On the different kinds of separability of the space of Borel functions

TL;DR

This paper investigates various strong forms of separability in the space of Borel functions on real sets, analyzing their properties and implications.

Contribution

It introduces and studies new separability properties in Borel function spaces, extending understanding beyond traditional separability.

Findings

Identifies stronger separability conditions in B(X)

Analyzes implications of these properties for function space topology

Provides new insights into the structure of Borel function spaces

Abstract

In this paper we consider some properties of a space B(X) of Borel functions on a set of reals X, with pointwise topology, that are stronger than separability.

Full text

$\bullet$ a space of Borel functions $B(X)$ on a set of reals $X$, with pointwise topology, to be countably selective sequentially separable if and only if $X$ has the property $S_{1}(B_{\Gamma},B_{\Gamma})$;

$\bullet$ there exists a consistent example of sequentially separable selectively separable space which is not selective sequentially separable. This is an answer to the question of A. Bella, M. Bonanzinga and M. Matveev;

$\bullet$ there is a consistent example of a compact $T_{2}$ sequentially separable space which is not selective sequentially separable. This is an answer to the question of A. Bella and C. Costantini;

$\bullet$ $\min\{\mathfrak{b},\mathfrak{q}\}=\{\kappa:2^{\kappa}$ is not selective sequentially separable$\}$. This is a partial answer to the question of A. Bella, M. Bonanzinga and M. Matveev.

††journal: …

1 Introduction

In [12], Osipov and Pytkeev gave necessary and sufficient conditions for the space $B_{1}(X)$ of first Baire class functions on a Tychonoff space $X$, with pointwise topology, to be (strongly) sequentially separable. In this paper we consider some properties of a space $B(X)$ of Borel functions on a set of reals $X$, with pointwise topology, that are stronger than (sequentially) separability.

2 Main definitions and notation

Many topological properties are defined or characterized in terms of the following classical selection principles. Let $\mathcal{A}$ and $\mathcal{B}$ be sets consisting of families of subsets of an infinite set $X$. Then:

$S_{1}(\mathcal{A},\mathcal{B})$ is the selection hypothesis: for each sequence $(A_{n}:n\in\mathbb{N})$ of elements of $\mathcal{A}$ there is a sequence $(b_{n}:n\in\mathbb{N})$ such that for each $n$, $b_{n}\in A_{n}$, and $\{b_{n}:n\in\mathbb{N}\}$ is an element of $\mathcal{B}$.

$S_{fin}(\mathcal{A},\mathcal{B})$ is the selection hypothesis: for each sequence $(A_{n}:n\in\mathbb{N})$ of elements of $\mathcal{A}$ there is a sequence $(B_{n}:n\in\mathbb{N})$ of finite sets such that for each $n$, $B_{n}\subseteq A_{n}$, and $\bigcup_{n\in\mathbb{N}}B_{n}\in\mathcal{B}$.

$U_{fin}(\mathcal{A},\mathcal{B})$ is the selection hypothesis: whenever $\mathcal{U}_{1}$, $\mathcal{U}_{2},...\in\mathcal{A}$ and none contains a finite subcover, there are finite sets $\mathcal{F}_{n}\subseteq\mathcal{U}_{n}$, $n\in\mathbb{N}$, such that $\{\bigcup\mathcal{F}_{n}:n\in\mathbb{N}\}\in\mathcal{B}$.

An open cover $\mathcal{U}$ of a space $X$ is:

$\bullet$ an $\omega$-cover if $X$ does not belong to $\mathcal{U}$ and every finite subset of $X$ is contained in a member of $\mathcal{U}$.

$\bullet$ a $\gamma$-cover if it is infinite and each $x\in X$ belongs to all but finitely many elements of $\mathcal{U}$.

For a topological space $X$ we denote:

$\bullet$ $\Omega$ — the family of all countable open $\omega$-covers of $X$;

$\bullet$ $\Gamma$ — the family of all countable open $\gamma$-covers of $X$;

$\bullet$ $B_{\Omega}$ — the family of all countable Borel $\omega$-covers of $X$;

$\bullet$ $B_{\Gamma}$ — the family of all countable Borel $\gamma$-covers of $X$;

$\bullet$ $F_{\Gamma}$ — the family of all countable closed $\gamma$-covers of $X$;

$\bullet$ $\mathcal{D}$ — the family of all countable dense subsets of $X$;

$\bullet$ $\mathcal{S}$ — the family of all countable sequentially dense subsets of $X$.

A $\gamma$-cover $\mathcal{U}$ of co-zero sets of $X$ is $\gamma_{F}$-shrinkable if there exists a $\gamma$-cover $\{F(U):U\in\mathcal{U}\}$ of zero-sets of $X$ with $F(U)\subset U$ for every $U\in\mathcal{U}$.

