Closed embeddings of spaces and groups into pseudocompact $k_\mathbb {R}$-groups

Abstract

A well-known theorem of Noble states that each Tychonoff space X is homeomorphic to a closed subspace of a pseudocompact $k_\mathbb {R}$-space. We strengthen this result by showing that any Tychonoff space X is homeomorphic to a closed subspace of an abelian pseudocompact $k_\mathbb {R}$-group G such that $w(G)\le \aleph _1\cdot w(X)$, and if, in addition, X is a precompact group, then X is topologically isomorphic to a closed subgroup of G. It is constructed the first examples of pseudocompact groups $G_1$ and $G_2$ (in fact, they are even countably compact and of weight $\aleph _2$) such that $G_1$ is Ascoli but not a $k_\mathbb {R}$-space, and $G_2$ is a $k_\mathbb {R}$-space but not a k-space. Under $\mathrm {MA+\lnot CH}$, we show that any pseudocompact group of weight $\aleph _1$ is Ascoli. These results are proved using topological properties of pseudocompact spaces X of weight $\aleph _1$ and of $\Sigma$-products in products of compact spaces. Being motivated by these results and the countably compact part of Noble’s theorem, it is shown by a well-known technique that each countably compact infinite group has a separable countably compact subgroup of cardinality continuum.

1 Introduction

All spaces are assumed to be Tychonoff. We denote by $C_p(X)$ and $C_k(X)$ the space C(X) of all continuous real-valued functions on a space X endowed with the pointwise topology or the compact-open topology, respectively. The space X is a $k_\mathbb {R}$-space if every k-continuous function $f :X \rightarrow \mathbb {R}$ is continuous (recall that f is k-continuous if each restriction of f to any compact set $K\subseteq X$ is continuous). In [19], Noble proved that every $k_\mathbb {R}$-space satisfies the conclusion of the Ascoli theorem: if X is a $k_\mathbb {R}$-space, then every compact subset $\mathcal {K}$ of $C_k(X)$ is evenly continuous; that is, the map $X\times \mathcal {K}\ni (x,f)\mapsto f(x)$ is continuous. Following [3], a space X is called an Ascoli space if every compact subset of $C_k(X)$ is evenly continuous. In other words, X is Ascoli if and only if the compact-open topology of $C_k(X)$ is Ascoli in the sense of [18, p.45]. A simple method of constructing of pseudocompact spaces which are Ascoli but not $k_\mathbb {R}$-spaces is given in Proposition 3.11 of [13].

We shall say that a space X is closed embeddable into a space Y if X is homeomorphic to a closed subspace Z of Y.

The following important result was proved by Noble in Construction 2.3 of [20].

Theorem 1.1

([20]) Each space X is closed embeddable into a pseudocompact $k_\mathbb {R}$-space Y. If, in addition, X is countably compact, then the space Y can be also taken countably compact.

This remarkable result was complemented in Theorem 3.13 of [13] as follows:

Theorem 1.2

([13]) Any space X is closed embeddable into a pseudocompact Ascoli space Y such that

• (i) Y is not a $k_\mathbb {R}$-space;

• (ii) $w(Y)= \max \{w(X), \aleph _2\}$;

• (iii) if, in addition, X is $\omega$-bounded (sequentially compact, totally countably compact, near sequentially compact or countably compact), then so is Y.

All the aforementioned results concern topological spaces. The main purpose of the article is to consider their corresponding group versions.

Now we describe the content of the article. In Section 2 we recall some of the basic notions and theorems used in the article.

In Section 3 we show that subspaces of products of compact metrizable spaces containing $\Sigma$-products are often $k_\mathbb {R}$-spaces. It should be noted that such spaces are of independent interest because they appear in many natural situations, see for example [8].

The first example of a pseudocompact group without infinite compact subsets was constructed by Sirota [24]. In Section 4 we recall the construction of Sirota type groups and obtain some additional topological properties of such groups. The obtained results are summarized in Theorem 4.12.

In Proposition 6.1 of Section 6 we construct the first example of a pseudocompact group which is an Ascoli space but not a $k_\mathbb {R}$-space.

The main results of the article are contained in Section 5. The group version of Noble’s Theorem 1.1 and Theorem 1.2 can be considered only for precompact groups because any pseudocompact group is precompact and each subgroup of a precompact group is precompact as well. In Theorem 5.1 we show that each precompact group H is topologically isomorphic to a closed subgroup of a pseudocompact $k_\mathbb {R}$-group G such that $w(G)\le \aleph _1\cdot w(H)$. Using Theorem 5.1 we prove in Theorem 5.4 that each space X is closed embeddable into an abelian pseudocompact $k_\mathbb {R}$-group G such that $w(G)\le \aleph _1\cdot w(X)$. To prove Theorems 5.1 and 5.4 we considerably use the results of Section 3.

Being motivated by the countably compact case in Theorem 1.1, van Douwen’s result [6] (see Proposition 2.3) and (ii) and (iii) of Theorem 1.2, one can naturally consider the following problem: Is it true that any countably compact group G contains “small” countably compact subgroups H? Here “small” means that H has size continuum $\mathfrak {c}$ or weight $w(H)\le \aleph _1$. We partially answer this problem in Theorem 7.3 by showing the perhaps known fact that each countably compact infinite group G has a separable countably compact subgroup H of cardinality $\mathfrak {c}$.

Spaces and groups of small weight are studied in Section 8. In Theorem 8.3 we show that if a space X contains a dense $\omega$-bounded subspace Y and $w(X)\le \aleph _1$, then X is a $k_\mathbb {R}$-space. On the other hand, in Proposition 8.5 we provide a simple method of constructing pseudocompact $k_\mathbb {R}$-spaces which are not k-spaces. Applying the obtained results we give the first example of an abelian pseudocompact $k_\mathbb {R}$-group G of weight $w(G)= \aleph _1$ which is not a k-space, see Proposition 8.14. In Theorem 8.11 we show that each pseudocompact group G such that $w(G)<\mathfrak {p}$ is near sequentially compact.

2 Preliminary results

As usual we set $\mathbb {N}=\{1,2,\dots \}$ and $\omega =\mathbb {N}\cup \{0\}$. If $\Omega$ is a set, we denote by $[\Omega ]^{<\omega }$ and $[\Omega ]^{\le \omega }$ the family of all finite subsets of $\Omega$ and the family of all countable subsets of $\Omega$, respectively.

Let X be a space. The weight of X is denoted by w(X). Let us recall the most important types of pseudocompact spaces: X is called

$\bullet$:

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

$\bullet$:

$\omega$-bounded if every sequence in X has compact closure;

$\bullet$:

totally countably compact if every sequence in X has a subsequence with compact closure;

$\bullet$:

near sequentially compact if for any (pairwise disjoint) sequence $(U_n)_{n\in \omega }$ of open sets in X there exists a sequence $(x_n)_{n\in \omega }\in \prod _{n\in \omega }U_n$ containing a convergent subsequence $(x_{n_k})_{k\in \omega }$;

$\bullet$:

selectively pseudocompact if for any sequence $(U_n)_{n\in \omega }$ of open sets in X there exists a sequence $(x_n)_{n\in \omega }\in \prod _{n\in \omega }U_n$ which has a cluster point;

$\bullet$:

weakly selectively pseudocompact if for any sequence $(U_n)_{n\in \omega }$ of open sets in X there exists a sequence $(M_n)_{n\in \omega }$ of finite sets in X such that $M_n\subseteq U_n$ for each $n\in \omega$ and a sequence $(M_n)_{n\in \omega }$ has a cluster point, i.e., there exists $x\in X$ such that $\{n\in \omega : U\cap M_n\ne \emptyset \}$ is infinite for any neighborhood U of x;

$\bullet$:

selectively $\omega$-bounded if each infinite collection of disjoint open sets has an infinite subcollection each of which meets some fixed compact set;

$\bullet$:

countably compact if every sequence in X has a cluster point.

