Selective sequential pseudocompactness

TL;DR

This paper introduces the class of selectively sequentially pseudocompact (SSP) spaces, explores their properties, closure under various operations, and constructs examples of SSP group topologies, expanding understanding of generalized compactness.

Contribution

It defines the SSP class, proves its closure properties, relates it to existing classes, and constructs new SSP group topologies with many generators.

Findings

SSP spaces are closed under products and continuous images.

All omega-bounded groups are SSP, but not all compact spaces are.

Constructed SSP group topologies on free and free Abelian groups with continuum-many generators.

Abstract

We say that a topological space X is selectively sequentially pseudocompact (SSP for short) if for every sequence (U_n) of non-empty open subsets of X, one can choose a point x_n in U_n for every n in such a way that the sequence (x_n) has a convergent subsequence. We show that the class of SSP spaces is closed under taking arbitrary products and continuous images, contains the class of all dyadic spaces and forms a proper subclass of the class of strongly pseudocompact spaces introduced recently by Garc\'ia-Ferreira and Ortiz-Castillo. We investigate basic properties of this new class and its relations with known compactness properties. We prove that every omega-bounded (=the closure of which countable set is compact) group is SSP, while compact spaces need not be SSP. Finally, we construct SSP group topologies on both the free group and the free Abelian group with continuum-many generators.

Equations

$$\text{the set }N(W)=\{n\in \mathbb{N}:W\cap U_n\ne \varnothing \}\text{ is infinite for every open neighborhood }W\text{ of }x.$$

$$x\notin U_n\text{ for every }n\in \mathbb{N}\text{ satisfying }n\ge n_{-1}.$$

$$V(k)\subseteq U(s(k))=(U\circ s)(k)\text{ for every }k\in \mathbb{N}.$$

$$\mathcal{B}=\{N_W:W\text{ is an open neighborhood of }x\}\cup \{\mathbb{N}\setminus F:F\text{ is a finite subset of }\mathbb{N}\}$$

$$\Sigma (p,X)=\{f\in X:|\{i\in I:f(i)\ne p(i)\}|\le \omega \}$$

$$r_{\mathcal{V}}(G)=\sup \{|X|:X\text{ is a }\mathcal{V}\text{-independent subset of }G\}$$

$$R=\{r_{\alpha ,n}:\alpha <\mathfrak{c},n\in \mathbb{N}\}\subseteq L$$

$$\text{the sequence }\{r_{\alpha ,n}:n\in \mathbb{N}\}\text{ converges to }{r}_{\alpha ,\omega }\text{ for every }\alpha <\mathfrak{c}.$$

$$\mathcal{U}=\{\{U_{\alpha ,n}:n\in \mathbb{N}\}:\alpha <\mathfrak{c}\},$$

$$S_{\alpha }=\bigcup _{n\in \mathbb{N}}\mathrm{supp}(U_{\alpha ,n})$$

$$U_{\alpha ,n}=p_{\alpha }^{\leftarrow }(W_n).$$

$$y_{\alpha,n}(\beta)=\left\{\begin{array}[]{ll}x_{\alpha,n}(\beta)&\mbox{if }\beta\in S_{\alpha}\\ r_{\alpha,n}&\mbox{if }\beta\in\mathfrak{c}\setminus S_{\alpha}\end{array}\right.\ \ \ \mbox{for every }\beta<\mathfrak{c}.$$

$$X\subseteq \{y_{\alpha ,n}:(\alpha ,n)\in F\times (\omega +1)\}.$$

$$\{{\pi }_{\beta }(y_{\alpha ,n}):(\alpha ,n)\in F\times (\omega +1)\}=\{r_{\alpha ,n}:(\alpha ,n)\in F\times (\omega +1)\}.$$

Full text

The first listed author was supported by CONACyT of México: Estancias Posdoctorales al Extranjero propuesta No. 263464

The second listed author was partially supported by the Grant-in-Aid for Scientific Research (C) No. 26400091 by the Japan Society for the Promotion of Science (JSPS)

All topological spaces are considered to be Tychonoff and all topological groups Hausdorff.

The symbol $\mathbb{N}$ denotes the set of natural numbers, $\mathbb{R}$ denotes the set of real numbers, $\mathbb{T}$ denotes the circle group $\{e^{i\theta}:\theta\in\mathbb{R}\}\subseteq\mathbb{R}^{2}.$

If $f:X\to Y$ is a function and $A$ is a subset of $Y$, then we let $f^{\leftarrow}(A)=\{x\in X:f(x)\in A\}$.

For a subset $A$ of a topological space $X$, we use cl${}_{X}(A)$ to denote the closure of $A$ in $X$.

Let us recall two well-known compactness-type properties.

• (i)

sequentially compact if every sequence in $X$ has a convergent subsequence;

• (ii)

pseudocompact if every continuous real-valued function defined on $X$ is bounded;

The notion of sequential compactness is well known; see, for example, [9, Section 3.10]. The notion of pseudocompactness is due to Hewitt [13].

• •

point-finite provided that the set $\{n\in\mathbb{N}:x\in U_{n}\}$ is finite for each point $x\in X$;

• •

locally finite provided that each point $x\in X$ has an open neighborhood $W$ such that the set $N(W)=\{n\in\mathbb{N}:W\cap U_{n}\not=\emptyset\}$ is finite.

Clearly, every locally finite sequence is point-finite.

The following fact is well known.

• (i)

$X$* is pseudocompact;*

• (ii)

every sequence of non-empty open subsets of $X$ is not locally finite;

• (ii)

every sequence of pairwise disjoint non-empty open subsets of $X$ is not locally finite.

1. Disjoint refinement principle

The next lemma is probably known. We include its proof here only for the reader’s convenience.

Lemma 1.1**.**

Let $\mathscr{U}=\{U_{n}:n\in\mathbb{N}\}$ be a point-finite family of non-empty open subsets of a topological space $X$ which is not locally finite. Then there exist a sequence $\mathscr{V}=\{V_{k}:k\in\mathbb{N}\}$ of pairwise disjoint non-empty open subsets of $X$ and a strictly increasing sequence $\{n_{k}:k\in\mathbb{N}\}\subseteq\mathbb{N}$ such that $V_{k}\subseteq U_{n_{k}}$ for all $k\in\mathbb{N}$.

Proof.

Since $\mathscr{U}$ is not locally finite, there exists $x\in X$ such that

[TABLE]

Since $\mathscr{U}$ is point-finite, we can fix $n_{-1}\in\mathbb{N}$ such that

[TABLE]

By induction on $k\in\mathbb{N}$ we shall select $n_{k}\in\mathbb{N}$ and a non-empty open subset $V_{k}$ of $X$ satisfying the following conditions:

• (ik)

$n_{k-1}<n_{k}$,

• (iik)

$\mathrm{cl}_{X}(V_{k})\subseteq U_{n_{k}}$,

• (iiik)

if $k\geq 1$, then $V_{k}\cap V_{j}=\emptyset$ whenever $j\in\mathbb{N}$ and $j<k$.

We let $n_{0}=n_{-1}+1$ and use regularity of $X$ to fix a non-empty open subset $V_{0}$ of $U_{n_{0}}$ satisfying (ii0). Condition (i0) is clear and condition (iii0) is vacuous.

Suppose now that $k\in\mathbb{N}$, $k\geq 1$ and $n_{k-1}\in\mathbb{N}$ and a non-empty open subset $V_{k-1}$ of $X$ satisfying (ik-1)–(iiik-1) have already been chosen.

