#  Iterations and unions of star selection properties on topological
spaces111The first-listed author was supported for this research by Consejo
Nacional de Ciencia y Tecnología (CONACYT, México), Scholarship 769010.

JAVIER CASAS-DE LA ROSA, WILLIAM CHEN-MERTENS, SERGIO GARCIA-BALAN

###### Abstract

In this paper, we investigate what selection principles properties are
possessed by small (with respect to the bounding and dominating numbers)
unions of spaces with certain (star) selection principles.. Furthermore, we
give several results about iterations of these properties and weaker
properties than paracompactness. In addition, we study the behaviour of these
iterated properties on $\Psi$-spaces. Finally, we show that, consistently,
there is a normal star-Menger space that is not strongly star-Menger; this
example answers a couple of questions posed in [7].

_Key words._ Menger, star Menger, strongly star Menger, Hurewicz, star
Hurewicz, strongly star Hurewicz, star selection principles, $\Psi$-spaces,
iterated stars.

_2020 Mathematics Subject Classification_ : Primary 54D20; Secondary 54A35.

##  1 Introduction

###  1.1 Notation and terminology

Let $X$ be a set and let $\mathcal{U}$ be a collection of subsets of $X$. If
$A$ is a subset of $X$, then the star of $A$ with respect to $\mathcal{U}$,
denoted by $St(A,\mathcal{U})$, is the set $\bigcup\\{U\in\mathcal{U}:U\cap
A\neq\emptyset\\}$; for $A=\\{x\\}$ with $x\in X$, we write
$St(x,\mathcal{U})$ instead of $St(\\{x\\},\mathcal{U})$. We denote by
$[X]^{<\omega}$ the collection of all finite subsets of $X$. Throughout this
paper, all spaces are assumed to be regular, unless a specific separation
axiom is indicated. For notation and terminology, we refer to [11].

We recall some classical star covering properties following the terminology of
[10]. A space $X$ is said to be strongly starcompact (strongly star-Lindelöf),
briefly $SSC$ ($SSL$), if for every open cover $\mathcal{U}$ of $X$ there
exists a finite (countable) subset $F$ of $X$ such that $St(F,\mathcal{U})=X$.
A space $X$ is starcompact (star-Lindelöf), briefly $SC$ ($SL$), if for every
open cover $\mathcal{U}$ of $X$ there exists a finite (countable) subset
$\mathcal{V}$ of $\mathcal{U}$ such that
$St(\bigcup\mathcal{V},\mathcal{U})=X$. It is well-known that countable
compactness and strongly starcompactness are equivalent for Hausdorff spaces
(see [10]). We refer the reader to the survey of Matveev [17] for a more
detailed treatment of these star covering properties.

Recall that a space $X$ is said to be _metacompact_ (_metaLindelöf_) if every
open cover $\mathcal{U}$ of $X$ has a point-finite (point-countable) open
refinement $\mathcal{V}$. Further, a space $X$ is said to be _paracompact_
(_paraLindelöf_) if every open cover $\mathcal{U}$ of $X$ has a locally-finite
(locally-countable) open refinement $\mathcal{V}$. For more information about
the relationships among these covering properties (and others), we refer the
reader to [5].

Recall that for $f,g\in\omega^{\omega}$, $f\leq^{*}g$ means that $f(n)\leq
g(n)$ for all but finitely many $n$ (and $f\leq g$ means that $f(n)\leq g(n)$
for all $n$). A subset $B$ of $\omega^{\omega}$ is _bounded_ if there is
$g\in\omega^{\omega}$ such that $f\leq^{*}g$ for each $f\in B$. A subset $D$
of $\omega^{\omega}$ is _dominating_ if for each $g\in\omega^{\omega}$ there
is $f\in D$ such that $g\leq^{*}f$. The minimal cardinality of an unbounded
subset of $\omega^{\omega}$ is denoted by $\mathfrak{b}$, and the minimal
cardinality of a dominating subset of $\omega^{\omega}$ is denoted by
$\mathfrak{d}$. The family of all meager subsets of $\mathbb{R}$ is denoted by
$\mathcal{M}$ and the minimum of the cardinalities of subfamilies
$\mathcal{U}\subset\mathcal{M}$ such that $\bigcup\mathcal{U=\mathbb{R}}$ is
denoted by $cov(\mathcal{M})$.

Recall that a family $\mathcal{A}$ of infinite subsets of $\omega$ is almost
disjoint (a.d., for short) if the intersection of any two distinct sets in
$\mathcal{A}$ is finite. Let $\mathcal{A}$ be an a.d. family, we consider
$\Psi(\mathcal{A})=\mathcal{A}\cup\omega$ with the following topology: the
points of $\omega$ are isolated and a basic neighbourhood of a point
$a\in\mathcal{A}$ is of the form $\\{a\\}\cup(a\setminus F)$, where $F$ is a
finite subset of $\omega$. Then $\Psi(\mathcal{A})$ is called a $\Psi$-space
(see [13]).

###  1.2 Classical (star) selection principles

Likely, among classical selection principles, the most well-known are the
Menger, Rothberger and Hurewicz properties. Let us recall those notions and
its star versions. Given a topological space $X$, we denote by $\mathcal{O}$
the collection of all open covers of $X$ and by $\Gamma$ the collection of all
$\gamma$-covers of $X$. Recall that an open cover $\mathcal{U}$ of $X$ is a
$\gamma$-cover if it is infinite and each $x\in X$ belongs to all but finitely
many elements of $\mathcal{U}$. A space $X$ is Menger ($M$) if for each
sequence $\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there is a
sequence $\\{\mathcal{V}_{n}:n\in\omega\\}$ such that for each $n\in\omega$,
$\mathcal{V}_{n}$ is a finite subset of $\mathcal{U}_{n}$ and
$\\{\bigcup\mathcal{V}_{n}:n\in\omega\\}$ is an open cover of $X$ (see [18]).
A space $X$ is Rothberger ($R$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there is a sequence
$\\{U_{n}:n\in\omega\\}$ such that for each $n\in\omega$,
$U_{n}\in\mathcal{U}_{n}$ and $\\{U_{n}:n\in\omega\\}$ is an open cover of $X$
(see [20]). A space $X$ is Hurewicz ($H$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there is a sequence
$\\{\mathcal{V}_{n}:n\in\omega\\}$ such that for each $n\in\omega$,
$\mathcal{V}_{n}$ is a finite subset of $\mathcal{U}_{n}$ and for each $x\in
X$, $x\in\bigcup\mathcal{V}_{n}$ for all but finitely many $n$ (see [14]). The
following star versions for the cases Menger and Rothberger were introduced in
[15] and the star versions for the Hurewicz case were defined in [2].

######  Definition 1.1.

A space $X$ is:

  1. 1.

star-Menger ($SM$) if for each sequence $\\{\mathcal{U}_{n}:n\in\omega\\}$ of
open covers of $X$, there is a sequence $\\{\mathcal{V}_{n}:n\in\omega\\}$
such that for each $n\in\omega$, $\mathcal{V}_{n}$ is a finite subset of
$\mathcal{U}_{n}$ and
$\\{St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover
of $X$.

  2. 2.

strongly star-Menger ($SSM$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there exists a
sequence $\\{F_{n}:n\in\omega\\}$ of finite subsets of $X$ such that
$\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.

  3. 3.

star-Rothberger ($SR$) if for each sequence $\\{\mathcal{U}_{n}:n\in\omega\\}$
of open covers of $X$, there are $U_{n}\in\mathcal{U}_{n}$, $n\in\omega$, such
that $\\{St(U_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.

  4. 4.

strongly star-Rothberger ($SSR$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there exists a
sequence $\\{x_{n}:n\in\omega\\}$ of elements of $X$ such that
$\\{St(x_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.

  5. 5.

star-Hurewicz ($SH$) if for each sequence $\\{\mathcal{U}_{n}:n\in\omega\\}$
of open covers of $X$, there is a sequence $\\{\mathcal{V}_{n}:n\in\omega\\}$
such that for each $n\in\omega$, $\mathcal{V}_{n}$ is a finite subset of
$\mathcal{U}_{n}$ and for each $x\in X$, $x\in
St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})$ for all but finitely many $n$.

  6. 6.

strongly star-Hurewicz ($SSH$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$, there exists a
sequence $\\{F_{n}:n\in\omega\\}$ of finite subsets of $X$ such that for each
$x\in X$, $x\in St(F_{n},\mathcal{U}_{n})$ for all but finitely many $n$.

It is worth to mention that for paracompact Hausdorff spaces the three Menger-
type properties, $SM$, $SSM$ and $M$ are equivalent and the same situation
holds for the three Rothberger-type properties and the three Hurewicz-type
properties (see [15] and [2]). Even more, those equivalences still true for
paraLindelöf spaces (see [7]).

The following diagram shows the relationships among these properties (in the
diagram $C$ and $L$ are used to denote compactness and the Lindelöf property,
respectively). We mention that none of the arrows in the following diagram
reverse. We refer the reader to [16] to see the current state of knowledge
about these relationships with others.

${C}$${H}$${M}$${R}$${L}$${SSC}$${SSH}$${SSM}$${SSR}$${SSL}$${SC}$${SH}$${SM}$${SR}$${SL}$
Figure 1: Star selection principles.

##  2 Small unions of some star spaces

In [27] Tall proved that if a space $X$ is Lindelöf and it can be written as a
union of less than $\mathfrak{d}$ compact spaces, then $X$ is Menger. It turns
out that we can replace “compact” by “star-Hurewicz” in Tall’s result (see
Proposition 2.3 below)222In [8], the authors also use the idea of unions of
size less than $\mathfrak{d}$ many Hurewicz-type spaces to obtain some results
about star-Scheepers spaces., or we can replace “$\mathfrak{d}$” and “compact”
by “$\mathfrak{b}$” and “star-Menger” (Proposition 2.7). Furthermore, a
Lindelöf space that can be written as a union of less than $\mathfrak{b}$
star-Hurewicz spaces, is Hurewicz (Proposition 2.4). These results are
contained in Theorem 2.2 below.

In addition, we investigated what happens if instead of starting with a
Lindelöf space that can be written as some small union, we consider a star-
Lindelöf space or a strongly star-Lindelöf space or an absolutely strongly
star-Lindelöf space (see Definition 2.17 below). Some other interesting
relationships were obtained and they are described in Theorem 2.10 and Theorem
2.19 below. Let us first introduce some notation that allows to present these
results in an organized manner333Preliminary versions of some results in this
section are contained in the PhD Dissertation of the third listed author (see
[12])..

######  Definition 2.1.

Let $X$ be any space, $A$ and $B$ denote some properties and $\kappa$ is some
cardinal.

$\big{(}A,(<\kappa,B)\big{)}$

stands for “$X$ satisfies property $A$ and it can be written as a union of
less than $\kappa$ spaces each of them satisfying property $B$”.

For instance, if $L,C$ and $M$ denote Lindelöf, compact and Menger,
respectively, then Tall’s result can be written as
“$\big{(}L,(<\mathfrak{d},C)\big{)}\rightarrow M$”. More in general, we have:

######  Theorem 2.2.

For any space $X$ the following holds:

$\big{(}L,(<\mathfrak{d},SH)\big{)}$

$\big{(}L,(<\mathfrak{b},SH)\big{)}$

$H$

$M$

$\big{(}L,(<\mathfrak{b},SM)\big{)}$

The proof is divided as Propositions 2.3, 2.4 and 2.7.

######  Proposition 2.3.

If $X$ is a Lindelöf space and $X$ is the union of less than $\mathfrak{d}$
star-Hurewicz spaces, then $X$ is Menger.

###### Proof.

Let $\kappa$ be a cardinal smaller than $\mathfrak{d}$ and put
$X=\bigcup_{\alpha<\kappa}Y_{\alpha}$ with each $Y_{\alpha}$ being a star-
Hurewicz space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open
covers of $X$. Since $X$ is Lindelöf, we can assume that for each
$n\in\omega$, $\mathcal{U}_{n}$ is countable and put
$\mathcal{U}_{n}=\\{U_{n}^{i}:i\in\omega\\}$. Since each $Y_{\alpha}$ is star-
Hurewicz, for each $\alpha<\kappa$, there exists a finite subset
$\mathcal{V}_{n}^{\alpha}$ of $\mathcal{U}_{n}$ such that
$\\{St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n}):n\in\omega\\}$ is a
$\gamma$-cover of $Y_{\alpha}$. Define, for each $\alpha<\kappa$, a function
$f_{\alpha}$ as follows: for each $n\in\omega$, let
$f_{\alpha}(n)=min\\{i\in\omega:\mathcal{V}_{n}^{\alpha}\subseteq\\{U_{n}^{j}:j\leq
i\\}\\}$. Since the collection $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less
than $\mathfrak{d}$, there exists $g\in\omega^{\omega}$ such that for every
$\alpha<\kappa$, $g\nleq^{*}f_{\alpha}$. For each $n\in\omega$, let
$\mathcal{W}_{n}=\\{U_{n}^{i}:i\leq g(n)\\}$.
_Claim:_ $\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an
open cover of $X$.
Let $x\in X$. Then, there exists $\alpha<\kappa$ such that $x\in Y_{\alpha}$.
Hence, there is $n_{0}\in\omega$ so that for every $n\geq n_{0}$, $x\in
St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n})$. Since
$g\nleq^{*}f_{\alpha}$, we can take $n>n_{0}$ such that $g(n)>f_{\alpha}(n)$.
Then $x\in St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(\bigcup_{j\leq f_{\alpha}(k)}U_{n}^{j},\mathcal{U}_{n})\subseteq
St(\bigcup_{j\leq
g(n)}U_{n}^{j},\mathcal{U}_{n})=St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n})$.
Therefore, the collection
$\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover
of $X$. Thus, $X$ is star-Menger. Finally, since $X$ is Lindelöf, $X$ is a
paracompact space and this allow us to conclude that $X$ is Menger.  ∎

######  Proposition 2.4.

If $X$ is a Lindelöf space and $X$ is the union of less than $\mathfrak{b}$
star-Hurewicz spaces, then $X$ is Hurewicz.

