We
restrict our consideration to _sets of reals_ , i.e., spaces homeomorphic to
subsets of the real line. In the realm of sets of reals, we have the following
implications.

$\text{countability }\longrightarrow\text{ Rothberger }\longrightarrow\text{
strong measure zero}$

It is consistent that these properties hold only for countable spaces and are,
thus, equivalent, e.g., in the Laver model for the consistency of Borel’s
conjecture [12]. On the other hand, the Continuum Hypothesis implies that the
Rothberger property is strictly in between countablility and strong measure
zero.

A space $X$ is _Menger_ if, for every sequence
$\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$ of open covers
of the space $X$, there are finite subsets
$\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$
such that the family $\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$ covers $X$.
Menger’s property is a generalization of Rothberger’s property, and, in the
realm of sets of reals, Menger’s property is strictly in between
$\sigma$-compactness and the Lindelöf property.

The above covering properties, are central in the _selection principles
theory_ [23].

###  1.2. Luzin sets

A _Luzin_ set is an uncountable set of reals whose intersection with every
meager set is at most countable; its existence, is independent with ZFC. A
Luzin set is Rothberger (and thus Menger), and it is not $\sigma$-compact.
Thus, a Luzin set is a counterexample to Menger’s conjecture which says that,
in the realm of sets of reals, the Menger property and $\sigma$-compactness,
are equivalent [13, 10]. A Luzin set also solves Hurewicz’s problem, whether
the Menger property is equivalent to the Hurewicz property, a formally
stronger than Menger, a covering property; every Luzin set is Menger, but not
Hurewicz [10]. A detailed analysis of Luzin sets, gives a good intuition how
to proceed, using combinatorics, in order to construct Menger sets of reals,
with no any extra assumptions [7, 4].

###  1.3. Products of Luzin sets

Let $\operatorname{cov}(\mathcal{M})$ be the minimal cardinality of a family
of meager subsets of the real line, that covers the real line. Let
$\operatorname{cof}(\mathcal{M})$ be the minimal cardinality of a _cofinal_
family of meager subsets of the real line, i.e., every meager subset of the
real line is contained in a member of the family. A set of reals $X$ is a
_$\operatorname{cov}(\mathcal{M})$ -Luzin_ set if
$\left|X\right|\geq\operatorname{cov}(\mathcal{M})$ and, for every meager set
$M$, we have $\left|X\cap M\right|<\operatorname{cov}(\mathcal{M})$. Every
$\operatorname{cov}(\mathcal{M})$-Luzin set is hereditary Rothberger and,
assuming that
$\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$, a direct
construction of such a set is possible.

Luzin sets and $\operatorname{cov}(\mathcal{M})$-Luzin sets, play an important
role in investigations of products spaces with the Rothberger, or Menger
properties. Just, Miller, Scheepers, and Szeptycki proved that, under the
Continuum Hypothesis, there is a Luzin set, whose square is not Menger [9,
Lemma 2.6]; it was the first example showing that products of Rothberger sets,
may not be Menger. Let $\mathfrak{c}$ be the cardinality of the real line.
Assuming the Continuum Hypothesis or the equality
$\operatorname{cov}(\mathcal{M})=\mathfrak{c}$, independently, many authors
proved that there are two $\operatorname{cov}(\mathcal{M})$-Luzin sets in all
finite powers Rothberger, with a non-Menger product ([21, Proposition 3.1],
[9, Page 205], [16, Theorem 13], [11, Section 3], [3, Theorem 4]). Category
theoretic methods used there, have substantial limitations, which prevent to
weaken assumptions. A natural question arises, whether the existence of such
$\operatorname{cov}(\mathcal{M})$-Luzin sets can be proven under
$\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$. We show
that, assuming that the cardinal number $\operatorname{cov}(\mathcal{M})$ is
regular, the answer is positive. In particular, we prove the existence of such
sets, e.g., in the Random model, where
$\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})=\aleph_{1}<\mathfrak{c}$.
In this model a category theoretic approach fails.

In the second part of the paper, we use already known methods, to construct,
for any Luzin-type set $X$, another Luzin type-set $Y$ with some additional
properties (as being a topological group, or whose all finite powers are
Rothberger) such that the product space $X\times Y$ is not Menger.

We apply obtained results to products of function spaces with some local
properties.

##  2\. Luzin-type sets via combinatorial approach

Let $\mathbb{N}$ be the set of natural numbers and
${\operatorname{P}(\mathbb{N})}$ be the family of subsets of $\mathbb{N}$.
Identify every element of the set ${\operatorname{P}(\mathbb{N})}$ with its
characteristic function, an element of the Cantor cube
${\\{0,1\\}^{\mathbb{N}}}$. Using this identification, we introduce a topology
in ${\operatorname{P}(\mathbb{N})}$. Since the Cantor cube is homeomorphic to
Cantor’s set, every subspace of ${\operatorname{P}(\mathbb{N})}$ is a set of
reals. Let ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be the set of all infinite
subsets of $\mathbb{N}$. We identify each set
$a\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ with its increasing enumeration,
an element of the Baire space ${\mathbb{N}^{\mathbb{N}}}$. For each natural
number $n$, by $a(n)$ denote the $n$-th smallest element of the set $a$. Thus,
we have
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\subseteq{\mathbb{N}^{\mathbb{N}}}$.
Relative topologies in the set ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, from
the Baire space ${\mathbb{N}^{\mathbb{N}}}$ and from the Cantor space
${\operatorname{P}(\mathbb{N})}$, are equal. For elements
$x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, we write $x\leq^{*}y$, if $x$
_is dominated_ by $y$, i.e., if the set $\\{\,n:y(n)<x(n)\,\\}$ is finite. A
subset of ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ is _dominating_ if every
element in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ is dominated by an element
from this set. Analogously, we define relation $\leq^{*}$ in
${\mathbb{N}^{\mathbb{N}}}$, and a dominating subset of
${\mathbb{N}^{\mathbb{N}}}$. Let $\mathfrak{d}$ be the minimal cardinality of
a dominating set in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ (equivalently in
${\mathbb{N}^{\mathbb{N}}}$). For elements
$x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, define
$\max\\{x,y\\}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that
$\max\\{x,y\\}(n):=\max\\{x(n),y(n)\\}$, for all natural numbers $n$. A set
$X\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ is _2-dominating_ , if the
set $\\{\,\max\\{x,y\\}:x,y\in X\,\\}$ is dominating.

The cardinal numbers $\operatorname{cov}(\mathcal{M})$ and
$\operatorname{cof}(\mathcal{M})$ do not change if, in their definitions,
consider families of meager subsets of ${\operatorname{P}(\mathbb{N})}$,
instead of the real line.

######  Theorem 2.1.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There
are $\operatorname{cov}(\mathcal{M})$-Luzin sets
$X,Y\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that:

  1. (1)

all finite powers of $X$ and $Y$ are Rothberger,

  2. (2)

the product space $X\times Y$ is not Menger,

  3. (3)

the union $X\cup Y$ is $2$-dominating.

In order to prove Theorem 2.1 and later discussion, we need the following
auxiliary results and notations. A set
$r\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ _reaps_ a family
$A\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ if, for each set $a\in A$,
both sets $a\cap r$ and $a\setminus r$ are infinite. Let $\mathfrak{r}$ be the
minimal cardinality of a family
$A\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ that no set $r$ reaps. For
elements $x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, we write
$x\leq^{\infty}y$ if $y\not\leq^{*}x$, i.e., if the set
$\\{\,n:x(n)<y(n)\,\\}$ is infinite. For a set
$Z\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and an element
$x\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, we write $Z\leq^{\infty}x$ if
$z\leq^{\infty}x$ for all elements $z\in Z$. We use this convention to any
binary relation in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. For a set
$x\in{\operatorname{P}(\mathbb{N})}$, define $x^{\text{\tt
c}}:=\mathbb{N}\setminus x$. For an element
$z\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ with $z(1)\neq 1$, define
$\tilde{z}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that
$\tilde{z}(1):=z(1)$, and $\tilde{z}(n+1):=z(\tilde{z}(n))$, for all natural
numbers $n$. For natural numbers $j,k$ with $j<k$, let
$[j,k):=\\{\,j,j+1,\dotsc,k-1\,\\}$. For a set
$a\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and a function
$f\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, define

$a/f:=\\{\,n\in\mathbb{N}:a\cap[f(n),f(n+1))\neq\emptyset\,\\}.$

The following combinatorial characterization of meager subsets of
${\operatorname{P}(\mathbb{N})}$ is an important tool, in our constructions.