For a topological space $X$ we denote:

$\bullet$ $\Gamma_{F}$ — the family of all countable $\gamma_{F}$-shrinkable $\gamma$-covers of $X$.

We will use the following notations.

$\bullet$ $C_{p}(X)$ is the set of all real-valued continuous functions $C(X)$ defined on a space $X$, with pointwise topology.

$\bullet$ $B_{1}(X)$ is the set of all first Baire class functions $B_{1}(X)$ i.e., pointwise limits of continuous functions, defined on a space $X$, with pointwise topology.

$\bullet$ $B(X)$ is the set of all Borel functions, defined on a space $X$, with pointwise topology.

If $X$ is a space and $A\subseteq X$, then the sequential closure of $A$, denoted by $[A]_{seq}$, is the set of all limits of sequences from $A$. A set $D\subseteq X$ is said to be sequentially dense if $X=[D]_{seq}$. If $D$ is a countable sequentially dense subset of $X$ then $X$ call sequentially separable space.

Call $X$ strongly sequentially separable if $X$ is separable and every countable dense subset of $X$ is sequentially dense.

A space $X$ is $($countably$)$ selectively separable (or $M$-separable, [3]) if for every sequence $(D_{n}:n\in\mathbb{N})$ of (countable) dense subsets of $X$ one can pick finite $F_{n}\subset D_{n}$, $n\in\mathbb{N}$, so that $\bigcup\{F_{n}:n\in\mathbb{N}\}$ is dense in $X$.

In [3] the authors started to investigate a selective version of sequential separability.

A space $X$ is $($countably$)$ selectively sequentially separable (or $M$-sequentially separable, [3]) if for every sequence $(D_{n}:n\in\mathbb{N})$ of (countable) sequentially dense subsets of $X$, one can pick finite $F_{n}\subset D_{n}$, $n\in\mathbb{N}$, so that $\bigcup\{F_{n}:n\in\mathbb{N}\}$ is sequentially dense in $X$.

In Scheeper’s terminology [16] countably selectively separability equivalently to the selection principle $S_{fin}(\mathcal{D},\mathcal{D})$, and countably selective sequentially separability equivalently to the $S_{fin}(\mathcal{S},\mathcal{S})$.

Recall that the cardinal $\mathfrak{p}$ is the smallest cardinal so that there is a collection of $\mathfrak{p}$ many subsets of the natural numbers with the strong finite intersection property but no infinite pseudo-intersection. Note that $\omega_{1}\leq\mathfrak{p}\leq\mathfrak{c}$.

For $f,g\in\mathbb{N}^{\mathbb{N}}$, let $f\leq^{*}g$ if $f(n)\leq g(n)$ for all but finitely many $n$. $\mathfrak{b}$ is the minimal cardinality of a $\leq^{*}$-unbounded subset of $\mathbb{N}^{\mathbb{N}}$. A set $B\subset[\mathbb{N}]^{\infty}$ is unbounded if the set of all increasing enumerations of elements of $B$ is unbounded in $\mathbb{N}^{\mathbb{N}}$, with respect to $\leq^{*}$. It follows that $|B|\geq\mathfrak{b}$. A subset $S$ of the real line is called a $Q$-set if each one of its subsets is a $G_{\delta}$. The cardinal $\mathfrak{q}$ is the smallest cardinal so that for any $\kappa<\mathfrak{q}$ there is a $Q$-set of size $\kappa$. (See [7] for more on small cardinals including $\mathfrak{p}$).

3 Properties of a space of Borel functions

Theorem 3.1**.**

For a set of reals $X$, the following statements are equivalent:

$B(X)$* satisfies $S_{1}(\mathcal{S},\mathcal{S})$ and $B(X)$ is sequentially separable;* 2. 2.

$X$* satisfies $S_{1}(B_{\Gamma},B_{\Gamma})$;* 3. 3.

$B(X)\in S_{fin}(\mathcal{S},\mathcal{S})$* and $B(X)$ is sequentially separable;* 4. 4.

$X$* satisfies $S_{fin}(B_{\Gamma},B_{\Gamma})$;* 5. 5.

$B_{1}(X)$* satisfies $S_{1}(\mathcal{S},\mathcal{S})$;* 6. 6.

$X$* satisfies $S_{1}(F_{\Gamma},F_{\Gamma})$;* 7. 7.

$B_{1}(X)$* satisfies $S_{fin}(\mathcal{S},\mathcal{S})$.*

Proof.

It is obviously that $(1)\Rightarrow(3)$.