###### Proof.

Let $\kappa$ be a cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}Y_{\alpha}$ with each $Y_{\alpha}$ being a star-
Hurewicz space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open
covers of $X$. Since $X$ is Lindelöf, we can assume that for each
$n\in\omega$, $\mathcal{U}_{n}$ is countable and put
$\mathcal{U}_{n}=\\{U_{n}^{i}:i\in\omega\\}$. For each $\alpha<\kappa$, there
exists a finite subset $\mathcal{V}_{n}^{\alpha}$ of $\mathcal{U}_{n}$ such
that $\\{St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n}):n\in\omega\\}$ is
a $\gamma$-cover of $Y_{\alpha}$. Define, for each $\alpha<\kappa$, a function
$f_{\alpha}$ as follows: for each $n\in\omega$, let
$f_{\alpha}(n)=min\\{i\in\omega:\mathcal{V}_{n}^{\alpha}\subseteq\\{U_{n}^{j}:j\leq
i\\}\\}$. Since the collection $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less
than $\mathfrak{b}$, there exists $g\in\omega^{\omega}$ such that for every
$\alpha<\kappa$, $f_{\alpha}\leq^{*}g$. For each $n\in\omega$, let
$\mathcal{W}_{n}=\\{U_{n}^{i}:i\leq g(n)\\}$.
_Claim:_ $\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is a
$\gamma$-cover of $X$.
Let $x\in X$. Then, there exists $\alpha<\kappa$ such that $x\in Y_{\alpha}$.
Hence, there is $n_{0}\in\omega$ so that for every $n\geq n_{0}$, $x\in
St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n})$. Since
$f_{\alpha}\leq^{*}g$, there is $n_{1}\in\omega$ such that for every $n\geq
n_{1}$, $f_{\alpha}(n)\leq g(n)$. Put $m=max\\{n_{0},n_{1}\\}$. Hence, for
each $k\geq m$, $x\in St(\bigcup\mathcal{W}_{k},\mathcal{U}_{k})$. Indeed, let
$k\geq m$. Then $x\in
St(\bigcup\mathcal{V}_{k}^{\alpha},\mathcal{U}_{k})\subseteq St(\bigcup_{j\leq
f_{\alpha}(k)}U_{k}^{j},\mathcal{U}_{k})\subseteq St(\bigcup_{j\leq
g(k)}U_{k}^{j},\mathcal{U}_{k})=St(\bigcup\mathcal{W}_{k},\mathcal{U}_{k})$.
Therefore, the collection
$\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is a
$\gamma$-cover of $X$. Thus, $X$ is star-Hurewicz. Finally, since $X$ is
Lindelöf, $X$ is a paracompact space and we conclude that $X$ is Hurewicz.  ∎

In this article, by large cover we mean the following:

######  Definition 2.5.

A cover $\mathcal{U}=\\{U_{\alpha}:\alpha<\kappa\\}$ of a space $X$ it’s
called large if for every $\alpha<\kappa$,
$\\{U_{\beta}:\alpha\leq\beta<\kappa\\}$ is a cover of $X$. We denote the
class of large covers of $X$ by $\mathcal{L}(X)$.

Observe that when we consider countable covers, the previous definition and
the one given in [22] coincide.

For the following lemma we recall some classical notation of (star) selection
principles introduced by M. Scheepers (Kočinac) in [22] ([15]). Let
$\mathcal{A}$ and $\mathcal{B}$ be collections of families of sets.

$S_{fin}(\mathcal{A},\mathcal{B})$: For each sequence $\\{A_{n}:n\in\omega\\}$
of elements of $\mathcal{A}$ there is a sequence $\\{B_{n}:n\in\omega\\}$ such
that for each $n\in\omega$, $B_{n}\in[A_{n}]^{<\omega}$ and
$\bigcup\\{B_{n}:n\in\omega\\}$ is an element of $\mathcal{B}$.

$U_{fin}(\mathcal{A},\mathcal{B})$: For each sequence $\\{A_{n}:n\in\omega\\}$
of elements of $\mathcal{A}$ there is a sequence $\\{B_{n}:n\in\omega\\}$ such
that for each $n\in\omega$, $B_{n}\in[A_{n}]^{<\omega}$ and $\\{\bigcup
B_{n}:n\in\omega\\}$ is an element of $\mathcal{B}$.

$S^{*}_{fin}(\mathcal{A},\mathcal{B})$: For each sequence
$\\{A_{n}:n\in\omega\\}$ of elements of $\mathcal{A}$ there exists a sequence
$\\{B_{n}:n\in\omega\\}$ such that for each $n\in\omega$,
$B_{n}\in[A_{n}]^{<\omega}$ and $\\{St(\bigcup B_{n},A_{n}):n\in\omega\\}$ is
an element of $\mathcal{B}$.

The following Lemma will be useful to clarify some steps in the proofs of
Propositions 2.7 and Lemma 2.9.

######  Lemma 2.6 (Folklore).

For any space $X$:

  1. 1.

$S_{fin}(\mathcal{O},\mathcal{O})\leftrightarrow
U_{fin}(\mathcal{O},\mathcal{L})$.

  2. 2.

$S^{*}_{fin}(\mathcal{O},\mathcal{O})\leftrightarrow
S^{*}_{fin}(\mathcal{O},\mathcal{L})$.

###### Proof.

Let $X$ be any space. Observe that
$U_{fin}(\mathcal{O},\mathcal{L})\rightarrow S_{fin}(\mathcal{O},\mathcal{O})$
and $S^{*}_{fin}(\mathcal{O},\mathcal{L})\rightarrow
S^{*}_{fin}(\mathcal{O},\mathcal{O})$ are immediate. Now, assume
$S_{fin}(\mathcal{O},\mathcal{O})$ ($S^{*}_{fin}(\mathcal{O},\mathcal{O})$
respectively) holds. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of
open covers of $X$ and let $m\in\omega$. Since the collection
$\\{\mathcal{U}_{n}:m\leq n<\omega\\}$ is a sequence of open covers of $X$,
then for each $n\geq m$ there exists a finite subset $\mathcal{V}_{n}^{m}$ of
$\mathcal{U}_{n}$ such that $\bigcup\\{\mathcal{V}_{n}^{m}:m\leq n<\omega\\}$
($\\{St(\bigcup\mathcal{V}_{n}^{m},\mathcal{U}_{n}):m\leq n<\omega\\}$, resp.)
is an open cover of $X$. So, for each $n\in\omega$ we define
$\mathcal{W}_{n}=\bigcup_{m\leq n}\mathcal{V}_{n}^{m}$. Hence, for each
$m\in\omega$ the collection $\bigcup\\{\mathcal{W}_{n}:m\leq n<\omega\\}$
($\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):m\leq n<\omega\\}$, resp.) is
an open cover of $X$. Furthermore, for each $m\in\omega$ the collection
$\\{\bigcup\mathcal{W}_{n}:m\leq n<\omega\\}$ is an open cover of $X$. Thus,
for each $n$, $\mathcal{W}_{n}$ is a finite subset of $\mathcal{U}_{n}$ and
the collection $\\{\bigcup\mathcal{W}_{n}:n<\omega\\}$
($\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n<\omega\\}$, resp.) is a
large cover of $X$. Hence, $U_{fin}(\mathcal{O},\mathcal{L})$
($S^{*}_{fin}(\mathcal{O},\mathcal{L})$, resp.) holds.  ∎

######  Proposition 2.7.

If $X$ is a Lindelöf space and $X$ is the union of less than $\mathfrak{b}$
star-Menger spaces, then $X$ is Menger.

###### Proof.

Let $\kappa$ be a cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}Y_{\alpha}$ with each $Y_{\alpha}$ being a star-
Menger space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open
covers of $X$. Since $X$ is Lindelöf, we can assume that for each
$n\in\omega$, $\mathcal{U}_{n}$ is countable and put
$\mathcal{U}_{n}=\\{U_{n}^{i}:i\in\omega\\}$. Since or each $\alpha<\kappa$,
$Y_{\alpha}$ is star-Menger, by Lemma 2.6, for each $\alpha<\kappa$ there is
$\mathcal{V}_{n}^{\alpha}$ finite subset of $\mathcal{U}_{n}$ such that for
every $m\in\omega$,
$\\{St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n}):m\leq n<\omega\\}$ is
an open cover of $Y_{\alpha}$. Define, for each $\alpha<\kappa$, a function
$f_{\alpha}$ as follows: for each $n\in\omega$, let
$f_{\alpha}(n)=min\\{i\in\omega:\mathcal{V}_{n}^{\alpha}\subseteq\\{U_{n}^{j}:j\leq
i\\}\\}$. Since the collection $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less
than $\mathfrak{b}$, there exists $g\in\omega^{\omega}$ such that for every
$\alpha<\kappa$, $f_{\alpha}\leq^{*}g$. For each $n\in\omega$, let
$\mathcal{W}_{n}=\\{U_{n}^{i}:i\leq g(n)\\}$.
_Claim:_ $\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an
open cover of $X$.
Let $x\in X$ and fix $\alpha<\kappa$ such that $x\in Y_{\alpha}$. Hence, there
is $m\in\omega$ so that for every $n\geq m$, $f_{\alpha}(n)\leq g(n)$. Let
$n\geq m$ such that $x\in
St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n})$. Observe that
$St(\bigcup\mathcal{V}_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(\bigcup_{j\leq f_{\alpha}(n)}U_{n}^{j},\mathcal{U}_{n})\subseteq
St(\bigcup_{j\leq
g(n)}U_{n}^{j},\mathcal{U}_{n})=St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n})$.
Therefore, the collection
$\\{St(\bigcup\mathcal{W}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover
of $X$. Thus, $X$ is star-Menger. Finally, since $X$ is Lindelöf, $X$ is a
paracompact space and this allow us to conclude that $X$ is Menger.  ∎

Following [4] (see also [19]), we recall some modifications of the Menger and
Hurewicz properties, called $E^{*}_{\omega}$ and $E^{**}_{\omega}$ properties,
respectively. We say that a space $X$ has the property $E^{*}_{\omega}$
($E^{**}_{\omega}$), if for every sequence $\\{\mathcal{U}_{n}:n\in\omega\\}$
of countable open covers of $X$, there exists a sequence
$\\{\mathcal{V}_{n}:n\in\omega\\}$ such that for each $n\in\omega$,
$\mathcal{V}_{n}$ is a finite subset of $\mathcal{U}_{n}$ and
$\\{\bigcup\mathcal{V}_{n}:n\in\omega\\}$ is an open cover ($\gamma$-cover) of
$X$. If we recall the definition of countably compact space (every countable
cover has a finite subcover), properties $E^{*}_{\omega}$ and
$E^{**}_{\omega}$ could be called countably Menger and countably Hurewicz,
respectively. As pointed out in [19], in the class of Lindelöf spaces, the
properties Menger and $E^{*}_{\omega}$ are the same. However, this fact is not
true in general. The space $\omega_{1}$ (with the order topology) has the
property $E^{*}_{\omega}$ and it is not a Menger space.

######  Lemma 2.8.

If $X$ can be written as a union of less than $\mathfrak{d}$ many Hurewicz
spaces, then $X$ is $E^{*}_{\omega}$.

###### Proof.

Assume $X=\bigcup_{\alpha<\kappa}H_{\alpha}$ so that $\kappa<\mathfrak{d}$ and
each $H_{\alpha}$ is Hurewicz. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a
sequence of countable open covers of $X$. For each $n\in\omega$ let
$\mathcal{U}_{n}=\\{U^{i}_{n}:i\in\omega\\}$. For each $\alpha<\kappa$ and
$n\in\omega$, there exists
$\mathcal{F}^{\alpha}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ so that
$\\{\bigcup\mathcal{F}^{\alpha}_{n}:n\in\omega\\}$ is a $\gamma$-cover of
$H_{\alpha}$. For each $\alpha<\kappa$ and $n\in\omega$, let
$f_{\alpha}(n)=min\\{m\in\omega:\mathcal{F}^{\alpha}_{n}\subseteq\\{U^{i}_{n}:i\leq
m\\}\\}$. Given that $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less than
$\mathfrak{d}$, there exists $g\in\omega^{\omega}$ so that for each
$\alpha<\kappa$, $g\not\leq^{*}f_{\alpha}$. For each $n\in\omega$, let
$G_{n}=\\{U^{i}_{n}:i\leq g(n)\\}$. Let us check that $\\{\bigcup
G_{n}:n\in\omega\\}$ is an open cover of $X$. Let $x\in X$, then there is
$\alpha<\kappa$ so that $x\in H_{\alpha}$. Thus, there is $n_{0}\in\omega$ so
that for each $n\geq n_{0}$, $x\in\bigcup\mathcal{F}^{\alpha}_{n}$. Pick
$m\geq n_{0}$ with $f_{\alpha}(m)<g(m)$. Hence,
$x\in\bigcup\mathcal{F}^{\alpha}_{m}\subseteq\bigcup G_{m}$. Therefore, $X$ is
$E^{*}_{\omega}$.  ∎

Analogous to Lemma 2.8, the following also holds:

######  Lemma 2.9.

  1. 1.