######  Lemma 2.2 (Bartoszyński, Judah [2, Theorem 2.2.4]).

Let $M\subseteq{\operatorname{P}(\mathbb{N})}$ be a meager set. There are a
set $a\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and a function
$f\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that the set $M$ is contained
in the meager set

$\\{\,x\in{\operatorname{P}(\mathbb{N})}:x\cap[f(n),f(n+1))\neq
a\cap[f(n),f(n+1))\text{ for all but finitely many }n\,\\}.$

Let $\mathfrak{d}^{\prime}$ be the minimal cardinality of a set
$A\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that, for each function
$b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, there is a set $a\in A$ such
that, the set $\\{\,n:\left|a\cap[b(n),b(n+1))\right|\geq 2\,\\}$ is finite.
The following Lemma, used in our main result, is essentially due to Blass [6,
Theorem 2.10]. We provide a proof, for the sake of completeness.

######  Lemma 2.3.

We have $\mathfrak{d}=\mathfrak{d}^{\prime}$.

###### Proof.

($\mathfrak{d}\leq\mathfrak{d}^{\prime}$): Let
$A\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be a set from the definition
of the number $\mathfrak{d}^{\prime}$, of cardinality $\mathfrak{d}^{\prime}$,
closed under finite modifications of its elements.

The set $A$ is dominating in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$: Fix an
element $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. There is a set $a\in A$
such that the set $\\{\,n:\left|a\cap[b(n),b(n+1))\right|\geq 2\,\\}$ is
finite. Since the set $A$ is closed under finite modifications of its
elements, we may assume that, this finite set is empty, and $b(1)<a(1)$. Fix a
natural number $n\in\mathbb{N}$, and let $b(n)\leq a(n)$. If $a(n+1)<b(n+1)$,
then $b(n)\leq a(n)<a(n+1)<b(n+1)$, and thus
$\left|a\cap[b(n),b(n+1))\right|\geq 2$, a contradiction. It follows that
$b(n+1)\leq a(n+1)$. We have $b\leq^{*}a$, and thus the set $A$ is dominating
in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$.

($\mathfrak{d}^{\prime}\leq\mathfrak{d}$): Let
$D\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be a dominating set, of
cardinality $\mathfrak{d}$, closed under finite modifications of its elements.

The set $\\{\,\tilde{d}:d\in D\,\\}$ satisfies the condition from the
definition of the number $\mathfrak{d}^{\prime}$: Fix an element
$b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. There exists en element $d\in D$
such that $b(n)<d(n)$, for all natural numbers $n$. Fix a natural number $n$.
If there is a natural number $k$ with
$b(n)\leq\tilde{d}(k)<\tilde{d}(k+1)<b(n+1)$, then

$d(\tilde{d}(k))=\tilde{d}(k+1)<b(n+1)\leq b(\tilde{d}(k)),$

a contradiction. Thus, $\\{\,n:|\,\tilde{d}\cap[b(n),b(n+1))\,|\geq
2\,\\}=\emptyset$. ∎

We have the following inequalities between considered cardinals [6]:

$\operatorname{cov}(\mathcal{M})\leq\mathfrak{d}\leq\operatorname{cof}(\mathcal{M})\text{
and }\operatorname{cov}(\mathcal{M})\leq\mathfrak{r}.$

######  Lemma 2.4.

Let $\mathcal{M}^{\prime}$ be a family of meager sets in
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ with
$\left|\mathcal{M}^{\prime}\right|<\operatorname{cov}(\mathcal{M})$, and
$Z\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be a set with
$\left|Z\right|<\operatorname{cov}(\mathcal{M})$. Let
$d\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. There are elements
$x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\setminus\bigcup\mathcal{M}^{\prime}$
such that $Z\leq^{\infty}x,Z\leq^{\infty}y$, and $d\leq^{*}\max\\{x^{\text{\tt
c}},y^{\text{\tt c}}\\}$.

###### Proof.

Let $\mathcal{M}^{\prime}=\\{\,M_{\beta}:\beta<\alpha\,\\}$, for some ordinal
number $\alpha<\operatorname{cov}(\mathcal{M})$. Fix an ordinal number
$\beta<\alpha$. By Lemma 2.2, there are a set
$a_{\beta}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and a function
$f_{\beta}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that the set
$M_{\beta}$ is contained in the meager set

$\\{\,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}:y\cap[f_{\beta}(n),f_{\beta}(n+1))\neq
a_{\beta}\cap[f_{\beta}(n),f_{\beta}(n+1))\text{ for all but finitely many
}n\,\\}.$

Thus, we may assume that the set $M_{\beta}$ is equal to the above meager set.
Since $\alpha<\operatorname{cov}(\mathcal{M})$, there is a set
$a\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\setminus\bigcup_{\beta<\alpha}M_{\beta}$.
Then the sets

$f^{\prime}_{\beta}:=\\{\,f_{\beta}(n):a\cap[f_{\beta}(n),f_{\beta}(n+1))=a_{\beta}\cap[f_{\beta}(n),f_{\beta}(n+1)),n\in\mathbb{N}\,\\}$

are infinite, for all ordinal numbers $\beta<\alpha$.

Assume that $Z=\\{\,z_{\beta}:\beta<\alpha\,\\}$. We have
$\alpha<\operatorname{cov}(\mathcal{M})=\mathfrak{d}$. By Lemma 2.3, there is
a function $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that the sets

$I_{\beta}:=\\{\,n:|\,\tilde{z}_{\beta}/\tilde{d}\cap(b(n),b(n+1))\,|\geq
2\,\\},J_{\beta}:=\\{\,n:|\,f^{\prime}_{\beta}/\tilde{d}\cap(b(n),b(n+1))\,|\geq
2\,\\}$

are infinite for all ordinal numbers $\beta<\alpha$.

Since $\alpha<\operatorname{cov}(\mathcal{M})\leq\mathfrak{r}$, there is a
decomposition of the set $\mathbb{N}$ into infinite sets $r,s,t,u$, each of
them reaps the family $\\{\,I_{\beta},J_{\beta}:\beta<\alpha\,\\}$. Define

$\displaystyle x:=\bigcup_{n\in r\cup
u}\bigl{[}\tilde{d}(b(n)),\tilde{d}(b(n+1)+1)\bigr{)}\cup\bigcup_{n\in
s}a\cap\bigl{[}\tilde{d}(b(n)),\tilde{d}(b(n+1)+1))\bigr{)},$ $\displaystyle
y:=\bigcup_{n\in t\cup
s}\bigl{[}\tilde{d}(b(n)),\tilde{d}(b(n+1)+1)\bigr{)}\cup\bigcup_{n\in
u}a\cap\bigl{[}\tilde{d}(b(n)),\tilde{d}(b(n+1)+1))\bigr{)}.$

Fix an ordinal number $\beta<\alpha$ and a natural number $n\in s\cap
J_{\beta}$. There is a natural number $i$ such that

$f^{\prime}_{\beta}(i),f^{\prime}_{\beta}(i+1)\in[\tilde{d}(b(n)),\tilde{d}(b(n+1)))$

Thus,

$x\cap[f^{\prime}_{\beta}(i),f^{\prime}_{\beta}(i+1))=a\cap[f^{\prime}_{\beta}(i),f^{\prime}_{\beta}(i+1))=a_{\beta}\cap[f^{\prime}_{\beta}(i),f^{\prime}_{\beta}(i+1)).$

Let $j$ be a natural number such that $f_{\beta}(j)=f^{\prime}_{\beta}(i).$ We
have

$x\cap[f_{\beta}(j),f_{\beta}(j+1))=a_{\beta}\cap[f_{\beta}(j),f_{\beta}(j+1)).$

Since the set $s\cap J_{\beta}$ is infinite, the set

$\\{\,n\in\mathbb{N}:x\cap[f_{\beta}(j),f_{\beta}(j+1))=a_{\beta}\cap[f_{\beta}(j),f_{\beta}(j+1))\,\\}$

is infinite, too. Thus, $x\notin\bigcup\mathcal{M}^{\prime}.$ Analogously, we
get $y\notin\bigcup\mathcal{M}^{\prime}$.

Fix an ordinal number $\beta<\alpha$ and a natural number $n\in t\cap
I_{\beta}$. Since

$|\,\tilde{z}_{\beta}/\tilde{d}\cap(b(n),b(n+1))\,|\geq 2,$

there is a natural number $i$ such that

$\tilde{z}_{\beta}(i),\tilde{z}_{\beta}(i+1)\in[\tilde{d}(b(n)+1),\tilde{d}(b(n+1))).$

Since $x\cap[\tilde{d}(b(n)+1),\tilde{d}(b(n+1)))=\emptyset$, we have

$z(\tilde{z}_{\beta}(i))=\tilde{z}_{\beta}(i+1)<\tilde{d}(b(n+1))\leq
x(\tilde{z}_{\beta}(i)),$

for all natural numbers $n\in t$. The set $t\cap I_{\beta}$ is infinite, and
thus $z\leq^{\infty}x$. Analogously, we get $z\leq^{\infty}y$.