We introduce the notion of weakly selectively pseudocompact spaces being motivated by the following unexpected Theorem 1.2 of [14].

Theorem 2.1

([14]) Each pseudocompact group is weakly selectively pseudocompact.

Clearly, any selectively pseudocompact space is weakly selectively pseudocompact. In [14] a pseudocompact group is constructed that is not selectively pseudocompact. It follows from Theorem 2.1 that this pseudocompact group is a weakly selectively pseudocompact space that is not selectively pseudocompact. There is a pseudocompact space in which every countable set is discrete and closed [23]. This pseudocompact space is not weakly selectively pseudocompact, for another such an example see Example 8.13 below.

Taking into account Theorem 3.2 of [9] and Theorem 1.2 of [10], we have the following relationships between the above-defined notions and note that none of these implications is reversible:

A subset A of a space X is k-closed if $A\cap K$ is closed in K for every compact subset K of X. The space X is called a k-space if any k-closed subset of X is closed.

Let $\{X_i\}_{i\in I}$ ba a nonempty family of spaces, $X=\prod _{i\in I}X_i$, and $x=(x_i)\in X$. Then the subset $\Sigma (X,x)$ of X defined by

$$ \Sigma (X,x):=\big \{ y=(y_i)\in X: \{i\in I: y_i\not = x_i\} \text{ is } \text{ countable }\big \} $$

is called a $\Sigma$-product in X (around x). In the partial case when G is a topological group with the identity e and $\tau$ is an infinite cardinal, the subgroup

$$ G_\tau :=\Sigma \big (G^\tau ,(e_i)\big )=\big \{ (g_i)\in G^\tau : \text{ the } \text{ support } \{i\in \tau : g_i\not =e\} \text{ of } (g_i) \text{ is } \text{ countable }\big \} $$

of $G^\tau$ is called the $\Sigma$-product in $G^\tau$.

The local character $\chi (x,X)$ of a space X at a point x is the least cardinality of a local base at x. The character $\chi (X)$ is the least upper bound of the local characters. Recall that the small cardinal $\mathfrak {p}$ is defined as follows

$$\begin{aligned} \mathfrak {p}=\big \{ |\mathcal {B}|:&\;\mathcal {B} \text{ is } \text{ a } \text{ subbase } \text{ for } \text{ a } \text{ free } \text{ filter } \text{ on } \omega \text{, } \text{ and } \text{ there } \text{ is } \text{ no } \text{ infinite } A \\&\; \text{ such } \text{ that } A\subseteq ^*B \text{ for } \text{ all } B\in \mathcal {B}\big \}, \end{aligned}$$

where $A\subseteq ^*B$ means that $A{\setminus }B$ is finite and $B{\setminus }A$ is infinite. Recall that $\omega _1<\mathfrak {p}$ is consistent with ZFC since it follows from $MA+\lnot CH$.

We shall use the following interesting result, see Lemma 1.11 of [21].

Proposition 2.2

([21]) Each countable space X such that $\chi (X)<\mathfrak {p}$ is Fréchet–Urysohn.

The following important result is due to van Douwen, see Proposition 1.3 of [6]. We give its simple and independent proof.

Proposition 2.3

([6]) Let X be a pseudocompact space.

• (i) If X is infinite and homogeneous, then X has no isolated points.

• (ii) If X has no isolated points, then $|X|\ge \mathfrak {c}$.

Proof

(i) is clear.

(ii) Since X has no isolated points, $\beta X$ is not scattered. Therefore there is a continuous mapping p from $\beta X$ onto [0, 1]. Hence p(X) is a pseudocompact dense subspace of [0, 1] and, therefore, $p(X)=[0,1]$. Thus $|X|\ge \mathfrak {c}$. $\square$

3 Spaces containing dense $\Sigma$-products

Now we consider subspaces of products containing $\Sigma$-products. Following [12], a subset Z of a topological space X is k-dense if for every $x\in X$, there is $M\subseteq Z$ such that $x\in \overline{M}$ and $\overline{M}$ is compact.

Proposition 3.1

Let $X= \prod _{\alpha \in \mathcal {A}} X_\alpha$ be the product of a nonempty family $\{X_\alpha \}_{\alpha \in \mathcal {A}}$ of spaces, $x=(x_\alpha )\in X$ be a point in X, and let $S:=\Sigma (X,x)$. If Y is a subspace of X such that $S\subseteq Y\subseteq X$, then S is k-dense in Y if one of the following conditions is satisfied:

• (i) $\prod _{\alpha \in \mathcal {A}} \{x_\alpha ,y_\alpha \}\subseteq Y$ for each $y=(y_\alpha )\in Y$;

• (ii) $Y=X$;

• (iii) $|\mathcal {A}|=\omega _1$.

Proof

(i) Let $y=(y_\alpha )\in Y$ be arbitrary. Set $P:=\prod _{\alpha \in \mathcal {A}} \{x_\alpha ,y_\alpha \}$ and $M:=P\cap S$. Then P is compact in Y and $y\in \overline{M}=P$. Thus S is k-dense in Y.

(ii) immediately follows from (i).

(iii) Let $y=(y_\alpha )\in Y$ be arbitrary. For every $\beta \in \mathcal {A}$, set

$$ z_{\alpha ,\beta }:= {\left\{ \begin{array}{ll} y_\beta ,& \text { if } \beta <\alpha , \\ x_\beta ,& \text { if } \beta \ge \alpha , \end{array}\right. } $$

and $z_\alpha :=(z_{\alpha ,\beta })_\beta$. Put $M=\{z_\alpha :\alpha <\omega _1\}\subseteq S$. It is easy to see that $\overline{M}^{\, X}=M\cup \{y\}$ and $\overline{M}$ is compact. Thus S is k-dense in Y. $\square$

Proposition 3.2

Let a subspace Z of a space X be C-embedded in X, k-dense in X and a $k_\mathbb {R}$-space. Then X is a $k_\mathbb {R}$-space.

Proof

Let $f :X \rightarrow \mathbb {R}$ be a k-continuous function. Then the restriction $g=f{\restriction }_Z$ of f to Z is also k-continuous. Since Z is a $k_\mathbb {R}$-space, we obtain that g is continuous. As Z is C-embedded in X, g can be extended to a continuous function $h :X \rightarrow \mathbb {R}$. It remains to show that $h=f$. To this end, fix an arbitrary point $x\in X$. Since Z is k-dense in X, there is $M\subseteq Z$ such that $x\in \overline{M}$ and $\overline{M}$ is compact. Taking into account that f and h are continuous on $\overline{M}$ since $\overline{M}$ is compact, the equality $f{\restriction }_M=h{\restriction }_M$ and the inclusion $x\in \overline{M}$ imply that $f(x)=h(x)$, as desired. $\square$

Proposition 3.3

Let $\{X_\alpha \}_{\alpha \in \mathcal {A}}$ be a nonempty family of compact spaces such that for every $\alpha \in \mathcal {A}$, $X_\alpha =\prod _{i\in I_\alpha } K_{i,\alpha }$ is the product of a nonempty family $\{K_{i,\alpha }\}_{i\in I_\alpha }$ of compact metrizable spaces. If S is a $\Sigma$-product in the product $X=\prod _{\alpha \in \mathcal {A}}X_\alpha$, then S is an $\omega$-bounded $k_\mathbb {R}$-space.