If $X$ can be written as a union of less than $\mathfrak{b}$ many Menger
spaces, then $X$ is $E^{*}_{\omega}$.

  2. 2.

If $X$ can be written as a union of less than $\mathfrak{b}$ many Hurewicz
spaces, then $X$ is $E^{**}_{\omega}$.

###### Proof.

$(1)$ Assume $X=\bigcup_{\alpha<\kappa}M_{\alpha}$ so that
$\kappa<\mathfrak{b}$ and each $M_{\alpha}$ is Menger. Let
$\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of countable open covers of
$X$. For each $n\in\omega$ let $\mathcal{U}_{n}=\\{U^{i}_{n}:i\in\omega\\}$.
By Lemma 2.6, for each $\alpha<\kappa$ and $n\in\omega$, there exists
$\mathcal{F}^{\alpha}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ so that
$\\{\bigcup\mathcal{F}^{\alpha}_{n}:n\in\omega\\}$ is a large cover of
$M_{\alpha}$. For each $\alpha<\kappa$ and $n\in\omega$, let
$f_{\alpha}(n)=min\\{m\in\omega:\mathcal{F}^{\alpha}_{n}\subseteq\\{U^{i}_{n}:i\leq
m\\}\\}$. Given that $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less than
$\mathfrak{b}$, there exists $g\in\omega^{\omega}$ so that for each
$\alpha<\kappa$, $f_{\alpha}\leq^{*}g$. For each $n\in\omega$, let
$G_{n}=\\{U^{i}_{n}:i\leq g(n)\\}$. Let us check that $\\{\bigcup
G_{n}:n\in\omega\\}$ is an open cover of $X$. Let $x\in X$, then there is
$\alpha<\kappa$ so that $x\in M_{\alpha}$. Thus, there is $n_{0}\in\omega$ so
that for each $n\geq n_{0}$, $f_{\alpha}(n)\leq g(n)$. Since
$\\{\bigcup\mathcal{F}^{\alpha}_{n}:n\in\omega\\}$ is a large cover of
$M_{\alpha}$, there is $n\geq n_{0}$ such that
$x\in\bigcup\mathcal{F}^{\alpha}_{n}\subseteq\bigcup_{i\leq
f_{\alpha}(n)}U^{i}_{n}\subseteq\bigcup_{i\leq g(n)}U^{i}_{n}=\bigcup G_{n}$.
Therefore, $X$ is $E^{*}_{\omega}$.

$(2)$ Assume $X=\bigcup_{\alpha<\kappa}H_{\alpha}$ so that
$\kappa<\mathfrak{b}$ and each $H_{\alpha}$ is Hurewicz. Let
$\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of countable open covers of
$X$. For each $n\in\omega$ let $\mathcal{U}_{n}=\\{U^{i}_{n}:i\in\omega\\}$.
For each $\alpha<\kappa$ and $n\in\omega$, there exists
$\mathcal{F}^{\alpha}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ so that
$\\{\bigcup\mathcal{F}^{\alpha}_{n}:n\in\omega\\}$ is $\gamma$-cover of
$H_{\alpha}$. For each $\alpha<\kappa$ and $n\in\omega$, let
$f_{\alpha}(n)=min\\{m\in\omega:\mathcal{F}^{\alpha}_{n}\subseteq\\{U^{i}_{n}:i\leq
m\\}\\}$. Given that $\\{f_{\alpha}:\alpha<\kappa\\}$ has size less than
$\mathfrak{b}$, there exists $g\in\omega^{\omega}$ so that for each
$\alpha<\kappa$, $f_{\alpha}\leq^{*}g$. For each $n\in\omega$, let
$G_{n}=\\{U^{i}_{n}:i\leq g(n)\\}$. Let us check that $\\{\bigcup
G_{n}:n\in\omega\\}$ is a $\gamma$-cover of $X$. Let $x\in X$, then there is
$\alpha<\kappa$ so that $x\in H_{\alpha}$. Thus, there is $n_{0}\in\omega$ so
that for each $n\geq n_{0}$, $f_{\alpha}(n)\leq g(n)$. Since
$\\{\bigcup\mathcal{F}^{\alpha}_{n}:n\in\omega\\}$ is a $\gamma$-cover of
$H_{\alpha}$, we can fix $n_{1}\geq n_{0}$ such that for each $n\geq n_{1}$,
$x\in\bigcup\mathcal{F}^{\alpha}_{n}\subseteq\bigcup_{i\leq
f_{\alpha}(n)}U^{i}_{n}\subseteq\bigcup_{i\leq g(n)}U^{i}_{n}=\bigcup G_{n}$.
Therefore, $X$ is $E^{**}_{\omega}$.  ∎

######  Theorem 2.10.

For any space $X$ the following holds:

$\big{(}SSL,(<\mathfrak{d},H)\big{)}$

$\big{(}SSL,(<\mathfrak{b},H)\big{)}$

$SSH$

$SSM$

$\big{(}SSL,(<\mathfrak{b},M)\big{)}$

$\big{(}SL,(<\mathfrak{d},H)\big{)}$

$\big{(}SL,(<\mathfrak{b},H)\big{)}$

$SH$

$SM$

$\big{(}SL,(<\mathfrak{b},M)\big{)}$

The proof is divided as Propositions 2.11, 2.12, 2.13, 2.14, 2.15 and 2.16.

######  Proposition 2.11.

If $X$ is a strongly star-Lindelöf space and $X$ is the union of less than
$\mathfrak{d}$ Hurewicz spaces, then $X$ is strongly star-Menger.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{d}$ and put
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is strongly star-Lindelöf, for each $n\in\omega$ there exists
$C_{n}\in[X]^{\leq\omega}$ such that $St(C_{n},\mathcal{U}_{n})=X$. For each
$n\in\omega$, put $C_{n}=\\{x_{n}^{i}:i\in\omega\\}$. Note that
$St(C_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(x_{n}^{i},\mathcal{U}_{n})$
for each $n\in\omega$. So, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(x_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is a
countable open cover of $X$. By Lemma 2.8, $X$ is $E^{*}_{\omega}$ and then we
can get finite subcollections $\mathcal{F}_{n}$ of $\mathcal{W}_{n}$ so that
$\\{\bigcup\mathcal{F}_{n}:n\in\omega\\}$ is a cover of $X$. Equivalently, we
get finite subsets $F_{n}$ of $X$ such that the collection
$\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$. Thus,
$X$ is strongly star-Menger.  ∎

######  Proposition 2.12.

If $X$ is a strongly star-Lindelöf space and $X$ is the union of less than
$\mathfrak{b}$ Hurewicz spaces, then $X$ is strongly star-Hurewicz.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is strongly star-Lindelöf, for each $n\in\omega$ there exists
$C_{n}\in[X]^{\leq\omega}$ such that $St(C_{n},\mathcal{U}_{n})=X$. For each
$n\in\omega$, put $C_{n}=\\{x_{n}^{i}:i\in\omega\\}$. Note that
$St(C_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(x_{n}^{i},\mathcal{U}_{n})$
for each $n\in\omega$. So, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(x_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is a
countable open cover of $X$. By Lemma 2.9 (2), $X$ is $E^{**}_{\omega}$ and
then we can get finite subcollections $\mathcal{F}_{n}$ of $\mathcal{W}_{n}$
so that $\\{\bigcup\mathcal{F}_{n}:n\in\omega\\}$ is a $\gamma$-cover of $X$.
Equivalently, we get finite subsets $F_{n}$ of $X$ such that the collection
$\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is a $\gamma$-cover of $X$. Thus,
$X$ is strongly star-Hurewicz.  ∎

######  Proposition 2.13.

If $X$ is a strongly star-Lindelöf space and $X$ is the union of less than
$\mathfrak{b}$ Menger spaces, then $X$ is strongly star-Menger.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}M_{\alpha}$ with each $M_{\alpha}$ being a Menger
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is strongly star-Lindelöf, for each $n\in\omega$ there exists
$C_{n}\in[X]^{\leq\omega}$ such that $St(C_{n},\mathcal{U}_{n})=X$. For each
$n\in\omega$, put $C_{n}=\\{x_{n}^{i}:i\in\omega\\}$. Observe that for each
$n\in\omega$,
$St(C_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(x_{n}^{i},\mathcal{U}_{n})$.
Thus, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(x_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is a
countable open cover of $X$. By Lemma 2.9 (1), $X$ is $E^{*}_{\omega}$ and
then, we can get finite collections $\mathcal{F}_{n}$ of $\mathcal{W}_{n}$
such that the collection $\\{\bigcup\mathcal{F}_{n}:n\in\omega\\}$ is an open
cover of $X$. Equivalently, we get finite subsets $F_{n}$ of $X$ such that
$\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$. Thus,
$X$ is strongly star-Menger.  ∎

######  Proposition 2.14.

If $X$ is a star-Lindelöf space and $X$ is the union of less than
$\mathfrak{d}$ Hurewicz spaces, then $X$ is star-Menger.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{d}$ and put
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is star-Lindelöf, for each $n\in\omega$ there exists
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{\leq\omega}$ such that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=X$. For each $n\in\omega$, put
$\mathcal{V}_{n}=\\{V_{n}^{i}:i\in\omega\\}$. Note that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(V_{n}^{i},\mathcal{U}_{n})$
for each $n\in\omega$. So, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(V_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is a
countable open cover of $X$. By Lemma 2.8, $X$ is $E^{*}_{\omega}$ and then we
can get finite subcollections $\mathcal{H}_{n}$ of $\mathcal{W}_{n}$ so that
$\\{\bigcup\mathcal{H}_{n}:n\in\omega\\}$ is an open cover of $X$. That is, we
get finite sets $\mathcal{F}_{n}$ of $\mathcal{U}_{n}$ such that the
collection $\\{St(\bigcup\mathcal{F}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an
open cover of $X$. Thus, $X$ is star-Menger.  ∎

######  Proposition 2.15.

If $X$ is a star-Lindelöf space and $X$ is the union of less than
$\mathfrak{b}$ Hurewicz spaces, then $X$ is star-Hurewicz.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is star-Lindelöf, for each $n\in\omega$ there exists
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{\leq\omega}$ such that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=X$. For each $n\in\omega$, put
$\mathcal{V}_{n}=\\{V_{n}^{i}:i\in\omega\\}$. Note that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(V_{n}^{i},\mathcal{U}_{n})$
for each $n\in\omega$. So, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(V_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is a
countable open cover of $X$. By Lemma 2.9 (2), $X$ is $E^{**}_{\omega}$ and
then we can get finite subcollections $\mathcal{H}_{n}$ of $\mathcal{W}_{n}$
so that $\\{\bigcup\mathcal{H}_{n}:n\in\omega\\}$ is a $\gamma$-cover of $X$.
That is, we get finite sets $\mathcal{F}_{n}$ of $\mathcal{U}_{n}$ such that
the collection $\\{St(\bigcup\mathcal{F}_{n},\mathcal{U}_{n}):n\in\omega\\}$
is a $\gamma$-cover of $X$. Thus, $X$ is star-Hurewicz.  ∎

######  Proposition 2.16.

If $X$ is a star-Lindelöf space and $X$ is the union of less than
$\mathfrak{b}$ Menger spaces, then $X$ is star-Menger.

###### Proof.

Let $\kappa$ be any cardinal smaller than $\mathfrak{b}$ and put
$X=\bigcup_{\alpha<\kappa}M_{\alpha}$ with each $M_{\alpha}$ being a Menger
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of
$X$. Since $X$ is star-Lindelöf, for each $n\in\omega$ there exists
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{\leq\omega}$ such that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=X$. For each $n\in\omega$, put
$\mathcal{V}_{n}=\\{V_{n}^{i}:i\in\omega\\}$. Note that
$St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})=\bigcup_{i\in\omega}St(V_{n}^{i},\mathcal{U}_{n})$
for each $n\in\omega$. So, for each $n\in\omega$, the collection
$\mathcal{W}_{n}=\\{St(V_{n}^{i},\mathcal{U}_{n}):i\in\omega\\}$ is an
countable open cover of $X$. By Lemma 2.9 (1), $X$ is $E^{*}_{\omega}$ and
then we can get finite subcollections $\mathcal{H}_{n}$ of $\mathcal{W}_{n}$
so that $\\{\bigcup\mathcal{H}_{n}:n\in\omega\\}$ is an open cover of $X$.
Equivalently, we obtain finite subcollections $\mathcal{F}_{n}$ of
$\mathcal{U}_{n}$ so that the collection
$\\{St(\bigcup\mathcal{F}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover
of $X$. Thus, $X$ is star-Menger.  ∎

Another interesting fact is that we also have theorems of same structure for
the selective versions of the star selection principles. We recall the
necessary definitions to state the analogous theorem for these selective
versions.

