Finite powers and products of Menger sets

TL;DR

This paper constructs specific examples of Menger sets with particular properties regarding their finite powers and products, and extends these results to a broader set-theoretic model using forcing techniques.

Contribution

It introduces new constructions of Menger sets that are not Scheepers and explores their properties in different set-theoretic contexts.

Findings

Existence of a Menger set not Scheepers under mild hypotheses

Construction of Menger sets with non-Menger products in all finite powers

Results hold in the Blass–Shelah model for various ultrafilter and dominating numbers

Abstract

We construct, using mild combinatorial hypotheses, a real Menger set that is not Scheepers, and two real sets that are Menger in all finite powers, with a non-Menger product. By a forcing-theoretic argument, we show that the same holds in the Blass--Shelah model for arbitrary values of the ultrafilter and dominating number.

Equations

$$\text{All finite powers are Menger}⟶\text{Scheepers}⟶\text{Menger}$$

$$\mathrm{maxfin}[X]:=\{\max [F]:F\text{ is a finite subset of }X\}$$

$$a/h:=\{n\in \mathbb{N}:a\cap [h(n),h(n+1))\ne \varnothing \}.$$

$$I_{z}:={\bigl{\{}\,n:|\,{\tilde{z}}/{\tilde{d}}\cap(b(n),b(n+1))\,|\geq 2\,\bigr{\}}}$$

$$x:=\bigcup _{n\in r}[\tilde{d}(b(n)),\tilde{d}(b(n+1)+1))\text{ and }y:=\bigcup _{n\in r^{\text{c}}}[\tilde{d}(b(n)),\tilde{d}(b(n+1)+1)).$$

$$z(\tilde{z}(i))=\tilde{z}(i+1)<\tilde{d}(b(n+1))\le x(\tilde{z}(i)).$$

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

$$\mathrm{cl}_{h}(u):=\bigcup{\bigl{\{}\,[h(n),h(n+1)):u\cap[h(n),h(n+1))\neq\emptyset\,\bigr{\}}}\in{\tilde{U}}.$$

$$Y_{\alpha }:=$$

$${\tilde{Y}}_{\alpha }:=$$

$$\bigcup _{\beta <\alpha }{F}_{\beta }\cup \left\{[s\le x_{\alpha }]:s\in \mathrm{maxfin}[D_{\alpha }\cup \{x_{\beta }:\beta <\alpha \}]\right\}$$

$$l\in [h(n),h(n+1))\cap [d\le x_{\alpha }]\text{ and }k\in [h(n),h(n+1))\cap [d\le x_{\alpha }^{\text{c}}].$$

$$2h(n+1)<d(h(n))\le d(l)\le x_{\alpha }(l)<x_{\alpha }(h(n+1))$$

$$2h(n+1)<d(h(n))\le d(k)\le x_{\alpha }^{\text{c}}(k)<x_{\alpha }^{\text{c}}(h(n+1)).$$

$$[1,2h(n+1)+1]\subseteq x_{\alpha }([1,h(n+1)])\cup x_{\alpha }^{\text{c}}([1,h(n+1)]),$$

$$\displaystyle 2h(n+1)+1=\bigl{|}\,[1,2h(n+1)+1]\,\bigr{|}\leq\bigl{|}\,x_{\alpha}([1,h(n+1)])\,\bigr{|}+\bigl{|}\,x_{\alpha}^{\text{\tt c}}([1,h(n+1)])\,\bigr{|}=$$

$$h(n+1)+h(n+1)=2h(n+1),$$

$$\bigcup _{n\in I}[d(n),d(n+1))\in U,\bigcup _{n\in J}[d(n),d(n+1))\in \tilde{U},\text{ and }\left|i-j\right|>3,\text{ for all }i\in I,j\in J.$$

$$\bigcup _{n\in I}[\tilde{d}(n),\tilde{d}(n+1))\in U,\bigcup _{n\in J}[\tilde{d}(n),\tilde{d}(n+1))\in \tilde{U},\text{ and }\left|i-j\right|>3$$

$$x^{\text{c}}:=\bigcup _{n\in I}[\tilde{d}(n),\tilde{d}(n+2)),\text{ and }{y}^{\text{c}}:=\bigcup _{n\in J}[\tilde{d}(n),\tilde{d}(n+2)).$$

$$a\uplus b:=(2a)\cup (2b+1)=\left\{2k:k\in a\right\}\cup \left\{2k+1:k\in b\right\}.$$

$$Y:=\bigcup _{\alpha <\kappa }(a_{\alpha }\uplus {\tilde{M}}_{\alpha })\cup \mathrm{Fin}$$

$$a_{\alpha }(n)<2a_{\alpha }(n)+1\le (a_{\alpha }\uplus a_{\alpha })(2n)\le (a_{\alpha }\uplus y)(2n)\le b(2n)=b^{\prime }(n)$$

$$Y^{\prime}:=Y\setminus U\subseteq\bigl{(}\bigcup_{\alpha<\kappa}a_{\alpha}\uplus\tilde{M}_{\alpha}\bigr{)}\cap\{\,x\in{[\mathbb{N}]^{\mbox{\tiny$\infty$}}}:x\leq b\,\}$$

$$2a_{\alpha }\oplus (a_{\alpha }\uplus y)=2y+1\in (2X)\oplus Y.$$

$$\{a_{\beta }:\beta <\alpha \}\cup M_{\alpha }{<}^{\infty }{a}_{\alpha }.$$

$$\{x\in X:\Phi (x)(n)=m\}=\{x\in X:\Phi (x)(n)\le m\}\setminus \{x\in X:\Phi (x)(n)<m\},$$

$$Y:=\bigl{(}(X\cup\mathrm{Fin})^{n+1}\times M\bigr{)}\setminus\bigl{(}X_{\xi}^{n+1}\times M\bigr{)}$$

$$Y_j^i:=\{\begin{array}{ll}X\cup \mathrm{Fin},& \text{ if }i\ne j,\\ \mathrm{Fin},& \text{ if }i=j,\end{array}$$

$${\tilde{X}}_n:=\{x_1\oplus \cdots \oplus x_n:x_1,\dots ,x_n\in X\},$$

Full text

1. Introduction

By space we mean a topological space. A space is Menger if, for each sequence $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dotsc$ of open covers of that space, there are finite sets $\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$ such that the family $\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$ covers the space. If, in this definition, we request that each finite subset of the space is contained in a set $\bigcup\mathcal{F}_{n}$, then the space is Scheepers. We have the following implications between considered properties of spaces.

[TABLE]

The first of these implications follows from the results of Just, Miller, Scheepers and Szeptycki [11, Theorem 3.9, implications in Figure 3].

A set of reals is a space homeomorphic to a subspace of the real line, with the standard topology. We restrict our consideration to the realm of sets of reals. The three properties are consistently equivalent [33]. In other words, for these properties to differ, special set theoretic hypotheses are necessary. The natural hypotheses to this end concern cardinal characteristics of the continuum [7]; the necessary definitions are provided in the next sections. Assuming $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$, there is a Menger set of reals that is not Scheepers ([24, Theorem 32], [11, Theorem 2.8]). The methods used for that are category theoretic, and do not lend themselves for generalizations. We obtain the same result by a purely combinatorial approach, using the far milder hypothesis $\mathfrak{d}\leq\mathfrak{r}$ (Theorem 2.1).

Assuming $\operatorname{cov}(\mathcal{M})=\mathfrak{c}$ or $\mathfrak{b}=\mathfrak{d}$, there are sets of reals with all finite powers Menger, whose product is not Menger ([29, Theorem 3.3], [20, Proposition 3.4]). The assumptions $\operatorname{cov}(\mathcal{M})=\mathfrak{c}$ and $\mathfrak{b}=\mathfrak{d}$ each imply $\mathfrak{d}\leq\mathfrak{r}$. The inequality $\mathfrak{d}\leq\mathfrak{r}$ has recently played a central role in the construction of two Menger sets of reals whose product is not Menger [25]. We refine these methods to establish the mentioned stronger result from the same hypothesis $\mathfrak{d}\leq\mathfrak{r}$, assuming that the cardinal number $\mathfrak{d}$ is regular (Theorem 2.5). This additional assumption follows from the earlier hypothesis $\mathfrak{b}=\mathfrak{d}$.

These results, that are optimal for the used methods, suggest the question of the necessity of the hypothesis $\mathfrak{d}\leq\mathfrak{r}$ [25]. Arguing directly in the Blass–Shelah model [2], we answer this question in the negative.

Finally, we apply our results to products of function spaces with the topology of pointwise convergence.