Fix a natural number $k\geq\tilde{d}(b(1))$. There is a natural number $n$
such that $k\in[\tilde{d}(b(n)),\tilde{d}(b(n+1)))$. Assume that $n\in r\cup
u$. Since $x^{\text{\tt
c}}\cap[\tilde{d}(b(n)),\tilde{d}(b(n+1)+1))=\emptyset$, we have

$d(k)<d(\tilde{d}(b(n+1)))=\tilde{d}(b(n+1)+1)\leq x^{\text{\tt c}}(k).$

Analogously, if $n\in t\cup s$, then $d(k)<y^{\text{\tt c}}(k)$. Thus,
$d\leq^{*}\max\\{x^{\text{\tt c}},y^{\text{\tt c}}\\}$. ∎

A _filter_ is a subset of ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ with empty
intersection, that is closed under finite intersections and taking supersets.
An _ultrafilter_ is a maximal filter. For elements
$x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, let $[x\leq y]:=\\{\,n:x(n)\leq
y(n)\,\\}$. For a set $F\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and
elements $x,y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, we write $x\leq_{F}y$,
if $[x\leq y]\in F$. If the above set $F$ is a filter, then the relation
$\leq_{F}$ is transitive. Let $U$ be an ultrafilter. A set
$X\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ is _$\leq_{U}$ -unbounded_
if, for each element $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, there is an
element $x\in X$ with $b\leq_{U}x$. Let $\mathfrak{b}(U)$ be the minimal
cardinality of a $\leq_{U}$-unbounded subset of
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$.

######  Definition 2.5 (Szewczak, Tsaban [19, Definition 4.1]).

Let $U$ be an ultrafilter. A set
$X\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ with
$\left|X\right|\geq\operatorname{\mathfrak{b}}(U)$ is a _$U$ -scale_ if, for
each element $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, we have

$\left|\\{\,x\in X:x\leq_{U}b\,\\}\right|<\mathfrak{b}(U).$

######  Proposition 2.6.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There
are ultrafilters $U,\tilde{U}$ and sets
$X,Y\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that:

  1. (1)

the set $X$ is a $U$-scale and a $\operatorname{cov}(\mathcal{M})$-Luzin set,

  2. (2)

the set $Y$ is a $\tilde{U}$-scale and a
$\operatorname{cov}(\mathcal{M})$-Luzin set,

  3. (3)

the set $\\{\,\max\\{x^{\text{\tt c}},y^{\text{\tt c}}\\}:x\in X,y\in Y\,\\}$
is dominating in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$.

###### Proof.

Let $\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a
dominating set in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and
$\\{\,M_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a cofinal
family of meager sets in ${\operatorname{P}(\mathbb{N})}$.

Let $F_{0}=\tilde{F}_{0}=\\{\mathbb{N}\\}$ and
$x_{0},y_{0}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be elements such that
$d_{0}\leq^{*}\max\\{{x}_{0}^{\text{\tt c}},{y}_{0}^{\text{\tt c}}\\}$. Fix an
ordinal number $\alpha<\operatorname{cov}(\mathcal{M})$. We show that there
are elements $x_{\alpha},y_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$
and sets
$F_{\alpha},\tilde{F}_{\alpha}\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$
such that:

  1. (i)

the sets $F_{\alpha},\tilde{F}_{\alpha}$ are closed under finite
intersections,

  2. (ii)

$\bigcup_{\beta<\alpha}F_{\beta}\subseteq F_{\alpha}$,
$\bigcup_{\beta<\alpha}\tilde{F}_{\beta}\subseteq\tilde{F}_{\alpha}$,

  3. (iii)

$\left|F_{\alpha}\right|,|\,\tilde{F}_{\alpha}\,|<\operatorname{cov}(\mathcal{M})$,

  4. (iv)

$\\{\,d_{\beta},x_{\beta}:\beta<\alpha\,\\}\leq_{F_{\alpha}}x_{\alpha},\\{\,d_{\beta},y_{\beta}:\beta<\alpha\,\\}\leq_{\tilde{F}_{\alpha}}y_{\alpha}$,

  5. (v)

$x_{\alpha},y_{\alpha}\notin\bigcup_{\beta<\alpha}M_{\beta}$,

  6. (vi)

$d_{\alpha}\leq^{*}\max\\{{x}_{\alpha}^{\text{\tt c}},{y}_{\alpha}^{\text{\tt
c}}\\}$:

Let

$F=\bigcup_{\beta<\alpha}F_{\beta}\text{ and
}\tilde{F}=\bigcup_{\beta<\alpha}\tilde{F}_{\beta}.$

The cardinal number $\operatorname{cov}(\mathcal{M})$ is regular, and thus
$\left|F\right|,|\,\tilde{F}\,|<\operatorname{cov}(\mathcal{M})$. Define

$S:=\operatorname{maxfin}\\{\,d_{\beta},x_{\beta}:\beta<\alpha\,\\}\text{ and
}\tilde{S}:=\operatorname{maxfin}\\{\,d_{\beta},y_{\beta}:\beta<\alpha\,\\}.$

We have $|\,S\,|,|\,\tilde{S}\,|<\operatorname{cov}(\mathcal{M})$. For
elements $s,f\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, let $s\circ
f\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be an element such that $(s\circ
f)(n)=s(f(n))$, for all natural numbers $n$. By Lemma 2.4, there are elements
$x_{\alpha},y_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that :

  1. (vii)

$\\{\,s\circ f:s\in S\cup\tilde{S},f\in
F\cup\tilde{F}\,\\}\leq^{\infty}x_{\alpha},y_{\alpha}$,

  2. (viii)

$x_{\alpha},y_{\alpha}\notin\bigcup_{\beta<\alpha}M_{\beta}$,

  3. (ix)

$d_{\alpha}\leq^{*}\max\\{{x}_{\alpha}^{\text{\tt c}},{y}_{\alpha}^{\text{\tt
c}}\\}$.

The intersection of finitely many elements of the set
$F\cup{\bigl{\\{}\,[s\leq x_{\alpha}]:s\in S\,\bigr{\\}}}$ is infinite: Since
the set $F$ is closed under finite intersections, it is enough to show that
for an element $f\in F$ and a finite subset $S^{\prime}\subseteq S$, the
intersection $f\cap\bigcap_{s\in S^{\prime}}[s\leq x_{\alpha}]$ is infinite.
Let $f\in F$ and $S^{\prime}\subseteq S$ be a finite set. We have

$\bigcap_{s\in S^{\prime}}[s\leq x_{\alpha}]=\bigcap_{s\in
S^{\prime}}\\{\,n:s(n)\leq
x_{\alpha}(n)\,\\}\supseteq{\bigl{\\{}\,n:\max_{s\in S^{\prime}}s(n)\leq
x_{\alpha}(n)\,\bigr{\\}}}=\bigl{[}\max[S^{\prime}]\leq x_{\alpha}\bigr{]}.$

By the definition of the set $S$, we have $\max[S^{\prime}]\in S$. By (vii),
we have $\max[S^{\prime}]\circ f\leq^{\infty}x_{\alpha}$, and thus

$\max[S^{\prime}](f(n))\leq x_{\alpha}(n)\leq x_{\alpha}(f(n)),$

for infinitely many natural numbers $n$. It follows that the intersection
$f\cap\bigl{[}\max[S^{\prime}]\leq x_{\alpha}\bigr{]}$ is infinite, and the
intersection $f\cap\bigcap_{s\in S^{\prime}}[s\leq x_{\alpha}]$ is infinite,
too.

Let $F_{\alpha}$ be the set $F\cup\\{\,[s\leq x_{\alpha}]:s\in S\,\\}$, closed
under finite intersections. Since $\mathbb{N}\in F$, we have
$d_{\beta},x_{\beta}\in S$, for all ordinal numbers $\beta<\alpha$. Thus,
$\\{\,d_{\beta},x_{\beta}:\beta<\alpha\,\\}\leq_{F_{\alpha}}x_{\alpha}$.
Analogously, define the set $\tilde{F}_{\alpha}$ and show, that
$\\{\,d_{\beta},y_{\beta}:\beta<\alpha\,\\}\leq_{\tilde{F}_{\alpha}}y_{\alpha}$.
By the construction, we have
$\left|F_{\alpha}\right|,|\,\tilde{F}_{\alpha}\,|<\operatorname{cov}(\mathcal{M})$.

Let $U$ and $\tilde{U}$ be ultrafilters containing the sets
$\bigcup_{\alpha<\operatorname{cov}(\mathcal{M})}F_{\alpha}$ and
$\bigcup_{\alpha<\operatorname{cov}(\mathcal{M})}\tilde{F}_{\alpha}$,
respectively. Define

$X:=\\{\,x_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}\text{ and
}Y:=\\{\,y_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}.$

The set $X$ is a $U$-scale: By the construction, we have
$\left|X\right|=\operatorname{cov}(\mathcal{M})$. Fix an element
$b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. There is an ordinal number
$\beta<\operatorname{cov}(\mathcal{M})$ such that $b\leq^{*}d_{\beta}$. For
all ordinal numbers $\alpha$ with
$\beta<\alpha<\operatorname{cov}(\mathcal{M})$, we have
$d_{\beta}\leq_{U}x_{\alpha}$. Thus, $\left|\\{\,x\in
X:x\leq_{U}b\,\\}\right|<\operatorname{cov}(\mathcal{M}).$ Analogously, the
set $Y$ is a $\tilde{U}$-scale.