The following version of the strongly star-Lindelöf property was introduced
and studied by Bonanzinga in [1]:

######  Definition 2.17.

A space $X$ is absolutely strongly star-Lindelöf ($aSSL$) if for any open
cover $\mathcal{U}$ of $X$ and any dense subset $D$ of $X$, there is a
countable set $C\subseteq D$ such that $St(C,\mathcal{U})=X$.

The selective versions below are stronger properties than the classical star
selection principles (see for instance [6] for more information of these
properties444The Rothberger case and some other interesting properties are
also given in [6]).

######  Definition 2.18.

We say that a space $X$ is:

  1. 1.

selectively strongly star-Menger ($selSSM$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$ and each sequence
$\\{D_{n}:n\in\omega\\}$ of dense sets of $X$, there exists a sequence
$\\{F_{n}:n\in\omega\\}$ of finite sets such that $F_{n}\subseteq D_{n}$,
$n\in\omega$, and $\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open
cover of $X$ (see [9]).

  2. 2.

selectively strongly star-Hurewicz ($selSSH$) if for each sequence
$\\{\mathcal{U}_{n}:n\in\omega\\}$ of open covers of $X$ and each sequence
$\\{D_{n}:n\in\omega\\}$ of dense sets of $X$, there exists a sequence
$\\{F_{n}:n\in\omega\\}$ of finite sets such that $F_{n}\subseteq D_{n}$,
$n\in\omega$, and $\\{St(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is a
$\gamma$-cover of $X$.

######  Theorem 2.19.

For any space $X$ the following holds:

$\big{(}aSSL,(<\mathfrak{d},H)\big{)}$

$\big{(}aSSL,(<\mathfrak{b},H)\big{)}$

$SelSSH$

$SelSSM$

$\big{(}aSSL,(<\mathfrak{b},M)\big{)}$

The proof is divided as Propositions 2.20, 2.21, 2.22.

######  Proposition 2.20.

If $X$ is an absolutely strongly star-Lindelöf space and $X$ is the union of
less than $\mathfrak{d}$ Hurewicz spaces, then $X$ is selectively strongly
star-Menger.

###### Proof.

Assume $X$ is $aSSL$ and let $\kappa<\mathfrak{d}$ such that
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of open covers
and let $\\{D_{n}:n\in\omega\\}$ be any sequence of dense subsets of $X$. For
each $n\in\omega$ fix
$E_{n}=\\{d_{q}^{n}:q\in\omega\\}\in[D_{n}]^{\leq\omega}$ such that
$St(E_{n},\mathcal{U}_{n})=X$. Thus, for each $n\in\omega$, the collection
$\\{St(d_{q}^{n},\mathcal{U}_{n}):q\in\omega\\}$ is a countable open cover of
$X$. In a similar fashion as Proposition 2.11, we use Lemma 2.8 to conclude
the proof.  ∎

######  Proposition 2.21.

If $X$ is an absolutely strongly star-Lindelöf space and $X$ is the union of
less than $\mathfrak{b}$ Hurewicz spaces, then $X$ is selectively strongly
star-Hurewicz.

###### Proof.

Assume $X$ is $aSSL$ and let $\kappa<\mathfrak{b}$ such that
$X=\bigcup_{\alpha<\kappa}H_{\alpha}$ with each $H_{\alpha}$ being a Hurewicz
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of open covers
and let $\\{D_{n}:n\in\omega\\}$ be any sequence of dense subsets of $X$. For
each $n\in\omega$ fix
$E_{n}=\\{d_{q}^{n}:q\in\omega\\}\in[D_{n}]^{\leq\omega}$ such that
$St(E_{n},\mathcal{U}_{n})=X$. Thus, for each $n\in\omega$, the collection
$\\{St(d_{q}^{n},\mathcal{U}_{n}):q\in\omega\\}$ is a countable open cover of
$X$. In a similar fashion as Proposition 2.12, we use Lemma 2.9 (2) to
complete the proof.  ∎

######  Proposition 2.22.

If $X$ is an absolutely strongly star-Lindelöf space and $X$ is the union of
less than $\mathfrak{b}$ Menger spaces, then $X$ is selectively strongly star-
Menger.

###### Proof.

Assume $X$ is $aSSL$ and let $\kappa<\mathfrak{b}$ such that
$X=\bigcup_{\alpha<\kappa}M_{\alpha}$ with each $M_{\alpha}$ being a Menger
space. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of open covers
and let $\\{D_{n}:n\in\omega\\}$ be any sequence of dense subsets of $X$. For
each $n\in\omega$ fix
$E_{n}=\\{d_{q}^{n}:q\in\omega\\}\in[D_{n}]^{\leq\omega}$ such that
$St(E_{n},\mathcal{U}_{n})=X$. Thus, for each $n\in\omega$, the collection
$\\{St(d_{q}^{n},\mathcal{U}_{n}):q\in\omega\\}$ is a countable open cover of
$X$. In a similar way as Proposition 2.13, we use Lemma 2.9 (1) to conclude
the proof.  ∎

##  3 Iterated stars

A well-known study about iterations of star versions of Lindelöf properties
was made in [10] (see also [17]). The previous section motivated a similar
study for star selection properties. We start giving some results that involve
refinements of open covers.

In [10] the properties $n$-star Lindelöf and strongly $n$-star Lindelöf are
defined and the authors show that every $n$-star Lindelöf space is strongly
$n+1$-star-Lindelöf (Theorem 3.1.1 (3)). In the class of metaLindelöf spaces
the converse holds:

######  Proposition 3.1.

If $X$ is metaLindelöf and strongly $n+1$-star Lindelöf then $X$ is $n$-star
Lindelöf.

###### Proof.

Let $\mathcal{U}$ be an open cover of $X$. Since $X$ is metaLindelöf we can
assume that $\mathcal{U}$ is point-countable. Let $C\in[X]^{\omega}$ so that
$St^{n}\big{(}St(C,\mathcal{U}),\mathcal{U}\big{)}=X$. Since $\mathcal{U}$ is
point-countable, there exists $W\in[\mathcal{U}]^{\omega}$ so that
$St(C,\mathcal{U})\subseteq\bigcup W$. Hence,
$St^{n+1}(C,\mathcal{U})\subseteq St^{n}(\bigcup W,\mathcal{U})$ i.e., $X$ is
$n$-star-Lindelöf.  ∎

######  Lemma 3.2 (Folklore).

Let $\mathcal{U}$ be an open cover of a topological space $X$. If $A\subseteq
B\subseteq X$ and $\mathcal{V}$ is a refinement of $\mathcal{U}$, then for
each $n\in\omega$, $St^{n}(A,\mathcal{V})\subseteq St^{n}(B,\mathcal{U})$.

######  Proposition 3.3.

Every Paralindelöf and $n$-star Lindelöf space $X$ is $n$-strongly star
Lindelöf for each $n\in\omega$.

###### Proof.

Assume $X$ is a Paralindelöf and $n$-star Lindelöf space for some
$n\in\omega$. Let $\mathcal{U}$ be an open cover of $X$. Without loss of
generality we can assume that $\mathcal{U}$ is locally countable. For each
$x\in X$, let $U_{x}$ be an open set so that $|\\{U\in\mathcal{U}:U_{x}\cap
U\neq\emptyset\\}|\leq\omega$ and $\mathcal{V}:=\\{U_{x}:x\in X\\}$ refines
$\mathcal{U}$. Since $X$ is $n$-star Lindelöf, there exists
$C\in[X]^{\leq\omega}$ so that $St^{n}(\bigcup_{x\in C}U_{x},\mathcal{V})=X$.
For $x\in C$, let $\mathcal{W}_{x}=\\{U\in\mathcal{U}:U_{x}\cap
U\neq\emptyset\\}$. Then $|\mathcal{W}_{x}|\leq\omega$. If
$\mathcal{W}=\bigcup_{x\in C}\mathcal{W}_{x}$, then $|\mathcal{W}|\leq\omega$.
Thus, for each $W\in\mathcal{W}$, fix $y_{W}\in W$.
Claim: $St(\bigcup_{x\in C}U_{x},\mathcal{V})\subseteq
St(\\{y_{W}:W\in\mathcal{W}\\},\mathcal{U})$.
Indeed, let $y\in St(\bigcup_{x\in C}U_{x},\mathcal{V})$, hence, there exists
$V\in\mathcal{V}$ so that $y\in V$ and $V\cap\bigcup_{x\in
C}U_{x}\neq\emptyset$. Let $U\in\mathcal{U}$ so that $V\subseteq U$. Then,
$U\in\mathcal{W}$. Thus, $y\in St(y_{U},\mathcal{U})\subseteq
St(\\{y_{W}:W\in\mathcal{W}\\},\mathcal{U})$.
Since $\mathcal{V}$ is a refinement o $\mathcal{U}$, and $St(\bigcup_{x\in
C}U_{x},\mathcal{V})\subseteq St(\\{y_{W}:W\in\mathcal{W}\\},\mathcal{U})$, by
Lemma 3.2, $X=St^{n}(\bigcup_{x\in C}U_{x},\mathcal{V})\subseteq
St^{n}(\\{y_{W}:W\in\mathcal{W}\\},\mathcal{U})$. That is, $X$ is $n$-strongly
star Lindelöf.  ∎

Observe that if we have a paraLindelöf space $X$ that is either $n$-star
Lindelöf or $n$-strongly star Lindelöf for some $n\in\omega$, we can use
Proposition 3.3 and Proposition 3.1, as many times as needed to get that $X$
is star-Lindelöf. Since paraLindelöf, star-Lindelöf spaces are Lindelöf (check
Theorem 2.6 in [7] or Theorem 2.24 in [25]), we have the following:

######  Corollary 3.4.

In the class of paraLindelöf spaces, for each $n\in\omega$, the following
properties are equivalent:

  1. (i)

Lindelöf;

  2. (ii)

strongly $n$-star Lindelöf;

  3. (iii)

$n$-star Lindelöf.

In Proposition 53 of [17], Matveev shows that every metaLindelöf strongly
$2$-star Lindelöf is absolutely strongly $2$-star Lindelöf (he calls star
Lindelöf what we call strongly star Lindelöf). Thus, we can ask the general
case:

######  Question 3.5.

Is it true that every metaLindelöf strongly $n$-star Lindelöf space is
absolutely strongly $n$-star Lindelöf?

Additionally, since both properties, absolutely strongly $n+1$-star Lindelöf
and $n$-star Lindelöf are stronger than strongly $n+1$-star Lindelöf, it is
worth to investigate the following:

######  Question 3.6.

What is the relationship between the properties absolutely strongly $n+1$-star
Lindelöf and $n$-star Lindelöf?

We introduce similar definitions for the star versions of the Menger property:

######  Definition 3.7.

  1. 1.

A space $X$ is called $n$-star-Menger if for every sequence of open covers
$\\{\mathcal{U}_{n}:n\in\omega\\}$ there exists a sequence
$\\{\mathcal{V}_{n}:n\in\omega\\}$ so that for each $n\in\omega$
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ and
$\\{St^{n}(\bigcup\mathcal{V}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open
cover of $X$.

  2. 2.

A space $X$ is called strongly $n$-star-Menger if for every sequence of open
covers $\\{\mathcal{U}_{n}:n\in\omega\\}$ there exists a sequence
$\\{F_{n}:n\in\omega\\}$ so that for each $n\in\omega$ $F_{n}\in[X]^{<\omega}$
and $\\{St^{n}(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.

Observe that $1$-star-Menger and strongly $1$-star-Menger spaces are precisely
star-Menger and strongly star-Menger spaces, respectively.

It follows immediately from Proposition 3.1 that if $X$ is metaLindelöf and
strongly $n+1$-star Menger then $X$ is $n$-star Lindelöf. In addition, the
next proposition is the Menger version of Theorem 3.1.1(3) in [10].

######  Proposition 3.8.

If $X$ is $n$-star-Menger, then $X$ is strongly $n+1$-star-Menger.

###### Proof.

Assume $X$ is $n$-star-Menger. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a
sequence of open covers of $X$. For each $n\in\omega$, let
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ so that
$\\{St^{n}(\bigcup\mathcal{V}_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open
cover of $X$. Put $\mathcal{V}_{n}=\\{V_{n}^{i}:i\in I_{n}\\}$. Take for each
$n\in\omega$ and $i\in I_{n}$, $x_{n}^{i}\in V_{n}^{i}$ and define for each
$n\in\omega$, $F_{n}=\\{x_{n}^{i}:i\in I_{n}\\}$. Observe that for each
$n\in\omega$, $\bigcup\mathcal{V}_{n}\subseteq St(F_{n},\mathcal{U}_{n})$.
Hence,

$St^{n}(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})\subseteq
St^{n}(St(F_{n},\mathcal{U}_{n}),\mathcal{U}_{n})=St^{n+1}(F_{n},\mathcal{U}_{n}).$

Thus, $\\{St^{n+1}(F_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of
$X$. That is, $X$ is strongly $n+1$-star-Menger.  ∎

In the class of metacompact spaces the properties $n$-star-Menger and strongly
$n+1$-star-Menger coincide:

######  Proposition 3.9.

If $X$ is strongly $n+1$-star-Menger and metacompact, then $X$ is $n$-star-
Menger.