Proof

Since all spaces $X_\alpha$ are compact, it follows that the $\Sigma$-product S is $\omega$-bounded. To show that S is a $k_\mathbb {R}$-space, fix $x=(x_\alpha )\in X$ such that $S=\Sigma (X,x)$. For every $\alpha \in \mathcal {A}$, let $Z_\alpha :=\Sigma (X_\alpha ,x_\alpha )\subseteq X_\alpha$ be the $\Sigma$-product in $X_\alpha$ around the $\alpha$-th coordinate $x_\alpha \in X_\alpha$ of x. Consider the subspace

$$ Z:=\Sigma \Big (\prod _{\alpha \in \mathcal {A}} \prod _{i\in I_\alpha } K_{i,\alpha }, x\Big ) $$

of the product $X=\prod _{\alpha \in \mathcal {A}} \prod _{i\in I_\alpha } K_{i,\alpha }$ of metrizable compact spaces. It is clear that $Z\subseteq S$. Being a $\Sigma$-product, the space Z is Fréchet–Urysohn (see [8, 3.10.D]). Since Z is $\omega$-bounded and dense in the compact space X, it follows that Z is C-embedded in $X=\beta Z$ (the equality $X=\beta Z$ follows from Corollary 3.6.9 of [8]). Therefore Z is C-embedded in S as well.

We show that Z is k-dense in S. Let $y=(y_\alpha )\in S$. Then the set

$$ \delta (x,y):=\{\alpha \in \mathcal {A}:x_\alpha \ne y_\alpha \} $$

is at most countable. It follows from Proposition 3.1(ii) that for every $\alpha \in \delta (x,y)$, there exists $M_\alpha \subseteq Z_\alpha$ such that $\overline{M_\alpha }$ is compact and $y_\alpha \in \overline{M_\alpha }$. If $\alpha \in \mathcal {A}\setminus \delta (x,y)$, we set $M_\alpha =\{x_\alpha \}$. Define $M:=\prod _{\alpha \in \mathcal {A}}M_\alpha \subseteq S$. Then $y\in \overline{M}$ and $\overline{M}$ is compact. Thus Z is k-dense in S.

Since Z is a Fréchet–Urysohn space, it is a $k_\mathbb {R}$-space. Therefore Z is C-embedded in S, k-dense in S and is a $k_\mathbb {R}$-space. Thus, by Proposition 3.2, S is a $k_\mathbb {R}$-space. $\square$

In the next theorem we provide a method for constructing $k_\mathbb {R}$-spaces.

Theorem 3.4

Let $Q=\prod _{\alpha \in \mathcal {A}}X_\alpha$ be the product of a nonempty family $\{X_\alpha \}_{\alpha \in \mathcal {A}}$ of compact metrizable spaces, and let S be a $\Sigma$-product in $Q^{\omega _1}$. If X is a subspace of $Q^{\omega _1}$ such that $S\subseteq X$, then X is a pseudocompact $k_\mathbb {R}$-space.

Proof

Since S is $\omega$-bounded and dense in X, it follows that X is a pseudocompact space. To show that X is a $k_\mathbb {R}$-space, we recall that, by Proposition 3.3, S is a $k_\mathbb {R}$-space and, by Proposition 3.1(iii), S is k-dense in X. As S is $\omega$-bounded and dense in $Q^{\omega _1}=\beta S$ it follows that S is C-embedded in $Q^{\omega _1}$. Therefore S is C-embedded in X, too. Thus, by Proposition 3.2, X is a $k_\mathbb {R}$-space. $\square$

4 Sirota’s groups

Let M be a subset of a group G. Denote by $\left\langle M\right\rangle$ the algebraic hull of M.

Let G be a Boolean group. Then G is (algebraically) the direct sum of a family of the abelian two-element group $\mathbb {Z}(2)$. A subset M of G is called independent if $g\notin \left\langle M\setminus \{g\}\right\rangle$ for every g in M. In other words, M is independent if $\left\langle M_0\right\rangle \cap \left\langle M_1\right\rangle =\{0\}$ for every disjoint subsets $M_0, M_1 \subseteq M$. Note that the group $\mathbb {Z}(2)$ carries the structure of the two-element field (which in the field theory is denoted usually by $\mathbb {F}_2$). Therefore independent sets in Boolean groups are exactly linearly independent sets over the field $\mathbb {Z}(2)$.

The following two facts are well-known (they were also noticed in [24]).

Fact 4.1

Every infinite subset of a Boolean group contains an infinite independent subset.

Fact 4.2

Let G be a Boolean group, and let $M\subseteq G$ be an independent subset. Then any map $f :M\rightarrow \mathbb {Z}(2)$ can be extended to a homomorphism $\hat{f} :G\rightarrow \mathbb {Z}(2)$.

For the best of our knowledge, the first example of a pseudocompact group without convergent sequences was constructed by Sirota in 1969, see Theorem 5 of [24]. A topological Boolean group G is called a Sirota group if the following condition holds:

(S):

for every countable independent set $M\subseteq G$, any mapping $f :M\rightarrow \mathbb {Z}(2)$ can be extended to a continuous homomorphism $\hat{f} :G\rightarrow \mathbb {Z}(2)$.

Let us briefly recall the main points in the construction of Sirota groups. We use the following assertion proved in Proposition 4 of [22].

Fact 4.3

([22]) Let X be a space, and let $M\subseteq X$. A set M is discrete and $C^*$-embedded in X if and only if any two disjoint subsets of M are functionally separated in X.

Following [24], a set $X\subseteq \mathbb {Z}(2)^\kappa$ is called disjointed if for any countable disjoint subsets $M_0,M_1\subseteq X$ there exists $\alpha <\kappa$ such that $\pi _\alpha (M_0)=\{0\}$ and $\pi _\alpha (M_1)=\{1\}$, where $\pi _\alpha : \mathbb {Z}(2)^\kappa \rightarrow \mathbb {Z}(2)$ is the projection onto the $\alpha$-th coordinate.

Proposition 4.4

If $X\subseteq \mathbb {Z}(2)^\kappa$ is disjointed, then X is independent and each countable subset of X is discrete, closed and $C^*$-embedded in X.

Proof

Assuming that X is not independent we could find $M\subseteq X$ and $g\in M$ such that $g\in \left\langle M\setminus \{g\}\right\rangle$. Then $g\in \left\langle M_0\right\rangle$ for some finite $M_0\subseteq M\setminus \{g\}$. Then, for $M_1:=\{g\}$, we have $\pi _\alpha \big (\left\langle M_0\right\rangle \big )=\pi _\alpha (M_0)=\{0\}$ but $\pi _\alpha (M_1)=\{1\}$, a contradiction.

From the definition of disjointed sets and Fact 4.3 it follows that each countable subset of X is discrete and $C^*$-embedded in X. From the fact that each countable subset of X is discrete it follows that each countable subset of X is closed. $\square$

The following statement is contained in the proof of Theorem 3 of [24].

Proposition 4.5

If $X\subseteq \mathbb {Z}(2)^\kappa$ is disjointed, then $G:=\left\langle X\right\rangle$ is a Sirota group.

Proof

Let $M\subseteq G$ be a countable independent set, and let $f :M\rightarrow \mathbb {Z}(2)$ be a mapping. Choose a countable $Y\subseteq X$ such that $M\subseteq \left\langle Y\right\rangle$. It follows from Fact 4.2 that f can be extended to a homomorphism $f_1: \left\langle Y\right\rangle \rightarrow \mathbb {Z}(2)$. For every $i\in \mathbb {Z}(2)$, set

$$ M_i:=\{m\in M: f(m)=i\} \;\; \text{ and } \;\; Y_i:=\{y\in Y: f(y)=i\} $$

and observe that $M_i \subseteq Y_i$ since $f_1(m)=f(m)$ for every $m\in M$.