The set $X$ is a $\operatorname{cov}(\mathcal{M})$-Luzin set: Let
$M\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be a meager set. There is an
ordinal number $\alpha<\operatorname{cov}(\mathcal{M})$ such that $M\subseteq
M_{\alpha}$. By (v), we have

$M\cap X\subseteq M_{\alpha}\cap X\subseteq\\{\,x_{\beta}:\beta<\alpha\,\\},$

and thus $\left|M\cap X\right|<\operatorname{cov}(\mathcal{M})$. Analogously,
the set $Y$ is a $\operatorname{cov}(\mathcal{M})$-Luzin set.

By (vi), the set $\\{\,\max\\{{x_{\alpha}}^{\text{\tt
c}},{y_{\alpha}}^{\text{\tt
c}}\\}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ is a dominating subset of
$\\{\,\max\\{{x}^{\text{\tt c}},{y}^{\text{\tt c}}\\}:x,y\in X\cup Y\,\\}$. ∎

Let $\mathrm{Fin}$ be the set of all finite subsets of $\mathbb{N}$.

###### Proof of Theorem 2.1.

Let $X,Y\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be
$\operatorname{cov}(\mathcal{M})$-Luzin sets from Proposition 2.6. All finite
powers of the sets $X\cup\mathrm{Fin}$, and $Y\cup\mathrm{Fin}$ are Rothberger
[20, Lemma 2.20]. Let
$\tau\colon{\operatorname{P}(\mathbb{N})}\to{\operatorname{P}(\mathbb{N})}$ be
a homeomorphism such that $\tau(x):=x^{\text{\tt c}}$. The sets
$\tau[X\cup\mathrm{Fin}]$ and $\tau[Y\cup\mathrm{Fin}]$ are homeomorphic
copies of the $\operatorname{cov}(\mathcal{M})$-Luzin sets $X\cup\mathrm{Fin}$
and $Y\cup\mathrm{Fin}$, respectively. Thus, the sets
$\tau[X\cup\mathrm{Fin}]$ and $\tau[Y\cup\mathrm{Fin}]$ are
$\operatorname{cov}(\mathcal{M})$-Luzin subsets of
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, whose all finite powers are
Rothberger.

By Proposition 2.6(3), a continuous image of the product space
$\tau[X\cup\mathrm{Fin}]\times\tau[Y\cup\mathrm{Fin}]$ into
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, under the map
$(z,t)\mapsto\max\\{z,t\\}$, contains a dominating set
$\\{\,\max\\{x^{\text{\tt c}},y^{\text{\tt c}}\\}:x\in X,y\in Y\,\\}$. Thus,
the product space $\tau[X\cup\mathrm{Fin}]\times\tau[Y\cup\mathrm{Fin}]$ is
not Menger.

By Proposition 2.6(3), the union
$\tau[X\cup\mathrm{Fin}]\cup\tau[Y\cup\mathrm{Fin}]$ is a $2$-dominating
subset of ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. ∎

An open family of proper subsets of a space is an _$\omega$ -cover_ of the
space if, each finite subset of the space is contained in a member of the
family. A space $X$ is _Scheepers_ if, for every sequence of open covers
$\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$ of the space
$X$, there are finite subsets
$\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$
such that the family $\\{\,\bigcup\mathcal{F}_{n}:n\in\mathbb{N}\,\\}$ is an
$\omega$-cover of $X$. By the result of Tsaban and Zdomskyy, it is consistent
with ZFC that, the Menger and Scheepers properties are equivalent [22, Theorem
3.7]. Assuming that $\mathfrak{d}\leq\mathfrak{r}$, there is a Menger set of
reals that is not Scheepers [20, Theorem 2.1]. A separation of these
properties is also possible with Luzin type-sets. Assuming that the cardinal
number $\operatorname{cov}(\mathcal{M})$ is regular, the following theorem is
a generalization of the result of Bartoszyński, Shelah, and Tsaban [3, Theorem
3.1].

######  Theorem 2.7.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There is
a $\operatorname{cov}(\mathcal{M})$-Luzin set that is not Scheepers.

###### Proof.

Let $X$ and $Y$ be sets from Theorem 2.1. The set $X\cup Y$ is a
$\operatorname{cov}(\mathcal{M})$-Luzin set, a $2$-dominating subset of
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. No $2$-dominating subset of
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ is Scheepers [18, Theorem 2.1]. ∎

######  Corollary 2.8.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There is
a hereditary Rothberger set of reals that is not Scheepers.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. It
follows from Proposition 2.6 that, in particular, there is an ultrafilter $U$
with $\operatorname{\mathfrak{b}}(U)=\operatorname{cov}(\mathcal{M})$ and a
$\operatorname{cov}(\mathcal{M})$-Luzin set that is a $U$-scale. The below
Proposition 2.9 generalizes this result.

######  Proposition 2.9.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$.
Let $U\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be an ultrafilter with
$\mathfrak{b}(U)=\operatorname{cov}(\mathcal{M})$. There is a
$\operatorname{cov}(\mathcal{M})$-Luzin set that is a $U$-scale.

We need the following Lemma.

######  Lemma 2.10.

Let $U$ be an ultrafilter with
$\mathfrak{b}(U)=\operatorname{cov}(\mathcal{M})$. Let $\mathcal{M}^{\prime}$
be a family of meager sets in ${\operatorname{P}(\mathbb{N})}$ with
$\left|\mathcal{M}^{\prime}\right|<\operatorname{cov}(\mathcal{M})$ and
$Z\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ be a set with
$\left|Z\right|<\operatorname{cov}(\mathcal{M})$. There is an element
$x\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\setminus\bigcup\mathcal{M}^{\prime}$
such that $Z\leq_{U}x$.

###### Proof.

Since
$\left|Z\right|<\operatorname{cov}(\mathcal{M})=\operatorname{\mathfrak{b}}(U)$,
there is an element $y\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that
$Z\leq_{U}y$. Let $\mathcal{M}^{\prime}=\\{\,M_{\beta}:\beta<\alpha\,\\}$ for
some ordinal number $\alpha<\operatorname{cov}(\mathcal{M})$. Fix an ordinal
number $\beta<\alpha$. By Lemma 2.2, there are a set
$a_{\beta}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and a function
$f_{\beta}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that the set
$M_{\beta}$ is contained in the meager set

$\\{\,x\in{\operatorname{P}(\mathbb{N})}:x\cap[f_{\beta}(n),f_{\beta}(n+1))\neq
a_{\beta}\cap[f_{\beta}(n),f_{\beta}(n+1))\text{ for all but finitely many
}n\,\\}.$

Thus, we may assume that the set $M_{\beta}$ is equal to the above meager set.
Since $\alpha<\operatorname{cov}(\mathcal{M})$, there is a set
$a\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\setminus\bigcup\mathcal{M}^{\prime}$.
Then the sets

$f_{\beta}^{\prime}:=\\{\,f_{\beta}(n):a\cap[f_{\beta}(n),f_{\beta}(n+1))=a_{\beta}\cap[f_{\beta}(n),f_{\beta}(n+1))\,\\}$

are infinite, for all ordinal numbers $\beta<\alpha$.

Since $\alpha<\operatorname{cov}(\mathcal{M})\leq\mathfrak{d}$, by Lemma 2.3,
there is a function $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that the
sets

$I_{\beta}:=\\{\,n:\left|f^{\prime}_{\beta}\cap[b(n),b(n+1))\right|\geq
2\,\\}$

are infinite, for all ordinal numbers $\beta<\alpha$. We may assume that
$y(b(n))<b(n+1)$, for all natural numbers $n$.

Since $\alpha<\operatorname{cov}(\mathcal{M})\leq\mathfrak{r}$, there is a
decomposition of the set $\mathbb{N}$ into infinite sets $r_{1},r_{2},r_{3}$,
each of them reaps the family $\\{\,I_{\beta}:\beta\leq\alpha\,\\}$. Since $U$
is an ultrafilter, for the only one of the above sets $r_{i}$, we have
$\bigcup_{n\in r_{i}}[b(n),b(n+1))\in U$. Let say that for $r_{1}$. Define

$\displaystyle r_{1}^{2}:=$ $\displaystyle\\{\,n\in r_{1}:\min\\{\,k:k\geq
n,k\notin r_{1}\,\\}\in r_{2}\,\\},$ $\displaystyle r_{1}^{3}:=$
$\displaystyle\\{\,n\in r_{1}:\min\\{\,k:k\geq n,k\notin r_{1}\,\\}\in
r_{3}\,\\}.$

We have $r_{1}=r_{1}^{2}\cup r_{1}^{3}$ and $r_{1}^{2}\cap
r_{1}^{3}=\emptyset$. Since $U$ is an ultrafilter, for the only one of the
above sets $r_{1}^{j}$, we have $\bigcup_{n\in r_{1}^{j}}[b(n),b(n+1))\in U$.
Let say say that for $r_{1}^{2}$. Let

$x:=\bigcup_{n\in r_{3}}[b(n),b(n+1))\cap f.$

For a natural number $k\in\bigcup_{n\in r_{1}^{2}}[b(n),b(n+1))$, we have

$y(k)<y(b(n+1))<b(n+2)\leq x(k).$

Then $y\leq_{U}x$, and thus $Z\leq_{U}x$.