###### Proof.

Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence of open covers of $X$.
Since $X$ is metacompact we can assume that for each $n\in\omega$,
$\mathcal{U}_{n}$ is point-finite.
Hence, for each $n\in\omega$ there exists $F_{n}\in[X]^{<\omega}$ so that

$\big{\\{}St^{n+1}(F_{n},\mathcal{U}_{n}):n\in\omega\big{\\}}$

is an open cover of $X$. For each $n\in\omega$, let
$W_{n}=\\{U\in\mathcal{U}_{n}:F_{n}\cap U\neq\emptyset\\}$. Note
$|W_{n}|<\omega$ and $St(F_{n},\mathcal{U}_{n})=\bigcup W_{n}$. Then for each
$n\in\omega$
$St^{n}\big{(}St(F_{n},\mathcal{U}_{n}),\mathcal{U}_{n}\big{)}=St^{n}(\bigcup
W_{n},\mathcal{U}_{n})$. Thus, $\\{St^{n}(\bigcup
W_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$. That is, $X$ is
$n$-star-Menger.  ∎

######  Corollary 3.10.

If $X$ is strongly $2$-star-Menger and paracompact, then $X$ is Menger.

###### Proof.

Since paracompact implies metacompact, if $X$ is strongly $2$-star-Menger and
paracompact, using the previous result, $X$ is star-Menger. In addition,
paracompact star-Menger spaces are Menger (see [15]).  ∎

Actually something stronger holds. In [7], the authors showed that for
paraLindelöf spaces, the three Menger-type properties, strongly star-Menger,
star-Menger and Menger, are equivalent. The following result shows that this
fact is still true when taking iterations of those properties.

######  Theorem 3.11.

In the class of paraLindelöf spaces, given any $n\in\omega$ the properties
$n$-star-Menger, strongly $n$-star-Menger and Menger, are equivalent.

###### Proof.

Fix $m\in\omega$ and let $X$ be a paraLindelöf, $m$-star-Menger space. By
Corollary 3.4, $X$ is Lindelöf and, in particular, paracompact.
Let us recall that a cover $\mathcal{B}$ is said to be a star refinement of a
cover $\mathcal{U}$ ($\mathcal{B}\prec^{*}\mathcal{U}$), if for each
$B\in\mathcal{B}$, there is some $U\in\mathcal{U}$ so that
$St(B,\mathcal{B})\subseteq U$. A space is paracompact if every open cover has
an open star refinement (check, for instance, Theorem 5.1.12 in [11]).
Now let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of open covers of
$X$. For each $i\leq m$ define $\mathcal{B}_{n}^{i}$ open cover of $X$ so that
$\mathcal{B}_{n}^{m}\prec^{*}\mathcal{B}_{n}^{m-1}\prec^{*}\cdots\prec^{*}\mathcal{B}_{n}^{0}\prec^{*}\mathcal{U}_{n}$.
Claim: For each $n\in\omega$ and for each $W\in\mathcal{B}_{n}^{m}$, there is
$U_{W}\in\mathcal{U}_{n}$ so that $St^{m}(W,\mathcal{B}_{n}^{m})\subseteq
U_{W}$. Indeed, fix $n\in\omega$ and $W_{m}\in{\mathcal{B}_{n}^{m}}$, since
$\mathcal{B}_{n}^{m}\prec^{*}\mathcal{B}_{n}^{m-1}$, there is
$W_{m-1}\in\mathcal{B}_{n}^{m-1}$ so that
$St(W_{m},\mathcal{B}_{n}^{m})\subseteq W_{m-1}$. By Lemma 3.2,

$St^{2}(W_{m},\mathcal{B}_{n}^{m})=St(St(W_{m},\mathcal{B}_{n}^{m}),\mathcal{B}_{n}^{m})\subseteq
St(W_{m-1},\mathcal{B}_{n}^{m-1}).$

Now, since $\mathcal{B}_{n}^{m-1}\prec^{*}\mathcal{B}_{n}^{m-2}$, there is
$W_{m-2}\in\mathcal{B}_{n}^{m-2}$ so that
$St(W_{m-1},\mathcal{B}_{n}^{m-1})\subseteq W_{m-2}$. Then,
$St^{2}(W_{m},\mathcal{B}_{n}^{m})\subseteq W_{m-2}$. It is possible to repeat
this process $m-2$ more times to get $U_{W_{m}}\in\mathcal{U}_{n}$ with
$St^{m}(W_{m},\mathcal{B}_{n}^{m})\subseteq U_{W_{m}}$.
Since $X$ is $m$-star-Menger, let
$\mathcal{W}_{n}\in[\mathcal{B}_{n}^{m}]^{<\omega}$ so that
$\\{St^{m}(\bigcup\mathcal{W}_{n},\mathcal{B}_{n}^{m}):n\in\omega\\}$ is an
open cover of $X$.
For each $n\in\omega$ and each $W\in\mathcal{W}_{n}$ fix
$U_{W}\in\mathcal{U}_{n}$ such that $St^{m}(W,\mathcal{B}_{n}^{m})\subseteq
U_{W}$. For all $n\in\omega$, let
$\mathcal{V}_{n}=\\{U_{W}:W\in\mathcal{W}_{n}\\}$. Each
$\mathcal{V}_{n}\in[\mathcal{U}_{n}]^{<\omega}$ and
$\\{\bigcup\mathcal{V}_{n}:n\in\omega\\}$ is an open cover of $X$. Thus, $X$
is Menger.  ∎

##  4 $\Psi$-spaces

A natural question in this context is whether every $2$-star-Menger is star-
Menger. This is not the case, in fact, Tree ([28]) built a 2-star compact
(hence $2$-star-Menger), space which is not strongly 2-star Lindelöf (in
particular, not star Lindelöf and therefore, not star-Menger). Since, every
star-Menger space is both 2-star-Menger and star-Lindelöf, it is worth asking
whether the converse holds true, i.e., Is it true that every 2-star-Menger,
star-Lindelöf space is star-Menger?. The answer is no, at least consistently
(see Example 4.4 below). For this, we use a _Luzin family_ $\mathcal{A}$. That
is, $\mathcal{A}=\\{a_{\alpha}:\alpha<\omega_{1}\\}$ is an almost disjoint
family with the property that for each $\alpha<\omega_{1}$ and each
$n\in\omega$ the set $\\{\beta<\alpha:a_{\beta}\cap a_{\alpha}\subseteq n\\}$
is finite. First of all, we introduce the analogous definitions to Definition
3.7 for the Rothberger property:

######  Definition 4.1.

  1. 1.

A space $X$ is called $n$-star-Rothberger if for every sequence of open covers
$\\{\mathcal{U}_{n}:n\in\omega\\}$ there exists a sequence
$\\{U_{n}:n\in\omega\\}$ so that for each $n\in\omega$,
$U_{n}\in\mathcal{U}_{n}$ and $\\{St^{n}(U_{n},\mathcal{U}_{n}):n\in\omega\\}$
is an open cover of $X$.

  2. 2.

A space $X$ is called strongly $n$-star-Rothberger if for every sequence of
open covers $\\{\mathcal{U}_{n}:n\in\omega\\}$ there exists a sequence
$\\{x_{n}:n\in\omega\\}$ so that for each $n\in\omega$, $x_{n}\in X$ and
$\\{St^{n}(x_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.

Observe that $1$-star-Rothberger and strongly $1$-star-Rothberger spaces are
precisely star-Rothberger and strongly star-Rothberger spaces, respectively.

It is surprising that, regardless of the size of the continuum, $\Psi$-spaces
induced by a Luzin family, are always strongly 2-star-Rothberger.

######  Proposition 4.2.

If $\mathcal{A}$ is Luzin, then $\Psi(\mathcal{A})$ is strongly 2-star
Rothberger.

###### Proof.

Let $\mathcal{A}=\\{a_{\alpha}:\alpha<\omega_{1}\\}$ be a Luzin family. Let
$\\{\mathcal{U}_{n}:n\in\omega\\}$ be any sequence of open covers of
$\Psi(\mathcal{A})$. For each $\alpha<\omega_{1}$, fix
$U_{\alpha}^{0}\in\mathcal{U}_{0}$ such that $a_{\alpha}\in U_{\alpha}^{0}$.
For $\alpha<\omega_{1}$, let $f(a_{\alpha})\in a_{\alpha}\cap\omega$ so that
$a_{\alpha}\setminus f(a_{\alpha})\subseteq U_{\alpha}^{0}$.
For each $k\in\omega$, let $W_{k}=\\{\alpha<\omega_{1}:f(a_{\alpha})=k\\}$.
Observe $\bigcup_{k\in\omega}W_{k}=\omega_{1}$. Furthermore, there is
$A_{0}\subseteq\omega_{1}$ stationary and $n_{0}\in\omega$ such that for each
$\alpha\in A_{0}$, $f(a_{\alpha})=n_{0}$. Now, let $A_{1}\subseteq A_{0}$ and
$n_{1}\in\omega$ so that
$A_{1}=\\{\alpha<\omega_{1}:minU_{\alpha}^{0}=n_{1}\\}$ is stationary. For
each $k\in\omega$ and each $\alpha\in A_{1}$, let
$B_{\alpha,k}=\\{\beta<\alpha:a_{\beta}\cap a_{\alpha}\subseteq
max\\{k,n_{0},n_{1}\\}\\}$. Since $\mathcal{A}$ is Luzin, each $B_{\alpha,k}$
is finite.
Claim: For each $k\in\omega$, $G_{k}=\bigcup_{\alpha\in A_{1}}(\alpha\setminus
B_{\alpha,k})$ is cofinite.
Fix $k\in\omega$, there is $A^{k}\subseteq A_{1}$ stationary and
$B\in[\omega]^{<\omega}$ so that for each $\alpha\in A^{k}$, $B_{\alpha,k}=B$.
Then, $\omega_{1}\setminus B=\bigcup_{\alpha\in A^{k}}(\alpha\setminus
B)=\bigcup_{\alpha\in A^{k}}(\alpha\setminus
B_{\alpha,k})\subseteq\bigcup_{\alpha\in A_{1}}(\alpha\setminus
B_{\alpha,k})=G_{k}$.
Since for each $k\in\omega$, $G_{k}$ is cofinite, then $W_{k}\setminus G_{k}$
is finite. Hence, $\bigcup_{k\in\omega}W_{k}\cap G_{k}$ is cocountable. Now we
show that $\\{a_{\beta}:\beta\in\bigcup_{k\in\omega}W_{k}\cap
G_{k}\\}\subseteq St^{2}(n_{1},\mathcal{U}_{0})$. Fix $k\in\omega$ and let
$\beta\in W_{k}\cap G_{k}$, then $f(a_{\beta})=k$ and there is $\alpha\in
A^{k}$ such that $\beta\in\alpha\setminus B_{\alpha,k}$. Thus, $a_{\beta}\cap
a_{\alpha}\not\subseteq max\\{k,n_{0},n_{1}\\}$. Fix $m\in a_{\beta}\cap
a_{\alpha}$ with $m>max\\{k,n_{0},n_{1}\\}$. Hence, $m\in U_{\alpha}^{0}$,
$m\in U_{\beta}^{0}$ and $m>n_{1}=minU_{\alpha}^{0}$. Therefore, $m\in
St(n_{1},\mathcal{U}_{0})$ and $a_{\beta}\in U_{\beta}^{0}\subseteq
St(m,\mathcal{U}_{0})\subseteq St^{2}(n_{1},\mathcal{U}_{0})$.

Therefore, $St^{2}(n_{1},\mathcal{U}_{0})$ contains all but countably many
members of $\Psi(\mathcal{A})$, whence $\Psi(\mathcal{A})$ is strongly 2-star-
Rothberger.  ∎

Another fact, which is also interesting, is that $\Psi$-spaces induced by a
maximal almost-disjoint family can be characterized in terms of the $strongly$
2-$starcompact$ property (see [10] for information about iterated (strongly)
starcompact property).

######  Proposition 4.3 ([10]).

For an a.d. family $\mathcal{A}$, the following are equivalent:

  1. 1.

$\Psi(\mathcal{A})$ is strongly 2-starcompact.

  2. 2.

$\Psi(\mathcal{A})$ is strongly $k$-starcompact for every $k\geq 2$.

  3. 3.

$\Psi(\mathcal{A})$ is 2-starcompact.

  4. 4.

$\Psi(\mathcal{A})$ is $k$-starcompact for every $k\geq 2$.

  5. 5.

$\Psi(\mathcal{A})$ is $k$-starcompact for some $k\geq 2$.

  6. 6.

$\mathcal{A}$ is maximal.

It is worth to mention that the equivalences $(3)-(6)$ were showed in [10] for
spaces in general (by using pseudocompactness instead of a maximal almost
disjoint family) and the proof for $(6)\Rightarrow(1)$ is contained in the
proof of Example 2.2.5 in same article. For convenience of the reader, we
outline the proof of these equivalences for $\Psi$-spaces below.

###### Proof.

Since the implications
$(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)\Rightarrow(5)$ always hold, we
just show $(5)\Rightarrow(6)$ and $(6)\Rightarrow(1)$.