Let $j\in\mathbb{N}$ and $j<k$. Since (il) holds for all $l\in\mathbb{N}$ with $l\leq j$, we have $n_{-1}<n_{0}<\dots<n_{j}$. Combining this with (2), we conclude that $x\not\in U_{n_{j}}$. Finally, from this and (iij), we get $x\not\in\mathrm{cl}_{X}(V_{j})$. Since this holds for every $j\in\mathbb{N}$ with $j<k$, it follows that $W=X\setminus\bigcup_{j<k}\mathrm{cl}_{X}(V_{j})$ is an open neighborhood of $x$. Since the set $N(W)$ is infinite by (1), we can choose $n_{k}\in N(W)$ satisfying $n_{k-1}<n_{k}$. Then $W\cap U_{n_{k}}\not=\emptyset$ by the definition of $N(W)$, so we can choose a point $z\in W\cap U_{n_{k}}$ and its open neighborhood $V_{k}$ such that $\mathrm{cl}_{X}(V_{k})\subseteq U_{n_{k}}$. Clearly, conditions (ik)–(iiik) are satisfied. The inductive step has been completed.

The sequence $\{n_{k}:k\in\mathbb{N}\}$ is strictly increasing, as (ik) holds for every $k\in\mathbb{N}$. Since (iiik) holds for all $k\in\mathbb{N}$, we conclude that $\mathscr{V}=\{V_{k}:k\in\mathbb{N}\}$ is a sequence of non-empty pairwise disjoint open subsets of $X$. Finally, $V_{k}\subseteq U_{n_{k}}$ by (iik). ∎

Definition 1.2**.**

Let $X$ be a topological space.

• (i)

$\mathbb{U}_{X}$ denotes the family of all sequences $U=\{U(n):n\in\mathbb{N}\}$ of non-empty open subsets of $X$.

• (ii)

We say that $U\in\mathbb{U}_{X}$ is pairwise disjoint if $U(n)\cap U(m)=\emptyset$ whenever $m,n\in\mathbb{N}$ and $m\not=n$.

• (iii)

If $U\in\mathbb{U}_{X}$ and $s:\mathbb{N}\to\mathbb{N}$ is a monotonically increasing function, then the function $U\circ s\in\mathbb{U}_{X}$ shall be called a subsequence of $U$.

Proposition 1.3**.**

Let $X$ be a pseudocompact space and let $\mathbf{P}:\mathbb{U}_{X}\to\{0,1\}$ be a function satisfying the following conditions:

• (a)

$\mathbf{P}(U)=1$* whenever $U\in\mathbb{U}_{X}$ is not point-finite;*

• (b)

if $\mathbf{P}(V)=1$ for some subsequence $V$ of $U\in\mathbb{U}_{X}$, then $\mathbf{P}(U)=1$.

• (c)

if $U,V\in\mathbb{U}_{X}$, $V(n)\subseteq U(n)$ for every $n\in\mathbb{N}$ and $\mathbf{P}(V)=1$, then $\mathbf{P}(U)=1$.

• (d)

if $V\in\mathbb{U}_{X}$ is pairwise disjoint, then $\mathbf{P}(V)=1$.

Then $\mathbf{P}(U)=1$ for all $U\in\mathbb{U}_{X}$.

Proof.

Let $U\in\mathbb{U}_{X}$ be arbitrary. By (a), we may assume, without loss of generality, that $U$ is point-finite. Since $X$ is pseudocompact, $U$ cannot be locally finite by Fact 0.3. By Lemma 1.1, there exist a pairwise disjoint $V\in\mathbb{U}_{X}$ and a monotonically increasing function $s:\mathbb{N}\to\mathbb{N}$ such that

[TABLE]

Since $V\in\mathbb{U}_{X}$ is pairwise disjoint, $\mathbf{P}(V)=1$ by (d). From this, (3) and (c), we get $\mathbf{P}(U\circ s)=1$. Since $U\circ s$ is a subsequence of $U$, (b) implies $\mathbf{P}(U)=1$. ∎

Corollary 1.4**.**

Let $X$ be a pseudocompact space and let $\mathbf{P}:\mathbb{U}_{X}\to\{0,1\}$ be a function satisfying conditions (a), (b), (c) from Proposition 1.3. Then the following conditions are equivalent:

• (i)

$\mathbf{P}(U)=1$* for every $U\in\mathbb{U}_{X}$;*

• (ii)

$\mathbf{P}(V)=1$* for each pairwise disjoint $V\in\mathbb{U}_{X}$.*

We demonstrate the usefulness of this corollary on the following property which is perhaps not so familiar to the reader.

Definition 1.5**.**

A topological space $X$ is called sequentially pseudocompact if for every family $\{U_{n}:n\in\mathbb{N}\}$ of non-empty open subsets of $X$, there exists an infinite set $J\subseteq\mathbb{N}$ and a point $x\in X$ such that the set $\{n\in J:W\cap U_{n}=\emptyset\}$ is finite for every open neighborhood $W$ of $x$.

This notion appeared in [8, Definition 1.4] under the name sequentially feebly compact. A formally weaker property obtained by requiring the conclusion of Definition 1.5 to hold only for the families $\{U_{n}:n\in\mathbb{N}\}$ consisting of pairwise disjoint non-empty open subsets of $X$ was defined earlier in [3, Definition 1.8]. It was proved in [15, Proposition 1] that these two versions of Definition 1.5 are in fact equivalent. This result can be easily derived from the disjoint refinement principle.

Corollary 1.6**.**

[15, Proposition 1]** For every topological space $X$, the following conditions are equivalent:

• (i)

$X$* is sequentially pseudocompact;*

• (ii)

for every sequence $\{U_{n}:n\in\mathbb{N}\}$ of pairwise disjoint non-empty open subsets of $X$, there exists an infinite set $J\subseteq\mathbb{N}$ and a point $x\in X$ such that the set $\{n\in J:W\cap U_{n}=\emptyset\}$ is finite for every open neighborhood $W$ of $x$.

Proof.

The implication (i)$\Rightarrow$(ii) is trivial. Let us prove that (ii)$\Rightarrow$(i). Note that (ii) implies that every pairwise disjoint sequence of non-empty open subsets of $X$ is not locally finite. Therefore, $X$ is pseudocompact by Fact 0.3.

Define the function $\mathbf{P}:\mathbb{U}_{X}\to\{0,1\}$ as follows. For $U\in\mathbb{U}_{X}$, we let $\mathbf{P}(U)=1$ if there exists an infinite set $J\subseteq\mathbb{N}$ and a point $x\in X$ such that the set $\{n\in J:W\cap U(n)=\emptyset\}$ is finite for every open neighborhood $W$ of $x$, and we define $\mathbf{P}(U)=0$ otherwise. The function $\mathbf{P}$ clearly satisfies conditions (a), (b), (c) of Proposition 1.3. Now the conclusion follows from Corollary 1.4. ∎

2. Two “selective” pseudocompactness-type properties

The (equivalent) properties from the next theorem can be considered “selective” properties, as they all involve a selection of a point from each set of a countable sequence of non-empty open sets in such a way that the resulting sequence satisfies some condition agreed in advance.

Theorem 2.1**.**

For every topological space $X$, the following conditions are equivalent:

• (i)

for each sequence $\{U_{n}:n\in\mathbb{N}\}$ of pairwise disjoint non-empty open subsets of $X$, one can choose a point $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$ such that the set $\{x_{n}:n\in\mathbb{N}\}$ is not closed in $X$;

• (ii)

for each sequence $\{U_{n}:n\in\mathbb{N}\}$ of pairwise disjoint non-empty open subsets of $X$, one can choose a point $x_{n}\in U_{n}$ for each $n\in\mathbb{N}$ such that $\mathrm{cl}_{X}(\{x_{n}:n\in\mathbb{N}\})\setminus\bigcup_{n\in\mathbb{N}}U_{n}\not=\emptyset$;

• (iii)

for each sequence $\{U_{n}:n\in\mathbb{N}\}$ of non-empty open subsets of $X$, one can choose a point $x\in X$ and a point $x_{n}\in U_{n}$ for each $n\in\mathbb{N}$ such that the set $\{n\in\mathbb{N}:x_{n}\in W\}$ is infinite for every open neighborhood $W$ of $x$;

• (iv)

for each sequence $\{U_{n}:n\in\mathbb{N}\}$ of non-empty open subsets of $X$ on can choose a point $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$, select a free ultrafilter111Recall that an ultrafilter $p$ on $\mathbb{N}$ is called free if $\bigcap p=\emptyset.$* $p$ on $\mathbb{N}$ and find a point $x\in X$ such that $\{n\in\mathbb{N}:x_{n}\in W\}\in p$ for every open neighborhood $W$ of $x$.*

Proof.