Fix an ordinal number $\beta<\alpha$. Since the set $r_{3}$ reaps the family
$\\{\,I_{\beta}:\beta\leq\alpha\,\\}$, the set $I_{\beta}\cap r_{3}$ is
infinite. We have

$x\cap[f^{\prime}_{\beta}(n),f^{\prime}_{\beta}(n+1))=a\cap[f^{\prime}_{\beta}(n),f^{\prime}_{\beta}(n+1)),$

for infinitely many natural numbers $n$. By the definition of the set
$a^{\prime}_{\beta}$, we have

$x\cap[f_{\beta}(n),f_{\beta}(n+1))=a\cap[f_{\beta}(n),f_{\beta}(n+1))=a_{\beta}\cap[f_{\beta}(n),f_{\beta}(n+1)),$

for infinitely many natural numbers $n$. Thus, $x\notin M_{\beta}$. ∎

###### Proof of Proposition 2.9.

Let $\\{\,M_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a cofinal
family of meager sets in ${\operatorname{P}(\mathbb{N})}$ and
$\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a dominating
set in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$.

Let $x_{0}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. Fix an ordinal number
$\alpha<\operatorname{cov}(\mathcal{M})$. By Lemma 2.10, applied for the
family $\\{\,M_{\beta}:\beta<\alpha\,\\}$ of meager sets, and the set
$\\{\,d_{\beta},x_{\beta}:\beta<\alpha\,\\}$, there is an element
$x_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\setminus\bigcup_{\beta<\alpha}M_{\beta}$
such that $\\{\,d_{\beta},x_{\beta}:\beta<\alpha\,\\}\leq_{U}x_{\alpha}$.

Let $X:=\\{\,x_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$. By the
construction, we have
$\left|X\right|=\operatorname{cov}(\mathcal{M})=\operatorname{\mathfrak{b}}(U)$.
Fix an element $b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. There is an
ordinal number $\beta<\operatorname{cov}(\mathcal{M})$ such that
$b\leq^{*}d_{\beta}$. For every ordinal number $\alpha$ with
$\beta<\alpha<\operatorname{cov}(\mathcal{M})$, we have
$d_{\beta}\leq_{U}x_{\alpha}$. Then $\left|\\{\,x\in
X:x\leq_{U}b\,\\}\right|<\operatorname{cov}(\mathcal{M})=\mathfrak{b}(U)$, and
thus the set $X$ is a $U$-scale.

By the construction, the set $X$ is a $\operatorname{cov}(\mathcal{M})$-Luzin
set. ∎

Let $U$ be an ultrafilter. A space $X$ is _$U$ -Menger_ [19] if, for every
sequence $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$ of open
covers of the space $X$, there are finite subsets
$\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$
such that $\\{\,n:x\in\bigcup\mathcal{F}_{n}\,\\}\in U$, for all points $x\in
X$.

######  Corollary 2.11.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$.
Let $U$ be an ultrafilter with
$\mathfrak{b}(U)=\operatorname{cov}(\mathcal{M})$. There is a $U$-Menger
$\operatorname{cov}(\mathcal{M})$-Luzin set whose all finite powers are
Rothberger.

###### Proof.

By Proposition 2.9, there is a $\operatorname{cov}(\mathcal{M})$-Luzin set
$X$, that is a $U$-scale. Then $X\cup\mathrm{Fin}$ is a $U$-Menger
$\operatorname{cov}(\mathcal{M})$-Luzin set [19, Corollary 4.6]. By the result
of Tsaban, Zdomskyy, and the first named author [20, Lemma 2.21], all finite
powers of this set are Rothberger. ∎

##  3\. Luzin-type sets via topological approach

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cf}(\mathfrak{d})$
or $\operatorname{cov}(\mathcal{M})=\mathfrak{d}$. For every
$\operatorname{cov}(\mathcal{M})$-Luzin set $X$, there is a Rothberger set of
reals $Y$ such that the product space $X\times Y$ is not Menger [19, Corollary
2.11]. Under various assumptions, we construct, for every
$\operatorname{cov}(\mathcal{M})$-Luzin set $X$, another
$\operatorname{cov}(\mathcal{M})$-Luzin set (with some extra properties) such
that the product space $X\times Y$ is not Menger.

###  3.1. Products of Luzin-type sets in ${{\mathbb{Z}}^{\mathbb{N}}}$

Let $\mathbb{Z}$ be the set of integers. Let
$f,g\in{{\mathbb{Z}}^{\mathbb{N}}}$. Define functions
$(f+g),|f+g|,(f-g)\in{{\mathbb{Z}}^{\mathbb{N}}}$ as follows:

$\displaystyle(f+g)(n)$ $\displaystyle:=f(n)+g(n),$ $\displaystyle|f+g|(n)$
$\displaystyle:=|f(n)+g(n)|,$ $\displaystyle(f-g)(n)$
$\displaystyle:=f(n)-g(n),$

for all natural numbers $n$. For a function $f\in{{\mathbb{Z}}^{\mathbb{N}}}$
and a set $X\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$, let

$f-X:=\\{\,f-g:g\in X\,\\}.$

In the set $\mathbb{Z}$, consider the discrete topology. The space
${{\mathbb{Z}}^{\mathbb{N}}}$, with the Tychonoff topology, is homeomorphic to
the Baire space ${\mathbb{N}^{\mathbb{N}}}$ (and thus also to
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$), and ${{\mathbb{Z}}^{\mathbb{N}}}$
with the group operation $+$, is a topological group. The cardinal numbers
$\operatorname{cov}(\mathcal{M})$ and $\operatorname{cof}(\mathcal{M})$ do not
change if, in their definitions, consider families of meager subsets of
${{\mathbb{Z}}^{\mathbb{N}}}$, instead of the real line.

######  Theorem 3.1.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. For
every $\operatorname{cov}(\mathcal{M})$-Luzin set
$X\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$, there is a
$\operatorname{cov}(\mathcal{M})$-Luzin set
$Y\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$ such that the product space $X\times
Y$ is not Menger.

###### Proof.

Let $\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a
dominating set in ${\mathbb{N}^{\mathbb{N}}}$ and
$\\{\,M_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a cofinal
family of meager sets in ${{\mathbb{Z}}^{\mathbb{N}}}$. Let

$y_{0}\in d_{0}-X.$

Fix an ordinal number $\alpha<\operatorname{cov}(\mathcal{M})$. The set
$d_{\alpha}-X$ is a $\operatorname{cov}(\mathcal{M})$-Luzin set. Since the set
$\bigcup_{\beta<\alpha}M_{\beta}\cup\\{\,y_{\beta}:\beta<\alpha\,\\}$ is a
union of less than $\operatorname{cov}(\mathcal{M})$ meager sets, and the
cardinal number $\operatorname{cov}(\mathcal{M})$ is regular, this union
cannot cover the set $d_{\alpha}-X$. Thus, there is a function

$y_{\alpha}\in(d_{\alpha}-X)\setminus\bigl{(}\bigcup_{\beta<\alpha}M_{\beta}\cup\\{\,y_{\beta}:\beta<\alpha\,\\}\bigr{)}.$

The set $Y:=\\{\,y_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ is a
$\operatorname{cov}(\mathcal{M})$-Luzin set: By the construction, we have
$\left|Y\right|=\operatorname{cov}(\mathcal{M})$. Let
$M\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$ be a meager set. There is an ordinal
number $\alpha<\operatorname{cov}(\mathcal{M})$ such that $M\subseteq
M_{\alpha}$. Since $M\cap Y\subseteq M_{\alpha}\cap
Y\subseteq\\{\,y_{\beta}:\beta<\alpha\,\\}$, we have $\left|M\cap
Y\right|<\operatorname{cov}(\mathcal{M})$.

The product space $X\times Y$ is not Menger: For each ordinal number
$\alpha<\operatorname{cov}(\mathcal{M})$, we have $y_{\alpha}\in
d_{\alpha}-X$, and thus there is a function $x_{\alpha}\in X$ such that
$x_{\alpha}+y_{\alpha}=d_{\alpha}$. Then the continuous image of the product
space $X\times Y$ in ${\mathbb{N}^{\mathbb{N}}}$, under the map
$(x,y)\mapsto|x+y|$, contains a dominating set
$\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$. Thus, the
product space $X\times Y$ is not Menger [14, Proposition 3]. ∎

######  Theorem 3.2.

Assume that $\operatorname{cov}(\mathcal{M})=\mathfrak{c}$ and the cardinal
number $\mathfrak{c}$ is regular. For every $\mathfrak{c}$-Luzin set
$X\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$, there is a $\mathfrak{c}$-Luzin set
$Y\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$, whose all finite powers are
Rothberger, and the product space $X\times Y$ is not Menger.