$(2)\Leftrightarrow(4)$. By Theorem 1 in [15], $U_{fin}(B_{\Gamma},B_{\Gamma})=S_{1}(B_{\Gamma},B_{\Gamma})=S_{fin}(B_{\Gamma},B_{\Gamma})$.

$(3)\Rightarrow(2)$. Let $\{\mathcal{F}_{i}\}\subset B_{\Gamma}$ and $\mathcal{S}=\{h_{m}\}_{m\in\mathbb{N}}$ be a countable sequentially dense subset of $B(X)$. For each $i\in\mathbb{N}$ we consider a countable sequentially dense subset $\mathcal{S}_{i}$ of $B(X)$ and $\mathcal{F}_{i}=\{F^{m}_{i}\}_{m\in\mathbb{N}}$ where

$\mathcal{S}_{i}=\{f^{m}_{i}\}:=\{f^{m}_{i}\in B(X):f^{m}_{i}\upharpoonright F^{m}_{i}=h_{m}$ and $f^{m}_{i}\upharpoonright(X\setminus F^{m}_{i})=1$ for $m\in\mathbb{N}\}$.

Since $\mathcal{F}_{i}=\{F^{m}_{i}\}_{m\in\mathbb{N}}$ is a Borel $\gamma$-cover of $X$ and $\mathcal{S}$ is a countable sequentially dense subset of $B(X)$, we have that $\mathcal{S}_{i}$ is a countable sequentially dense subset of $B(X)$ for each $i\in\mathbb{N}$. Indeed, let $h\in B(X)$, there is a sequence $\{h_{s}\}_{s\in\mathbb{N}}\subset\mathcal{S}$ such that $\{h_{s}\}_{s\in\mathbb{N}}$ converge to $h$. We claim that $\{f^{s}_{i}\}_{s\in\mathbb{N}}$ converge to $h$. Let $K=\{x_{1},...,x_{k}\}$ be a finite subset of $X$, $\epsilon>0$ and let $W=\langle h,K,\epsilon\rangle:=\{g\in B(X):|g(x_{j})-h(x_{j})|<\epsilon$ for $j=1,...,k\}$ be a base neighborhood of $h$, then there is $m_{0}\in\mathbb{N}$ such that $K\subset F^{m}_{i}$ for each $m>m_{0}$ and $h_{s}\in W$ for each $s>m_{0}$. Since $f^{s}_{i}\upharpoonright K=h_{s}\upharpoonright K$ for every $s>m_{0}$, $f^{s}_{i}\in W$ for every $s>m_{0}$. It follows that $\{f^{s}_{i}\}_{s\in\mathbb{N}}$ converge to $h$.

Since $B(X)$ satisfies $S_{fin}(\mathcal{S},\mathcal{S})$, there is a sequence $(F_{i}=\{f^{m_{1}}_{i},...,f^{m_{s(i)}}_{i}\}:i\in\mathbb{N})$ such that for each $i$, $F_{i}\subset\mathcal{S}_{i}$, and $\bigcup_{i\in\mathbb{N}}F_{i}$ is a countable sequentially dense subset of $B(X)$.

For ${\bf 0}\in B(X)$ there is a sequence $\{f^{m_{s(i_{j})}}_{i_{j}}\}_{j\in\mathbb{N}}\subset\bigcup_{i\in\mathbb{N}}F_{i}$ such that $\{f^{m_{s(i_{j})}}_{i_{j}}\}_{j\in\mathbb{N}}$ converge to ${\bf 0}$. Consider a sequence $(F^{m_{s(i_{j})}}_{i_{j}}:j\in\mathbb{N})$.

(1) $F^{m_{s(i_{j})}}_{i_{j}}\in\mathcal{F}_{i_{j}}$.

(2) $\{F^{m_{s(i_{j})}}_{i_{j}}:j\in\mathbb{N}\}$ is a $\gamma$-cover of $X$.

Indeed, let $K$ be a finite subset of $X$ and $U=\langle{\bf 0},K,\frac{1}{2}\rangle$ be a base neighborhood of ${\bf 0}$, then there is $j_{0}\in\mathbb{N}$ such that $f^{m_{s(i_{j})}}_{i_{j}}\in U$ for every $j>j_{0}$. It follows that $K\subset F^{m_{s(i_{j})}}_{i_{j}}$ for every $j>j_{0}$. We thus get $X$ satisfies $U_{fin}(B_{\Gamma},B_{\Gamma})$, and, hence, by Theorem 1 in [15], $X$ satisfies $S_{1}(B_{\Gamma},B_{\Gamma})$.