2. The main results

2.1. Separation of the Menger and Scheepers properties

Let $\mathbb{N}$ be the set of positive natural numbers, and {[\mathbb{N}]^{\mbox{\tiny\infty}}} be the set of 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 $a$. We have {[\mathbb{N}]^{\mbox{\tiny\infty}}}\subseteq{\mathbb{N}^{\mathbb{N}}}, and thus every set from {[\mathbb{N}]^{\mbox{\tiny\infty}}} is viewed as a function. Let x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. The function $x$ is dominated by the function $y$, denoted $x\leq^{*}y$, if the set $\{\,n:y(n)<x(n)\,\}$ is finite. A subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is dominating if each function in {[\mathbb{N}]^{\mbox{\tiny\infty}}} is dominated by some function from this set. Let $\mathfrak{d}$ be the minimal cardinality of a dominating subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. For a finite subset $F$ of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, define $\max[F]:=\{\,\max\{\,x(n):x\in F\,\}:n\in\mathbb{N}\,\}$, an element of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. A subset $X$ of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is finitely dominating if the set

[TABLE]

is dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. We identify the Cantor cube ${\{0,1\}^{\mathbb{N}}}$ with the family ${\operatorname{P}(\mathbb{N})}$ of all subsets of $\mathbb{N}$, via characteristic functions. Since the Cantor cube is homeomorphic to Cantor’s set, every subspace of ${\operatorname{P}(\mathbb{N})}$ is a set of reals. The topologies in the set {[\mathbb{N}]^{\mbox{\tiny\infty}}}, induced from the Cantor space ${\operatorname{P}(\mathbb{N})}$, and the Baire space ${\mathbb{N}^{\mathbb{N}}}$, are equivalent. We use the following combinatorial characterizations of the Menger and Scheepers properties. A space is Menger (Scheepers) if and only if no continuous image of the space into {[\mathbb{N}]^{\mbox{\tiny\infty}}} is dominating [19, Proposition 3](finitely dominating [15, Theorem 7.4(2)]).

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. A property $\mathbf{P}$ of spaces is productive if the product space of two spaces with the property $\mathbf{P}$, has the property $\mathbf{P}$.

Theorem 2.1**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$.

1. (1)

There is a Menger set of reals that is not Scheepers. 2. (2)

The Menger property is not productive.

In order to prove Theorem 2.1 and later discussion, we introduce several notions and auxiliary results. Fix functions x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} and a set Z\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}}. We write $x<^{\infty}y$ if $y\nleq^{*}x$ and $Z<^{\infty}x$ if $z<^{\infty}x$ for all functions $z\in Z$. We use the latter convention to any binary relation. Let $\max\{x,y\}:=\{\,\max\{x(n),y(n)\}:n\in\mathbb{N}\,\}$, an element of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. If $x(1)\neq 1$, then define an element {\tilde{x}}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that ${\tilde{x}}(1):=x(1)$ and ${\tilde{x}}(n+1):=x({\tilde{x}}(n))$, for all natural numbers $n$. For $x\in{\operatorname{P}(\mathbb{N})}$, define $x^{\text{\tt c}}:=\mathbb{N}\setminus x$, and let [\mathbb{N}]^{\infty,\infty}:=\{\,a\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}:a^{\text{\tt c}}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}\,\}. For natural numbers $n,m$, with $n<m$, define $[n,m):=\{\,i\in\mathbb{N}:n\leq i<m\,\}$. For elements a,h\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, let

[TABLE]

Lemma 2.2**.**

Let $Z$ be a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} with $\left|Z\right|<\min\{\mathfrak{d},\mathfrak{r}\}$, and d\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There are elements $x,y\in[\mathbb{N}]^{\infty,\infty}$ such that $Z<^{\infty}x,y$, and $d\leq^{*}\max\{x^{\text{\tt c}},y^{\text{\tt c}}\}$.

Proof.

For natural numbers $n,m$ with $n<m$, let $(n,m):=\{\,i\in\mathbb{N}:n<i<m\,\}$. Since $\left|Z\right|<\mathfrak{d}$, there is a function b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that the sets

[TABLE]

are infinite for all elements $z\in Z$ [7, Theorem 2.10]. Since $\left|Z\right|<\mathfrak{r}$, there is a set $r\in[\mathbb{N}]^{\infty,\infty}$ that reaps the family $\{\,I_{z}:z\in Z\,\}$. Let

[TABLE]

Fix an element $z\in Z$ and a natural number $n\in r^{\text{\tt c}}\cap I_{z}$. Since $|\,{\tilde{z}}/{\tilde{d}}\cap(b(n),b(n+1))\,|\geq 2$, there is a natural number $i$ such that $\tilde{z}(i),\tilde{z}(i+1)\in[\tilde{d}(b(n)+1),\tilde{d}(b(n+1)))$. Since $x\cap[\tilde{d}(b(n)+1),\tilde{d}(b(n+1)))=\emptyset$, we have

[TABLE]

As the set $r^{\text{\tt c}}\cap I_{z}$ is infinite, we have $z<^{\infty}x$. Similarly, $z<^{\infty}y$.

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 $n\in r$. Since $x^{\text{\tt c}}\cap[\tilde{d}(b(n)),\tilde{d}(b(n+1))+1)=\emptyset$, we have

[TABLE]

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

Let $\kappa$ be an uncountable cardinal number. A subset $A$ of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is $\kappa$-unbounded [25, Definition 2.1] if $\left|A\right|\geq\kappa$ and, for each function x\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, we have $\left|\{\,a\in A:a\leq^{*}x\,\}\right|<\kappa$. Let $\tau\colon{\operatorname{P}(\mathbb{N})}\to{\operatorname{P}(\mathbb{N})}$ be a homeomorphism defined by $\tau(x):=x^{\text{\tt c}}$. Let $\mathrm{Fin}$ be the set of finite subsets of $\mathbb{N}$. Then {\operatorname{P}(\mathbb{N})}=\mathrm{Fin}\cup{[\mathbb{N}]^{\mbox{\tiny\infty}}}.

Proof of Theorem 2.1.

(1) Let $\{\,d_{\alpha}:\alpha<\mathfrak{d}\,\}$ be a dominating set in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Fix an ordinal number $\alpha<\mathfrak{d}$. Let $x_{\alpha},y_{\alpha}\in[\mathbb{N}]^{\infty,\infty}$ be elements obtained from Lemma 2.2, applied to the set $\{\,d_{\beta}:\beta<\alpha\,\}$ and to the function $d_{\alpha}$. Since the set $X:=\{\,x_{\alpha},y_{\alpha}:\alpha<\mathfrak{d}\,\}$ is $\mathfrak{d}$-unbounded, the set $X\cup\mathrm{Fin}$ is Menger [25, Corollary 2.4]. The continuous image $\tau[X\cup\mathrm{Fin}]$ of the set $X\cup\mathrm{Fin}$ is a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Since $\operatorname{maxfin}[\tau[X\cup\mathrm{Fin}]]$ contains a dominating set $\{\,\max\{{x_{\alpha}}^{\text{\tt c}},{y_{\alpha}}^{\text{\tt c}}\}:\alpha<\mathfrak{d}\,\}$, the image $\tau[X\cup\mathrm{Fin}]$ is finitely dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. A set of reals is Scheepers if no continuous image of this set into {[\mathbb{N}]^{\mbox{\tiny\infty}}} is finitely dominating [28, Theorem 2.1]. Thus, the set $X\cup\mathrm{Fin}$ is not Scheepers.

(2) Since the continuous image of the product space $(X\cup\mathrm{Fin})^{2}$ under the map $(x,y)\mapsto\max\{x^{\text{\tt c}},y^{\text{\tt c}}\}$ is dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, the product space $(X\cup\mathrm{Fin})^{2}$ is not Menger. ∎

Remark 2.3*.*

Let $\operatorname{cov}(\mathcal{M})$ be the minimal cardinality of a family of meager subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}} that covers {[\mathbb{N}]^{\mbox{\tiny\infty}}} and $\operatorname{cof}(\mathcal{M})$ be the minimal cardinality of a cofinal family of meager sets in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, i.e., every meager subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is contained in a member of the family. Assuming $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$, there is a Menger set of reals that is not Scheepers [24, Theorem 32]. Since $\operatorname{cov}(\mathcal{M})\leq\mathfrak{d}\leq\operatorname{cof}(\mathcal{M})$ and $\operatorname{cov}(\mathcal{M})\leq\mathfrak{r}$, the equality $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$ implies that $\mathfrak{d}\leq\mathfrak{r}$. Thus, Theorem 2.1(1) is a substantial extension of the mentioned result.

Remark 2.4*.*

It is always the case that, if there is a Menger set of reals that is not Scheepers, then the Menger property is not productive: Let $X$ be such a set. Some finite power of $X$ is not Menger [11, Theorem 3.9] and let $n$ be the minimal natural number with this property. The product space of Menger sets $X$ and $X^{n-1}$ is not Menger.