###### Proof.

Let $Q$ be a countable dense subset of ${{\mathbb{Z}}^{\mathbb{N}}}$. Let
${{\mathbb{Z}}^{\mathbb{N}}}=\\{\,d_{\alpha}:\alpha<\mathfrak{c}\,\\}$, and
$\\{\,M_{\alpha}:\alpha<\mathfrak{c}\,\\}$ be a cofinal family of meager sets
in ${{\mathbb{Z}}^{\mathbb{N}}}$. Let
$\\{\,(\mathcal{U}^{\alpha}_{1},\allowbreak\mathcal{U}^{\alpha}_{2},\allowbreak\dots):\alpha<\mathfrak{c}\,\\}$
be a set of all countable sequences of countable open families in
${{\mathbb{Z}}^{\mathbb{N}}}$.

By transfinite induction on ordinal numbers $\alpha<\mathfrak{c}$, choose
elements $y_{\alpha}\in{{\mathbb{Z}}^{\mathbb{N}}}$, sets
$Y_{\alpha}\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$ and sets
$U^{\alpha}_{1},\allowbreak
U^{\alpha}_{2},\allowbreak\dots\subseteq{{\mathbb{Z}}^{\mathbb{N}}}$ such
that:

  1. (i)

$y_{\alpha}\in(d_{\alpha}-X)\setminus\bigcup_{\beta<\alpha}M_{\beta}$,

  2. (ii)

$Y_{\alpha}:=Q\cup\\{\,y_{\beta}:\beta<\alpha\,\\}$

  3. (iii)

If
$\mathcal{U}^{\alpha}_{1},\allowbreak\mathcal{U}^{\alpha}_{2},\allowbreak\dots$
is a sequence of $\omega$-covers of the set $Y_{\alpha}$, then the sets
$U^{\alpha}_{1}\in\mathcal{U}^{\alpha}_{1},U^{\alpha}_{2}\in\mathcal{U}^{\alpha}_{2},\dotsc$
formulate an $\omega$-cover $\\{\,U^{\alpha}_{n}:n\in\mathbb{N}\,\\}$, of the
set $Y_{\alpha}\cup\\{y_{\alpha}\\}$.

  4. (iv)

If the condition from (iii) does not hold, then
$U^{\alpha}_{n}={{\mathbb{Z}}^{\mathbb{N}}}$, for all natural numbers $n$.

  5. (v)

If
$\\{\,U^{\beta}_{n}:n\in\mathbb{N}\,\\}\neq\\{{{\mathbb{Z}}^{\mathbb{N}}}\\}$,
then it is an $\omega$-cover of the set $Y_{\alpha}\cup\\{y_{\alpha}\\}$, for
all ordinal numbers $\beta<\alpha$.

Let $Y_{0}:=Q$. Fix an ordinal number $\alpha<\mathfrak{c}$, and let
$Y_{\alpha}:=\\{\,y_{\beta}:\beta<\alpha\,\\}\cup Q$. If the sequence
$\mathcal{U}^{\alpha}_{1},\allowbreak\mathcal{U}^{\alpha}_{2},\allowbreak\dots$
is not a sequence of $\omega$-covers of the set $Y_{\alpha}$, then
$U^{\alpha}_{n}:={{\mathbb{Z}}^{\mathbb{N}}}$, for all natural numbers $n$.
Assume that
$\mathcal{U}^{\alpha}_{1},\allowbreak\mathcal{U}^{\alpha}_{2},\allowbreak\dots$
is a sequence of $\omega$-covers of the set $Y_{\alpha}$. Since
$\left|Y_{\alpha}\right|<\mathfrak{c}=\operatorname{cov}(\mathcal{M})$, there
are sets
$U^{\alpha}_{1}\in\mathcal{U}^{\alpha}_{1},U^{\alpha}_{2}\in\mathcal{U}^{\alpha}_{2},\dotsc$
such that the family $\\{\,U^{\alpha}_{n}:n\in\mathbb{N}\,\\}$ is an
$\omega$-cover of the set $Y_{\alpha}$ [9, Theorem 4.8]. Assume that, for each
ordinal number $\beta<\alpha$, the family
$\\{\,U^{\beta}_{n}:n\in\mathbb{N}\,\\}$ is an $\omega$-cover of the set
$Y_{\alpha}$, or
$\\{\,U^{\beta}_{n}:n\in\mathbb{N}\,\\}=\\{{{\mathbb{Z}}^{\mathbb{N}}}\\}$.
For each finite set $F\subseteq Y_{\alpha}$ and each ordinal number
$\beta<\alpha$, define

$G^{F,\beta}:=\bigcup{\left\\{\,U^{\beta}_{n}:n\in\mathbb{N},F\subseteq
U^{\beta}_{n}\,\right\\}}.$

The set $G^{F,\beta}$ is open and dense in ${{\mathbb{Z}}^{\mathbb{N}}}$.
Define

$Z_{\alpha}:=\bigcup_{\beta<\alpha}M_{\beta}\cup\bigcup{\left\\{\,{{\mathbb{Z}}^{\mathbb{N}}}\setminus
G^{F,\beta}:\beta\leq\alpha,F\text{ is a finite subset of
}X_{\alpha}\,\right\\}}.$

The set $Z_{\alpha}$ is a union of less than $\mathfrak{c}$ meager sets. Since
the cardinal number $\mathfrak{c}$ is regular, the set $Z_{\alpha}$ cannot
cover the Luzin set $d_{\alpha}-X$, and thus there is a function

$y_{\alpha}\in(d_{\alpha}-X)\setminus Z_{\alpha}.$

Fix an ordinal number $\beta\leq\alpha$. Assume that
$\\{\,U^{\beta}_{n}:n\in\mathbb{N}\,\\}\neq\\{{{\mathbb{Z}}^{\mathbb{N}}}\\}$.
The family $\\{\,U^{\beta}_{n}:n\in\mathbb{N}\,\\}$ is an $\omega$-cover of
the set $Y_{\alpha}\cup\\{y_{\alpha}\\}$: For every finite set $F\subseteq
Y_{\alpha}$, we have $y_{\alpha}\in G^{F,\beta}$. Thus, there is an ordinal
number $n$ with $F\cup\\{y_{\alpha}\\}\subseteq U^{\beta}_{n}$.

Let $Y:=\bigcup_{\alpha<\mathfrak{c}}Y_{\alpha}$. By the construction, the set
$Y$ is a $\mathfrak{c}$-Luzin set. For each ordinal number
$\alpha<\mathfrak{c}$, we have $y_{\alpha}\in d_{\alpha}-X$, and thus there is
a function $x_{\alpha}\in X$ such that $x_{\alpha}+y_{\alpha}=d_{\alpha}$.
Then ${{\mathbb{Z}}^{\mathbb{N}}}$ is a continuous image of the product space
$X\times Y$, under the map $(x,y)\mapsto x+y$. Since the Menger property is
preserved under continuous maps, and the space ${{\mathbb{Z}}^{\mathbb{N}}}$
is not Menger (it is homeomorphic to ${\mathbb{N}^{\mathbb{N}}}$), the product
space $X\times Y$ is not Menger as well.

All finite powers of the set $Y$ are Rothberger: It is enough to show that,
for every sequence
$\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$ of
$\omega$-covers of the set $Y$, there are sets
$U_{1}\in\mathcal{U}_{1},U_{2}\in\mathcal{U}_{2},\dotsc$ such that the family
$\\{\,U_{n}:n\in\mathbb{N}\,\\}$ is an $\omega$-cover of $Y$ [9, Theorem 3.9].
Let $\mathcal{U}_{1},\mathcal{U}_{2},\dotsc$ be a sequence of $\omega$-covers
of the set $Y$, that are families of open sets in
${{\mathbb{Z}}^{\mathbb{N}}}$. There is an ordinal number
$\alpha<\mathfrak{c}$ such that the sequence
$\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dots$ is equal to the
sequence
$\mathcal{U}^{\alpha}_{1},\allowbreak\mathcal{U}^{\alpha}_{2},\allowbreak\dots$.
By the construction, the sets
$U^{\alpha}_{1}\in\mathcal{U}^{\alpha}_{1},U^{\alpha}_{2}\in\mathcal{U}^{\alpha}_{2},\dotsc$
formulates an $\omega$-cover $\\{\,U^{\alpha}_{n}:n\in\mathbb{N}\,\\}$ of the
set $Y_{\beta}$, for all ordinal numbers $\beta<\mathfrak{c}$. Thus, the
family $\\{\,U^{\alpha}_{n}:n\in\mathbb{N}\,\\}$ is an $\omega$-cover of the
set $Y$. ∎

###  3.2. Products of Luzin-type groups in ${\operatorname{P}(\mathbb{N})}$

A _$\operatorname{cov}(\mathcal{M})$ -Luzin group_, is a
$\operatorname{cov}(\mathcal{M})$-Luzin set that is a topological group.