(iv)$\Rightarrow$(ii) is clear.

(ii)$\Rightarrow$(i) Let $\{U_{n}:n\in\mathbb{N}\}$ be a sequence of non-empty pairwise disjoint open subsets of $X$. Let $\{x_{n}:n\in\mathbb{N}\}$ be the sequence as in the conclusion of item (ii). Since $\{x_{n}:n\in\mathbb{N}\}\subseteq\bigcup_{n\in\mathbb{N}}U_{n}$, yet $\mathrm{cl}_{X}(\{x_{n}:n\in\mathbb{N}\})\setminus\bigcup_{n\in\mathbb{N}}U_{n}\not=\emptyset$, the set $\{x_{n}:n\in\mathbb{N}\}$ is not closed in $X$.

(i)$\Rightarrow$(iii) First, note that the condition (i) implies that an infinite sequence of pairwise disjoint non-empty open subsets of $X$ cannot be locally finite. Applying Fact 0.3, we conclude that $X$ is pseudocompact.

Define the function $\mathbf{P}:\mathbb{U}_{X}\to\{0,1\}$ as follows. For $U\in\mathbb{U}_{X}$, we let $\mathbf{P}(U)=1$ if one can chose a point $x\in X$ and a point $x_{n}\in U(n)$ for each $n\in\mathbb{N}$ such that the set $\{n\in\mathbb{N}:x_{n}\in W\}$ is infinite for every open neighborhood $W$ of $x$, and we define $\mathbf{P}(U)=0$ otherwise. The function $\mathbf{P}$ clearly satisfies conditions (a), (b), (c) of Proposition 1.3.

Let use check the condition (d) of this proposition. Suppose that $V\in\mathbb{U}_{X}$ is pairwise disjoint. Apply (i) to choose a point $x_{k}\in V(k)$ for every $k\in\mathbb{N}$ such that the set $\{x_{k}:k\in\mathbb{N}\}$ is not closed in $X$. Let $x\in\mathrm{cl}_{X}(\{x_{k}:k\in\mathbb{N}\})\setminus\{x_{k}:k\in\mathbb{N}\}$. One can easily check now that $\{k\in W:x_{k}\in W\}$ is infinite for every neighborhood $W$ of $x$.

We have checked that $\mathbf{P}$ satisfies all the assumptions of Proposition 1.3, which implies that $\mathbf{P}(U)=1$ for all $U\in\mathbb{U}_{X}$. This means that (iii) holds.

(iii)$\Rightarrow$(iv) Let $\{U_{n}:n\in\mathbb{N}\}$ be a sequence of non-empty open subsets of $X$. By (iii), we can chose a point $x\in X$ and a point $x_{n}\in U_{n}$ for each $n\in\mathbb{N}$ such that the set $N_{W}=\{n\in\mathbb{N}:x_{n}\in W\}$ is infinite for every open neighbourhood $W$ of $x$. Therefore,

[TABLE]

is a base of a free filter on $\mathbb{N}.$ Let $p$ be an ultrafilter on $\mathbb{N}$ satisfying $\mathcal{B}\subseteq p$; such an ultrafilter exists by Zorn’s lemma. One easily checks that $p$ satisfies (iv). ∎

Definition 2.2**.**

A topological space $X$ satisfying any (and then all) of the equivalent conditions in Theorem 2.1 will be called selectively pseudocompact.

The property from item (iv) of Theorem 2.1 appeared recently in [11, 12] under the name “strong pseudocompactness”. The property from item (ii) of Theorem 2.1 appeared under the same name in the abstract of [12]. Since the term “strongly pseudocompact” is used in [1, 5] to denote two different properties, Definition 2.2 proposes a new name for this property reflecting its“selective” nature. The proposed new name also matches the name of another property introduced in Definition 2.3 below.

To the best of our knowledge, the following “selective” property is new.

Definition 2.3**.**

We shall call a topological space $X$ selectively sequentially pseudocompact if for every family $\{U_{n}:n\in\mathbb{N}\}$ of non-empty open subsets of $X,$ one can choose a point $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$ in such a way that the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a convergent subsequence.

Proposition 2.4**.**

For every topological space $X$, the following conditions are equivalent:

• (i)

$X$* is selectively sequentially pseudocompact;*

• (ii)

for every sequence $\{U_{n}:n\in\mathbb{N}\}$ of pairwise disjoint non-empty open subsets of $X$, one can choose $x_{n}\in U_{n}$ for all $n\in\mathbb{N}$ such that the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a convergent subsequence.

Proof.

(i)$\Rightarrow$(ii) trivially follows from Definition 2.3.

(ii)$\Rightarrow$(i) One easily sees that (ii) implies that every sequence pairwise disjoint non-empty open subsets of $X$ fails to be locally finite. Therefore, $X$ is pseudocompact by Fact 0.3.

Define the function $\mathbf{P}:\mathbb{U}_{X}\to\{0,1\}$ as follows. For $U\in\mathbb{U}_{X}$, we let $\mathbf{P}(U)=1$ if one can chose a point $x_{n}\in U(n)$ for each $n\in\mathbb{N}$ such that the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a convergent subsequence, and we define $\mathbf{P}(U)=0$ otherwise. The function $\mathbf{P}$ clearly satisfies conditions (b) and (c) of Proposition 1.3.

Let us check also condition (a). If $U\in\mathbb{U}_{X}$ is not point-finite, then there exists $x\in X$ such that the set $N=\{n\in\mathbb{N}:x\in U(n)\}$ is infinite. Define $x_{n}=x$ for all $n\in N$ and choose $x_{n}\in U(n)$ arbitrarily for each $n\in\mathbb{N}\setminus N$. Now the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a constant (thus, convergent) subsequence.

Finally, (ii) implies that $\mathbf{P}$ satisfies condition (d) as well.

We have checked that $\mathbf{P}$ satisfies all the assumptions of Proposition 1.3, which implies that $\mathbf{P}(U)=1$ for all $U\in\mathbb{U}_{X}$. This means that (i) holds. ∎

The following diagram summarizes connections between two “selective” pseudocompactness-type properties and other compactness-like properties.

compactsequentially compact$\scriptstyle{1}$countably compact$\scriptstyle{2}$selectively sequentially pseudocompact$\scriptstyle{3}$selectively pseudocompact$\scriptstyle{4}$sequentially pseudocompactpseudocompact

Diagram 1.

Example 2.5**.**

A compact space need not be sequentially pseudocompact. Indeed, the Stone-Čech compactification $\beta\omega$ of $\omega$ is not sequentially pseudocompact [8, Example 2.9]. Hence, none of the properties on the right side of Diagram 1 imply any of the properties on the left side.

Example 2.6**.**

Examples of selectively pseudocompact spaces which are not countably compact were constructed in [12]. Therefore, arrow 2 is not reversible.

Example 2.7**.**

Let $X$ be a pseudocompact space such that all its countable subsets are closed constructed in [17, Theorem 2]. By Theorem 2.1 (i), $X$ is not selectively pseudocompact. Hence, arrow 4 is not reversible.

In Example 5.4 below we shall show that arrow 1 is not reversible. Arrow 3 is not reversible either; see Example 5.5 below.

3. Basic properties of selective (sequential) pseudocompactness

In this section we collect basic properties of the class of selectively (sequentially) pseudocompact spaces.