$(2)\Rightarrow(1)$. Let $\{S_{i}\}\subset\mathcal{S}$ and $S=\{d_{n}:n\in\mathbb{N}\}\in\mathcal{S}$. Consider the topology $\tau$ generated by the family $\mathcal{P}=\{f^{-1}(G):G$ is an open set of $\mathbb{R}$ and $f\in S\cup\bigcup\limits_{i\in\mathbb{N}}S_{i}\}$. Since $P=S\cup\bigcup\limits_{i\in\mathbb{N}}S_{i}$ is a countable dense subset of $B(X)$ and $X$ is Tychonoff, we have that the space $Y=(X,\tau)$ is a separable metrizable space. Note that a function $f\in P$, considered as mapping from $Y$ to $\mathbb{R}$, is a continuous function i.e. $f\in C(Y)$ for each $f\in P$. Note also that an identity map $\varphi$ from $X$ on $Y$, is a Borel bijection. By Corollary 12 in [6], $Y$ is a $QN$-space and, hence, by Corollary 20 in [17], $Y$ has the property $S_{1}(B_{\Gamma},B_{\Gamma})$. By Corollary 21 in [17], $B(Y)$ is an $\alpha_{2}$ space.

Let $q:\mathbb{N}\mapsto\mathbb{N}\times\mathbb{N}$ be a bijection. Then we enumerate $\{S_{i}\}_{i\in\mathbb{N}}$ as $\{S_{q(i)}\}_{q(i)\in\mathbb{N}\times\mathbb{N}}$. For each $d_{n}\in S$ there are sequences $s_{n,m}\subset S_{n,m}$ such that $s_{n,m}$ converges to $d_{n}$ for each $m\in\mathbb{N}$. Since $B(Y)$ is an $\alpha_{2}$ space, there is $\{b_{n,m}:m\in\mathbb{N}\}$ such that for each $m$, $b_{n,m}\in s_{n,m}$, and, $b_{n,m}\rightarrow d_{n}$ $(m\rightarrow\infty)$. Let $B=\{b_{n,m}:n,m\in\mathbb{N}\}$. Note that $S\subset[B]_{seq}$.

Since $X$ is a $\sigma$-set (that is, each Borel subset of $X$ is $F_{\sigma}$)(see [17]), $B_{1}(X)=B(X)$ and $\varphi(B(Y))=\varphi(B_{1}(Y))\subseteq B(X)$ where $\varphi(B(Y)):=\{p\circ\varphi:p\in B(Y)\}$ and $\varphi(B_{1}(Y)):=\{p\circ\varphi:p\in B_{1}(Y)\}$.

Since $S$ is a countable sequentially dense subset of $B(X)$, for any $g\in B(X)$ there is a sequence $\{g_{n}\}_{n\in\mathbb{N}}\subset S$ such that $\{g_{n}\}_{n\in\mathbb{N}}$ converge to $g$. But $g$ we can consider as a mapping from $Y$ into $\mathbb{R}$ and a set $\{g_{n}:n\in\mathbb{N}\}$ as subset of $C(Y)$. It follows that $g\in B_{1}(Y)$. We get that $\varphi(B(Y))=B(X)$.

We claim that $B\in\mathcal{S}$, i.e. that $[B]_{seq}=B(X)$. Let $f\in B(Y)$ and $\{f_{k}:k\in\mathbb{N}\}\subset S$ such that $f_{k}\rightarrow f$ ($k\rightarrow\infty$). For each $k\in\mathbb{N}$ there is $\{f^{n}_{k}:n\in\mathbb{N}\}\subset B$ such that $f^{n}_{k}\rightarrow f_{k}$ ($n\rightarrow\infty$). Since $Y$ is a $QN$-space (Theorem 16 in [6]), there exists an unbounded $\beta\in\mathbb{N}^{\mathbb{N}}$ such that $\{f^{\beta(k)}_{k}\}$ converge to $f$ on $Y$. It follows that $\{f^{\beta(k)}_{k}:k\in\mathbb{N}\}$ converge to $f$ on $X$ and $[B]_{seq}=B(X)$.

$(5)\Rightarrow(6)$. By Velichko’s Theorem ([18]), a space $B_{1}(X)$ is sequentially separable for any separable metric space $X$.

Let $\{\mathcal{F}_{i}\}\subset F_{\Gamma}$ and $\mathcal{S}=\{h_{m}\}_{m\in\mathbb{N}}$ be a countable sequentially dense subset of $B_{1}(X)$.

Similarly implication $(3)\Rightarrow(2)$ we get $X$ satisfies $U_{fin}(F_{\Gamma},F_{\Gamma})$, and, hence, by Lemma 13 in [17], $X$ satisfies $S_{1}(F_{\Gamma},F_{\Gamma})$.