Since X is disjointed, also Y is disjointed. Therefore there exists $\alpha <\kappa$ such that $\pi _\alpha (Y_0)=\{0\}$ and $\pi _\alpha (Y_1)=\{1\}$. It is clear that the continuous homomorphism $\hat{f}=\pi _\alpha {\restriction }_{G}$ extends $f_1{\restriction }_{Y}$. Taking into account the inclusions $M_i \subseteq Y_i$ for $i\in \mathbb {Z}(2)$, we see that $\hat{f}$ extends f. $\square$

The following theorem is Theorem 4 from [24].

Theorem 4.6

([24]) Let $\kappa$ be a cardinal such that $\kappa =\kappa ^\omega$. Then $\mathbb {Z}(2)^\kappa$ has a dense pseudocompact disjointed subset X.

It follows from Theorem 4.6 and Proposition 4.4 that there exist infinite pseudocompact spaces X in which every countable subset is discrete, closed, and $C^*$-embedded in X. Such spaces were later constructed in [23].

The existence of pseudocompact Sirota groups is proved in Theorem 5 of [24].

Theorem 4.7

([24]) Let $\kappa$ be a cardinal such that $\kappa =\kappa ^\omega$. Then $\mathbb {Z}(2)^\kappa$ has a dense pseudocompact Sirota group of weight $\kappa$.

Proof

Using Theorem 4.6, fix a dense pseudocompact disjointed subset X of $\mathbb {Z}(2)^\kappa$. Then the group G (with the topology induced from $\mathbb {Z}(2)^\kappa$ is pseudocompact and has weight $\kappa$. By Proposition 4.5, G is a Sirota group.

$\square$

In fact, the conclusion of Proposition 4.4 holds true for Sirota groups. This was noticed in Remarks 4 and 5 of [24]. Let us clarify this fact.

Proposition 4.8

If $M\subseteq G$ is a countable independent subset of a Sirota group G, then M is a discrete and closed subset of G which is $C^*$-embedded in G.

Proof

To show that M is a discrete and $C^*$-embedded in G subset of G, by Fact 4.3, it suffices to prove that if $M_0,M_1\subseteq M$ is a partition of M into two sets, then $M_0$ and $M_1$ are functionally separated. Let $f :M\rightarrow \mathbb {Z}(2)$ be a function equal to 0 on $M_0$ and 1 on $M_1$. It follows from (S) that f can be extended to a continuous homomorphism $\hat{f} :G\rightarrow \mathbb {Z}(2)$. Therefore, $M_0$ and $M_1$ are functionally separated.

We prove that M is a closed subset of G. Assuming the converse fix a point $g\in \overline{M}\setminus M$. Since the set $M\cup \{g\}$ is not discrete, then, as we proved above, the set $M\cup \{g\}$ is not independent. As M is independent, then $g=g_1+g_2+...+g_n$ for some $g_1,g_2, \ldots ,g_n\in M$. Then the set $L=\{g\}\cup M \setminus \{g_1,g_2, \ldots ,g_n\}$ is independent. Therefore, L is discrete. A contradiction, since $g\in \overline{L\setminus \{g\}}$. $\square$

Fact 4.1 and Proposition 4.8 imply the following assertion.

Proposition 4.9

If G is an Sirota group, then each infinite set $M\subseteq G$ contains a countable closed discrete $C^*$-embedded subset.

Corollary 4.10

Each countably compact subspace of a Sirota group G is finite.

Proposition 2.1 of [1] states that every compact subset of a Sirota group is finite.

Below we consider topological properties of Sirota groups.

Proposition 4.11

If G is a pseudocompact Sirota group, then G is not a selectively pseudocompact space.

Proof

The completion $\widehat{G}$ of G is a compact Boolean group. Therefore, by Theorem 25.9 of [16], $\widehat{G}$ is topologically isomorphic to $\mathbb {Z}(2)^\kappa$ for some cardinal $\kappa$. Consequently, we can assume that G is a dense subgroup of $\mathbb {Z}(2)^\kappa$. For every $n\in \mathbb {N}$, we set

$$ U_n=\{ g\in G: g(n)=1 \text { and } g(i)=0 \text { for } i<n\}. $$

Then $U_n$ is an open nonempty subset of G and if $g_n\in U_n$, then the sequence $(g_n)_n$ is an independent set in G. It follows from Proposition 4.8 that $(g_n)_n$ is a discrete closed subset of G. Thus, G is not a selectively pseudocompact space. $\square$

In the next theorem we summarize the results obtained in Theorem 4.7, Propositions 4.9 and 4.11, and Corollary 4.10.

Theorem 4.12

Let $\kappa$ be a cardinal such that $\kappa =\kappa ^\omega$. Then there exists a pseudocompact group G for which the following conditions are satisfied:

• (i) $w(G)=\kappa$;

• (ii) each infinite set $M\subseteq G$ contains a countable closed discrete $C^*$-embedded subset;

• (iii) each countably compact subspace of G is finite;

• (iv) G is not a selectively pseudocompact space.

Corollary 4.13

([14]) There exists a pseudocompact not selectively pseudocompact group.

Corollary 4.14

([1]) There exists an infinite pseudocompact group, in which each compact subspace is finite.

5 Embeddings of spaces and precompact groups into pseudocompact $k_\mathbb {R}$-groups

The following theorem is a group analogue of Theorems 1.1 and 1.2.

Theorem 5.1

Let H be an (abelian) precompact group. Then:

• (i) H is topologically isomorphic to a closed subgroup of an (abelian) pseudocompact $k_\mathbb {R}$-group G such that $w(G)\le w(H)\cdot \aleph _1$. Consequently, if $w(H)\ge \aleph _1$, then $w(G)= w(H)$.

• (ii) If H is an abelian and $\omega$-bounded (totally countably compact, sequentially compact, countably compact), then the abelian pseudocompact $k_\mathbb {R}$-group G from (i) can be also chosen $\omega$-bounded (resp., totally countably compact, countably compact).

Proof

(i) Let $\overline{H}$ be the completion of H. By Corollary 5.2.4 of [2], we have $w(H)=\chi (H)=\chi (\overline{H})=w(\overline{H})$. This equalities and Theorem 3.4.21 of [2] (applied to the class $\mathcal {P}$ of (abelian) compact metrizable groups) imply that $\overline{H}$ is topologically isomorphic to a subgroup of the product Q of a family $\{K_i\}_{i<w(H)}$ of (abelian) compact metrizable groups. Hence also H is topologically isomorphic to a subgroup of $Q=\prod _{i<w(H)} K_i$. Let

$$ L:= \big \{ (h_i)\in Q^{\,\omega _1}: \text{ the } \text{ support } \{i<\omega _1: h_i\not =e\} \text{ of } (h_i) \text{ is } \text{ countable }\big \} $$

be the $\Sigma$-product in $Q ^{\,\omega _1}$. If S is a countable subset of L, then S is contained in a subgroup $L_S$ of L of the form $Q ^{\,\lambda _S}\times \textbf{e}_S$, where $\lambda _S$ is a countable subset of $\omega _1$ and $\textbf{e}_S$ is the identity element of the group $Q ^{\,\omega _1{\setminus }\lambda _S}$. Thus L is an $\omega$-bounded dense subgroup of $Q ^{\,\omega _1}$.

Let $T: H\rightarrow Q ^{\,\omega _1}$ be the diagonal embedding defined by

$$ T(h):=(h_i), \; \text{ where } h_i=h \text{ for } \text{ every } i<\omega _1. $$

Denote by G the subgroup of $Q ^{\,\omega _1}$ generated by T(H) and L. Since T(H) is a closed subgroup of $Q ^{\,\omega _1}$, T(H) is also closed in G. As T is an embedding, $T:H\rightarrow G$ is a topological isomorphism.

It follows from Theorem 3.4 that G is a $k_\mathbb {R}$-space.