Proposition 3.1**.**

Every infinite selectively sequentially pseudocompact space has a non-trivial convergent sequence.

Proof.

Let $X$ be an infinite selectively sequentially pseudocompact space. Since $X$ is Hausdorff, there exists a sequence $\{U_{n}:n\in\mathbb{N}\}$ consisting of pairwise disjoint non-empty open subsets of $X$. By Proposition 2.4, one can choose $x_{n}\in U_{n}$ for all $n\in\mathbb{N}$ such that the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a convergent subsequence. Since $U_{n}\cap U_{m}=\emptyset$ whenever $m,n\in\mathbb{N}$ and $m\not=n$, this subsequence is non-trivial. ∎

Proposition 3.2**.**

Let $f:X\to Y$ be a continuous function from a topological space $X$ onto a topological space $Y.$ If $X$ is selectively (sequentially) pseudocompact, then so is $Y$.

Proof.

Let $\{V_{n}:n\in\mathbb{N}\}$ be a sequence of pairwise disjoint non-empty open subsets of $Y$. For every $n\in\mathbb{N}$ define $U_{n}=f^{\leftarrow}(V_{n})$. Since $f$ is continuous and onto, $\{U_{n}:n\in\mathbb{N}\}$ is a sequence of pairwise disjoint non-empty open subsets of $X.$ Now the proof splits into two cases.

If $X$ is selectively sequentially pseudocompact, one can choose a point $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$ such that the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a convergent subsequence. Clearly $f(x_{n})\in V_{n}$ for every $n\in\mathbb{N}.$ Since $f$ is continuous, the sequence $\{f(x_{n}):n\in\mathbb{N}\}$ has a convergent subsequence in $Y$. Applying Proposition 2.4 to $Y$, we conclude that $Y$ is selectively sequentially pseudocompact.

If $X$ is selectively pseudocompact, by Theorem 2.1(ii), we can choose $x_{n}\in U_{n}$ for each $n\in\mathbb{N}$ and find $x\in X$ such that $x\in\mathrm{cl}_{X}(\{x_{n}:n\in\mathbb{N}\})\setminus\bigcup_{n\in\mathbb{N}}V_{n}$. Then $f(x_{n})\in V_{n}$ for all $n\in\mathbb{N}$ and $f(x)\in\mathrm{cl}_{X}(\{f(x_{n}):n\in\mathbb{N}\})\setminus\bigcup_{n\in\mathbb{N}}V_{n}$. Applying Theorem 2.1(ii) to $Y$, we conclude that $Y$ is selectively pseudocompact. ∎

Lemma 3.3**.**

Suppose that $X$ is a topological space having the following property: For every countable family $\{U_{n}:n\in\mathbb{N}\}$ of non-empty open subsets of $X$, there exists a selectively (sequentially) pseudocompact subspace $Y$ of $X$ such that $U_{n}\cap Y\not=\emptyset$ for all $n\in\mathbb{N}$. Then $X$ is selectively (sequentially) pseudocompact.

Proof.

Let $\{U_{n}:n\in\mathbb{N}\}$ be a sequence of pairwise disjoint non-empty open subsets of $X.$ By our hypothesis, $X$ contains a selectively (sequentially) pseudocompact subspace $Y$ such that $V_{n}=U_{n}\cap Y\not=\emptyset$ for every $n\in\mathbb{N}$. Since each $V_{n}$ is open in $Y$ and $Y$ is selectively (sequentially) pseudocompact, we can choose a point $x_{n}\in V_{n}\subseteq U_{n}$ for every $n\in\mathbb{N}$ such that the set $\{x_{n}:n\in\mathbb{N}\}$ is not closed in $Y$, and thus also in $X$ (the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a subsequence converging to some element of $Y$, respectively). This shows that $X$ is selectively (sequentially) pseudocompact. ∎

Corollary 3.4**.**

If some dense subspace of a topological space $X$ is selectively (sequentially) pseudocompact, then $X$ itself is selectively (sequentially) pseudocompact.

Proposition 3.5**.**

If every countable subset of a topological space $X$ is contained in a selectively (sequentially) pseudocompact subspace of $X$, then $X$ is selectively (sequentially) pseudocompact.

Proof.

Let $\{U_{n}:n\in\mathbb{N}\}$ be a family of non-empty open subsets of $X.$ Take $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$. By our hypothesis, the countable subset $\{x_{n}:n\in\mathbb{N}\}$ of $X$ is contained in a selectively (sequentially) pseudocompact subspace $Y$ of $X$. Since $x_{n}\in U_{n}\cap Y\not=\emptyset$ for every $n\in\mathbb{N}$, the conclusion follows from Lemma 3.3. ∎

Proposition 3.6**.**

A clopen subspace of a selectively (sequentially) pseudocompact space is selectively (sequentially) pseudocompact.

Proof.

Let $Y$ be a clopen subspace of a selectively (sequentially) pseudocompact space $X$. Let $\{U_{n}:n\in\mathbb{N}\}$ be a sequence of non-empty open subsets of $Y.$ Since $Y$ is open in $X,$ each $U_{n}$ is also open in $X.$ Since $X$ is selectively (sequentially) pseudocompact, we can use item (iii) of Theorem 2.1 to choose a point $x_{n}\in U_{n}$ for every $n\in\mathbb{N}$ and $x\in X$ such that the set $\{n\in\mathbb{N}:x_{n}\in W\}$ is infinite for every open neighborhood $W$ of $x$ (respectively, the sequence $\{x_{n}:n\in\mathbb{N}\}$ has a subsequence converging to $x\in X$). Since $\{x_{n}:n\in\mathbb{N}\}\subseteq Y$ and $Y$ is closed in $X$, it follows that $x\in Y$. This shows that $Y$ is selectively (sequentially) pseudocompact. ∎

4. selective (sequential) pseudocompactness in ($\Sigma$-)products

Lemma 4.1**.**

A countable product of selectively sequentially pseudocompact spaces is selectively sequentially pseudocompact.

Proof.

Suppose that $X_{i}$ is a selectively sequentially pseudocompact space for every $i\in\mathbb{N}$, and let $X=\prod_{i\in\mathbb{N}}X_{i}$ be the Tychonoff product of the sequence $\{X_{i}:i\in\mathbb{N}\}$. Let $\{U_{n}:n\in\mathbb{N}\}$ be a sequence of nonempty open subsets of $X$. Without loss of generality, we may assume that each $U_{n}$ has the form $U_{n}=\prod_{i\in\mathbb{N}}U_{n,i}$, where $U_{n,i}$ is a nonempty open subset of $X_{i}$.

Let $J_{-1}=\mathbb{N}$. Since each $X_{i}$ is selectively sequentially pseudocompact, by a straightforward induction on $i\in\mathbb{N}$, one can choose $x_{n,i}\in U_{n,i}$ for every $n\in\mathbb{N}$ and select a point $x_{i}\in X$ and an infinite set $J_{i}\subseteq J_{i-1}$ such that the sequence $\{x_{n,i}:n\in J_{i}\}$ converges to $x_{i}$.

Clearly, $y_{n}=(x_{n,i})_{i\in\mathbb{N}}\in\prod_{i\in\mathbb{N}}U_{n,i}=U_{n}$ for every $n\in\mathbb{N}.$

Let $j_{0}\in J_{0}$ be arbitrary. For every $i\geq 1$, choose $j_{i}\in J_{i}\setminus\{j_{k}:k\leq i-1\}$; this can be done as $J_{i}$ is infinite. By our construction, the set $J_{\omega}=\{j_{i}:i\in\mathbb{N}\}$ is infinite, while the set $J_{\omega}\setminus J_{i}$ is finite for every $i\in\mathbb{N}$.