$(6)\Rightarrow(5)$. By Corollary 20 in [17], $X$ satisfies $S_{1}(B_{\Gamma},B_{\Gamma})$. Since $X$ is a $\sigma$-set (see [17]), $B_{1}(X)=B(X)$ and, by implication $(2)\Rightarrow(1)$, we get $B_{1}(X)$ satisfies $S_{1}(\mathcal{S},\mathcal{S})$.

∎

In [16] (Theorem 13) M. Scheepers proved the following result

Theorem 3.2**.**

(Scheepers) For $X$ a separable metric space, the following are equivalent:

$C_{p}(X)$* satisfies $S_{1}(\mathcal{D},\mathcal{D})$;* 2. 2.

$X$* satisfies $S_{1}(\Omega,\Omega)$.*

We claim the theorem for a space $B(X)$ of Borel functions.

Theorem 3.3**.**

For a set of reals $X$, the following are equivalent:

$B(X)$* satisfies $S_{1}(\mathcal{D},\mathcal{D})$;* 2. 2.

$X$* satisfies $S_{1}(B_{\Omega},B_{\Omega})$.*

Proof.

$(1)\Rightarrow(2)$. Let $X$ be a set of reals satisfying the hypotheses and $\beta$ be a countable base of $X$. Consider a sequence $\{\mathcal{B}_{i}\}_{i\in\mathbb{N}}$ of countable Borel $\omega$-covers of $X$ where $\mathcal{B}_{i}=\{W^{j}_{i}\}_{j\in\mathbb{N}}$ for each $i\in\mathbb{N}$.

Consider a topology $\tau$ generated by the family $\mathcal{P}=\{W^{j}_{i}\cap A:i,j\in\mathbb{N}$ and $A\in\beta\}\bigcup\{(X\setminus W^{j}_{i})\cap A:i,j\in\mathbb{N}$ and $A\in\beta\}$.

Note that if $\chi_{P}$ is a characteristic function of $P$ for each $P\in\mathcal{P}$, then a diagonal mapping $\varphi=\Delta_{P\in\mathcal{P}}\chi_{P}:X\mapsto 2^{\omega}$ is a Borel bijection. Let $Z=\varphi(X)$.

Note that $\{\mathcal{B}_{i}\}$ is countable open $\omega$-cover of $Z$ for each $i\in\mathbb{N}$. Since $B(Z)$ is a dense subset of $B(X)$, then $B(Z)$ also has the property $S_{1}(\mathcal{D},\mathcal{D})$. Since $C_{p}(Z)$ is a dense subset of $B(Z)$, $C_{p}(Z)$ has the property $S_{1}(\mathcal{D},\mathcal{D})$, too.

By Theorem 3.2, the space $Z$ has the property $S_{1}(\Omega,\Omega)$. It follows that there is a sequence $\{W^{j(i)}_{i}\}_{i\in\mathbb{N}}$ such that $W^{j(i)}_{i}\in\mathcal{B}_{i}$ and $\{W^{j(i)}_{i}:i\in\mathbb{N}\}$ is an open $\omega$-cover of $Z$. It follows that $\{W^{j(i)}_{i}:i\in\mathbb{N}\}$ is Borel $\omega$-cover of $X$.

$(2)\Rightarrow(1)$. Assume that $X$ has the property $S_{1}(B_{\Omega},B_{\Omega})$. Let $\{D_{k}\}_{k\in\mathbb{N}}$ be a sequence countable dense subsets of $B(X)$ and $D_{k}=\{f^{k}_{i}:i\in\mathbb{N}\}$ for each $k\in\mathbb{N}$. We claim that for any $f\in B(X)$ there is a sequence $\{f_{k}\}\subset B(X)$ such that $f_{k}\in D_{k}$ for each $k\in\mathbb{N}$ and $f\in\overline{\{f_{k}:k\in\mathbb{N}\}}$. Without loss of generality we can assume $f=\bf{0}$. For each $f^{k}_{i}\in D_{k}$ let $W^{k}_{i}=\{x\in X:-\frac{1}{k}<f^{k}_{i}(x)<\frac{1}{k}\}$.

If for each $j\in\mathbb{N}$ there is $k(j)$ such that $W^{k(j)}_{i(j)}=X$, then a sequence $f_{k(j)}=f^{k(j)}_{i(j)}$ uniformly converge to $f$ and, hence, $f\in\overline{\{f_{k(j)}:j\in\mathbb{N}\}}$.

We can assume that $W^{k}_{i}\neq X$ for any $k,i\in\mathbb{N}$.

(a). $\{W^{k}_{i}\}_{i\in\mathbb{N}}$ a sequence of Borel sets of $X$.

(b). For each $k\in\mathbb{N}$, $\{W^{k}_{i}:i\in\mathbb{N}\}$ is a $\omega$-cover of $X$.