Let $a,b\in{\operatorname{P}(\mathbb{N})}$ and
$A,B\subseteq{\operatorname{P}(\mathbb{N})}$. Define

$\displaystyle a\oplus b:=$ $\displaystyle(a\cup b)\setminus(a\cap b),$
$\displaystyle a\oplus B:=$ $\displaystyle\\{\,a\oplus b:b\in B\,\\},$
$\displaystyle A\oplus B:=$ $\displaystyle\\{\,a\oplus b:a\in A,b\in B\,\\}.$

The space ${\operatorname{P}(\mathbb{N})}$, with the group operation $\oplus$,
is a topological group. For $Y\subseteq{\operatorname{P}(\mathbb{N})}$, and
$y\in{\operatorname{P}(\mathbb{N})}$, let $\langle Y\rangle$ be the group
generated by the set $Y$, and $\langle A,x\rangle:=\langle
A\cup\\{x\\}\rangle$.

######  Theorem 3.3.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$.
There exist $\operatorname{cov}(\mathcal{M})$-Luzin groups
$X,Y\subseteq{\operatorname{P}(\mathbb{N})}$ such that the product space
$X\times Y$ is not Menger.

###### Proof.

Let $\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a
dominating set in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. Let
$\\{\,M_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a cofinal
family of meager sets in ${\operatorname{P}(\mathbb{N})}$, and
$\tilde{M}_{\alpha}:=\bigcup_{\beta<\alpha}M_{\beta}\cup\\{\,x^{\text{\tt
c}}:x\in\mathrm{Fin}\,\\}$, for all ordinal numbers
$\alpha<\operatorname{cov}(\mathcal{M})$.

By transfinite induction on ordinal numbers
$\alpha<\operatorname{cov}(\mathcal{M})$, pick elements $x_{\alpha}$,
$y_{\alpha}\in{\operatorname{P}(\mathbb{N})}$, construct increasing groups
$X_{\alpha},Y_{\alpha}$ with
$\left|X_{\alpha}\right|,\left|Y_{\alpha}\right|<\operatorname{cov}(\mathcal{M})$
and groups $\tilde{X}_{\alpha}:=\bigcup_{\beta<\alpha}X_{\beta}$,
$\tilde{Y}_{\alpha}:=\bigcup_{\beta<\alpha}Y_{\beta}$ such that

  1. (i)

$x_{\alpha},y_{\alpha}\notin\tilde{M}_{\alpha}\oplus(\tilde{X}_{\alpha}\cup\tilde{Y}_{\alpha})$,

  2. (ii)

$d_{\alpha}\leq^{*}\max\\{{x}_{\alpha}^{\text{\tt c}},{y}_{\alpha}^{\text{\tt
c}}\\}$,

  3. (iii)

$(X_{\alpha}\setminus\tilde{X}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset,(Y_{\alpha}\setminus\tilde{Y}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset$:

Fix an ordinal number $\alpha<\operatorname{cov}(\mathcal{M})$. By Lemma 2.4,
applied to the set
$\tilde{M}_{\alpha}\oplus(\tilde{X}_{\alpha}\cup\tilde{Y}_{\alpha})$, a union
of less than $\operatorname{cov}(\mathcal{M})$ meager sets, and to the
function $d_{\alpha}$, there are elements
$x_{\alpha},y_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ which
satisfies (i) and (ii). Let
$X_{\alpha}:=\langle\tilde{X}_{\alpha},x_{\alpha}\rangle$ and
$Y_{\alpha}:=\langle\tilde{Y}_{\alpha},y_{\alpha}\rangle$. We have
$(X_{\alpha}\setminus\tilde{X}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset$:
Suppose not. Then there are elements $x\in\tilde{X}_{\alpha}$ and
$m\in\tilde{M}_{\alpha}$ such that $x\oplus x_{\alpha}=m$. We have
$x_{\alpha}=x\oplus m$, and thus
$x_{\alpha}\in\tilde{X}_{\alpha}\oplus\tilde{M}_{\alpha}$, a contradiction.
Similarly, we have
$(Y_{\alpha}\setminus\tilde{Y}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset$.

By (iii), the sets
$X:=\bigcup_{\alpha<\operatorname{cov}(\mathcal{M})}X_{\alpha}$ and
$Y:=\bigcup_{\alpha<\operatorname{cov}(\mathcal{M})}Y_{\alpha}$ are
$\operatorname{cov}(\mathcal{M})$-Luzin groups.

The product space $X\times Y$ is not Menger: By (iii) and the definition of
the sets $\tilde{M}_{\alpha}$ we have $x^{\text{\tt c}}\notin\mathrm{Fin}$,
for all elements $x\in X\cup Y$. Thus, the continuous image of the product
space $X\times Y$, under the map $(x,y)\mapsto\max\\{{x}^{\text{\tt
c}},{y}^{\text{\tt c}}\\}$, is in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$. By
(ii), the image contains a dominating set $\\{\,\max\\{{x_{\alpha}}^{\text{\tt
c}},{y_{\alpha}}^{\text{\tt
c}}\\}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$. Thus, the product space
$X\times Y$ is not Menger [14, Proposition 3]. ∎

######  Theorem 3.4.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. For
every $\operatorname{cov}(\mathcal{M})$-Luzin group
$X\subseteq{\operatorname{P}(\mathbb{N})}$, there is a
$\operatorname{cov}(\mathcal{M})$-Luzin group
$Y\subseteq{\operatorname{P}(\mathbb{N})}$ such that the product space
$X\times Y$ is not Menger.

###### Proof.

Let $\\{\,d_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a
dominating set in ${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ such that
$d_{\alpha}^{\text{\tt c}}\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$, for all
ordinal numbers $\alpha<\operatorname{cov}(\mathcal{M})$. The sets
$B_{\alpha}:=\\{\,b\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}:d_{\alpha}\leq^{*}b\,\\}$
are meager sets of cardinality $\mathfrak{c}$, for all ordinal numbers
$\alpha<\operatorname{cov}(\mathcal{M})$. Let
$\\{\,M_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ be a cofinal
family of meager sets in ${\operatorname{P}(\mathbb{N})}$, and
$\tilde{M}_{\alpha}:=\bigcup_{\beta<\alpha}M_{\beta}$, for all ordinal numbers
$\alpha<\operatorname{cov}(\mathcal{M})$.

By transfinite induction on ordinal numbers
$\alpha<\operatorname{cov}(\mathcal{M})$, pick elements $b_{\alpha}$,
$y_{\alpha}\in{\operatorname{P}(\mathbb{N})}$ and construct increasing groups
$Y_{\alpha}$ with $\left|Y_{\alpha}\right|<\operatorname{cov}(\mathcal{M})$,
and $\tilde{Y}_{\alpha}:=\bigcup_{\beta<\alpha}Y_{\beta}$ such that:

  1. (i)

$b_{\alpha}\in B_{\alpha}\setminus\tilde{Y}_{\alpha}\oplus\mathrm{Fin}\oplus
X$,

  2. (ii)

$y_{\alpha}\in(b_{\alpha}\oplus
X)\setminus\bigl{(}\tilde{Y}_{\alpha}\oplus\tilde{M}_{\alpha}\cup\\{\,y_{\alpha}:\beta<\alpha\,\\}\bigr{)}$

  3. (iii)

$(Y_{\alpha}\setminus\tilde{Y}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset$.

Since the set $B_{0}$ is a meager set of cardinality $\mathfrak{c}$, and
$\mathrm{Fin}\oplus X$ is a $\operatorname{cov}(\mathcal{M})$-Luzin set, there
is an element

$b_{0}\in B_{0}\setminus(\mathrm{Fin}\oplus X).$

Let

$y_{0}\in b_{0}\oplus X,$

and $Y_{0}:=\langle y_{0}\rangle$. Fix an ordinal number
$\alpha<\operatorname{cov}(\mathcal{M})$. Let
$\tilde{Y}_{\alpha}:=\bigcup_{\beta<\alpha}Y_{\beta}$. Since the cardinal
number $\operatorname{cov}(\mathcal{M})$ is regular, we have
$|\,\tilde{Y}_{\alpha}\,|<\operatorname{cov}(\mathcal{M})$. Thus, the set
$\tilde{Y}_{\alpha}\oplus\mathrm{Fin}\oplus L$, a union of less than
$\operatorname{cov}(\mathcal{M})$ sets that are
$\operatorname{cov}(\mathcal{M})$-Luzin sets, is a
$\operatorname{cov}(\mathcal{M})$-Luzin set. Since $B_{\alpha}$ is a meager
set of cardinality $\mathfrak{c}$, there is an element

$b_{\alpha}\in B_{\alpha}\setminus(\tilde{Y}_{\alpha}\oplus\mathrm{Fin}\oplus
X).$

Since the set $b_{\alpha}\oplus X$ is a
$\operatorname{cov}(\mathcal{M})$-Luzin set and the cardinal number
$\operatorname{cov}(\mathcal{M})$ is regular, there is an element

$y_{\alpha}\in(b_{\alpha}\oplus
X)\setminus\bigl{(}\tilde{Y}_{\alpha}\oplus\tilde{M}_{\alpha}\cup\\{\,x_{\beta}:\beta<\alpha\,\\}\bigr{)}.$

We have $b_{\alpha}\notin\tilde{Y}_{\alpha}\oplus\mathrm{Fin}\oplus X$, the
set $X$ is a group, and thus $(y_{\alpha}\oplus
X)\cap(\tilde{Y}_{\beta}\oplus\mathrm{Fin}\oplus X)=\emptyset$. Let
$Y_{\alpha}:=\langle\tilde{Y}_{\alpha},y_{\alpha}\rangle$.