Let $i\in\mathbb{N}$ be arbitrary. Since the sequence $\{x_{n,i}:n\in J_{i}\}$ converges to $x_{i}$ and $J_{\omega}\setminus J_{i}$ is finite, it follows that the sequence $\{x_{n,i}:n\in J_{\omega}\}$ converges to $x_{i}$ as well. Since this holds for every $i\in\mathbb{N}$, we conclude that the sequence $\{y_{n}:n\in J_{\omega}\}$ converges to $y=(x_{i})_{i\in\mathbb{N}}\in X.$ ∎

Definition 4.2**.**

If $\{X_{i}:i\in I\}$ is a family of sets, $X=\prod\{X_{i}:i\in I\}$ is their product and $p$ is a point in $X$, then the subset

[TABLE]

of $X$ is called the $\Sigma$-product of $\{X_{i}:i\in I\}$ with the basis point $p\in X.$

The standard proof of the following lemma is omitted.

Lemma 4.3**.**

Let $\{X_{i}:i\in I\}$ be a family of topological spaces and $X=\prod\{X_{i}:i\in I\}$ be its product and $p\in X$. Then every countable subset of $\Sigma(p,X)$ is contained in a subspace of $\Sigma(p,X)$ homeomorphic to the product $\prod_{i\in J}X_{i}$ for some at most countable set $J\subseteq I$.

Theorem 4.4**.**

Let $\{X_{i}:i\in I\}$ be a family of topological spaces and $X=\prod\{X_{i}:i\in I\}$ be its product and $p\in X$.

• (i)

If all $X_{i}$ are selectively sequentially pseudocompact, then so is $\Sigma(p,X)$.

• (ii)

If $\prod_{i\in J}X_{i}$ is selectively sequentially pseudocompact for every at most countable set $J\subseteq I$, then so is $\Sigma(p,X)$.

Proof.

Let $C$ be a countable subset of $\Sigma(p,X)$. By (4), for every $c\in C$, the set $J_{c}=\{i\in I:f(i)\not=p(i)\}$ is at most countable. Since $C$ is countable, the set $J=\bigcup_{c\in C}J_{c}$ is at most countable as well. Note that $C\subseteq Y$, where $Y=\{f\in X:\{i\in I:f(i)\not=p(i)\}\subseteq J\}$ is a subset of $\Sigma(p,X)$ homeomorphic to $X_{J}=\prod\{X_{i}:i\in J\}$.

(i) Since $X_{J}$ is selectively sequentially pseudocompact by Lemma 4.1, from Proposition 3.5 we conclude that $\Sigma(p,X)$ is selectively sequentially pseudocompact.

(ii) Since $X_{J}$ is selectively pseudocompact by our assumption, applying Proposition 3.5 we conclude that $\Sigma(p,X)$ is selectively pseudocompact. ∎

Corollary 4.5**.**

A product of topological spaces is selectively sequentially pseudocompact if and only if each factor is selectively sequentially pseudocompact.

Proof.

$(\Rightarrow)$ Follows from Proposition 3.2.

$(\Leftarrow)$ Follows from Theorem 4.4 (i) and Corollary 3.4, as each $\Sigma$-product is dense in the product. ∎

Corollary 4.6**.**

Every dyadic space (in particular, every compact group) is selectively sequentially pseudocompact.

Proof.

Let $X$ be a dyadic space. Then $X$ is a continuous image of $\{0,1\}^{\tau}$ for a suitable cardinal $\tau$. Since $\{0,1\}$ is obviously selectively sequentially pseudocompact, Corollary 4.5 implies that $\{0,1\}^{\tau}$ is selectively sequentially pseudocompact. Now selectively sequentially pseudocompactness of $X$ follows from Proposition 3.2.

The “in particular” clause follows from the well-known fact that compact groups are dyadic [2, Theorem 4.1.7]. ∎

Example 4.7**.**

The topological space $\beta\omega$ can be embedded in $D^{\mathfrak{c}},$ where $D=\{0,1\}$ is the two point discrete space. By Corollary 4.6, $D^{\mathfrak{c}}$ is selectively sequentially pseudocompact. Since $\beta\omega$ is not selectively sequentially pseudocompact by [8, Example 2.9], a compact (and thus, closed) subspace of a selectively sequentially pseudocompact space is not necessarily selectively sequentially pseudocompact.

Example 4.8**.**

Let $X$ be a countably compact space such that its square $X^{2}$ is not pseudocompact [9, Example 3.10.9]. Since countably compact spaces are selectively pseudocompact and selectively pseudocompact spaces are pseudocompact (see Diagram 1), this shows that selective pseudocompactness is not a productive property.

5. selective (sequential) pseudocompactness in topological groups

One may expect that in the class of topological groups additional implications would hold that are not present for general topological spaces in Diagram 1. The following result of this type was proved in [3]:

Fact 5.1**.**

Every pseudocompact group is sequentially pseudocompact.

In this section we establish another result of the same flavour.

Definition 5.2**.**

A topological space $X$ is said to be $\omega$-bounded if the closure of any countable subset of $X$ is compact.

Theorem 5.3**.**

Every $\omega$-bounded group is selectively sequentially pseudocompact.

Proof.

Suppose that $X$ is an $\omega$-bounded group. Let $S$ be a countable subset of $X.$ The subgroup $Z$ generated by $S$ is countable. Hence, $Y=$ cl${}_{X}(Z)$ is a compact group. By Corollary 4.6, $Y$ is selectively sequentially pseudocompact. By Proposition 3.5, $X$ is selectively sequentially pseudocompact. ∎

In topological groups we have the following diagram.

compact\textstyle{\omega\text{-bounded}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{5}$$\scriptstyle{6}selectively sequentially pseudocompact$\scriptstyle{7}$countably compact$\scriptstyle{2}$selectively pseudocompact$\scriptstyle{4}$sequentially pseudocompactpseudocompact

Diagram 2.

In this diagram, double arrows denote implications that hold only in the class of topological groups and do not hold for general topological spaces.

Example 5.4**.**

By Corollary 4.6, the Cantor cube $D^{\mathfrak{c}}$ is selectively sequentially pseudocompact. By [9, Theorem 3.10.33], $D^{\mathfrak{c}}$ is not sequentially compact. Therefore, the compact group $G=D^{\mathfrak{c}}$ is selectively sequentially pseudocompact but not sequentially pseudocompact. Hence, arrow 1 is not reversible even for compact (Abelian) groups.

Example 5.5**.**

There is a pseudocompact group $G$ which is not selectively pseudocompact [12, Example 2.4]. By [3, Proposition 1.10], $G$ is sequentially pseudocompact. Therefore, arrows 3 and 4 of Diagram 1 are not reversible even for topological groups.

If $H$ is a pseudocompact group, then $H$ is a $G_{\delta}$-dense subgroup of its Raikov completion $\rho H.$ By [2, Theorem 3.7.16], $\rho H$ is compact and by Corollary 4.6, $\rho H$ is selectively sequentially pseudocompact. Hence, Example 5.5 proves that selectively sequentially pseudocompactness is not hereditary with respect to $G_{\delta}$-dense subgroups.

Example 5.6**.**

By the Hewitt-Marczewski-Pondiczery theorem, $D^{\mathfrak{c}}$ is separable. Therefore, we can fix a countable dense subset $S$ of $D^{\mathfrak{c}}$. By a standard closing off argument, we can construct a countably compact subgroup $G$ of $D^{\mathfrak{c}}$ such that $S\subseteq G$ and $|G|=\mathfrak{c}$. Since $|D^{\mathfrak{c}}|=2^{\mathfrak{c}}>\mathfrak{c}$, we conclude that $G$ is a proper subgroup of $D^{\mathfrak{c}}$. Since $S$ is dense in $D^{\mathfrak{c}}$, this implies that the closure of the countable set $S$ in $G$ is not compact. Thus, $G$ is not $\omega$-bounded. This example shows that arrow 6 is not reversible even for topological groups.