By (2), $X$ has the property $S_{1}(B_{\Omega},B_{\Omega})$, hence, there is a sequence $\{W^{k}_{i(k)}\}_{k\in\mathbb{N}}$ such that $W^{k}_{i(k)}\in\{W^{k}_{i}\}_{i\in\mathbb{N}}$ for each $k\in\mathbb{N}$ and $\{W^{k}_{i(k)}\}_{k\in\mathbb{N}}$ is a $\omega$-cover of $X$.

Consider $\{f^{k}_{i(k)}\}$. We claim that $f\in\overline{\{f^{k}_{i(k)}:k\in\mathbb{N}\}}$. Let $K$ be a finite subset of $X$, $\epsilon>0$ and $U=\langle f,K,\epsilon\rangle$ be a base neighborhood of $f$, then there is $k_{0}\in\mathbb{N}$ such that $\frac{1}{k_{0}}<\epsilon$ and $K\subset W^{k_{0}}_{i(k_{0})}$. It follows that $f^{k_{0}}_{i(k_{0})}\in U$.

Let $D=\{d_{n}:n\in\mathbb{N}\}$ be a dense subspace of $B(X)$. Given a sequence $\{D_{i}\}_{i\in\mathbb{N}}$ of dense subspace of $B(X)$, enumerate it as $\{D_{n,m}:n,m\in\mathbb{N}\}$. For each $n\in\mathbb{N}$, pick $d_{n,m}\in D_{n,m}$ so that $d_{n}\in\overline{\{d_{n,m}:m\in\mathbb{N}\}}$. Then $\{d_{n,m}:m,n\in\mathbb{N}\}$ is dense in $B(X)$.

∎

In [16] (Theorem 35) and [4] (Corollary 2.10) proved the following result

Theorem 3.4**.**

(Scheepers) For $X$ a separable metric space, the following are equivalent:

$C_{p}(X)$* satisfies $S_{fin}(\mathcal{D},\mathcal{D})$;* 2. 2.

$X$* satisfies $S_{fin}(\Omega,\Omega)$.*

Then for the space $B(X)$ we have analogous result.

Theorem 3.5**.**

For a set of reals $X$, the following are equivalent:

$B(X)$* satisfies $S_{fin}(\mathcal{D},\mathcal{D})$;* 2. 2.

$X$* satisfies $S_{fin}(B_{\Omega},B_{\Omega})$.*

Proof.

It is proved similarly to the proof of Theorem 3.3. ∎

4 Question of A. Bella, M. Bonanzinga and M. Matveev

In [3], Question 4.3, it is asked to find a sequentially separable selectively separable space which is not selective sequentially separable.

The following theorem answers this question.

Theorem 4.1**.**

$(CH)$* There is a consistent example of a space $Z$, such that $Z$ is sequentially separable selectively separable, not selective sequentially separable.*

Proof.

By Theorem 40 and Corollary 41 in [15], there is a $\mathfrak{c}$-Lusin set $X$ which has the property $S_{1}(B_{\Omega},B_{\Omega})$, but $X$ does not have the property $U_{fin}(\Gamma,\Gamma)$.

Consider a space $Z=C_{p}(X)$.

By Velichko’s Theorem ([18]), a space $C_{p}(X)$ is sequentially separable for any separable metric space $X$.

(a). $Z$ is sequentially separable.

Since $X$ is Lindel$\ddot{o}$f and $X$ satisfies $S_{1}(B_{\Omega},B_{\Omega})$, $X$ has the property $S_{1}(\Omega,\Omega)$.

By Theorem 3.2, $C_{p}(X)$ satisfies $S_{1}(\mathcal{D},\mathcal{D})$, and, hence, $C_{p}(X)$ satisfies $S_{fin}(\mathcal{D},\mathcal{D})$.

(b). $Z$ is selectively separable.

By Theorem 4.1 in [11], $U_{fin}(\Gamma,\Gamma)=U_{fin}(\Gamma_{F},\Gamma)$ for Lindel$\ddot{o}$f spaces.

Since $X$ does not have the property $U_{fin}(\Gamma,\Gamma)$, $X$ does not have the property $S_{fin}(\Gamma_{F},\Gamma)$. By Theorem 8.11 in [9], $C_{p}(X)$ does not have the property $S_{fin}(\mathcal{S},\mathcal{S})$.

(c). $Z$ is not selective sequentially separable.

∎

Theorem 4.2**.**

$(CH)$* There is a consistent example of a space $Z$, such that $Z$ is countable sequentially separable selectively separable, not countable selective sequentially separable.*

Proof.