(ii) Since Q is a compact group, the group L is $\omega$-bounded. It easy to see that then the group $G=L+T(H)$ is $\omega$-bounded (resp., totally countably compact, countably compact). $\square$

In the case $w(H)= \aleph _0$, the following proposition complements Theorem 5.1 and shows that the multiplication by $\aleph _1$ is essential.

Proposition 5.2

A metrizable precompact group H is topologically isomorphic to a closed subgroup of a metrizable pseudocompact group G if and only if H is a compact group.

Proof

Assume that H is closed embeddable into a metrizable pseudocompact group G. Since metrizable pseudocompact spaces are compact (see Theorem 3.10.1 and Theorem 3.10.21 of [8]), the group G is compact. Hence its closed subgroup H is compact as well.

The converse assertion is trivial if we set $G:=H$. $\square$

In Theorem 4.3 of [5], Comfort and van Mill proved that every space X is embedded into a pseudocompact group G as a closed subspace such that $\langle X\rangle$ is dense in G. We provide an independent proof of the existence of an embedding with closed image and with the additional property that $w(X)=w(G)$.

Theorem 5.3

Each space X is closed embeddable into an abelian precompact group G such that $w(X)=w(G)$.

Proof

Let K be a compactification of X such that $w(X)=w(K)$, see Theorem 2.3.23 of [8], and let A(K) be the free abelian topological group over K (for the definition and numerous properties of A(K) we refer the reader to the book [2]). Choose a sequence $(q_n)_n\subseteq (0,1)$ of irrational numbers that is linearly independent over $\mathbb {Q}$ and, for simplicity of notations, set $\kappa :=w(X)$.

Claim 1. There is $D\subseteq C(K,[0,1))$ of cardinality $\le \kappa$ such that for each distinct $x_1,x_2, \ldots ,x_n\in K$, there is $f\in D$ such that

$$ f(x_1)=q_1, \; f(x_2)=q_2, \dots , \; f(x_n)=q_n. $$

Indeed, since K is embedded into $[0,1]^\kappa$ we can assume that $K=[0,1]^\kappa$. Let $\mathcal {F}_n$ be the family of all distinct elements $F=\{x_1,x_2, \ldots ,x_n\}$ in K containing exactly n elements. For each $F=\{x_1,x_2, \ldots ,x_n\}\in \mathcal {F}_n$, choose a finite subset $I_F$ of $\kappa$ such that the projection $p_{I_F}(F)$ of F onto the coordinates $I_F$ contains exactly n elements.

Fix $n\in \mathbb {N}$. For every $m\in \mathbb {N}$, let $\mathcal {P}_{m,n}$ be the partition of the cube $[0,1]^n$ by subcubes of the form

$$ \big [\tfrac{i_1}{m},\tfrac{i_1+1}{m}\big ]\times \cdots \times \big [\tfrac{i_n}{m},\tfrac{i_n+1}{m}\big ], $$

where $0\le i_k\le m-1$ for every $1\le k\le n$. If a family $\mathcal {K}=\{K_1,\dots , K_n\}$ in $\mathcal {P}_{m,n}$ is disjoint, choose a continuous function $f_{\mathcal {K},m,n} :[0,1]^n\rightarrow [0,1)$ such that

$$ f_{\mathcal {K},m,n}{\restriction }_{K_i}=q_i\; \text{ for } \text{ every } 1\le i\le n $$

and $f_{\mathcal {K},m,n}{\restriction }_{K}=0$ for every subcube such that $K\cap \bigcup \mathcal {K}=\emptyset$.

Define

$$ D:=\big \{ f_{\mathcal {K},m,n}\circ p_{I_F}: n,m\in \mathbb {N}, \; F\in \mathcal {F}_n \text{ and } \mathcal {K}\in \mathcal {P}_{m,n} \big \} $$

Since $I_F$ is a finite subset of $\kappa$, we obtain that $|D|\le \kappa$. It is easy to see that D is as desired that proves Claim 1.

We identify the circle group $\mathbb {T}=\mathbb {R}/\mathbb {Z}$ as a set with the set [0, 1). Let

$$ \varphi =\bigtriangleup D: K\rightarrow \mathbb {T}^D, \quad \varphi (x):=\big (d(x)\big )_{d\in D} \;\; (x\in K). $$

Clearly, $\varphi$ is continuous. Therefore, by the definition of A(K), $\varphi$ can be extended to the unique continuous homomorphism $\hat{\varphi }: A(K)\rightarrow \mathbb {T}^D$.

Claim 2. $\hat{\varphi }$ is a monomorphism.

Indeed, assume that $\hat{\varphi }(\chi )=\hat{\varphi }(\eta )$ for some

$$ \chi =a_1 z_1 +\cdots + a_k z_k \; \text{ and } \; \eta =b_1 t_1 +\cdots + b_j t_j $$

where $\{z_1,\dots ,z_k\}\in \mathcal {F}_k$ and $\{t_1,\dots ,t_j\}\in \mathcal {F}_j$ and $a_1,\dots ,a_k,b_1,\dots ,b_j\in \mathbb {Z}{\setminus }\{0\}$. If $\{z_1,\dots ,z_k\}\not =\{t_1,\dots ,t_j\}$, we can assume that $t_j\not \in \{z_1,\dots ,z_k\}$. Then, by the construction of D in Claim 1 choosing m sufficiently large, there is $d\in D$ such that

$$ d(z_1)= q_1, \dots , d(z_k)=q_k \; \text{ and } \; d(t_j)=q_{k+1}, \text{ and } \; d(t_i)=0 \text{ if } t_i\not \in \{z_1,\dots ,z_k\}. $$

Since $q_1, \dots , q_k, q_{k+1}$ are rationally independent, for the d-th coordinate, we obtain that

$$ \hat{\varphi }(\chi )_d =a_1 q_1 +\cdots + a_k q_k (\textrm{mod}\; 1) \not = b_1 q_1 +\cdots + b_j q_j (\textrm{mod}\; 1) =\hat{\varphi }(\eta )_d, $$

and hence $\hat{\varphi }(\chi )\not =\hat{\varphi }(\eta )$ that is impossible. Therefore we can assume that $k=j$ and $z_1=t_1,\dots ,z_k=t_k$. If $a_i\not = b_i$ for some $1\le i\le k$, for $n=k$ and choosing m sufficiently large, the construction of D in Claim 1 implies that there is $d\in D$ such that $d(z_i)=q_1$ and $d(z_j)=0$ if $j\not = i$. Then, for the d-th coordinate we obtain (recall that $q_1$ is irrational)

$$ \hat{\varphi }(\chi )_d = a_i q_1 \ \ (\textrm{mod}\; 1) \not = b_i q_1 \ \ (\textrm{mod}\; 1). $$

Therefore also in this case we obtain $\hat{\varphi }(\chi )\not =\hat{\varphi }(\eta )$ that is also impossible. Hence the equality $\hat{\varphi }(\chi )=\hat{\varphi }(\eta )$ implies $\chi =\eta$. Thus $\hat{\varphi }$ is a monomorphism and Claim 2 is proved.

Let $G:=\hat{\varphi }\big (A(X)\big )$ with the topology induced from $\mathbb {T}^D$. It is clear that G is abelian, precompact and $w(G)\le w(\mathbb {T}^D)=|D|\le w(X)$.

Claim 3. X is homeomorphic to $\hat{\varphi }(X)$ and $\hat{\varphi }(X)$ is closed in G. Indeed, since K is compact and $\varphi$ is injective (Claim 2), Theorem 3.1.13 of [8] implies that $\varphi$ is a homeomorphism onto its image $\varphi (K)$ in $\mathbb {T}^D$. This fact and the injectivity of $\hat{\varphi }$ (Claim 2) imply that X is homeomorphic to $\hat{\varphi }(X)$. Since $\varphi (X)=\varphi (K)\cap G$ and $\varphi (K)$ is closed in $\mathbb {T}^D$, it follows that $\hat{\varphi }(X)$ is a closed subspace of G. Claim 3 is proved.