We have
$(Y_{\alpha}\setminus\tilde{Y}_{\alpha})\cap\tilde{M}_{\alpha}=\emptyset$:
Suppose not. Then there are elements $y\in\tilde{Y}_{\alpha}$ and
$m\in\tilde{M}_{\alpha}$ such that $y\oplus y_{\alpha}=m$. We have
$y_{\alpha}=y\oplus m$, and thus
$y_{\alpha}\in\tilde{Y}_{\alpha}\oplus\tilde{M}_{\alpha}$, a contradiction.

By (iii), the set
$Y:=\bigcup_{\alpha<\operatorname{cov}(\mathcal{M})}Y_{\alpha}$ is a
$\operatorname{cov}(\mathcal{M})$-Luzin set, and since
$\\{\,Y_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$ is an increasing
sequence of groups, it is a group.

The product space $X\times Y$ is not Menger: Let
$Z\subseteq{\operatorname{P}(\mathbb{N})}$ be the image of the product space
$X\times Y$, under the map $(x,y)\mapsto x\oplus y$. Suppose that there are
elements $x\in X$, $y\in Y$ such that $x\oplus
y\in\mathrm{Fin}\setminus\\{\emptyset\\}$. Then there is an ordinal number
$\alpha<\operatorname{cov}(\mathcal{M})$ such that $y\in Y_{\alpha}$. Thus,
there is an element $y^{\prime}\in\tilde{Y}_{\alpha}$ such that
$y=y^{\prime}\oplus y_{\alpha}$. Since $y_{\alpha}\in b_{\alpha}\oplus X$,
there is an element $x^{\prime}\in X$ such that $y_{\alpha}=b_{\alpha}\oplus
x^{\prime}$. We have

$x\oplus y=x\oplus y^{\prime}\oplus y_{\alpha}=x\oplus y^{\prime}\oplus
b_{\alpha}\oplus x^{\prime},$

and thus

$b_{\alpha}=y^{\prime}\oplus(x\oplus y)\oplus(x\oplus
x^{\prime})\in\tilde{Y}_{\alpha}\oplus\mathrm{Fin}\oplus X,$

a contradiction with (i). Thus, we have
$Z\subseteq{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\cup\\{\emptyset\\}$. Assume
that the product space $X\times Y$ is Menger. Since the Menger property is
preserved under continuous mappings, the set $Z$ is Menger. Since
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}\cup\\{\emptyset\\}$ is a set of reals,
the set $Z\setminus\\{\emptyset\\}$ is an $F_{\sigma}$ subset of $Z$, and thus
it is Menger as well. Since the set $Z\setminus\\{\emptyset\\}$ is a subset of
${[\mathbb{N}]^{\mbox{\tiny$\infty$}}}$ and it contains a dominating set
$\\{\,b_{\alpha}:\alpha<\operatorname{cov}(\mathcal{M})\,\\}$, it cannot be
Menger, a contradiction. ∎

##  4\. Applications to function spaces

For a set of reals $X$, let $\operatorname{C}_{\mathrm{p}}(X)$ be the space of
all continuous real-valued functions on the spaces $X$ with the topology of
pointwise convergence. Properties of the set $X$ can describe local properties
of the space $\operatorname{C}_{\mathrm{p}}(X)$, and vice versa. A space $Y$
has _countable fan tightness_ [1] if, for each point $y\in Y$ and for every
sequence $U_{1},\allowbreak U_{2},\allowbreak\dots$ of subsets of the space
$Y$ with $y\in\bigcap_{n}\overline{U_{n}}$, there are finite sets
$F_{1}\subseteq U_{1},F_{2}\subseteq U_{2},\dotsc$ such that
$y\in\overline{\bigcup_{n}F_{n}}$. If we request that the above sets
$F_{1},\allowbreak F_{2},\allowbreak\dots$ are singletons, then the space $Y$
has _countable strong fan tightness_ [15]. A space is _M-separable_ [5] if,
for every sequence $D_{1},\allowbreak D_{2},\allowbreak\dots$ of dense subsets
of the space, there are finite sets $F_{1}\subseteq D_{1},F_{2}\subseteq
D_{2},\dotsc$ such that the union $\bigcup_{n}F_{n}$ is a dense subset of the
space. If we request that, the above sets $D_{1},\allowbreak
D_{2},\allowbreak\dots$ are singletons, then the space is _R-separable_ [5].

Let $X$ be a set of reals. By the results of Sakai [15], and Scheepers [17,
Theorem 13], the statements: all finite powers of the set $X$ are Rothberger,
the space $\operatorname{C}_{\mathrm{p}}(X)$ has countable strong fan
tightness, the space $\operatorname{C}_{\mathrm{p}}(X)$ is R-separable, are
equivalent. By the results of Just, Miller, Scheepers, and Szeptycki [17,
Theorem 3.9], and Scheepers [17, Theorem 35], the statements: all finite
powers of the set $X$ are Menger, the space $\operatorname{C}_{\mathrm{p}}(X)$
has countable fan tightness, the space $\operatorname{C}_{\mathrm{p}}(X)$ is
M-separable, are equivalent, too.

Let $X$ and $Y$ be sets of reals, and $X\sqcup Y$ be a topological sum of
these sets. The product space $X\times Y$ is a closed subspace of the product
space $(X\sqcup Y)^{2}$. Thus, if $X\times Y$ is not Menger, then $(X\sqcup
Y)^{2}$ is not Menger, too. Since the product space
$\operatorname{C}_{\mathrm{p}}(X)\times\operatorname{C}_{\mathrm{p}}(Y)$ is
homeomorphic to the space $\operatorname{C}_{\mathrm{p}}(X\sqcup Y)$, we have
the following corollaries from Theorem 2.1.

######  Corollary 4.1.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There
are $\operatorname{cov}(\mathcal{M})$-Luzin sets $X$ and $Y$ such that the
spaces $\operatorname{C}_{\mathrm{p}}(X)$ and
$\operatorname{C}_{\mathrm{p}}(Y)$ have countable strong fan tightness (they
are R-separable) but their product space does not have countable fan tightness
(is not M-separable).

######  Corollary 4.2.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$
and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular. There
are hereditary Rothberger sets of reals $X$ and $Y$ such that the spaces
$\operatorname{C}_{\mathrm{p}}(X)$ and $\operatorname{C}_{\mathrm{p}}(Y)$ have
countable strong fan tightness (they are R-separable), but the product space
$\operatorname{C}_{\mathrm{p}}(X)\times\operatorname{C}_{\mathrm{p}}(Y)$ does
not have countable fan tightness (is not M-separable).

##  5\. Open problems

In the light of the above results, the following problems arises.

######  Problem 5.1.

Do Theorems 2.1, 2.7, 3.1, 3.2, 3.4 remain true if, we do not assume that the
cardinal number $\operatorname{cov}(\mathcal{M})$ is regular?

######  Problem 5.2.

Assume that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$.
Let $X$ be a $\operatorname{cov}(\mathcal{M})$-Luzin set. Is there a
$\operatorname{cov}(\mathcal{M})$-Luzin group $Y$ such that the product space
$X\times Y$ is not Menger? What if the cardinal number
$\operatorname{cov}(\mathcal{M})$ is regular? What if the Continuum Hypothesis
holds?

######  Problem 5.3.

Let $X$ be a $\operatorname{cov}(\mathcal{M})$-Luzin set. Is there a Menger
(Rothberger) set of reals $Y$, whose all finite powers are Menger
(Rothberger), but the product space $X\times Y$ is not Menger? What if the
Continuum Hypothesis holds?

######  Problem 5.4.

Assume that, the cardinal number $\operatorname{cov}(\mathcal{M})$ is
singular. Can a $\operatorname{cov}(\mathcal{M})$-Luzin set be a union of less
than $\operatorname{cov}(\mathcal{M})$ meager sets?

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