Assume that $\Psi(\mathcal{A})$ is $k$-starcompact for some $k\geq 2$. If
$\mathcal{A}$ is not maximal, there is some $B\subseteq\omega$ which is almost
disjoint from every member of $\mathcal{A}$. Consider the open cover
$\mathcal{U}$ consisting of singletons from $\omega$ together with sets of the
form $A\setminus B$ for $A\in\mathcal{A}$. Then for any
$\mathcal{W}\subseteq\mathcal{U}$, $St^{k}(\bigcup\mathcal{W},\mathcal{U})$
intersects $B$ just in those natural numbers whose singletons were chosen in
$\mathcal{W}$, as all other members of $\mathcal{U}$ are disjoint from $B$.
Therefore $\Psi(A)$ is not $k$-starcompact.

If $\mathcal{A}$ is not 2-strongly starcompact, then there is an open cover
$\mathcal{U}$ so that the star of every finite subcollection misses some
element of $\omega$, otherwise taking another star would give all of
$\Psi(\mathcal{A})$. Now we can choose an infinite subset $B$ with increasing
enumeration $\\{b_{i}:i<\omega\\}$ of $\omega$ recursively so that for every
$j<\omega$, $b_{j}\not\in St(\\{b_{i}:i<j\\},\mathcal{U})$. Now we claim that
$B$ is almost disjoint from every member of $\mathcal{A}$, and so
$\mathcal{A}$ cannot be maximal. To see this, suppose that there is
$A\in\mathcal{A}$ which intersects $B$ infinitely. Then the neighborhood of
$A$ in $\mathcal{U}$ contains two points $b_{j_{0}},b_{j_{1}}$ with
$j_{0}<j_{1}$. So this neighborhood is contained in
$St(b_{j_{0}},\mathcal{U})$, contradicting the choice of $j_{1}$.  ∎

######  Example 4.4.

$(\mathfrak{d}=\omega_{1})$ There is a Tychonoff, strongly 2-star-Rothberger,
strongly star-Lindelöf, not star-Menger space.

###### Proof.

Let $\mathcal{A}$ be a Luzin family. Since $\Psi(\mathcal{A})$ is separable,
it is strongly star-Lindelöf and, by Proposition 4.2, it is strongly 2-star-
Rothberger. Since $\mathfrak{d}=\omega_{1}$, by Proposition 9 in [3],
$\Psi(\mathcal{A})$ is not star-Menger.  ∎

######  Corollary 4.5.

If $\mathfrak{d}=\omega_{1}$ then there is a Tychonoff, 2-star Menger, star-
Lindelöf, not star-Menger space.

In contrast to $\Psi$-spaces induced from Luzin families, we also have
examples of $\Psi$-spaces that not hold any iteration of star-Menger property.

######  Example 4.6.

There is a $\Psi$-space which is not $m$-star Menger for any $m<\omega$.

###### Proof.

Identify $\omega^{<\omega}$ with $\omega$ and let $\mathcal{A}$ be the
branches of the tree $\omega^{<\omega}$. Now, for each $n$, let
$\mathcal{U}_{n}$ be the open cover consisting of the sets
$N_{s}=\\{a\in\Psi(\mathcal{A}):s\sqsubseteq a\\}$, where $s\in\omega^{n+1}$,
and $\\{x\\}$ for all $x\not\in\bigcup\\{N_{s}:s\in\omega^{n+1}\\}$. Then, for
each $n\in\omega$, $\mathcal{U}_{n}$ consists of pairwise disjoint clopen sets
and therefore, no elements are picked up when taking stars.

Suppose $\mathcal{V}_{n}\subseteq\mathcal{U}_{n}$ is finite. By induction,
define $x\in\mathcal{A}$ so that
$N_{x\upharpoonright(n+1)}\not\in\mathcal{V}_{n}$. It is clear that $x\not\in
St^{m}(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})$ for any $m,n<\omega$.  ∎

We can also define $\Psi$-spaces associated to maximal almost disjoint
families that do not satisfy any iteration of the star-Rothberger property.

######  Example 4.7.

There is a MAD family $\mathcal{A}$ so that $\Psi(\mathcal{A})$ is not
$k$-star Rothberger for any $k$.

###### Proof.

Enumerate all infinite subsets of $\omega$ as $\langle
b_{\alpha}:\alpha<\mathfrak{c},\alpha\textrm{ even}\rangle$ and all functions
in $\prod_{n}{}^{n}3$ as $\langle
s_{\alpha}:\alpha<\mathfrak{c},\alpha\textrm{ odd}\rangle$. Note that any
ordinal can be written as $\alpha+n$ for some $n<\omega$, and “odd” and “even”
above refer to the parity of this integer $n$.

Let $\\{B_{t}:t\in{}^{<\omega}3\\}$ be a sequence of infinite subsets of
$\omega$ so that $B_{\emptyset}=\omega$ and for all $t\in{}^{<\omega}2$,
$\\{B_{t^{\frown}0},B_{t^{\frown}1},B_{t^{\frown}2}\\}$ partition $B_{t}$.

We will define $\mathcal{A}=\\{a_{\alpha}:\alpha<\mathfrak{c}\\}$. Suppose
$a_{\beta}$, $\beta<\alpha$, have already been defined.

We will construct $a_{\alpha}$ infinite subsets of $\omega$ and $x_{\alpha}$
branches of the tree so that

  1. 1.

For $\beta<\alpha$, $a_{\alpha}$ is almost disjoint from $a_{\beta}$.

  2. 2.

For each $n$, $\\{z_{m}:m\geq n\\}\subseteq B_{x_{\alpha}\upharpoonright n}$,
where $\\{z_{m}:m<\omega\\}$ is the increasing enumeration of $a_{\alpha}$ (so
that in particular, $a_{\alpha}\subseteq^{*}B_{x_{\alpha}\upharpoonright n}$
for any $n$).

If $\alpha$ is even, then we will construct $a_{\alpha}$ so that

  1. 3’.

If $b_{\alpha}$ is almost disjoint from each $a_{\beta},$ $\beta<\alpha$, then
$a_{\alpha}\subseteq b_{\alpha}$.

Let us assume the case where $b_{\alpha}$ is almost disjoint from each
$a_{\beta},$ $\beta<\alpha$, and hence the third condition applies. Note that
if $3^{\prime}$ is satisfied, then so is the first condition. Let $x_{\alpha}$
be a branch through the tree so that $B_{x_{\alpha}\upharpoonright n}\cap
b_{\beta}$ is infinite for each $n$. Define $\\{z_{n}:n<\omega\\}$ to be an
increasing sequence of natural numbers so that $z_{n}\in
B_{x_{\alpha}\upharpoonright n}\cap b_{\alpha}$.

If $\alpha$ is odd, then we will construct $a_{\alpha}$ so that

  1. 3”.

For each $n$, $\\{z_{m}:m\geq n\\}\cap B_{s_{\alpha}(n+1)}=\emptyset$.

This construction will proceed in $\omega$ steps. Let $S$ be the subtree of
${}^{<\omega}3$ of all $t$ so that for all $s\leq t$, $s\neq s_{\alpha}(n)$,
where $n$ is the length of $s$. The resulting subtree $S$ still splits at
every node.

Let $x_{\alpha}$ be a branch through this tree which is not $x_{\beta}$ for
any $\beta<\alpha$. This is possible since the tree has $\mathfrak{c}$
branches.

In step $n$, pick $z_{n}\in B_{x_{\alpha}\upharpoonright(n+1)}$ greater than
all previously chosen $z_{m}$, $m<n$. Let $a_{\alpha}=\\{z_{n}:n<\omega\\}$.
As $B_{x_{\alpha}\upharpoonright n}$ is $\subseteq$-decreasing along the
branch, this ensures that for each $n$, $\\{z_{m}:m\geq n\\}\subseteq
B_{x_{\alpha}\upharpoonright n}$.

If $\beta<\alpha$, then let $i$ be so that $x_{\alpha}(i)\neq x_{\beta}(i)$.
We have that $a_{\alpha}\subseteq^{*}B_{x_{\alpha}\upharpoonright(i+1)}$ and
$a_{\beta}\subseteq^{*}B_{x_{\beta}\upharpoonright(i+1)}$, yet
$B_{x_{\alpha}\upharpoonright(i+1)}$ is disjoint from
$B_{x_{\beta}\upharpoonright(i+1)}$. Therefore $a_{\alpha}$ is almost disjoint
from $a_{\beta}$.

Since $x_{\alpha}\upharpoonright(n+1)\neq s_{\alpha}(n+1)$ and $\\{z_{m}:m\geq
n\\}\subseteq B_{x_{\alpha}\upharpoonright n}$, we have that 3” holds.

For each $n$, let $a_{\alpha}^{(n)}$ be the set $a_{\alpha}$ with the least
$n$ elements removed and let $\mathcal{U}_{n}$ be the open cover of
$\Psi(\mathcal{A})$ consisting of singletons from $\omega$ together with
$\\{a_{\alpha}\\}\cup a_{\alpha}^{(n)}$. This sequence of covers shows that
$\Psi(\mathcal{A})$ is not $k$-star Rothberger. For any selection of sets
$\\{U_{n}:n<\omega\\}$, for each $n$ there is some $s(n)\in{}^{n}3$ so that
$U_{n}\subseteq B_{s(n)}$. Now take $\alpha$ so that $s=s_{\alpha}$. Then
$a_{\alpha}$ is disjoint from the $k$-iterated star of $U_{n}$ in
$\mathcal{U}_{n}$ for each $n$ by construction.  ∎

We finish this section giving a result on $\Psi$-spaces that combines the
style of the results in Section 2 with iterations of a star selection property
introduced in Section 3.

######  Theorem 4.8.

For any $k\in\omega$, if $X$ is the union of less than $\mathfrak{b}$
k-strongly star-Menger $\Psi$-spaces on $\omega$, then $X$ is k-strongly star-
Menger.

###### Proof.

Fix $k\in\omega$. Let $\kappa$ be a cardinal less than $\mathfrak{b}$. For
each $\alpha<\kappa$, let $\Psi_{\alpha}$ be a $\Psi$-space on $\omega$
defined by an a.d. family $\mathcal{A}_{\alpha}$ and put
$X=\bigcup_{\alpha<\kappa}\Psi_{\alpha}$ where each $\Psi_{\alpha}$ is
$k$-strongly star-Menger. Let $\\{\mathcal{U}_{n}:n\in\omega\\}$ be a sequence
of open covers of $X$ consisting of basic open sets (for each $\alpha<\kappa$,
$n\in\omega$ and $U\in\mathcal{U}_{n}$, $|U\cap\mathcal{A}_{\alpha}|\leq 1$).
For each $\alpha<\kappa$, let $F_{n}^{\alpha}\in[\Psi_{\alpha}]^{<\omega}$
such that for every $m\in\omega$,
$\\{St^{k}(F_{n}^{\alpha},\mathcal{U}_{n}):m\leq n<\omega\\}$ is an open cover
of $\Psi_{\alpha}$.

Fix $\alpha<\kappa$ and $n\in\omega$. For each $A\in
F_{n}^{\alpha}\cap\mathcal{A}_{\alpha}$, let $U_{A}$ be a member of
$\mathcal{U}_{n}$ such that $A\in U_{A}$. So, $U_{A}=\\{A\\}\cup A\setminus
F_{A}$ for some $F_{A}\in[A]^{<\omega}$. For each $A\in
F_{n}^{\alpha}\cap\mathcal{A}_{\alpha}$, fix $n_{A}\in A\setminus F_{A}$.
Thus, for each $\alpha<\kappa$ and for each $n\in\omega$, let

$G_{n}^{\alpha}=(F_{n}^{\alpha}\cap\omega)\cup\\{n_{A}:A\in
F_{n}^{\alpha}\cap\mathcal{A}_{\alpha}\\}\cup\\{\bigcup F_{A}:A\in
F_{n}^{\alpha}\cap\mathcal{A}_{\alpha}\\}.$

Then, for each $\alpha<\kappa$ and each $n\in\omega$, we have
$G_{n}^{\alpha}\in[\omega]^{<\omega}$ and
$St(F_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(G_{n}^{\alpha},\mathcal{U}_{n})$. Indeed, let $x\in
St(F_{n}^{\alpha},\mathcal{U}_{n})$. Then there exists $U\in\mathcal{U}_{n}$
such that $x\in U$ and $U\cap F_{n}^{\alpha}\neq\emptyset$. We have two cases:
If $U\cap(F_{n}^{\alpha}\cap\omega)\neq\emptyset$, then $U\cap
G_{n}^{\alpha}\neq\emptyset$ and therefore, $x\in U\subseteq
St(G_{n}^{\alpha},\mathcal{U}_{n})$.
If $U\cap(F_{n}^{\alpha}\cap\mathcal{A}_{\alpha})\neq\emptyset$, then
$U=\\{A\\}\cup A\setminus F$ for some $A\in
F_{n}^{\alpha}\cap\mathcal{A}_{\alpha}$ and for some $F\in[A]^{<\omega}$. Then
$x\in St(F_{A}\cup\\{n_{A}\\},\mathcal{U}_{n})\subseteq
St(G_{n}^{\alpha},\mathcal{U}_{n})$.

We conclude that $St(F_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(G_{n}^{\alpha},\mathcal{U}_{n})$.