2.2. Products of Menger sets with strong properties

A property of spaces $\mathbf{P}$ is additive if the union of any two spaces with the property $\mathbf{P}$, has the property $\mathbf{P}$.

Theorem 2.5**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$ and the cardinal number $\mathfrak{d}$ is regular.

1. (1)

There are two sets of reals whose all finite powers are Menger, but whose union is not Scheepers. 2. (2)

The Scheepers property is not additive. 3. (3)

There are two sets of reals whose all finite powers are Menger, but whose product space is not Menger. 4. (4)

The Scheepers property is not productive.

For elements x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, let $[x\leq y]:=\{\,n:x(n)\leq y(n)\,\}$ and $[x<y]:=\{\,n:x(n)<y(n)\,\}$. Let $A$ be a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, and x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. We write $x\leq_{A}y$ if $[x\leq y]\in A$, and $x<_{A}y$ if $[x<y]\in A$. A subset $X$ of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is $\leq_{A}$-bounded if there is a function b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that $X\leq_{A}b$. A subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is $\leq_{A}$-unbounded if it is not $\leq_{A}$-bounded.

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. Let $U$ be an ultrafilter, and $\operatorname{\mathfrak{b}}(U)$ be the minimal cardinality of a $\leq_{U}$-unbounded set in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. A set X\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}} is a $U$-scale [25, Definition 4.1] if $\left|X\right|\geq\operatorname{\mathfrak{b}}(U)$ and, for each function b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, we have $\left|\{\,x\in X:x\leq_{U}b\,\}\right|<\operatorname{\mathfrak{b}}(U)$. A space $X$ is $U\text{-Menger}$ if, for each sequence $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dotsc$ of open covers of the space $X$, there are finite sets $\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$ such that, for each point $x\in X$, we have $\{\,n:x\in\bigcup\mathcal{F}_{n}\,\}\in\mathcal{U}$. A set of reals is $U\text{-Menger}$ if every continuous image of the set into {[\mathbb{N}]^{\mbox{\tiny\infty}}} is $\leq_{U}$-bounded [25, Proposition 4.4]. Since no $\leq_{U}$-bounded set in {[\mathbb{N}]^{\mbox{\tiny\infty}}} is finitely dominating, every $U\text{-Menger}$ set is Scheepers [15, Theorem 7.4(2)].

An ultrafilter $U$ is near coherent to an ultrafilter ${\tilde{U}}$ if there is a function h\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that for each set $u\in U$, we have

[TABLE]

The near coherence relation is an equivalence relation. The near coherence of filters (NCF) is the statement introduced by Blass that any two ultrafilters are near coherent [8]. NCF is independent of ZFC [8] and it follows, e.g., from the inequality $\mathfrak{u}<\mathfrak{g}$ [13], where $\mathfrak{u}$ is the ultrafilter number and $\mathfrak{g}$ is the groupwise density number. It demonstrates that the forthcoming Lemma 2.6 needs a set theoretic assumption.

In order to prove Theorem 2.5, we need the following Lemmata. Let $\operatorname{cf}(\mathfrak{d})$ be the cofinality of the cardinal number $\mathfrak{d}$. A subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is centered if the finite intersections of its elements, are infinite. For natural numbers $n,m$ with $n\leq m$, let $[n,m]:=\{\,i\in\mathbb{N}:n\leq i\leq m\,\}$. For elements x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, we write $x\leq y$ if $[x\leq y]=\mathbb{N}$.

Lemma 2.6**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$. There are non near coherent ultrafilters $U$ and ${\tilde{U}}$ such that $\operatorname{\mathfrak{b}}(U)=\operatorname{\mathfrak{b}}({\tilde{U}})=\operatorname{cf}(\mathfrak{d})$.

Proof.

Let ${\langle\,D_{\alpha}:\alpha<\operatorname{cf}(\mathfrak{d})\,\rangle}$ be a strictly increasing sequence of subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}} of cardinality less than $\mathfrak{d}$ such that the set $\bigcup_{\alpha<\operatorname{cf}(\mathfrak{d})}D_{\alpha}$ is dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. We may assume that for each function x\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} there is a function $d$ from the dominating set with $x\leq d$. Proceed by transfinite induction on ordinal numbers $\alpha<\mathfrak{d}$. Define increasing sequences ${\langle\,F_{\alpha}:\alpha<\mathfrak{d}\,\rangle}$, ${\langle\,{\tilde{F}}_{\alpha}:\alpha<\mathfrak{d}\,\rangle}$ of subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, and a set $\{\,x_{\alpha}:\alpha<\operatorname{cf}(\mathfrak{d})\,\}\subseteq[\mathbb{N}]^{\infty,\infty}$ such that, for each ordinal number $\alpha<\mathfrak{d}$:

• •

the sets $F_{\alpha}$, ${\tilde{F}}_{\alpha}$ with $|\,F_{\alpha}\,|,|\,{\tilde{F}}_{\alpha}\,|<\mathfrak{d}$ are closed under finite intersections,

• •

$D_{\alpha}\cup\{\,x_{\beta}:\beta<\alpha\,\}\leq_{F_{\alpha}}x_{\alpha}$ and $D_{\alpha}\cup\{\,{x_{\beta}}^{\text{\tt c}}:\beta<\alpha\,\}\leq_{{\tilde{F}}_{\alpha}}{x_{\alpha}}^{\text{\tt c}}$.

For functions s,f\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, let $s\circ f$ be a function in {[\mathbb{N}]^{\mbox{\tiny\infty}}} such that $(s\circ f)(n):=s(f(n))$, for all natural numbers $n$. Fix an ordinal number $\alpha<\mathfrak{d}$. Let

[TABLE]

Since $|\,Y_{\alpha}\cup{\tilde{Y}}_{\alpha}\,|<\mathfrak{d}\leq\mathfrak{r}$, there is an element $x_{\alpha}\in[\mathbb{N}]^{\infty,\infty}$ such that $Y_{\alpha}\cup{\tilde{Y}}_{\alpha}<^{\infty}x_{\alpha}$ and $Y_{\alpha}\cup{\tilde{Y}}_{\alpha}<^{\infty}{x_{\alpha}}^{\text{\tt c}}$ [25, Lemma 3.4].

Fix elements $s\in\operatorname{maxfin}[D_{\alpha}\cup\{\,x_{\beta}:\beta<\alpha\,\}]$ and $f\in\bigcup_{\beta<\alpha}F_{\beta}$. Since the set $[(s\circ f)\leq x_{\alpha}]$ is infinite, the intersection $f\cap[s\leq x_{\alpha}]$ is infinite, too. Thus, the set

[TABLE]

is centered. Let $F_{\alpha}$ be the latter set closed under finite intersections. Then $\left|F_{\alpha}\right|<\mathfrak{d}$. Similarly, define a set ${\tilde{F}}_{\alpha}$.

The sets $F:=\bigcup_{\alpha<\operatorname{cf}(\mathfrak{d})}F_{\alpha}$ and ${\tilde{F}}:=\bigcup_{\alpha<\operatorname{cf}(\mathfrak{d})}{\tilde{F}}_{\alpha}$ are closed under finite intersections. Let $U$ and ${\tilde{U}}$ be ultrafilters containing the sets $F$ and ${\tilde{F}}$, respectively. We have $\operatorname{\mathfrak{b}}(U)=\operatorname{cf}(\mathfrak{d})$: Let $Y$ be a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} with $\left|Y\right|<\operatorname{cf}(\mathfrak{d})$. There is an ordinal number $\alpha<\operatorname{cf}(\mathfrak{d})$ such that, for each function $y\in Y$, there is a function $d_{y}\in D_{\alpha}$ with $y\leq^{*}d_{y}$. Since $D_{\alpha}\leq_{U}x_{\alpha}$, we have $Y\leq_{U}x_{\alpha}$.

The ultrafilters $U$ and $\tilde{U}$ are not near coherent: Fix h\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There are an ordinal number $\alpha<\mathfrak{d}$ and an element $d\in D_{\alpha}$ such that $2h(n+1)<d(h(n))$ for all natural numbers $n$. We may assume that $\mathbb{N}\in F_{0}\cap\tilde{F}_{0}$. Since $d\in D_{\alpha}\subseteq Y_{\alpha}$, we have $[d\leq x_{\alpha}]\in F_{\alpha}$ and $[d\leq x_{\alpha}^{\text{\tt c}}]\in\tilde{F}_{\alpha}$. Next, we show that $\mathrm{cl}_{h}([d\leq x_{\alpha}])\cap\mathrm{cl}_{h}([d\leq x_{\alpha}^{\text{\tt c}}])=\emptyset$. Suppose contrary that there is a natural number $n$ such that $[h(n),h(n+1))\cap[d\leq x_{\alpha}]\neq\emptyset$ and $[h(n),h(n+1))\cap[d\leq x_{\alpha}^{\text{\tt c}}]\neq\emptyset$. Pick

[TABLE]

We have

[TABLE]

and

[TABLE]

From the above equations, we get

[TABLE]

and thus

[TABLE]

a contradiction. ∎

Remark 2.7*.*

The non near coherence of the ultrafilters $U$ and $\tilde{U}$ in the proof of Lemma 2.6 could be also established with the help of the following shorter topological argument, which, however, is not self-contained.