It follows from Claim 3 that $w(X)\le w(G)$. Since also $w(G)\le w(X)$, we obtain $w(G)=w(X)$. This and Claim 3 finish the proof of the theorem. $\square$

The following theorem immediately follows from Theorems 5.3 and 5.1.

Theorem 5.4

Any space X is closed embeddable into an abelian pseudocompact $k_\mathbb {R}$-group G such that $w(G)\le \aleph _1\cdot w(X)$. Consequently, if $w(X)\ge \aleph _1$, then $w(G)=w(X)$.

6 A Boolean pseudocompact group which is Ascoli but not a $k_\mathbb {R}$-space and a Boolean pseudocompact group which is a $k_\mathbb {R}$-space but not a k-space

The main result of this section is the following example of a Boolean pseudocompact group which is Ascoli but not a $k_\mathbb {R}$-space.

Proposition 6.1

For a cardinal $\tau >\aleph _1$, let $H_\tau$ be the $\Sigma$-product in $\mathbb {Z}(2)^\tau$, and let $G_\tau := H_\tau \cup (1+H_\tau )\subseteq \mathbb {Z}(2)^\tau$, where $1=(1_i)\in \mathbb {Z}(2)^\tau$. Then the group $G_\tau$ is sequentially compact and $\omega$-bounded (hence Ascoli) but not a $k_\mathbb {R}$-space.

Proof

First we prove the following claim.

Claim 1. If K is a compact subset of $G_\tau$, then $F:=K\cap H_\tau$ and $G:=K\cap (1+H_\tau )$ are compact. Indeed, it suffices to show that the sets F and G are closed. By the symmetry we shall prove only that F is closed in $G_\tau$.

Assume that $x=(x_i)\in \overline{F}$, and suppose for a contradiction that $x\not \in H_\tau$ and hence $x\in 1+ H_\tau$. For every $\alpha<\omega _1 <\tau$, we consider the natural projection $\pi _\alpha :\mathbb {Z}(2)^\tau \rightarrow \mathbb {Z}(2)^\alpha$. Since $H_\tau$ is sequentially compact and K is compact, it follows that F is sequentially compact. As $\alpha$ is countable, we obtain that $\pi _\alpha (F)$ is a sequentially compact subset of the metrizable compact space $\mathbb {Z}(2)^\alpha$. Therefore $\pi _\alpha (F)$ is compact, and hence $\pi _\alpha (x)\in \pi _\alpha (F)=\pi _\alpha (\overline{F})$. Choose $q_\alpha \in F$ such that $\pi _\alpha (x)=\pi _\alpha (q_\alpha )$.

Consider the set $Q:=\{q_\alpha : \alpha <\omega _1\}\subseteq F$. As $Q\subseteq K$, the set Q has a cluster point $y=(y_i)\in K$. Taking into account that $\pi _\alpha (x)=\pi _\alpha (q_\alpha )$ for every $\alpha <\omega _1$, we obtain that $\pi _{\omega _1}(x)=\pi _{\omega _1}(y)$. Therefore y has at least $\omega _1$ non-zero coordinates and hence $y\not \in H_\tau$. On the other hand, the support A of Q has cardinality $\aleph _1$. As $\omega _1 <\tau$, we obtain that $y_i=0$ for every $i\in \tau {\setminus }A$. Therefore $y\not \in 1+H_\tau$. Hence $y\not \in G_\tau$, which is impossible. The claim is proved.

Consider the characteristic function $\textbf{1}_{H_\tau }$ of $H_\tau$. Since $H_\tau$ is dense in $G_\tau$, $\textbf{1}_{H_\tau }$ is discontinuous. On the other hand, it follows from Claim 1 that $\textbf{1}_{H_\tau }$ is k-continuous. Thus $G_\tau$ is not a $k_\mathbb {R}$-space.

To show that $G_\tau$ is sequentially compact and $\omega$-bounded, let $A\subseteq G_\tau$ be countable. Then $\overline{A}=\overline{A_0}\cup \overline{A_1}$, where $A_0=A\cap H_\tau$ and $A_0=A\cap (1+H_\tau )$. By symmetry, it suffices to prove that $\overline{A_0}$ is metrizable and compact in $H_\tau$. Since $A_0$ is countable, it sits in some compact metrizable subgroup of $H_\tau$ of the form $\mathbb {Z}(2)^\lambda \times \{0_i\}^{\tau {\setminus }\lambda }$, where $\lambda \subseteq \tau$ is countable. Thus $\overline{A_0}$ is a compact subset of $H_\tau$. $\square$

Now we construct the first example of a countably compact $k_\mathbb {R}$-group which is not a k-space.

Proposition 6.2

There exists an abelian $\omega$-bounded $k_\mathbb {R}$-group G of weight $w(G)=\aleph _2$ which is not a k-space.

Proof

Let H be an abelian $\omega$-bounded group such that $w(H)=\aleph _2$ and H is not a k-space (see Proposition 6.1). According to Theorem 5.1, the group H is topologically isomorphic to a closed subgroup of an abelian $\omega$-bounded $k_\mathbb {R}$-group G such that $w(G)=\aleph _2$. The group G is not a k-space because it contains a closed subspace H which is not a k-space. $\square$

The condition $\tau >\aleph _1$ in Proposition 6.1 is essential.

Proposition 6.3

Let $H_{\omega _1}$ be the $\Sigma$-product in $\mathbb {Z}(2)^{\omega _1}$, and let $G_{\omega _1}:= H_{\omega _1} \cup (1+H_{\omega _1})\subseteq \mathbb {Z}(2)^{\omega _1}$, where $1=(1_i)\in \mathbb {Z}(2)^{\omega _1}$. Then $G_{\omega _1}$ is an $\omega$-bounded, sequentially compact and non-compact group which is a k-space.

Proof

The group $G_{\omega _1}$ is $\omega$-bounded by the same reason as $G_\tau$ in Proposition 6.1. Since $H_{\omega _1}$ is sequentially compact it follows that also $G_{\omega _1}$ is sequentially compact. As $G_{\omega _1}$ is a dense subgroup of the compact group $\mathbb {Z}(2)^{\omega _1}$, we obtain that $G_{\omega _1}$ is not compact. Since $w(G_{\omega _1})\le w\big (\mathbb {Z}(2)^{\omega _1}\big )=\aleph _1$, the group $G_{\omega _1}$ is a k-space by Theorem 8.1. $\square$

7 Small countably compact subgroups of countably compact groups

Recall that a space X is crowded if it has no isolated points, and X is scattered if each nonempty subspace of X has an isolated point. Recall also that a compact space is either scattered or it contains a crowded compact subset.

Proposition 7.1

If a space X does not have non-trivial convergent sequences and $|X|< 2^{\omega _1}$, then any compact subset of X is finite.

Proof

Suppose for a contradiction that X has an infinite compact subset K. If K is scattered, then, by the proof of Theorem 2.1 of [15], K contains a non-trivial convergent sequence. Therefore, by the assumption of the proposition, K is not scattered. In this case K contains an infinite crowded compact subset L. But L is nowhere first countable since, otherwise, we would have a convergent sequence. Therefore, $\chi (x,L)\ge \aleph _1$ for every $x\in L$. Then, by the Čech–Pospišil theorem [4], we obtain $|L|\ge 2^{\aleph _1}$ which is impossible by the assumption of the proposition. $\square$

Corollary 7.2

If a compact space K has no non-trivial convergent sequences, then $|K|\ge 2^{\omega _1}$.

For a stronger result, see van Douwen [7, Theorem 6.3].

The next theorem shows that any countably compact infinite group G contains small (by cardinality) countably compact subgroups. Perhaps this is a known result, in any case, the method of proof is quite standard.