Now, we define, for each $\alpha<\kappa$, a function
$f_{\alpha}:\omega\rightarrow\omega$ as $f_{\alpha}(n)=max(G_{n}^{\alpha})$
for each $n\in\omega$. Since the collection $\\{f_{\alpha}:\alpha<\kappa\\}$
has size less than $\mathfrak{b}$, there exists $g\in\omega^{\omega}$ such
that for every $\alpha<\kappa$, $f_{\alpha}\leq^{*}g$. For each $n\in\omega$,
let $D_{n}=\\{i\in\omega:0\leq i\leq g(n)\\}$. Then each $D_{n}$ is a finite
subset of $\omega$ and it follows that
$\\{St^{k}(D_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.
Indeed, let $x\in X$. Then, there exists $\alpha<\kappa$ such that
$x\in\Psi_{\alpha}$. Since $f_{\alpha}\leq^{*}g$, there is $m\in\omega$ so
that for every $n\geq m$, $f_{\alpha}(n)\leq g(n)$. Furthermore, since the
collection $\\{St^{k}(F_{n}^{\alpha},\mathcal{U}_{n}):m\leq n<\omega\\}$ is an
open cover of $\Psi_{\alpha}$, let $n\geq m$ such that $x\in
St^{k}(F_{n}^{\alpha},\mathcal{U}_{n})$. We obtain that
$St(F_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(G_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St(\\{1,\ldots,f_{\alpha}(n)\\},\mathcal{U}_{n})\subseteq
St(\\{1,\ldots,g(n)\\},\mathcal{U}_{n})=St(D_{n},\mathcal{U}_{n})$. Thus,
$x\in St^{k}(F_{n}^{\alpha},\mathcal{U}_{n})\subseteq
St^{k}(D_{n},\mathcal{U}_{n})$. Therefore, the collection
$\\{St^{k}(D_{n},\mathcal{U}_{n}):n\in\omega\\}$ is an open cover of $X$.
Thus, $X$ is $k$-strongly star-Menger.  ∎

##  5 Normal star-Menger not strongly star-Menger not Dowker space

Recall that $X$ is a Dowker space if and only if $X$ is normal and its
Cartesian product with the closed unit interval $I$ is not normal.
Equivalently, $X$ is normal and not countably paracompact. In [7] the
following questions were posed:

######  Question 5.1 ([7] Question 2.4).

Is there a normal star-Menger space which is not strongly star-Menger?

######  Question 5.2 ([7] Question 2.21).

Are normal, countably paracompact star-Menger spaces strongly star-Menger?
I.e., if $X$ is normal, star-Menger, not strongly star-Menger, is $X$ a Dowker
Space?

In this section we present a consistent example (Example 5.4 below), of a
normal star Menger not strongly star-Menger not Dowker space. This space
answers consistently in the afirmative Question 5.1 and in the negative
Question 5.2.

In [26] Tall presented an example of a separable normal space with an
uncountable discrete subspace. Below we provide details of the construction of
such example for sake of completeness:

######  Example 5.3 ([26] Example E).

Assuming $2^{\aleph_{0}}=2^{\aleph_{1}}$ there exists a separable normal
$T_{1}$ space with an uncountable closed subspace.

Construction: Let $L$ be a set of cardinality $\aleph_{1}$ disjoint from
$\omega$. The existence of a strongly independent family $\mathcal{F}$555For
an infinite cardinal $\kappa$, a family
$\mathcal{F}\subseteq\mathcal{P}(\kappa)$ is called _independent_ if for all
pairs of disjoint $F,G\in[\mathcal{F}]^{<\omega}$ we have:
$C_{F,G}=\bigcap_{A\in F}A\cap\bigcap_{A\in G}(\kappa\smallsetminus
A)\neq\emptyset$ (Assume $\bigcap\emptyset=\kappa$). If in addition, for each
pair $(F,G)$ as above, $|C_{F,G}|=\kappa$, $\mathcal{F}$ is called _strongly
independent_. of subsets of $\aleph_{0}$ of size $2^{\aleph_{0}}=\mathfrak{c}$
is guaranteed by the Fichtenholz-Kantorovitch-Hausdorff Theorem666For every
infinite cardinal $\kappa$ there exists a strongly independent family
$\mathcal{F}\subseteq\mathcal{P}(\kappa)$ such that
$|\mathcal{F}|=2^{\kappa}$..

Write $\mathcal{F}=\\{A_{\alpha}:\alpha<\mathfrak{c}\\}$. Since
$|L|=\aleph_{1}$, $|\mathcal{P}(L)|=2^{\aleph_{1}}$. Assuming
$2^{\aleph_{0}}=2^{\aleph_{1}}$ it is possible to build a function
$f:\mathcal{P}(L)\to\\{A_{\alpha}:\alpha<\mathfrak{c}\\}\cup\\{\omega\smallsetminus
A_{\alpha}:\alpha<\mathfrak{c}\\}$ which is bijective and complement-
preserving (for each $B\subseteq L$, $f(L\smallsetminus
B)=\omega\smallsetminus f(B)$).
Now let $X=L\cup\omega$ with a subbase $\varphi$ for a topology defined by

  1. 1.

if $M\subseteq L$, then $M\cup f(M)\in\varphi$,

  2. 2.

if $n\in\omega$, then $\\{n\\}\in\varphi$,

  3. 3.

if $p\in X$, then $X\smallsetminus\\{p\\}\in\varphi$.

Observe that by condition (3) $X$ is $T_{1}$. By (2) $\omega$ is open,
therefore $L=X\smallsetminus\omega$ is closed and, by (1) for any $x\in L$,
$\\{x\\}\cup f(\\{x\\})$ is open such that $[\\{x\\}\cup f(\\{x\\})]\cap
L=\\{x\\}$, that is $L$ is discrete. $X$ is separable since $\omega$ is dense
in $X$: let $U$ be any nonempty basic open set, then

$U=\bigcap_{U\in F}U\cap\bigcap_{U\in G}U\cap\bigcap_{U\in H}U$

where $F,G,H$ are finite (possibly empty), each $U\in F$ is a subbasic open
set defined as in (1), each $U\in G$ is a subbasic open set defined as in (2),
and each $U\in H$ is a subbasic open set defined as in (3). To show
$U\cap\omega\neq\emptyset$ it is enough to observe that $|\big{(}\bigcap_{U\in
F}U\big{)}\cap\omega|=\omega$. This is always the case since $\mathcal{F}$ is
a strongly independent family. Now let $Y,Z$ be disjoint closed subsets of $X$
and observe:

$\displaystyle U_{Y}$ $\displaystyle=$ $\displaystyle\big{(}(Y\smallsetminus
L)\cup[(Y\cap L)\cup f(Y\cap L)]\big{)}\cap(X\smallsetminus Z)$
$\displaystyle=$ $\displaystyle\big{(}Y\cup f(Y\cap
L)\big{)}\cap(X\smallsetminus Z)$ $\displaystyle U_{Z}$ $\displaystyle=$
$\displaystyle\big{(}(Z\smallsetminus L)\cup[(L\smallsetminus Y)\cup
f(L\smallsetminus Y)]\big{)}\cap(X\smallsetminus Y)$

are open sets and $Y\subseteq U_{Y}$, $Z\subseteq U_{Z}$. Assume $x\in
U_{Y}\cap U_{Z}$, then $x\in X\smallsetminus(Y\cup Z)$ and $x\in f(Y\cap
L)\cap f(L\smallsetminus Y)$. But this is a contradiction since $f$ is
complement preserving: $f(L\smallsetminus Y)=f(L\smallsetminus(Y\cap
L))=\omega\smallsetminus(Y\cap L)$. Hence, $X$ is normal. $\blacksquare$

The following example presented by Song in [23] and [24] is a modification of
Example 5.3. Song proved, in particular, that this space is normal, star-
Lindelöf and not strongly star-Lindelöf (actually he showed something
stronger: there is $\mathcal{U}\in\mathcal{O}(X)$ such that for all
$L\subseteq X$ Lindelöf subspace of $X$, $St(L,\mathcal{U})\neq X$).

######  Example 5.4 ([23], [24]).

Assuming $2^{\aleph_{0}}=2^{\aleph_{1}}$ there exists a normal $T_{1}$ space
which is star-Lindelöf and not strongly star Lindelöf.

Construction: Let $X_{0}=L\cup\omega$ denote the space built in Example 5.3.
Let $X=L\cup(\omega_{1}\times\omega)$ and topologize it as follows, a basic
open set of

(i)

$x\in L$ is a set of the form $V^{U}_{\alpha}(x)=(U\cap
L)\cup\big{(}(\alpha,\omega_{1})\times(U\cap\omega)\big{)}$ where $U$ is a
neighbourhood of $x\in X_{0}$ and $\alpha<\omega_{1}$.

(ii)

$\langle\alpha,n\rangle\in(\omega_{1}\times\omega)$ is a set of the form
$V_{W}(\langle\alpha,n\rangle)=W\times\\{n\\}$ where $W$ is a neighbourhood of
$\alpha$ in $\omega_{1}$ with the usual topology.

Condition (i) guarantees that $X$ is $T_{1}$. Furthermore,
$\omega_{1}\times\omega$ is open in $X$ and for $x\in L$, if we let
$U=\\{x\\}\cup f(\\{x\\})$. then for any $\alpha<\omega_{1}$,
$V_{\alpha}^{U}(x)\cap L=\\{x\\}$. That is, $L$ is closed and discrete in $X$.

$X$ is normal: Let $Y,Z\subseteq X$ closed and disjoint. Define $Y_{L}=Y\cap
L$ and $Z_{L}=Z\cap L$ and for each $n\in\omega$,
$Y_{n}=Y\cap(\omega_{1}\times\\{n\\})$,
$Z_{n}=Z\cap(\omega_{1}\times\\{n\\})$. Since $Y\cap Z=\emptyset$ and
$\omega_{1}\times\\{n\\}$ is a copy of $\omega_{1}$ with the usual topology
(for each $n\in\omega$), then we can find clopen sets
$Y_{n}^{\prime},Z_{n}^{\prime}\subseteq\omega_{1}\times\\{n\\}$ such that
$Y_{n}^{\prime}\cap Z_{n}^{\prime}=\emptyset$, $Y_{n}\subseteq
Y_{n}^{\prime}$, $Z_{n}\subseteq Z_{n}^{\prime}$ and so that for each
$n\in\omega$, $Y_{n}^{\prime}$ is cofinal in $\omega_{1}\times\\{n\\}$ if and
only if $Y_{n}$ is cofinal in $\omega_{1}\times\\{n\\}$ and $Z_{n}^{\prime}$
is cofinal in $\omega_{1}\times\\{n\\}$ if and only if $Z_{n}$ is cofinal in
$\omega_{1}\times\\{n\\}$. This is possible since for each $n\in\omega$,
$Y_{n}$ and $Z_{n}$ cannot be both cofinal (otherwise $Y_{n}\cap
Z_{n}\neq\emptyset$). Let

$\mathcal{Y}=Y_{L}\cup\bigcup_{n\in\omega}Y_{n}^{\prime},\qquad\mathcal{Z}=Z_{L}\cup\bigcup_{n\in\omega}Z_{n}^{\prime}$

Observe $Y\subseteq\mathcal{Y}$, $Z\subseteq\mathcal{Z}$ and
$\mathcal{Y}\cap\mathcal{Z}=\emptyset$.
_Claim:_ $\mathcal{Y}$ and $\mathcal{Z}$ are closed in $X$.
Indeed, if
$\langle\alpha,m\rangle\in(\omega_{1}\times\omega)\smallsetminus\mathcal{Y}$,
since $Y_{m}^{\prime}$ is clopen in $\omega_{1}\times\\{m\\}$, then there is
$U$ open neighbourhood of $\langle\alpha,m\rangle$ in
$\omega_{1}\times\\{m\\}$ (and therefore open neighbourhood in $X$), such that
$U\cap Y_{m}^{\prime}=\emptyset$. Now, let $x\in L\smallsetminus\mathcal{Y}$
and assume that for each $U$ open neighbourhood of $x$ in $X_{0}$ and each
$\alpha<\omega_{1}$, $V_{\alpha}^{U}(x)\cap\mathcal{Y}\neq\emptyset$. This
implies that for each $U$ open neighbourhood of $x$ in $X_{0}$ and each
$\alpha<\omega_{1}$ there is some $n\in\omega$ such that
$V_{\alpha}^{U}(x)Y_{n}^{\prime}\neq\emptyset$ and $Y_{n}^{\prime}$ is cofinal
in $\omega_{1}\times\\{n\\}$. Then $Y_{n}$ is cofinal in
$\omega_{1}\times\\{n\\}$ and $V_{\alpha}^{U}(x)Y_{n}\neq\emptyset$. Hence,
$x\in\overline{Y}=Y$ which is a contradition. Thus, $\mathcal{Y}$ is closed. A
similar argument shows that $\mathcal{Z}$ is closed.
Since $Y_{L}$ and $Z_{L}$ are disjoint closed subsets of $X_{0}$ and $X_{0}$
is normal (recall $X_{0}$ is the space constructed in Example 5.3), then there
exist disjoint open sets $U_{Y}$, $U_{Z}$ in $X_{0}$ such that $Y_{L}\subseteq
U_{Y}$, $Z_{L}\subseteq U_{Z}$. Let

$V_{Y}=(U_{Y}\cap Y)\cup\bigcup_{n\in
U_{Y}\cap\omega}(\omega_{1}\times\\{n\\},\qquad V_{Z}=(U_{Z}\cap
Z)\cup\bigcup_{n\in U_{Z}\cap\omega}(\omega_{1}\times\\{n\\}.$

Observe that $V_{Y}$ and $V_{Z}$ are disjoint open subsets in $X$ and
$Y_{L}\subseteq V_{Y}$, $Z_{L}\subseteq V_{Z}$. Let
$W_{Y}=\mathcal{Y}\cup(V_{Y}\smallsetminus\mathcal{Z})$,
$W_{Z}=\mathcal{Z}\cup(V_{Z}\smallsetminus\mathcal{Y})$. Hence, $W_{Y}$ and
$W_{Z}$ are open sets in $X$, $W_{Y}\cap W_{Z}=\emptyset$, and $y\ \subseteq
W_{Y}$, $Z\subseteq W_{Z}$.