Let $\{\,x_{\alpha}:\alpha<\operatorname{cf}(\mathfrak{d})\,\}$ be defined as in the proof of Lemma 2.6. The set $\{\,x_{\alpha}:\alpha<\operatorname{cf}(\mathfrak{d})\,\}$ is a $U$-scale: Let b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There is an ordinal number $\alpha<\operatorname{cf}(\mathfrak{d})$ such that the function $b$ is dominated by some function from the set $D_{\alpha}$. For each ordinal number $\beta$ with $\alpha\leq\beta<\operatorname{cf}(\mathfrak{d})$, we have $D_{\alpha}\leq_{U}x_{\beta}$, and thus $b\leq_{U}x_{\beta}$. Analogously, we have $\operatorname{\mathfrak{b}}({\tilde{U}})=\operatorname{cf}(\mathfrak{d})$ and the set $\{\,{x_{\alpha}}^{\text{\tt c}}:\alpha<\operatorname{cf}(\mathfrak{d})\,\}$ is a ${\tilde{U}}$-scale. Let $X:=\{\,x_{\alpha}:\alpha<\mathfrak{d}\,\}$. Then the set $X\cup\mathrm{Fin}$ is $U\text{-Menger}$ [25, Theorem 5.3(2)]. The continuous image $\tau[X\cup\mathrm{Fin}]$ of the set $X\cup\mathrm{Fin}$ in {[\mathbb{N}]^{\mbox{\tiny\infty}}} contains a $\leq_{{\tilde{U}}}$-unbounded set $\{\,{x_{\alpha}}^{\text{\tt c}}:\alpha<\mathfrak{d}\,\}$, and thus the set $X\cup\mathrm{Fin}$ is not ${\tilde{U}}\text{-Menger}$ [25, Proposition 4.4]. We conclude that the ultrafilter $U$ is not near coherent to the ultrafilter ${\tilde{U}}$ [32, Theorem 2.32].

Lemma 2.8**.**

Let $U$ and ${\tilde{U}}$ be non near coherent ultrafilters, and d\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There are sets I,J\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that

[TABLE]

Proof.

Let $b_{k}:=\{\,8n+k:n\in\mathbb{N}\,\}$, where $k\in\{0,\dotsc,7\}$. Let $k_{0}$ be such that $x:=\bigcup_{i\in b_{k_{0}}}[d(i),d(i+1))\in U$. Define $d^{\prime}:=\{\,d(n):|n-i|>3\text{ for all }i\in b_{k_{0}}\,\}$, an element of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Since the ultrafilters $U$ and ${\tilde{U}}$ are non near coherent, there are sets $y\in U$ and $\tilde{y}\in{\tilde{U}}$ such that $\mathrm{cl}_{d^{\prime}}(y)\cap\mathrm{cl}_{d^{\prime}}(\tilde{y})=\emptyset$. We may assume that $y\subseteq x$. Let $I:=\{\,n:[d(n),d(n+1))\cap y\neq\emptyset\,\}$ and $J:=\{\,n:[d(n),d(n+1))\cap\tilde{y}\neq\emptyset\,\}$. Then $\bigcup_{n\in I}[d(n),d(n+1))\in U$ and $\bigcup_{n\in J}[d(n),d(n+1))\in{\tilde{U}}$. Fix $i\in I$ and $j\in J$. Since $y\subseteq x$, we have $i\in b_{k_{0}}$. Let $n,m$ be natural numbers such that $d(n),d(m)$ are consecutive elements of $d^{\prime}$ with $n\leq j<j+1\leq m$. By the definition of $d^{\prime}$, we have $|n-i|,|m-i|>3$. Since $[d(n),d(m))\cap y=\emptyset$, we have $i<n$ or $m\leq i$, and thus $|i-j|>3$. ∎

Lemma 2.9**.**

Let $U$ and ${\tilde{U}}$ be non near coherent ultrafilters, and d\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There are elements $x,y\in[\mathbb{N}]^{\infty,\infty}$ such that $d<_{U}x$, $d<_{\tilde{U}}y$, and $d\leq^{*}\max\{x^{\text{\tt c}},y^{\text{\tt c}}\}$.

Proof.

By Lemma 2.8, there are sets I,J\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that

[TABLE]

for all natural numbers $i\in I$, and $j\in J$. Let x,y\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} be elements such that

[TABLE]

Fix a natural number $n\in I$ and a natural number $k\in[{\tilde{d}}(n),{\tilde{d}}(n+1))$. Since $x\cap[{\tilde{d}}(n),{\tilde{d}}(n+2))=\emptyset$, we have $d(k)<d({\tilde{d}}(n+1))={\tilde{d}}(n+2)\leq x(k)$. Thus, $d<_{U}x$. Similarly, we have $d<_{\tilde{U}}y$.

Fix a natural number $n$ and a natural number $k\in[{\tilde{d}}(n),{\tilde{d}}(n+1))$. One of the sets $x^{\text{\tt c}}$, or $y^{\text{\tt c}}$ has empty intersection with the interval $[{\tilde{d}}(n),{\tilde{d}}(n+2))$. If $x^{\text{\tt c}}\cap[{\tilde{d}}(n),{\tilde{d}}(n+2))=\emptyset$, then $d(k)<d({\tilde{d}}(n+1))<{\tilde{d}}(n+2)\leq x^{\text{\tt c}}(k)$. If the set $y^{\text{\tt c}}$ has this property, then $d(k)\leq y^{\text{\tt c}}(k)$. Thus, $d\leq^{*}\max\{x^{\text{\tt c}},y^{\text{\tt c}}\}$ ∎

Let $\mathbf{P}$ be a property of spaces. A space $X$ is productively $\mathbf{P}$ if, for each space $Y$ with the property $\mathbf{P}$, the product space $X\times Y$, has the property $\mathbf{P}$.

Lemma 2.10**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$. Let $U$ be an ultrafilter with $\operatorname{\mathfrak{b}}(U)=\mathfrak{d}$. For any ultrafilter ${\tilde{U}}$ non near coherent to the ultrafilter $U$ with $\operatorname{\mathfrak{b}}({\tilde{U}})=\mathfrak{d}$, there are a $U$-scale X\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}}, and a ${\tilde{U}}$-scale Y\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that

1. (1)

the union $(X\cup\mathrm{Fin})\cup(Y\cup\mathrm{Fin})$ is not Scheepers, 2. (2)

the product space $(X\cup\mathrm{Fin})\times(Y\cup\mathrm{Fin})$ is not Menger.

Proof.

(1) Let $\{\,e_{\alpha}:\alpha<\mathfrak{d}\,\}$ be a dominating set in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Proceed by transfinite induction on ordinal numbers $\alpha<\mathfrak{d}$. Let $d_{0}=e_{0}$. Fix an ordinal number $\alpha<\mathfrak{d}$ and assume that elements d_{\beta}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} have been already defined for all ordinal numbers $\beta$ with $\beta<\alpha$. Since $\operatorname{\mathfrak{b}}(U)=\operatorname{\mathfrak{b}}({\tilde{U}})=\mathfrak{d}$, there is a function d_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that $\{\,d_{\beta}:\beta<\alpha\,\}\leq_{U}d_{\alpha}$, $\{\,d_{\beta}:\beta<\alpha\,\}\leq_{{\tilde{U}}}d_{\alpha}$ and $e_{\alpha}\leq^{*}d_{\alpha}$. Then the set $\{\,d_{\alpha}:\alpha<\mathfrak{d}\,\}$ is a dominating set in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, simultaneously a $U$-scale, and a ${\tilde{U}}$-scale.

For each ordinal number $\alpha<\mathfrak{d}$, let x_{\alpha},y_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} be elements obtained from Lemma 2.9, applied to the function $d_{\alpha}$. The sets $X:=\{\,x_{\alpha}:\alpha<\mathfrak{d}\,\}$, and $Y:=\{\,y_{\alpha}:\alpha<\mathfrak{d}\,\}$ are a $U$-scale, and a ${\tilde{U}}$-scale in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, respectively. Thus the sets $X\cup\mathrm{Fin}$, and $Y\cup\mathrm{Fin}$ are productively $U\text{-Menger}$, and productively ${\tilde{U}}\text{-Menger}$, respectively [25, Theorem 5.3(2)]. Let $Z:=X\cup Y\cup\mathrm{Fin}$. The set $\operatorname{maxfin}[\tau[Z]]$, a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, contains a dominating set $\{\,\max\{{x_{\alpha}}^{\text{\tt c}},{y_{\alpha}}^{\text{\tt c}}\}:\alpha<\mathfrak{d}\,\}$. The continuous image $\tau[Z]$ of the set $Z$ is finitely dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, and thus the set $Z$ is not Scheepers [28, Theorem 2.1].