Consider the $\mathfrak{c}$-Lusin set $X$ (see Theorem 40 and Corollary 41 in [15]), then $X$ has the property $S_{1}(B_{\Omega},B_{\Omega})$, but $X$ does not have the property $U_{fin}(\Gamma,\Gamma)$ and, hence, $X$ does not have the property $S_{fin}(B_{\Gamma},B_{\Gamma})$.

Consider a space $Z=B_{1}(X)$.

By Velichko’s Theorem in [18], a space $B_{1}(X)$ is sequentially separable for any separable metric space $X$.

(a). $Z$ is sequentially separable.

By Theorem 3.3, $B(X)$ satisfies $S_{1}(\mathcal{D},\mathcal{D})$. Since $Z$ is dense subset of $B(X)$ we have that $Z$ satisfies $S_{1}(\mathcal{D},\mathcal{D})$ and, hence, $Z$ satisfies $S_{fin}(\mathcal{D},\mathcal{D})$.

(b). $Z$ is countable selectively separable.

Since $X$ does not have the property $S_{fin}(B_{\Gamma},B_{\Gamma})$, by Theorem 3.1, $B_{1}(X)$ does not have the property $S_{fin}(\mathcal{S},\mathcal{S})$.

(c). $Z$ is not countable selective sequentially separable.

∎

5 Question of A. Bella and C.

Costantini.

In [5], Question 2.7, it is asked to find a compact $T_{2}$ sequentially separable space which is not selective sequentially separable.

The following theorem answers this question.

Theorem 5.1**.**

$(\mathfrak{b}<\mathfrak{q})$* There is a consistent example of a compact $T_{2}$ sequentially separable space which is not selective sequentially separable.*

Proof.

Let $D$ be a discrete space of size $\mathfrak{b}$. Since $\mathfrak{b}<\mathfrak{q}$, a space $2^{\mathfrak{b}}$ is sequentially separable (see Proposition 3 in [13]).

We claim that $2^{\mathfrak{b}}$ is not selective sequentially separable.

On the contrary, suppose that $2^{\mathfrak{b}}$ is selective sequentially separable. Since $non(S_{fin}(B_{\Gamma},B_{\Gamma}))=\mathfrak{b}$ (see Theorem 1 and Theorem 27 in [15]), there is a set of reals $X$ such that $|X|=\mathfrak{b}$ and $X$ does not have the property $S_{fin}(B_{\Gamma},B_{\Gamma})$. Hence there exists sequence $(A_{n}:n\in\mathbb{N})$ of elements of $B_{\Gamma}$ that for any sequence $(B_{n}:n\in\mathbb{N})$ of finite sets such that for each $n$, $B_{n}\subseteq A_{n}$, we have that $\bigcup_{n\in\mathbb{N}}B_{n}\notin B_{\Gamma}$.

Consider an identity mapping $id:D\mapsto X$ from the space $D$ onto the space $X$. Denote $C^{i}_{n}=id^{-1}(A^{i}_{n})$ for each $A^{i}_{n}\in A_{n}$ and $n,i\in\mathbb{N}$. Let $C_{n}=\{C^{i}_{n}\}_{i\in\mathbb{N}}$ (i.e. $C_{n}=id^{-1}(A_{n})$) and let $\mathcal{S}=\{h_{i}\}_{i\in\mathbb{N}}$ be a countable sequentially dense subset of $B(D,\{0,1\})=2^{\mathfrak{b}}$.

For each $n\in\mathbb{N}$ we consider a countable sequentially dense subset $\mathcal{S}_{n}$ of $B(D,\{0,1\})$ where

$\mathcal{S}_{n}=\{f^{i}_{n}\}:=\{f^{i}_{n}\in B(D,2):f^{i}_{n}\upharpoonright C^{i}_{n}=h_{i}$ and $f^{i}_{n}\upharpoonright(X\setminus C^{i}_{n})=1$ for $i\in\mathbb{N}\}$.

Since $C_{n}=\{C^{i}_{n}\}_{i\in\mathbb{N}}$ is a Borel $\gamma$-cover of $D$ and $\mathcal{S}$ is a countable sequentially dense subset of $B(D,\{0,1\})$, we have that $\mathcal{S}_{n}$ is a countable sequentially dense subset of $B(D,\{0,1\})$ for each $n\in\mathbb{N}$.