Example 5.7**.**

Under CH, there is a countably compact group $G$ without convergent sequences. Then, $G$ is not selectively sequentially pseudocompact. This example proves that, assuming CH, arrow 7 is not reversible even for topological groups.

We do not know if an additional set-theoretic assumption in this example can be omitted; see Question 8.3 below.

In Example 7.8 below we shall show that arrow 2 of Diagram 1 and arrow 5 of Diagram 2 are not reversible in the class of topological groups.

6. Building dense $\mathcal{V}$-independent sets in $\mathfrak{c}$-powers, for a fixed variety $\mathcal{V}$

We start this section with the following general lemma.

Lemma 6.1**.**

Let $H$ be a separable metric space. Then for every uncountable subset $Y$ of $H$ one can find a set $X\subseteq Y$ such that $X=\bigcup_{\alpha<|Y|}S_{\alpha}$, where:

• (i)

each $S_{\alpha}$ is a countably infinite compact set with a single non-isolated point (that is, $S_{\alpha}$ is a non-trivial convergent sequence taken together with its limit), and

• (ii)

$S_{\alpha}\cap S_{\beta}=\emptyset$* whenever $\alpha\not=\beta$.*

Proof.

First, note that every uncountable subset $Z$ of $H$ contains a countably infinite compact set $S$ with a single non-isolated point. Indeed, since $Z$ is a subspace of a separable metric space $H$, it has a countable base. On the other hand, $Z$ is uncountable, so it cannot be discrete. Therefore, $Z$ contains a non-isolated point $z$. Therefore, by induction on $n\in\mathbb{N}$, we can choose a point $x_{n}\in U(z,1/n)\setminus\{z,x_{0},\dots,x_{n-1}\}$, where $U(z,1/n)$ is the ball centered at $z$ of radius $1/n$. Now $S=\{x_{n}:n\in\mathbb{N}\}\cup\{z\}$ is the desired set.

By transfinite induction on $\alpha<|Y|$, we shall select a countably infinite subset $S_{\alpha}$ of $Y$ with a single non-isolated point such that $S_{\alpha}\cap\bigcup_{\beta<\alpha}S_{\beta}=\emptyset$.

For $\alpha=0$, we use the above observation (with $Z=Y$) to choose the required $S_{0}$.

Suppose now that the ordinal $\alpha$ satisfies $0<\alpha<|Y|$ and the family $\{S_{\beta}:\beta<\alpha\}$ has already been chosen. Note that $|\bigcup_{\beta<\alpha}S_{\beta}|\leq\alpha+\omega<|Y|$, so the set $Z=Y\setminus\bigcup_{\beta<\alpha}S_{\beta}$ satisfies $|Z|=|Y|$; in particular, $Z$ is uncountable. Applying the above observation to this $Z$, we get the desired $S_{\alpha}$. ∎

A variety of groups is a class of groups closed under Cartesian products, subgroups and quotients [16].

Definition 6.2**.**

Let $\mathcal{V}$ be a non-trivial variety of groups. A subset $X$ of a group $G$ is said to be $\mathcal{V}$-independent if

1. (i)

$\langle{X}\rangle\in\mathcal{V},$ and 2. (ii)

for each map $f:X\to H\in\mathcal{V}$ there exists a unique homomorphism $\tilde{f}:\langle{X}\rangle\to H$ extending $f.$

We shall need the following useful fact [6, Lemmas 2.3 and 2.4].

Lemma 6.3**.**

Let $\mathcal{V}$ be a variety of groups and $X$ be a subset of a group $G$.

• (i)

$X$* is $\mathcal{V}$-independent if and only if each finite subset of $X$ is $\mathcal{V}$-independent.*

• (ii)

If $H$ is a group and $f:G\to H$ is a homomorphism such that $f(X)$ is a $\mathcal{V}$-independent subset of $H$, $\langle X\rangle\in\mathcal{V}$ and $f\restriction_{X}:X\to H$ is an injection, then $X$ is a $\mathcal{V}$-independent subset of $G$.

Definition 6.4**.**

Let $\mathcal{V}$ be a non-trivial variety of groups. For a group $G$, the cardinal

[TABLE]

is called the $\mathcal{V}$-rank of $G$.

Lemma 6.5**.**

Let $\mathcal{V}$ be a non-trivial variety. If $H$ is a separable metric group such that $r_{\mathcal{V}}(H)\geq\omega$, then there exists a $\mathcal{V}$-independent subset $X\subseteq H^{\omega}$ such that $X$ contains $\mathfrak{c}$ many pairwise disjoint convergent closed sequences in $X$.

Proof.

Since $H$ is a separable metric group, so is $H^{\omega}$. By [6, Lemma 4.1], the group $H^{\omega}$ has a $\mathcal{V}$-independent subset $Y$ of cardinality $\mathfrak{c}$. Applying Lemma 6.1 to this $Y$, we can choose a subset $X$ of $Y$ as in the conclusion of this lemma. Since $X$ is a subset of the $\mathcal{V}$-independent set $Y$, it is also $\mathcal{V}$-independent by Lemma 6.3 (i). ∎

Theorem 6.6**.**

Suppose that $H$ is a compact metric group with $r_{\mathcal{V}}(H)\geq\omega$. Then $H^{\mathfrak{c}}$ contains a dense selectively sequentially pseudocompact $\mathcal{V}$-independent subset of size continuum.

Proof.

Define $L=H^{\omega}.$ By Lemma 6.5, there exists a $\mathcal{V}$-independent set

[TABLE]

such that

[TABLE]

We shall call a subset $V$ of $L^{\mathfrak{c}}$ a basic open set if $V=\prod_{\gamma<\mathfrak{c}}V_{\gamma}$, where each $V_{\gamma}$ is an open subset of $L$ and the set $\mathrm{supp}(V)=\{\gamma<\mathfrak{c}:V_{\gamma}\neq L\}$ is finite. Let $\mathcal{B}$ be a base for $L^{\mathfrak{c}}$ consisting of basic open subsets $L^{\mathfrak{c}}$ such that $|\mathcal{B}|=\mathfrak{c}$. Let $\mathcal{U}=[\mathcal{B}]^{\omega}.$ Since $|\mathcal{U}|=\mathfrak{c},$ we can enumerate

[TABLE]

where $U_{\alpha,n}\in\mathcal{B}$ for all $\alpha<\mathfrak{c}$ and $n\in\mathbb{N}$.

Let $\alpha<\mathfrak{c}.$ Since each $U_{\alpha,n}$ is a basic open set in $L^{\mathfrak{c}}$, the set $\mathrm{supp}(U_{\alpha,n})$ is finite, and so the set

[TABLE]

is at most countable. Let $p_{\alpha}:L^{\mathfrak{c}}\to L^{S_{\alpha}}$ be the natural projection.

Fix $n\in\mathbb{N}$. Since $U_{\alpha,n}$ is a non-empty basic open subset of $L^{\mathfrak{c}}$ and $\mathrm{supp}(U_{\alpha,n})\subseteq S_{\alpha}$ by (6), the set $W_{n}=p_{\alpha}(U_{\alpha,n})$ is a non-empty open subset of $L^{S_{\alpha}}$ such that

[TABLE]

Since $S_{\alpha}$ is countable and $L$ is a compact metric space, the space $L^{S_{\alpha}}$ is compact and metric. In particular, $L^{S_{\alpha}}$ is sequentially compact, and so also selectively sequentially pseudocompact; see Diagram 1. Therefore, we can choose a point $x_{\alpha,n}\in W_{n}$ for every $n\in\mathbb{N}$ in such a way that a suitable subsequence $\{x_{\alpha,n}:n\in J_{\alpha}\}$ of the sequence $\{x_{\alpha,n}:n\in\mathbb{N}\}$ converges to some element $x_{\alpha,\omega}\in L^{S_{\alpha}}$.