(2) The continuous image of the product space $(X\cup\mathrm{Fin})\times(Y\cup\mathrm{Fin})$, under the map $(x,y)\mapsto\max\{x^{\text{\tt c}},y^{\text{\tt c}}\}$, is dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Thus, this product space is not Menger. ∎

Proof of Theorem 2.5.

(1) Since the cardinal number $\mathfrak{d}$ is regular, there is an ultrafilter $U$ with $\operatorname{\mathfrak{b}}(U)=\mathfrak{d}$ [9]. By Lemma 2.6 and the fact that the near coherence relation is an equivalence relation, there is an ultrafilter ${\tilde{U}}$ non near coherent to the ultrafilter $U$ with $\operatorname{\mathfrak{b}}({\tilde{U}})=\mathfrak{d}$. Let $X$ and $Y$ be a $U$-scale, and a ${\tilde{U}}$-scale, respectively, obtained from Lemma 2.10. The sets $X\cup\mathrm{Fin}$, and $Y\cup\mathrm{Fin}$, are productively $U\text{-Menger}$, and productively ${\tilde{U}}\text{-Menger}$, respectively, and thus all finite powers of these sets are Menger. Apply Lemma 2.10(1).

(2) If all finite powers of a space are Menger, then the space is Scheepers. Apply (1).

(3) Apply sets from (1) and Lemma 2.10(2).

(4) If all finite powers of a space are Menger, then the space is Scheepers. Apply (3). ∎

It is consistent with ZFC, that for any ultrafilter $U$, the $U\text{-Menger}$ and Scheepers properties are equivalent [31, proof of Theorem 3.7]. In contrast to this result, we have the following corollary from Theorem 2.5.

Corollary 2.11**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$. For each ultrafilter $U$, there is a Scheepers set of reals that is not $U\text{-Menger}$.

Proof.

Assume that $\operatorname{\mathfrak{b}}(U)<\mathfrak{d}$. Since every set of reals of cardinality less than $\mathfrak{d}$ is Scheepers [28, Theorem 2.1], each $\leq_{U}$-unbounded subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} of cardinality $\operatorname{\mathfrak{b}}(U)$, is Scheepers, but not $U\text{-Menger}$. Assume that $\operatorname{\mathfrak{b}}(U)=\mathfrak{d}$. By Lemma 2.10(2), there are an ultrafilter ${\tilde{U}}$, and two sets of reals that are productively $U\text{-Menger}$, and productively ${\tilde{U}}\text{-Menger}$, respectively, whose product space is not Menger. One of these sets is ${\tilde{U}}\text{-Menger}$, and thus Scheepers [25, Proposition 4.4], [28, Theorem 2.1], but it is not $U\text{-Menger}$. ∎

2.3. Additivity and productivity of the Menger property.

Let $\operatorname{add}(\mathrm{Menger})$ be the minimal cardinality of a family of Menger sets of reals, whose union is not Menger. For every subset $X$ of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, containing a $\operatorname{cf}(\mathfrak{d})$-unbounded, or a $\mathfrak{d}$-unbounded set, there is a Menger set of reals $Y$ such the product space $X\times Y$ is not Menger [25, Theorem 2.7]. We prove the same statement, assuming that the set $X$ contains an $\operatorname{add}(\mathrm{Menger})$-unbounded set.

Theorem 2.12**.**

Let $X$ be a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} containing an $\operatorname{add}(\mathrm{Menger})$-unbounded set. There is a Menger set of reals $Y$ such that the product space $X\times Y$ is not Menger.

Recall the result from an earlier work [25], a main tool in the proof of Theorem 2.12.

Lemma 2.13** ([25, Lemma 2.6]).**

For sets $a,b\in{\operatorname{P}(\mathbb{N})}$, let

[TABLE]

Then:

1. (1)

For each set a\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} and each natural number $n$, we have $(a\uplus a)(2n)=2a(n)+1$. 2. (2)

For all sets a,b,c,d\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} with $a\leq b$ and $c\leq d$, we have $a\uplus c\leq b\uplus d$.∎

For sets $x,y$, let $x\oplus y:=(x\cup y)\setminus(x\cap y)$. The space ${\operatorname{P}(\mathbb{N})}$, with the group operation $\oplus$ and the standard topology (see the beginning of Subsection 2.1 for the explanation of what kind of topology on ${\operatorname{P}(\mathbb{N})}$ we consider) is a topological group. For sets $X,Y\subseteq{\operatorname{P}(\mathbb{N})}$, let $X\oplus Y:=\{\,x\oplus y:x\in X,y\in Y\,\}$. For an element x\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} let $2x:=\{\,2k:k\in x\,\}$, and $(x+1):=\{\,k+1:k\in x\,\}$. For a set X\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}} and an element a\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}, let $2X:=\{\,2x:x\in X\,\}$, $X+1:=\{\,x+1:x\in X\,\}$ and $a\uplus X:=\{\,a\uplus x:x\in X\,\}$.

Proof of Theorem 2.12.

Let $\kappa:=\operatorname{add}(\mathrm{Menger})$. There is a family $\{\,M_{\alpha}:\alpha<\kappa\,\}$ of Menger sets of reals whose union is not Menger. Then there is a continuous image of this union in {[\mathbb{N}]^{\mbox{\tiny\infty}}}, a dominating set. Since the Menger property is preserved by continuous mappings, we may assume that $\bigcup_{\alpha<\kappa}M_{\alpha}$ is a union of Menger subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Let $\{\,a_{\alpha}:\alpha<\kappa\,\}$ be a $\kappa$-unbounded subset of $X$. For each ordinal number $\alpha<\kappa$, the set $\tilde{M}_{\alpha}:=\{\,\max\{a_{\alpha},x\}:x\in M_{\alpha}\,\}$, a continuous image of the Menger set $M_{\alpha}$, is Menger.

The set

[TABLE]

is Menger: Let $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dotsc$ be a sequence of open covers of the set $Y$ which are open also in ${\operatorname{P}(\mathbb{N})}$. There are sets $U_{1}\in\mathcal{U}_{1},U_{2}\in\mathcal{U}_{2},\dotsc$ such that the set $U:=\bigcup_{n}U_{n}$ contains the set $\mathrm{Fin}$. Since ${\operatorname{P}(\mathbb{N})}\setminus U$ is a compact subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}, there is an element b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that ${\operatorname{P}(\mathbb{N})}\setminus U\leq b$. Define $b^{\prime}(n):=b(2n)$ for all natural numbers $n$. Fix an ordinal number $\alpha<\kappa$, and an element $y\in\tilde{M}_{\alpha}$. Assume that $a_{\alpha}\uplus y\leq b$. By Lemma 2.13, we have

[TABLE]

for all natural numbers $n$. Thus, $a_{\alpha}\leq b^{\prime}$. Since the set $\{\,a_{\alpha}:\alpha<\kappa\,\}$ is $\kappa$-unbounded, there is an ordinal number $\alpha^{\prime}<\kappa$ such that $\{\,\beta<\kappa:a_{\beta}\leq b^{\prime}\,\}\subseteq\alpha^{\prime}$. We conclude that the set

[TABLE]

is a closed subset of the Menger set $\bigcup_{\beta<\alpha^{\prime}}(a_{\beta}\uplus\tilde{M}_{\beta})$, and thus it is Menger. There are finite sets $\mathcal{F}_{1}\subseteq\mathcal{U}_{1},\mathcal{F}_{2}\subseteq\mathcal{U}_{2},\dotsc$ such that the family $\bigcup_{n}\mathcal{F}_{n}$ covers the set $Y^{\prime}$. Thus, the family $\bigcup_{n}(\mathcal{F}_{n}\cup\{U_{n}\})$ is a cover of the set $Y$.

The set $2X\oplus Y$ is dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}: By the definition of the operation $\uplus$, the set $2X\oplus Y$ is a subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Fix an ordinal number $\alpha<\kappa$, and an element $y\in\tilde{M}_{\alpha}$. We have

[TABLE]

Thus, the set $(2X)\oplus Y$ contains a dominating set $(2\bigcup_{\alpha<\kappa}\tilde{M}_{\alpha})+1$.