Theorem 7.3

Each countably compact infinite group G has a separable countably compact subgroup H of cardinality $\mathfrak {c}$.

Proof

Let $H_0$ be an arbitrary countable infinite subgroup of G (for example, the subgroup generated by a countably infinite subset of G). For every countable subset A of $H_0$, take a cluster point $g_A$ of A. Denote by $H_1$ the subgroup of G generated by the family

$$ \big \{ g_A: A\in [H_0]^{\le \omega }\big \}. $$

Since each $h\in H_0$ is the unique cluster point of the constant sequence (h) in $H_0$, we obtain that $H_0\subseteq H_1$. Observe also that $|H_1|=\mathfrak {c}$. Indeed, it is clear that $|H_1|\le \mathfrak {c}$. Let us show that $|H_1|\ge \mathfrak {c}$. Since $H_0$ is dense and relatively countably compact in $H_1$, $H_1$ is an infinite pseudocompact group. Now Proposition 2.3 implies that $|H_1|\ge \mathfrak {c}$, as desired.

Let $\alpha$ be a countable ordinal, and assume that for every countable ordinal $i<\alpha$, we have constructed a subgroup $H_i$ of G such that $H_i$ is generated by the set

$$ \Big \{ g_A: A\in \Big [\bigcup _{j<i} H_j\Big ]^{\le \omega }\Big \}, $$

where for every $A\in \big [\bigcup _{j<i} H_j\big ]^{\le \omega }$, $g_A$ is a cluster point of A in G. For every $A\in \big [\bigcup _{i<\alpha } H_i\big ]^{\le \omega }$, choose a cluster point $g_A$ of A in G. Denote by $H_\alpha$ the subgroup of G generated by the family

$$ \Big \{ g_A: A\in \Big [\bigcup _{i<\alpha } H_i\Big ]^{\le \omega }\Big \}. $$

Since the cardinality of $\big [\bigcup _{i<\alpha } H_i\big ]^{\le \omega }$ is $\mathfrak {c}^\omega =2^\omega$, it follows that the cardinality of $H_\alpha$ is $\mathfrak {c}$. Finally, we set $H=\bigcup _{\alpha <\omega _1} H_\alpha$. Then $|H|=\mathfrak {c}$. To see that H is countably compact, let A be a countable subset of H. Take $\alpha <\omega _1$ such that $A\subseteq H_\alpha$. Then A has a cluster point in $H_{\alpha +1}\subseteq H$.

Since $H_0$ is dense in H and $|H_0|=\omega$, then H is separable. $\square$

Proposition 7.4

Under $2^\omega <2^{\omega _1}$, let G be an infinite countably compact group without non-trivial convergent sequences. Then G contains a separable countably compact subgroup H of cardinality $\mathfrak {c}$ which does not have infinite compact subsets.

Proof

By Theorem 7.3, there exists a separable countably compact subgroup H of G of cardinality $\mathfrak {c}$. By $2^\omega <2^{\omega _1}$ and Proposition 7.1, H has no infinite compact subsets. $\square$

Problem 7.5

Does there exist a pseudocompact (or countably compact) group of cardinality $\mathfrak {c}$ without non-trivial convergent sequences which has infinite compact subspaces?

It is known (see for example [1] and Theorem 4.12 here) that there are pseudocompact abelian groups without infinite compact subsets. In [1], under Martin’s Axiom, it is shown that there are even countably compact abelian groups without infinite compact subsets. In Theorem 4.1 of [17] it is shown that there is a countably compact group G without non-trivial convergent sequences. These results motivate the following problems.

Problem 7.6

Does there exist in $\textrm{ZFC}$ a countably compact group without infinite compact subspaces?

Note that an infinite $\omega$-bounded group H contains non-trivial convergent sequences (indeed, take an infinite countable subgroup H, then its closure $\overline{H}$ is an infinite compact subgroup of G which is dyadic by the Ivanovskij–Kuz’minov theorem).

Problem 7.7

Does there exist a totally countably compact group without non-trivial convergent sequences?

Problem 7.8

Does there exist a countably compact group G such that any infinite subset $M\subseteq G$ contains a discrete countable subset $S\subseteq M$ that is $C^*$-embedded in G?

8 Spaces and groups of small weight

In this section we show that spaces containing dense $\omega$-bounded subsets or $\Sigma$-products are often $k_\mathbb {R}$-spaces. The obtained results are of independent interest because such spaces are appeared in many natural cases, see for example [8]. We also essentially use these assertions to prove the main results of the article.

Theorem 8.1

If X is an $\omega$-bounded space such that $w(X)\le \omega _1$, then X is a k-space.

Proof

By Theorem 2.3.23 of [8], we can assume that X is a subspace of $[0,1]^{\omega _1}$. To prove that X is a k-space we show that each k-closed subset F of X is closed. Suppose aiming for a contradiction that there is a k-closed subset F of X which is not closed. Take an arbitrary point $x=(w_i)\in \overline{F}\setminus F$. For every $\alpha <\omega _1$, let $p_\alpha : [0,1]^{\omega _1} \rightarrow [0,1]^{\alpha }$ be the projection onto the first $\alpha$-coordinates. Since F is k-closed and X is $\omega$-bounded, F is $\omega$-bounded as well. Hence, for every $\alpha <\omega _1$, $p_\alpha (F)$ being $\omega$-bounded is a compact subset of the compact metrizable space $[0,1]^\alpha$. Therefore, there exists $x_\alpha \in F$ such that $p_\alpha (x)=p_\alpha (x_\alpha )$.

Set $K:=\overline{\{x_\alpha :\alpha <\omega _1\}}$. The equalities $p_\alpha (x)=p_\alpha (x_\alpha )$ ($\alpha <\omega _1$) imply that $x\in K$. For every $\beta <\omega _1$, set $K_\beta :=\overline{\{x_\alpha :\alpha \le \beta \}}$. Since F is $\omega$-bounded, the set $K_\beta \subseteq K\cap F$ is compact. For every finite $B\subseteq \omega _1$, we also set $\beta (B):=\max (B)$.

Claim 1. K is a compact subset of X.

Indeed, let $\{U_i\}_{i\in I}$ be an open cover of K. Choose $i_0\in I$ such that $x\in U_{i_0}$. Choose a standard closed neighborhood $V_{B}$ of x which is contained in $U_{i_0}$, where B is a finite subset of $\omega _1$ and $V_{B}$ has a form

$$ V_B=\big \{ (z_i)\in X: z_b\in W_b \text{ for } b\in B\big \}, $$

where $W_b$ is a closed interval in [0, 1]. For the simplicity of notations set $\beta :=\beta (B)=\max (B)$. Then the equalities $p_\alpha (x)=p_\alpha (x_\alpha )$ ($\alpha <\omega _1$) imply that

$$ x_\alpha \in V_B\; \text{ for } \text{ every } \; \alpha >\beta . $$

Since $K_\beta$ is compact, there is a finite subset J of I such that $K_\beta \subseteq \bigcup _{j\in J} U_j$. Then

$$ K= \overline{\{x_\alpha :\alpha > \beta \}} \cup K_\beta \subseteq V_B \cup K_\beta \subseteq U_{i_0}\cup \bigcup _{j\in J} U_j, $$

which means that K is a compact subset of X.

Claim 2. $K{\setminus }F=\{x\}$.