$X$ is not strongly star-Lindelöf: List
$L=\\{x_{\alpha}:\alpha<\omega_{1}\\}$. Since $L$ is a closed discrete subset
of $X_{0}$, for $\alpha<\omega_{1}$ let $D_{\alpha}$ be an open neighbourhood
of $x_{\alpha}$ in $X_{0}$ such that $D_{\alpha}\cap L=\\{x_{\alpha}\\}$.
Hence,

$\mathcal{U}=\\{V_{\alpha}^{D_{\alpha}}(x_{\alpha}):\alpha<\omega_{1}\\}\cup\\{\omega_{1}\times\omega\\}\in\mathcal{O}(X).$

Assume $E\in[X]^{\omega}$, we show $St(E,\mathcal{U})\neq X$. Since $E$ is
countable, fix $\beta_{0},\beta_{1}<\omega_{1}$ such that
$sup\\{\alpha:x_{\alpha}\in E\cap L\\}<\beta_{0}$ and
$sup\\{\gamma:\langle\gamma,n\rangle\in E\textit{ for some
}n\in\omega\\}<\beta_{1}$. Let $\alpha=max\\{\beta_{0},\beta_{1}\\}$ and
observe $E\cap V_{\alpha}^{D_{\alpha}}(x_{\alpha})=\emptyset$ Since
$V_{\alpha}^{D_{\alpha}}(x_{\alpha})$ is the only element of $\mathcal{U}$
that contains $x_{\alpha}$, then $x_{\alpha}\notin St(E,\mathcal{U})$. Thus,
$X$ is not strongly star-Lindelöf.

$X$ is star-Lindelöf: Let $\mathcal{U}\in\mathcal{O}(X)$ and define

$M=\\{n\in\omega:(\exists
U\in\mathcal{U})(\exists\beta<\omega_{1})[(\beta,\omega_{1})\times\\{n\\}\subseteq
U]\\}.$

For each $n\in M$ fix $U_{n}\in\mathcal{U}$ and $\beta_{n}<\omega_{1}$ such
that $(\beta,\omega_{1})\times\\{n\\}\subseteq U_{n}$. Put
$\mathcal{V}^{\prime}=\\{U_{n}:n\in M\\}$.
_Claim:_ $L\subseteq St(\bigcup\mathcal{V}^{\prime},\mathcal{U})$.
Indeed, let $x\in L$, there is $U^{x}\in\mathcal{U}_{n}$ such that $x\in
U^{x}$ and therefore, there is $U$ open neighbourhood of $x$ in $X_{0}$ and
$\alpha<\omega_{1}$ such that $V_{\alpha}^{U}(x)\subseteq U^{x}$. Since
$V_{\alpha}^{U}(x)\cap(\omega_{1}\times\omega)=(\alpha,\omega_{1})\times(U\cap\omega)$
and $U=N\cup f(N)$ for some $N\subseteq L$, with $x\in N$, it holds true that
$n\in f(N)\rightarrow n\in M$. Then, for $n\in f(N)$,
$V_{\alpha}^{U}(x)\cap(\omega_{1}\times\\{n\\})\cap[(\beta_{n},\omega_{1})\times\\{n\\}]\neq\emptyset$.
Thus, $V_{\alpha}^{U}(x)\cap U_{n}\neq\emptyset$. Hence, $U^{x}\cap
U_{n}\neq\emptyset$. Therefore $x\in St(U_{n},\mathcal{U})\subseteq
St(\bigcup\mathcal{V}^{\prime},\mathcal{U})$. Now, $\omega_{1}\times\omega$ is
a countable union of strongly star compact spaces, then there is a countable
$\mathcal{V}^{\prime\prime}\subseteq\mathcal{U}$ such that
$\omega_{1}\times\omega\subseteq
St(\bigcup\mathcal{V}^{\prime\prime},\mathcal{U})$. If we let
$\mathcal{V}=\mathcal{V}^{\prime}\cup\mathcal{V}^{\prime\prime}$, then
$St(\bigcup\mathcal{V},\mathcal{U})=X$. $\blacksquare$

######  Proposition 5.5.

Assuming $2^{\aleph_{0}}=2^{\aleph_{1}}$ and $\aleph_{1}<\mathfrak{d}$ the
space $X$ built in Example 5.4 is normal, star-Menger, and is not either
strongly star-Menger nor Dowker.

###### Proof.

It has been shown that $X$ is normal and not strongly star-Lindelöf (in
particular, $X$ is not strongly star-Menger). It remains to show that it is
star-Menger and is not a Dowker space.

$X$ is star-Menger: let $(\mathcal{U}_{n}:n\in\omega)$ be any sequence of open
covers of $X$. Write $L=\\{x_{\alpha}:\alpha<\omega_{1}\\}$ and for each
$\alpha<\omega_{1}$ and each $n\in\omega$, let
$f_{\alpha}(n)=min\\{i\in\omega:(\exists
U\in\mathcal{U}_{n})(\exists\beta<\omega_{1})[x_{\alpha}\in
U\wedge(\beta,\omega_{1})\times\\{i\\}\subseteq U]\\}$. Observe that for each
$\alpha<\omega_{1}$, $f_{\alpha}:\omega\to\omega$ is well defined. Since
$\\{f_{\alpha}:\alpha<\omega_{1}\\}$ has size less than $\mathfrak{d}$, there
is a funtion $g\in\omega^{\omega}$ such that for all
$\alpha<\omega_{1}:g\not\leq^{*}f_{\alpha}$. For $n\in\omega$ let

$M_{n}=\\{i\in\omega:(\exists
U\in\mathcal{U}_{n})(\exists\beta<\omega_{1})[(\beta,\omega_{1})\times\\{i\\}\subseteq
U]\\}.$

Now, for each $n\in\omega$ and each $i\in M_{n}$, fix
$U_{n}^{i}\in\mathcal{U}_{n}$ and $\beta_{n}^{i}<\omega_{1}$ such that
$(\beta_{n}^{i},\omega_{1})\times\\{i\\}\subseteq U_{n}^{i}$ and let
$\mathcal{V}_{n}=\\{U_{n}^{i}:i\in M_{n}\cap g(n)\\}$.
_Claim:_
$L\subseteq\bigcup\\{St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n}):n\in\omega\\}$.
Indeed, fix $x_{\alpha}\in L$. There is $n\in\omega$ such that
$f_{\alpha}(n)<g(n)$. Hence, there are $U\in\mathcal{U}_{n}$ and
$\beta<\omega_{1}$ such that $x_{\alpha}\in U$ and
$(\beta,\omega_{1})\times\\{f_{\alpha}(n)\\}\subseteq U$. Thus,
$f_{\alpha}(n)\in M_{n}$ and $U_{n}^{f_{\alpha}(n)}\in\mathcal{V}_{n}$. In
addition, $U_{n}^{f_{\alpha}(n)}\cap U\neq\emptyset$. Hence, $x\in
St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n})\subseteq\bigcup\\{St(\bigcup\mathcal{V}_{n},\mathcal{U}_{n}):n\in\omega\\}$.

$X$ it is not a Dowker space: Let us recall the following characterization: A
normal space $D$ is a Dowker space (see [21]) if, and only if, $D$ has a
countable increasing open cover $\\{U_{n}:n\in\omega\\}$ such that there is no
closed cover $\\{F_{n}:n\in\omega\\}$ of $D$ with $F_{n}\subseteq U_{n}$ for
each $n\in\omega$. Hence, let $\\{U_{n}:n\in\omega\\}$ be any countable
increasing open cover ($U_{0}\subseteq U_{1}\subseteq\cdots$) of $X$, we must
find a countable cover of closed sets $\\{F_{n}:n\in\omega\\}$, such that for
each $n\in\omega$, $F_{n}\subseteq U_{n}$.

For each $i\in\omega$ define $n_{i}=min\\{n\in\omega:i\leq
n\wedge(\exists\gamma<\omega_{1})\big{[}[\gamma,\omega_{1})\times\\{i\\}\subseteq
U_{n}\big{]}\\}$. Observe that since $\\{U_{n}:n\in\omega\\}$ is a countable
cover of $X$, $n_{i}$ is well defined for each $i\in\omega$. In addition, for
each $n\in\omega$ and $i\in\omega$ with $i\leq n_{i}\leq n$ let

$\gamma_{i}^{n}=min\\{\gamma<\omega_{1}:[\gamma,\omega_{1})\times\\{i\\}\subseteq
U_{n}\\}\qquad(*)$

Since for each $n\in\omega$, $U_{n}\subseteq U_{n+1}$, then $\gamma_{i}^{n}$
is well defined. Now, for $n\in\omega$ let

$F_{n}=\big{(}\bigcup_{i\leq
n}\\{[\gamma_{i}^{n},\omega_{1})\times\\{i\\}:i\leq n_{i}\leq
n\\}\big{)}\cup(U_{n}\cap L).$

_Claim:_

(1)

For each $n\in\omega$, $F_{n}$ is closed,

(2)

For each $n\in\omega$, $F_{n}\subseteq U_{n}$,

(3)

$\bigcup_{n\in\omega}F_{n}=X$.

Indeed, to show (1), fix $n\in\omega$. First assume $x\in(X\smallsetminus
F_{n})\cap(\omega_{1}\times\omega)$. Hence $x=\langle\alpha,m\rangle$ for some
$\alpha<\omega_{1}$ and $m\in\omega$. If
$F_{n}\cap(\omega_{1}\times\\{m\\})=\emptyset$, any
$U\subseteq\omega_{1}\times\\{m\\}$ open neighbourhood of $x$ is disjoint from
$F_{n}$. If $F_{n}\cap(\omega_{1}\times\\{m\\})\neq\emptyset$, then
$\alpha<\gamma_{m}^{n}$ and for each $\beta<\alpha$,
$(\beta,\alpha]\times\\{m\\}$ is an open neighbourhood of $x$ disjoint from
$F_{n}$. Now, assume $x\in(X\smallsetminus F_{n})\cap L$, let $N\subseteq L$
such that $N\cap F_{n}=\emptyset$ and $x\in N$. Observe that $U=N\cup
f(N)\smallsetminus(n+1)=\big{(}N\cup f(N)\big{)}\cap\big{(}\bigcap_{j\leq
n+1}(X_{0}\smallsetminus\\{j\\})\big{)}$ is an open neighbourhood of $x$ in
$X_{0}$ (see condition (1) and (3) of Example 5.3). Hence, for any
$\alpha<\omega_{1}$, $V_{\alpha}^{U}(x)$ ( $=[U\cap
L]\cup[(\alpha,\omega_{1})\times(U\cap\omega)]$) is an open neighbourhood of
$x$ in $X$ such that $V_{\alpha}^{U}(x)\cap F_{n}=\emptyset$ since
$F_{n}\subseteq\omega_{1}\times[0,n]$ and
$V_{\alpha}^{U}(x)\cap(\omega_{1}\times[0,n])=\emptyset$. Thus, $F_{n}$ is
closed.
To show (2), fix $n\in\omega$. If $x\in F_{n}\cap L$, then $x\in U_{n}$. If
$x=\langle\alpha,m\rangle\in F_{n}\cap(\omega_{1}\times\omega)$, then there is
some $i\leq n_{i}\leq n$ such that
$\langle\alpha,m\rangle\in[\gamma_{i}^{n},\omega_{1})\times\\{i\\}$. Thus ,
$m=i$ and $[\gamma_{i}^{n},\omega_{1})\times\\{i\\}\subseteq U_{n}$. Hence
$F_{n}\subseteq U_{n}$.
Let us show (3). If $x\in X\cap L$, then there is some $n\in\omega$ such that
$x\in U_{n}$. Hence, $x\in U_{n}\cap L\subseteq F_{n}$. If $x\in
X\smallsetminus L$, there is some $i\in\omega$ such that
$x\in\omega_{1}\times\\{i\\}$. By $(*)$ and the fact that $U_{n}\subseteq
U_{n+1}$, $\\{\gamma_{i}^{n}:n\in\omega\\}$ is a decreasing sequence of
ordinals. Since $U_{n}:n\in\omega)$ covers $X$, there is some $m\in\omega$
such that $\gamma_{i}^{m}=0$. Thus, $x\in F_{m}$.  ∎

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Department of Mathematics and Statistics, York University, 4700 Keele St.
Toronto, ON M3J 1P3 Canada

Email address: J. Casas-de la Rosa<EMAIL_ADDRESS>
Email address: W. Chen-Mertens<EMAIL_ADDRESS>
Email address: S. Garcia-Balan<EMAIL_ADDRESS>