Since the set $(2X)\oplus Y$ is a continuous image of the product space $X\times Y$, the product space $X\times Y$ is not Menger. ∎

Theorem 2.14**.**

Assume that $\operatorname{add}(\mathrm{Menger})<\mathfrak{d}$. The Menger property is not productive.

We need the following Lemma.

Lemma 2.15**.**

There is an $\operatorname{add}(\mathrm{Menger})$-unbounded subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}}.

Proof.

Let $\kappa:=\operatorname{add}(\mathrm{Menger})$, and ${\langle\,M_{\alpha}:\alpha<\kappa\,\rangle}$ be an increasing sequence of Menger subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}} whose union is a dominating set. Fix an ordinal number $\alpha<\kappa$. The set $\{\,a_{\beta}:\beta<\alpha\,\}\cup M_{\alpha}$ is a union of two Menger sets, and thus it is not dominating in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. There is a function a_{\alpha}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that

[TABLE]

Let $A:=\{\,a_{\alpha}:\alpha<\kappa\,\}$, and b\in{[\mathbb{N}]^{\mbox{\tiny\infty}}}. There is an ordinal number $\alpha<\kappa$, and a function $d\in M_{\alpha}$ such that $b\leq^{*}d$. By the construction, we have $\{\,\beta<\kappa:a_{\beta}\leq^{*}b\,\}\subseteq\alpha$, and thus $\left|\{\,a\in A:a\leq^{*}b\,\}\right|<\kappa.$ ∎

Proof of Theorem 2.14.

The $\operatorname{add}(\mathrm{Menger})$-unbounded set, constructed in Lemma 2.15 has cardinality $\operatorname{add}(\mathrm{Menger})$, and thus it is Menger. Apply Theorem 2.12. ∎

2.4. Products of Menger sets in the Blass–Shelah model

By Theorem 2.1(2), assuming $\mathfrak{d}\leq\mathfrak{r}$, the Menger property is not productive. On the other hand, in the Miller model, where $\mathfrak{d}>\mathfrak{r}$, the Menger property is productive [33]. We show that the inequality $\mathfrak{d}\leq\mathfrak{r}$ is not necessary to prove that the Menger property is not productive. In this section, we shall work in the model constructed by Blass and Shelah [2] and use their notations. Let $\nu$ and $\delta$ be uncountable regular ordinal numbers in a model of ZFC $+$ GCH with $\nu<\delta$. Let $V(\delta,0)$ be the initial Cohen extension of the model of ZFC $+$ GCH, after adding $\delta$-many Cohen reals. For each ordinal number $\xi<\nu$, let $V(\delta,\xi)$ be the model obtained after $\xi$ stages in the Mathias forcing iteration with finite support, and $s_{\xi}$ be a Mathias real over the model $V(\delta,\xi)$ with respect to a certain ultrafilter $U_{\xi}$, in this model. Since $U_{\xi}$ is chosen to contain the set $\{\,s_{\zeta}:\zeta<\xi\,\}$ for all ordinal numbers $\zeta<\nu$, the sequence ${\langle\,s_{\xi}:\xi<\nu\,\rangle}$ is almost decreasing. Let $\mathfrak{u}$ be the minimal cardinality of a basis of an ultrafilter. In the model $V(\delta,\nu)$, the final extension of the model of ZFC $+$ GCH, we have $\mathfrak{r}=\mathfrak{u}=\nu$, $\mathfrak{d}=\delta$, and thus $\mathfrak{r}<\mathfrak{d}$ [2].

Theorem 2.16**.**

In the Blass–Shelah model $V(\delta,\nu)$, there are two sets of reals whose all finite powers are Menger, but their product space is not Menger.

Let $\mathfrak{b}$ be the minimal cardinality of a $\leq^{*}$-unbounded set in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. For each ordinal number $\xi<\nu$, let M_{\xi}:=V(\delta,\xi)\cap{[\mathbb{N}]^{\mbox{\tiny\infty}}}.

Lemma 2.17**.**

In the Blass–Shelah model, we have $\operatorname{add}(\mathrm{Menger})=\mathfrak{r}$. In particular, $\operatorname{add}(\mathrm{Menger})<\mathfrak{d}$.

Proof.

For each ordinal number $\xi<\nu$, the set $M_{\xi}$ is Scheepers in the model $V(\delta,\nu)$111This is not a direct application of [23, Theorem 11], but rather of its proof which gives: If $X$ is a ground model set of reals and $P$ is an iteration such that each real is contained in some intermediate model, and there is an unbounded real over every intermediate model, then $X$ is Menger in the final model.  [23, Theorem 11]. Thus, $\operatorname{add}(\mathrm{Menger})\leq\nu=\mathfrak{u}$. Since $\mathfrak{r}=\mathfrak{b}$ [14, Lemma 3.3] and $\mathfrak{b}\leq\operatorname{add}(\mathrm{Menger})$ [29, Corollary 2.2(2)], we have $\operatorname{add}(\mathrm{Menger})=\mathfrak{r}$. ∎

Remark 2.18*.*

Theorem 2.14 and Lemma 2.17 imply that, in the Blass–Shelah model, the Menger property is not productive.

Let $X$ be a space. A map \Phi\colon X\to{[\mathbb{N}]^{\mbox{\tiny\infty}}} is upper continuous if the sets $\{\,x\in X:\Phi(x)(n)\leq m\,\}$ are open for all natural numbers $n,m$.

Lemma 2.19**.**

Let $X$ be a space and \Phi\colon X\to{[\mathbb{N}]^{\mbox{\tiny\infty}}} be an upper continuous map. The map $\Phi$ is Borel.

Proof.

Let $n,m\in\mathbb{N}$. The set

[TABLE]

is a difference of two open sets, and thus is Borel. ∎

Proof of Theorem 2.16.

The family $\{\,[x<s_{\xi}]:x\in M_{\xi},\xi<\nu\,\}$ is centered: Fix a natural number $n$. Let ${\langle\,\xi_{i}:i\leq n\,\rangle}$ be a nondecreasing sequence of ordinal numbers smaller than $\nu$, and $x_{i}\in M_{\xi_{i}}$ for natural numbers $i\leq n$. For each ordinal number $\xi<\nu$, we have $\operatorname{maxfin}[M_{\xi}]=M_{\xi}$. Thus, assume that the sequence ${\langle\,\xi_{i}:i\leq n\,\rangle}$ is strictly increasing. For each ordinal number $\xi<\nu$, and functions $x,y\in M_{\xi}$, by the genericity of the function $s_{\xi}$, the set $[x<s_{\xi}]\cap y$ is infinite. Thus, the intersection $\bigcap_{i\leq n}[x_{i}<s_{\xi_{i}}]$ is infinite.

Let $U$ be an ultrafilter containing the family $\{\,[x<s_{\xi}]:x\in M_{\xi},\xi<\nu\,\}$. For each ordinal number $\xi<\nu,$ let $\tilde{M}_{\xi}$ be the set \{\,z_{\xi}(x):x\in M_{\xi}\,\}\subseteq{[\mathbb{N}]^{\mbox{\tiny\infty}}}, where $z_{\xi}(x)(n)=s_{\xi}(n)+1$ if $n\in x$ and $z_{\xi}(x)(n)=s_{\xi}(n)$ otherwise. Thus $\tilde{M}_{\xi}$ is a continuous image of $M_{\xi}$ under the map $x\mapsto z_{\xi}(x)$, and hence $\tilde{M}_{\xi}\in V_{\delta,\xi+1}$, which is going to be used later. Let $X:=\bigcup_{\xi<\nu}\tilde{M_{\xi}}$.

All finite powers of the set $X\cup\mathrm{Fin}$ are Menger, in the model $V(\delta,\nu)$: We prove a formally stronger statement that, for each natural number $n$, and each finite product $M$ of the sets $M_{\xi}$, the product space $(X\cup\mathrm{Fin})^{n}\times M$ is Menger. Each finite product of the sets $M_{\xi}$ is Menger [23, Theorem 11]. Proceed by induction. Fix a natural number $n$. Let $k$ be a natural number, and $\xi_{i}<\nu$ be ordinal numbers for natural numbers $i\leq k$. Let $M:=\prod_{i\leq k}M_{\xi_{i}}$, and \Psi\colon(X\cup\mathrm{Fin})^{n+1}\times M\to{[\mathbb{N}]^{\mbox{\tiny\infty}}} be a continuous map. There is an upper continuous map \Phi\colon M\to{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that, for elements $x=(x_{1},\dotsc,x_{n+1})\in(X\cup\mathrm{Fin})^{n+1}$, and $y=(y_{1},\dotsc,y_{k})\in M$, and a natural number $l$, we have:

If $\Phi(y)(l)<\min\{\,x_{j}(l):j\leq n+1\,\}$, then $\Psi(x,y)(l)<\Phi(y)(l)$ [25, Lemma 5.1].