For every $n\in\omega+1$ , define $y_{\alpha,n}\in L^{\mathfrak{c}}$ by

[TABLE]

Claim 1**.**

For every $\alpha<\mathfrak{c}$, the following holds:

• (i)

$y_{\alpha,n}\in U_{\alpha,n}$ for every $n\in\mathbb{N}$;

• (ii)

the sequence $\{y_{\alpha,n}:n\in J_{\alpha}\}$ converges to $y_{\alpha,\omega}.$

Proof.

Fix an arbitrary $\alpha<\mathfrak{c}$.

(i) By (8), $p_{\alpha}(y_{\alpha,n})=x_{\alpha,n}\in W_{n}$, so $y_{\alpha,n}\in p_{\alpha}^{\leftarrow}(W_{n})=U_{\alpha,n}$ by (7).

(ii) It suffices to show that the sequence $\{y_{\alpha,n}(\beta):n\in J_{\alpha}\}$ converges to $y_{\alpha,\omega}(\beta)$ for every $\beta<\mathfrak{c}$. In checking this,we consider two cases.

Case 1. By our choice of the set $J_{\alpha}$, the sequence $\{x_{\alpha,n}:n\in J_{\alpha}\}$ converges to $x_{\alpha,\omega}\in L^{S_{\alpha}}$ in $L^{S_{\alpha}}$. Since $\beta\in S_{\alpha}$, this implies that the sequence $\{y_{\alpha,n}(\beta):n\in J_{\alpha}\}$ converges to $y_{\alpha,\omega}(\beta)$.

Case 2. $\beta\in\mathfrak{c}\setminus S_{\alpha}$. By (8), $y_{\alpha,n}(\beta)=r_{\alpha,n}$ for every $n\in\omega+1$. Now it remains only to note that, by our construction, the sequence $\{r_{\alpha,n}:n\in\mathbb{N}\}$ converges to $r_{\alpha,\omega}$. ∎

Claim 2**.**

$Y=\{y_{\alpha,n}:\alpha<\mathfrak{c},n\in\omega+1\}\subseteq L^{\mathfrak{c}}$ is selectively sequentially pseudocompact and dense in $L^{\mathfrak{c}}$.

Proof.

Let $\mathcal{A}=\{A_{n}:n\in\mathbb{N}\}$ be a countable sequence of elements of $\mathcal{B}.$ There exists $\alpha<\mathfrak{c}$ such that $A_{n}=U_{\alpha,n}$ for every $n\in\mathbb{N}$; that is, $\mathcal{A}=\{U_{\alpha,n}:n\in\mathbb{N}\}$. By Claim 1, $y_{\alpha,n}\in U_{\alpha,n}$ for each $n\in\mathbb{N}$ and the sequence $\{y_{\alpha,n}:n\in J_{\alpha}\}$ converges to $y_{\alpha,\omega}$. Since $\{y_{\alpha,n}:n\in\omega+1\}\subseteq Y$, this proves that $Y$ is selectively sequentially pseudocompact.

Let $O$ be a non-empty open subset of $L^{\mathfrak{c}}$. Since $\mathcal{B}$ is a base of $L^{\mathfrak{c}}$, there exists $B\in\mathcal{B}$ such that $B\subseteq O$. Let $\mathcal{A}$ be the constant sequence all elements of which are equal to $B$. Then $\mathcal{A}=\{U_{\alpha,n}:n\in\mathbb{N}\}$ for some $\alpha<\mathfrak{c}$, and so $\{y_{\alpha,n}:n\in J_{\alpha}\}\subseteq B\cap Y\subseteq O\cap Y$. This shows that $Y$ is dense in $L^{\mathfrak{c}}.$ ∎

Claim 3**.**

$Y$ is $\mathcal{V}$-independent.

Proof.

By Lemma 6.3 (i), it suffices to show that every finite subset $X$ of $Y$ is $\mathcal{V}$-independent. Let $X$ be a finite subset of $Y$. There exists a finite subset $F$ of $\mathfrak{c}$ such that

[TABLE]

Since each set $S_{\alpha}$ is at most countable and $F$ is finite, the set $S=\bigcup_{\alpha\in F}S_{\alpha}$ is at most countable. Therefore, we can choose $\beta\in\mathfrak{c}\setminus S.$ Let $\pi_{\beta}:L^{\mathfrak{c}}\to L$ be the projection on the $\beta$th coordinate. To prove that $X$ is $\mathcal{V}$-independent, it suffices to check that $G=L^{\mathfrak{c}}$, $H=L$, $X$ and $f=\pi_{\beta}$ satisfy the assumptions of Lemma 6.3 (ii).

Since $\mathcal{V}$ is a variety of groups, it is closed under taking arbitrary products and subgroups. Since $L\in\mathcal{V}$ and $\langle X\rangle$ is a subgroup of $L^{\mathfrak{c}}$, this implies $\langle X\rangle\in\mathcal{V}$.

Let $(\alpha,n)\in F\times(\omega+1)$ be arbitrary. Then $\alpha\in F$, and so $S_{\alpha}\subseteq S$. Since $\beta\not\in S$, this implies $\beta\not\in S_{\alpha}$. Therefore, $\pi_{\beta}(y_{\alpha,n})=r_{\alpha,n}$ by (8). This shows that

[TABLE]

Since the latter set is a faithfully indexed subset of $R$, it follows from (9) that $f\restriction_{X}=\pi_{\beta}\restriction_{X}:X\to R$ is an injection. Since $f(X)=\pi_{\beta}(X)\subseteq R$ and $R$ is $\mathcal{V}$-independent, $f(X)$ is $\mathcal{V}$-independent by Lemma 6.3 (i). ∎

The conclusion of our theorem now follows from Claims  1, 2 and 3. ∎

7. selectively sequentially pseudocompact group topologies on $\mathcal{V}$-free groups

Definition 7.1**.**

Let $\mathcal{V}$ be a non-trivial variety of groups.

1. (i)

A subset $X$ of a group $G$ is a $\mathcal{V}$-base of $G$ if $X$ is $\mathcal{V}$-independent and $\langle{X}\rangle=G.$ 2. (ii)

A group is $\mathcal{V}$-free if it contains a $\mathcal{V}$-base.

If $\mathcal{V}$ is a non-trivial variety, then for every set $X$ there exists a unique (up to isomorphism) group $F_{X}(\mathcal{V})$ such that $X\subseteq F_{X}(\mathcal{V})$ and $X$ is a $\mathcal{V}$-base of $F_{X}(\mathcal{V}).$

Definition 7.2**.**

For every cardinal $\tau$, we use $F_{\tau}(\mathcal{V})$ to denote the unique (up to isomorphism) $\mathcal{V}$-free group having a $\mathcal{V}$-base of cardinality $\tau.$

Definition 7.3**.**

[6, Definition 5.2] A variety $\mathcal{V}$ is said to be precompact if there exists a compact zero-dimensional metric group $H\in\mathcal{V}$ with $r_{\mathcal{V}}(H)\geq\omega.$

Corollary 7.4**.**

For a precompact variety $\mathcal{V}$, the group $F_{\mathfrak{c}}(\mathcal{V})$ admits a selectively sequentially pseudocompact zero-dimensional group topology.

Proof.

By Theorem 6.6, $H^{\mathfrak{c}}$ contains a dense selectively sequentially pseudocompact $\mathcal{V}$-independent subset $X$ of size continuum. It follows from Definition 7.1 (i) that $X$ is a $\mathcal{V}$-base of $\langle{X}\rangle$, so $\langle{X}\rangle$ is algebraically isomorphic to $F_{\mathfrak{c}}(\mathcal{V})$. Since $X\subseteq\langle{X}\rangle\subseteq H^{\mathfrak{c}}$ and $X$ is dense in $H^{\mathfrak{c}}$, we conclude that $X$ is also dense in $\langle{X}\rangle$. Since $X$ is selectively sequentially pseudocompact, from Corollary 3.4 we conclude that $\langle{X}\rangle$ is selectively sequentially pseudocompact as well. Since $H$ is zero-dimensional, so is $H^{\mathfrak{c}}$ and $\langle{X}\rangle$. ∎

Since both the variety $\mathcal{G}$ of all groups and the variety $\mathcal{A}$ of all Abelian groups are precompact by [6, Lemma 5.3], from Corollary 7.4 we obtain the following corollary.