Indeed, let $h\in B(D,\{0,1\})$, there is a sequence $\{h_{s}\}_{s\in\mathbb{N}}\subset\mathcal{S}$ such that $\{h_{s}\}_{s\in\mathbb{N}}$ converge to $h$. We claim that $\{f^{s}_{n}\}_{s\in\mathbb{N}}$ converge to $h$. Let $K=\{x_{1},...,x_{k}\}$ be a finite subset of $D$, $\epsilon=\{\epsilon_{1},...,\epsilon_{k}\}$ where $\epsilon_{j}\in\{0,1\}$ for $j=1,...,k$, and $W=\langle h,K,\epsilon\rangle:=\{g\in B(D,\{0,1\}):|g(x_{j})-h(x_{j})|\in\epsilon_{j}$ for $j=1,...,k\}$ be a base neighborhood of $h$, then there is a number $m_{0}$ such that $K\subset C^{i}_{n}$ for $i>m_{0}$ and $h_{s}\in W$ for $s>m_{0}$. Since $f^{s}_{n}\upharpoonright K=h_{s}\upharpoonright K$ for each $s>m_{0}$, $f^{s}_{n}\in W$ for each $s>m_{0}$. It follows that a sequence $\{f^{s}_{n}\}_{s\in\mathbb{N}}$ converge to $h$.

Since $B(D,\{0,1\})$ is selective sequentially separable, there is a sequence $\{F_{n}=\{f^{i_{1}}_{n},...,f^{i_{s(n)}}_{n}\}:n\in\mathbb{N}\}$ such that for each $n$, $F_{n}\subset\mathcal{S}_{n}$, and $\bigcup_{n\in\mathbb{N}}F_{n}$ is a countable sequentially dense subset of $B(D,\{0,1\})$.

For $\bf{0}$ $\in B(D,\{0,1\})$ there is a sequence $\{f^{i_{j}}_{n_{j}}\}_{j\in\mathbb{N}}\subset\bigcup_{n\in\mathbb{N}}F_{n}$ such that $\{f^{i_{j}}_{n_{j}}\}_{j\in\mathbb{N}}$ converge to $\bf{0}$. Consider a sequence $\{C^{i_{j}}_{n_{j}}:j\in\mathbb{N}\}$.

(1) $C^{i_{j}}_{n_{j}}\in C_{n_{j}}$.

(2) $\{C^{i_{j}}_{n_{j}}:j\in\mathbb{N}\}$ is a $\gamma$-cover of $D$.

Indeed, let $K$ be a finite subset of $D$ and $U=\langle{\bf 0},K,\{0\}\rangle$ be a base neighborhood of ${\bf 0}$, then there is a number $j_{0}$ such that $f^{i_{j}}_{n_{j}}\in U$ for every $j>j_{0}$. It follows that $K\subset C^{i_{j}}_{n_{j}}$ for every $j>j_{0}$. Hence, $\{A^{i_{j}}_{n_{j}}=id(C^{i_{j}}_{n_{j}}):j\in\mathbb{N}\}\in B_{\Gamma}$ in the space $X$, a contradiction.

∎

Let $\mu=\min\{\kappa:2^{\kappa}$ is not selective sequentially separable$\}$. It is well-known that $\mathfrak{p}\leq\mu\leq\mathfrak{q}$ (see [3]).

Theorem 5.2**.**

$\mu=\min\{\mathfrak{b},\mathfrak{q}\}$.

Proof.

Let $\kappa<\min\{\mathfrak{b},\mathfrak{q}\}$. Then, by Proposition 3 in [13], $2^{\kappa}$ is a sequentially separable space.

Let $X$ be a set of reals such that $|X|=\kappa$ and $X$ be a $Q$-set.

Analogously proof of implication $(2)\Rightarrow(1)$ in Theorem 3.1, we can claim that $B(X,\{0,1\})=2^{X}=2^{\kappa}$ is selective sequentially separable.

It follows that $\mu\geq\min\{\mathfrak{b},\mathfrak{q}\}$.

Since $\mu\leq\mathfrak{q}$ we suppose that $\mu>\mathfrak{b}$ and $\mathfrak{b}<\mathfrak{q}$. Then, by Theorem 5.1, $2^{\mathfrak{b}}$ is not selective sequentially separable. It follows that $\mu=\min\{\mathfrak{b},\mathfrak{q}\}$. ∎

In [3], Question 4.12 : is it the case $\mu\in\{\mathfrak{p},\mathfrak{q}\}$ ?

A partial positive answer to this question is the existence of the following models of set theory (Theorem 8 in [1]):

1. $\mu=\mathfrak{p}=\mathfrak{b}<\mathfrak{q}$;

2. $\mathfrak{p}<\mu=\mathfrak{b}=\mathfrak{q}$.

and

3. $\mu=\mathfrak{p}=\mathfrak{q}<\mathfrak{b}$.

The author does not know whether, in general, the answer can be negative. In this regard, the following question is of interest.

Question. Is there a model of set theory in which $\mathfrak{p}<\mathfrak{b}<\mathfrak{q}$ ?