By Lemma 2.19, the map $\Phi$ is Borel.

Let $\xi<\nu$ be an ordinal number with $\xi>\max\{\,\xi_{i}:i\leq k\,\}$ such that the map $\Phi$ is coded in the model $V(\delta,\xi)$, and $X_{\xi}:=\bigcup_{\xi\leq\alpha<\nu}\tilde{M_{\alpha}}$. We have $\Phi[M]\subseteq M_{\xi}$, and thus $[\Phi(y)<x]\in U$ for all elements $y\in M$ and $x\in X_{\xi}$. It follows that $\Psi[X_{\xi}^{n+1}\times M]\leq_{U}s_{\xi}$.

Fix an ordinal number $\alpha<\xi$. By the inductive assumption, the set $(X\cup\mathrm{Fin})^{n}\times M_{\alpha}\times M$ is Menger, and thus its continuous image $(X\cup\mathrm{Fin})^{n}\times\tilde{M_{\alpha}}\times M$ is Menger, too. The set

[TABLE]

is a union of less than $\left|\xi\right|$-many sets with the Menger property. Since $\xi<\operatorname{add}(\mathrm{Menger})$, by Lemma 2.17, the set $Y$ is Menger. Let b^{\prime}\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} be a function such that $\Psi[Y]\leq^{\infty}b^{\prime}$. Thus, $\Psi[(X\cup\mathrm{Fin})^{n+1}\times M]\leq^{\infty}\max\{s_{\xi},b^{\prime}\}$.

Let $S:=\{\,s_{\xi}:\xi<\nu\,\}$. Since $\left|S\right|<\mathfrak{d}$, all finite powers of the set $S$ are Menger. The map $(x,s)\mapsto\{\,n:x(n)\neq s(n)\,\}$ from the product space $X\times S$ to the set {[\mathbb{N}]^{\mbox{\tiny\infty}}}, is continuous. Moreover, it is also surjective: Given $x\in[\mathbb{N}]^{\infty}$, let $\xi$ be such that $x\in V(\delta,\xi)$, i.e., $x\in M_{\xi}$. Then $x=\{\,n:z_{\xi}(x)(n)\neq s_{\xi}(n)\,\}$ by our definition of $z_{\xi}(n)$. Thus, the product space $X\times S$ is not Menger, because $[\mathbb{N}]^{\infty}$ is not Menger and the Menger property is preserved by continuous images. ∎

2.5. Unions and products of Rothberger sets with strong properties

A space is Rothberger if, for each sequence $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dotsc$ of open covers of that space, 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 the space.

Theorem 2.20**.**

Assume that $\operatorname{cov}(\mathcal{M})=\mathfrak{d}$ and the cardinal number $\mathfrak{d}$ is regular.

1. (1)

There are two sets of reals whose all finite powers are Rothberger, but whose union is not Scheepers. 2. (2)

There are two sets of reals whose all finite powers are Rothberger, but whose product space is not Menger.

In order to prove Theorem 2.20, we need the following Lemma. Let $\kappa$ be an uncountable cardinal number. A space $X$ is $\kappa$-concentrated, if there is a countable set $D\subseteq X$ such that, for any open set $U\subseteq X$ containing $D$, we have $\left|X\setminus U\right|<\kappa$.

Lemma 2.21**.**

Let $U$ be an ultrafilter such that $\operatorname{\mathfrak{b}}(U)\leq\operatorname{cov}(\mathcal{M})$, and $X$ be a $U$-scale in {[\mathbb{N}]^{\mbox{\tiny\infty}}}. All finite powers of the set $X\cup\mathrm{Fin}$ are Rothberger.

Proof.

Every set of cardinality smaller than $\operatorname{cov}(\mathcal{M})$ is Rothberger [11, Theorem 4.2], and therefore so is every $\operatorname{cov}(\mathcal{M})$-concentrated set. Since $\operatorname{\mathfrak{b}}(U)\leq\operatorname{cov}(\mathcal{M})$, the set $X\cup\mathrm{Fin}$ has cardinality smaller than $\operatorname{cov}(\mathcal{M})$ or it is $\operatorname{cov}(\mathcal{M})$-concentrated [25, Proposition 4.3]. Thus, in both cases, the set $X\cup\mathrm{Fin}$ is Rothberger [19, Proposition 4]. Fix a natural number $n>1$, and assume that the set $(X\cup\mathrm{Fin})^{n-1}$ is Rothberger. Let $\mathcal{U}_{1},\allowbreak\mathcal{U}_{2},\allowbreak\dotsc$ be a sequence of open covers of the space $(X\cup\mathrm{Fin})^{n}$. For natural numbers $i,j\leq n$, let

[TABLE]

and $\pi_{i}$ be the projection from $(X\cup\mathrm{Fin})^{n}$ onto the $i$-th coordinate. A countable union of Rothberger sets is Rothberger. The set $Y:=\bigcup_{i\leq n}\prod_{j\leq n}Y^{i}_{j}$ is Rothberger, and thus there are sets $U_{1}\in\mathcal{U}_{1},U_{2}\in\mathcal{U}_{2},\dotsc$ such that the set $\bigcup_{k}U_{k}$ covers the set $Y$. The set $Y^{\prime}:=(X\cup\mathrm{Fin})^{n}\setminus\bigcup_{k}U_{k}$ is a closed subset of the $U\text{-Menger}$ set $(X\cup\mathrm{Fin})^{n}$ [25, Theorem 5.3(2)], and thus it is $U\text{-Menger}$, too. For each natural number $i\leq n$, as the projection $\pi_{i}[Y^{\prime}]$ is a $U\text{-Menger}$ subset of a $U$-scale $X$, it has cardinality less than $\operatorname{\mathfrak{b}}(U)$. Since $Y^{\prime}$ is a subset of the set $\prod_{i\leq n}\pi_{i}[Y^{\prime}]$, a set of cardinality less than $\operatorname{cov}(\mathcal{M})$, it is Rothberger [11, Theorem 4.2]. There are sets $U_{1}^{\prime}\in\mathcal{U}_{1},U_{2}^{\prime}\in\mathcal{U}_{2},\dotsc$ such that $Y^{\prime}\subseteq\bigcup_{k}U_{k}^{\prime}$. Finally, $(X\cup\mathrm{Fin})^{n}=\bigcup_{n}(U_{n}\cup U_{n}^{\prime})$, and thus the product space $(X\cup\mathrm{Fin})^{n}$ is Rothberger [30, Theorem A.1]. ∎

Proof of Theorem 2.20.

(1) Since the cardinal number $\mathfrak{d}$ is regular, there is an ultrafilter $U$ with $\operatorname{\mathfrak{b}}(U)=\mathfrak{d}$ [9]. By Lemma 2.6 and the fact that the near coherence relation is an equivalence relation, there is an ultrafilter ${\tilde{U}}$ non near coherent to the ultrafilter $U$ with $\operatorname{\mathfrak{b}}({\tilde{U}})=\mathfrak{d}$. Since $\operatorname{cov}(\mathcal{M})=\mathfrak{d}$ and $\operatorname{cov}(\mathcal{M})\leq\mathfrak{r}$ [7, Theorem 5.19], we have $\mathfrak{d}\leq\mathfrak{r}$. Let $X$ and $Y$ be a $U$-scale, and a ${\tilde{U}}$-scale, respectively, obtained from Lemma 2.10. We have $\operatorname{\mathfrak{b}}(U)=\operatorname{\mathfrak{b}}({\tilde{U}})=\operatorname{cov}(\mathcal{M})$. By Lemma 2.21, all finite powers of the sets $X\cup\mathrm{Fin}$ and $Y\cup\mathrm{Fin}$ are Rothberger. By Lemma 2.10(1), the union $(X\cup\mathrm{Fin})\cup(Y\cup\mathrm{Fin})$ is not Scheepers.

(2) Apply sets from (1) and Lemma 2.10(2).∎

Remark 2.22*.*

Assuming that $\operatorname{cov}(\mathcal{M})=\operatorname{cof}(\mathcal{M})$ and the cardinal number $\operatorname{cov}(\mathcal{M})$ is regular (a stronger assumption than used in Theorem 2.20), there are two hereditarily Rothberger sets satisfying the statement of Theorem 2.20 [27].