Corollary 7.5**.**

Both the free group $F_{\mathfrak{c}}(\mathcal{G})$ with $\mathfrak{c}$-many generators and the free Abelian group $F_{\mathfrak{c}}(\mathcal{A})$ with $\mathfrak{c}$-many generators admit a selectively sequentially pseudocompact zero-dimensional group topology.

Corollary 7.6**.**

The group $\mathbb{T}^{\mathfrak{c}}$ contains the free Abelian group $F_{\mathfrak{c}}(\mathcal{A})$ with $\mathfrak{c}$-many generators as a dense selectively sequentially pseudocompact subgroup.

Proof.

The circle group $\mathbb{T}$ is a compact metric Abelian group. By [2, Lemma 7.1.6], $\mathbb{T}$ contains an $\mathcal{A}$-independent set $X$ of cardinality $\mathfrak{c}.$ By Theorem 6.6, the group $\mathbb{T}^{\mathfrak{c}}$ contains a dense $\mathcal{A}$-independent selectively sequentially pseudocompact subspace $X$ of size $\mathfrak{c}$. Arguing as in the proof of Corollary 7.4, we deduce that $\langle{X}\rangle$ is a dense selectively sequentially pseudocompact subgroup of $\mathbb{T}^{\mathfrak{c}}$ algebraically isomorphic to $F_{\mathfrak{c}}(\mathcal{A})$. ∎

Example 7.7**.**

Let $G=F_{\mathfrak{c}}(\mathcal{G})$ be the $\mathcal{G}$-free group of size $\mathfrak{c}$. By Corollary 7.5, $G$* admits a selectively sequentially pseudocompact group topology*. On the other hand, no non-trivial subgroup of $G$ admits a countably compact group topology. Indeed, if $H$ is a subgroup of $G$, then it is $\mathcal{G}$-free, and it is known that no non-trivial $\mathcal{G}$-free group admits a countably compact group topology. In particular, there exists a selectively sequentially pseudocompact group without non-trivial countably compact subgroups.

Example 7.8**.**

Let $G=F_{\mathfrak{c}}(\mathcal{A})^{\omega}$ be the countable power of the $\mathcal{A}$-free group of size $\mathfrak{c}$. Then $G$ is an Abelian group of size $\mathfrak{c}$. By Corollary 7.5, $F_{\mathfrak{c}}(\mathcal{A})$ admits a selectively sequentially pseudocompact group topology. Since the class of selectively sequentially pseudocompact spaces is closed under arbitrary products, $G$ also admits a selectively sequentially pseudocompact group topology. By [19, Theorem 17], this topology cannot be countably compact. Therefore, $G$* is an Abelian group (of size $\mathfrak{c}$) which admits a selectively sequentially pseudocompact group topology, yet does not admit a countably compact group topology.* This example proves that arrows 2 and 5 are not reversible even for topological groups.

8. Final remarks and open questions

García-Ferreira and Tomita asked whether every compact group contains a proper dense selectively pseudocompact subgroup [12, Question 2.10]. This question has a consistent positive answer. Indeed, assuming $2^{\omega_{1}}>\mathfrak{c}$, it was proved in [14] that every non-metrizable compact group has a proper dense countably compact subgroup, and countably compact spaces are selectively pseudocompact; see Diagram 1.

In ZFC, one can obtain a positive solution to even stronger version of the question of García-Ferreira and Tomita for large classes of topological groups.

Remark 8.1**.**

(i) It was proved in [14] that if a compact group $G$ can be mapped onto a product of $w(G)$-many nontrivial compact metric groups, then $G$ contains $|G|$-many dense, $\omega$-bounded proper subgroups. (Here $w(G)$ denotes the weight of $G$.)

(ii) The assumption of item (i) is satisfied when $G$ is either connected or Abelian [14]. Therefore, if a non-metrizable compact group $G$ is either connected or Abelian, then $G$ contains $|G|$-many dense, $\omega$-bounded proper subgroups.

(iii) Since $\omega$-bounded groups are selectively sequentially pseudocompact by Corollary 4.6, it follows from (ii) that if a non-metrizable compact group $G$ is either connected or Abelian, then $G$ contains $|G|$-many dense, selectively sequentially pseudocompact proper subgroups.

The following question is related to this remark.

Question 8.2**.**

Does every non-metrizable selectively (sequentially) pseudocompact Abelian group contains a proper dense selectively (sequentially) pseudocompact subgroup?

Example 5.7 justifies the following question.

Question 8.3**.**

(i) Is there a ZFC example of sequentially pseudocompact (Abelian) group which is not selectively sequentially pseudocompact?

(ii) Is there a ZFC example of a countably compact (Abelian) group which is not selectively sequentially pseudocompact?

It should be interesting to investigate how the new notion is behaving in $C_{p}(X,G)$-theory [18]. Let $G$ be a topological group. For a topological space $X$, we denote by $C_{p}(X,G)$ the group of all $G$-valued continuous maps from $X$ into $G$ endowed with the topology of pointwise convergence, i.e., the subspace topology $C(X,G)$ inherits from the Tychonoff product $X^{G}$. A space $X$ is said to be $b_{G}$-discrete if every countable subset $D$ of $X$ is discrete and each function from $D$ to $G$ can be extended to a continuous function from $X$ to $G$.

Remark 8.4**.**

If $G$ is a metric group and $X$ is a topological space such that $C_{p}(X,G)$ is pseudocompact and dense in $G^{X}$, then $G$ is compact and $X$ is $b_{G}$-discrete. Indeed, since $C_{p}(X,G)$ is pseudocompact and dense in $G^{X}$, the latter space is pseudocompact. Since $G$ is a continuous projection of the pseudocompact space $G^{X}$, it is pseudocompact. Now it remains only to note that a metric pseudocompact space is compact. As a dense pseudocompact subgroup of the compact group $G^{X}$, the group $C_{p}(X,G)$ is $G_{\delta}$-dense in $G^{X}$. Since $G$ is first countable, this is equivalent to $X$ being $b_{G}$-discrete [7, Proposition 4.6].

This remark motivates the compactness condition on $G$ in the following problem.

Problem 8.5**.**

Let $G$ be a compact metric group and $X$ be a $b_{G}$-space. Find necessary and sufficient conditions on a space $X$ for $C_{p}(X,G)$ to be selectively (sequentially) pseudocompact.

Let $G$ be a topological group. Recall that spaces $X$ and $Y$ are called $G$-equivalent if topological groups $C_{p}(X,G)$ and $C_{p}(Y,G)$ are topologically isomorphic [18].

Question 8.6**.**

Let $G$ be a topological group, and let $X$ and $Y$ be $G$-equivalent spaces such that $C_{p}(X,G)$ is dense in $G^{X}$ and $C_{p}(Y,G)$ is dense in $G^{Y}$. If $X$ is selectively (sequentially) pseudocompact, must $Y$ also be selectively (sequentially) pseudocompact?

Acknowledgement: The authors would like to thank Boaz Tsaban for his comment that inspired Problem 8.5 and Question 8.6. This paper was written during the first listed author’s stay at the Department of Mathematics of Faculty of Science of Ehime University (Matsuyama, Japan) in the capacity of Visiting Foreign Researcher under the support by CONACyT of México: Estancias Posdoctorales al Extranjero propuesta No. 263464. He would like to thank CONACyT for its support and the host institution for its hospitality.