3. Applications to function spaces

Let $X$ be a space, and $\operatorname{C}_{\mathrm{p}}(X)$ be the space of continuous real-valued functions on $X$, with the topology of pointwise convergence. Properties of the space $X$ can describe local properties of the space $\operatorname{C}_{\mathrm{p}}(X)$, and vice versa. E.g., the space $\operatorname{C}_{\mathrm{p}}(X)$ is metrizable (or just first countable) if and only if the space $X$ is countable. We apply results from the previous sections to products of function spaces. A space $Y$ has countable fan tightness [1] if, for each point $y\in Y$ and for every sequence $U_{1},\allowbreak U_{2},\allowbreak\dotsc$ 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 $F_{1},\allowbreak F_{2},\allowbreak\dotsc$ are singletons, then the space $Y$ has countable strong fan tightness [21]. A space is M-separable [5] if, for every sequence $D_{1},\allowbreak D_{2},\allowbreak\dotsc$ 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 $D_{1},\allowbreak D_{2},\allowbreak\dotsc$ are singletons, then the space is R-separable [5].

Proposition 3.1**.**

**

1. (1)

Assume that $\mathfrak{d}\leq\mathfrak{r}$ and the cardinal number $\mathfrak{d}$ is regular. There are sets of reals $X$, $Y$ such that the spaces $\operatorname{C}_{\mathrm{p}}(X)$, $\operatorname{C}_{\mathrm{p}}(Y)$ have countable fan tightness (are M-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). 2. (2)

In the Blass–Shelah model, there are sets of reals $X$, $Y$ such that the spaces $\operatorname{C}_{\mathrm{p}}(X)$, $\operatorname{C}_{\mathrm{p}}(Y)$ have countable fan tightness (are M-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). 3. (3)

Assume that $\operatorname{cov}(\mathcal{M})=\mathfrak{d}$ and the cardinal number $\mathfrak{d}$ is regular. There are sets of reals $X$, $Y$ such that the spaces $\operatorname{C}_{\mathrm{p}}(X)$, $\operatorname{C}_{\mathrm{p}}(Y)$ have countable strong fan tightness (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).

To prove Proposition 3.1, we need the below result of Scheepers [22, Theorem 35, Theorem 13].

Theorem 3.2** (Scheepers [22]).**

Let $X$ be a set of reals.

1. (1)

The space $\operatorname{C}_{\mathrm{p}}(X)$ has countable fan tightness if and only if the space $\operatorname{C}_{\mathrm{p}}(X)$ is M-separable. 2. (2)

The space $\operatorname{C}_{\mathrm{p}}(X)$ has countable strong fan tightness if and only if the space $\operatorname{C}_{\mathrm{p}}(X)$ is R-separable.∎

Proof of Proposition 3.1.

(1) By Theorem 2.5, there are sets of reals $X$, $Y$ whose all finite powers are Menger but their products space $X\times Y$ is not Menger. Let $X\sqcup Y$ be the topological sum of these sets. Since the product space $X\times Y$ is a closed subspace of the product space $(X\sqcup Y)^{2}$, this product is not Menger. The spaces $\operatorname{C}_{\mathrm{p}}(X)$ and $\operatorname{C}_{\mathrm{p}}(Y)$ have countable fan tightness [11, Theorem 3.9]. The space $\operatorname{C}_{\mathrm{p}}(X\sqcup Y)$ is homeomorphic to the product space $\operatorname{C}_{\mathrm{p}}(X)\times\operatorname{C}_{\mathrm{p}}(Y)$, and thus it does not have countable strong fan tightness. For M-separability, apply Theorem 3.2(1).

(2) Proceed as in (1), with the exception that the sets $X$ and $Y$ are as in Theorem 2.16.

(3) By Theorem 2.20(3), there are sets of reals $X$ and $Y$ whose all finite powers are Rothberger, but their product space $X\times Y$ is not Menger. The spaces $\operatorname{C}_{\mathrm{p}}(X)$, $\operatorname{C}_{\mathrm{p}}(Y)$ have countable strong fan tightness [21]. As in (1) the product space $\operatorname{C}_{\mathrm{p}}(X)\times\operatorname{C}_{\mathrm{p}}(Y)$ does not have countable fan tightness. For R-separability and M-separability, apply Theorem 3.2. ∎

4. Comments and open problems

Let $\operatorname{cov}(\mathcal{N})$ be the minimal cardinality of a family of Lebesgue null subsets of {[\mathbb{N}]^{\mbox{\tiny\infty}}} that covers {[\mathbb{N}]^{\mbox{\tiny\infty}}}, and $\operatorname{cof}(\mathcal{N})$ be the minimal cardinality of a cofinal family of Lebesgue null sets, i.e., every Lebesgue null subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is contained in a member of the family. Assuming $\operatorname{cov}(\mathcal{N})=\mathfrak{b}=\operatorname{cof}(\mathcal{N})$, there is a Scheepers set of reals whose square is not Menger [24, Theorem 43]. The Scheepers sets constructed here are each $U$-Menger for some ultrafilter $U$, and their finite powers are Menger.

Problem 4.1**.**

Assume that $\mathfrak{d}\leq\mathfrak{r}$

1. (1)

Is there a Scheepers set of reals with a non-Menger square? 2. (2)

Is there a Scheepers set of reals such that, for each ultrafilter $U$, the set is not $U\text{-Menger}$?

By Remark 2.4, if the Menger and Scheepers properties are different, then the Menger property is not productive.

Problem 4.2**.**

Assume that the Menger and Scheepers properties are equivalent. Does it follow that the Menger property is productive?

By Theorem 2.16, in the Blass-Shelah model the Menger property is not productive. Thus, it is natural to examine whether the Menger and Scheepers properties differ in the Blass–Shelah model.

A subset of {[\mathbb{N}]^{\mbox{\tiny\infty}}} is a semifilter if it closed under almost supersets. Semifilter trichotomy is the statement: for each semifilter $S$, there is a function h\in{[\mathbb{N}]^{\mbox{\tiny\infty}}} such that the set $S/h$ is either the filter of cofinite subsets of $\mathbb{N}$, or an ultrafilter, or the full semifilter {[\mathbb{N}]^{\mbox{\tiny\infty}}}. Semifilter trichotomy is equivalent to the statement $\mathfrak{u}<\mathfrak{g}$. Semifilter trichotomy implies that the Menger and Scheepers properties are equivalent. [31, Theorem 3.7]. In the Miller model (where semifilter trichotomy holds), the Menger property is productive. These results motivate the following problem.

Problem 4.3**.**

Assume semifilter trichotomy. Is the Menger property productive?

By Theorems 2.1(2) and 2.14, if $\mathfrak{d}\leq\mathfrak{r}$ or $\operatorname{add}(\text{Menger})<\mathfrak{d}$, then the Menger property is not productive.

Problem 4.4**.**

Is the assumption $\mathfrak{d}\leq\mathfrak{r}$ or $\operatorname{add}(\text{Menger})<\mathfrak{d}$ necessary for the the Menger property being not productive?

4.1. Products of Menger topological groups

Proposition 4.5**.**

Assume that the cardinal number $\mathfrak{d}$ is regular.

1. (1)

If $\mathfrak{d}\leq\mathfrak{r}$, then there are two topological groups whose all finite powers are Menger but their product space is not Menger. 2. (2)

If $\operatorname{cov}(\mathcal{M})=\mathfrak{d}$, then there are two topological groups whose all finite powers are Rothberger but their product space is not Menger.

Proof.

(1) There is a homeomorphic copy $C\subseteq{\operatorname{P}(\mathbb{N})}$ of the Cantor space, that is linearly independent [18]. By Lemma 2.10, there are ultrafilters $U$, $\tilde{U}$, and subsets $X,Y$ of the set $C$ that are productively $U\text{-Menger}$, and productively ${\tilde{U}}\text{-Menger}$, respectively, whose product space $X\times Y$ is not Menger. For each natural number $n$, the set

[TABLE]

a continuous image of a productively $U\text{-Menger}$ set $X^{n}$, is productively $U\text{-Menger}$. Thus the group $\tilde{X}=\bigcup_{n}\tilde{X}_{n}$ is productively $U\text{-Menger}$, too. Analogously, construct a $Y$ that is productively ${\tilde{U}}\text{-Menger}$. Thus, all finite powers of the groups $X$ and $Y$ are Menger. Since the set $C$ is linearly independent, the product space $\tilde{X}\times\tilde{Y}=(X\cap C)\times(Y\cap C)$ is a closed subset of the product space $X\times Y$. Since the product $X\times Y$ is not Menger, the product space $\tilde{X}\times\tilde{Y}$ is not Menger, too.

(2) Proceed as in (1). By Lemma 2.21, all finite powers of the groups $X$, $Y$ are Rothberger. ∎

Problem 4.6**.**

Is there, consistently, a Menger topological group whose square is not Menger?

Acknowledgments

The first and second named authors thank the third named author for his kind hospitality at the Kurt Gödel Research Center in fall 2016. We also thank the Center’s Director, researchers and staff for the excellent academic and friendly atmosphere. The third author would like to thank the Austrian Science Fund FWF (Grants I 2374-N35 and I 3709-N35) for generous support for this research. We are also grateful to the anonymous referee for careful reading of the manuscript.