Indeed, for every finite subset B of $\omega _1$ and each $n\in \mathbb {N}$, set

$$ V_{B,n}:=\big \{ (z_i)\in X: |z_b-w_b|\le \tfrac{1}{n} \text{ for } b\in B\big \}. $$

Then $V_{B,n}$ and hence also $V_{B,n}\cup K_{\beta (B)}$ are closed (recall that $K_{\beta (B)}$ is compact). For every $\alpha >\beta (B)$, we have $x_\alpha \in V_{B,n}$ and hence $K\subseteq V_{B,n}\cup K_{\beta (B)}$. Therefore

$$\begin{aligned} K&\subseteq \bigcap _{B\in [\omega _1]^{<\omega }, \, n\in \mathbb {N}} \big ( V_{B,n}\cup K_{\beta (B)}\big ) \\&\subseteq \left( \bigcup _{B\in [\omega _1]^{<\omega }} K_{\beta (B)} \right) \cup \left( \bigcap _{B\in [\omega _1]^{<\omega }, \, n\in \mathbb {N}} V_{B,n}\right) \\&= \left( \bigcup _{\beta <\omega _1} K_{\beta } \right) \cup \{x\} \subseteq F\cup \{x\}. \end{aligned}$$

Thus $K{\setminus }F=\{x\}$ and the claim is proved.

By Claims 1 and 2, for the compact set $K\subseteq X$, we have that $K\cap F$ is not closed in K. Therefore F is not k-closed, a contradiction. $\square$

Corollary 8.2

If G is an $\omega$-bounded group such that $w(G)=\aleph _1$, then G is a k-space.

Note that Proposition 6.1 implies that there is an $\omega$-bounded group G with $w(G)=\aleph _2$ that is not a $k_\mathbb {R}$-space.

Theorem 8.3

If a space X contains a dense $\omega$-bounded subset Y and $w(X)\le \omega _1$, then X is a $k_\mathbb {R}$-space.

Proof

For every $x\in X$, the subset $Y\cup \{x\}$ is $\omega$-bounded and satisfies $w\big (Y\cup \{x\}\big )\le w(X)\le \aleph _1$. Hence, by Theorem 8.1, $Y\cup \{x\}$ is a k-space. Therefore the family $\mathcal {M}=\{ Y\cup \{x\}: x\in X\}$ is a cover of X containing $k_\mathbb {R}$-spaces. For every $A,B\in \mathcal {M}$, the intersection $A\cap B$ is dense in X because it contains Y. Thus, by Theorem 6.3 of [12], X is a $k_\mathbb {R}$-space. $\square$

In Proposition 3.11 of [13] we provide a simple method for constructing pseudocompact Ascoli spaces which are not $k_\mathbb {R}$-spaces. In fact these spaces are subspaces of the compact space $[0,1]^\lambda$ for $\lambda >\omega _1$ consisting of $\Sigma$-products. For the case $\lambda =\omega _1$ we obtain the following assertion.

Corollary 8.4

Let X be a subspace of $[0,1]^{\omega _1}$ containing a $\Sigma$-product. Then X is a $k_\mathbb {R}$-space.

Proof

Each $\Sigma$-product in $[0,1]^{\omega _1}$ is $\omega$-bounded and $w(X)\le w\big ([0,1]^{\omega _1}\big )=\aleph _1$. Now Theorem 8.3 applies.

$\square$

In the next proposition we give a simple construction of $k_\mathbb {R}$-spaces which are not k-spaces.

Proposition 8.5

Let Z be a space of weight $\aleph _1$ which is not a k-space, and let $p:Z\rightarrow \big [\tfrac{1}{2},1\big ]^{\omega _1}$ be an embedding. Let $S:=\Sigma \big ([0,1]^{\omega _1},\textbf{0}\big )$ be the $\Sigma$-product in $[0,1]^{\omega _1}$ around the zero $\textbf{0}$. Then the space $X:=p(Z)\cup S$ is a $k_\mathbb {R}$-space of weight $\aleph _1$ which is not a k-space.

Proof

We identify Z with its image p(Z). It is clear that Z is a closed subspace of X. The space X being dense in $[0,1]^{\omega _1}$ has weight $\aleph _1$. The space X is a $k_\mathbb {R}$-space by Corollary 8.4. To show that X is not a k-space, let A be a k-closed subset of Z which is not closed in Z. Since Z is closed in X, for every compact subset K of X it follows that $A\cap K= A\cap (Z\cap K)$ is closed in K. Therefore A is k-closed in X. On the other hand, the choice of A shows that A is not closed in X. Thus X is not a k-space.

$\square$

Proposition 8.6

Let X be a space such that $\chi (X)<\mathfrak {p}$.

• (i) If X is countably compact, then X is sequentially compact.

• (ii) If X is weakly selectively pseudocompact, then X is near sequentially compact.

Proof

(i) Let $(x_n)_n\subseteq X$ be a sequence, and let x be a cluster point for $(x_n)_n$. It follows from Proposition 2.2 that the space $Y=\{x\}\cup \{x_n:n\in \omega \}$ is Fréchet–Urysohn. Therefore, one can choose a subsequence $(x_{n_k})_k$ converging to x.

(ii) Let $(U_n)_{n\in \omega }$ be a sequence of open sets in X. Choose a sequence $(M_n)_{n\in \omega }$ of finite sets in X such that $M_n\subseteq U_n$ for $n\in \omega$ and a sequence $(M_n)_{n\in \omega }$ has a cluster point x. Set $M:=\bigcup \{M_n:n\in \omega \}$. From Proposition 2.2 it follows that the space $Y=\{x\}\cup M$ is Fréchet–Urysohn. It follows that M contains a non-trivial convergent sequence S. Since $M_n$ is finite for every $n\in \omega$, it is clear that S contains a convergent subsequence $\{x_{n_k}\}_{k\in \omega }$ such that $x_{n_k}\in U_{n_k}$ for every $k\in \omega$. Thus X is a near sequentially compact space. $\square$

Corollary 8.7

Under $\omega _1<\mathfrak {p}$, if G is a countably compact group such that $w(G)\le \aleph _1$, then G is sequentially compact.

Problem 8.8

Is there a sequentially compact group G with $w(G)=\aleph _1$ that is not a k-space ($k_\mathbb {R}$-space)?

Note that Proposition 6.1 implies that there is a sequentially group G with $w(G)=\aleph _2$ that is not a $k_\mathbb {R}$-space.

Proposition 8.9

Under $\textrm{CH}$, there is an infinite separable countably compact abelian group G such that $w(G)=|G|=\mathfrak {c}$ and G does not have infinite compact subsets. The group G is not an Ascoli space.

Proof

Let $G'$ be some infinite countably compact abelian group without non-trivial convergent sequences, see [17]. It follows from Proposition 7.4 that $G'$ has some separable subgroup G of cardinality $\mathfrak {c}$ which does not have infinite compact subsets. Therefore, by Corollary 2.3 of [11], G is not an Ascoli space. Since G is separable, $w(G)\le \mathfrak {c}$. Since the group G is not metrizable, we have $w(G)=\mathfrak {c}$. $\square$

Problem 8.10

Does there exist in $\textrm{ZFC}$ a countably compact group with $w(G)=\aleph _1$ that is not k-space ($k_\mathbb {R}$-space)?

The following theorem follows from Theorem 2.1 and Proposition 8.6.

Theorem 8.11

If G is a pseudocompact group such that $w(G)<\mathfrak {p}$, then G is near sequentially compact and hence Ascoli.

Corollary 8.12

Under $\omega _1<\mathfrak {p}$, if G is a pseudocompact group such that $w(G)\le \aleph _1$, then G is near sequentially compact and hence Ascoli.

Below we construct a pseudocompact space which is not weakly selectively pseudocompact.

Example 8.13

Let S be the $\Sigma$-product in $[0,1]^{\omega _1}$ around a point $x\in [0,1]^{\omega _1}$, and let

$$ X = \left( \{\omega \} \times ([0,1]^{\omega _1}\setminus S)\right) \cup \left( \omega \times S\right) \subseteq (\omega +1)\times [0,1]^{\omega _1}. $$

Then X is a pseudocompact space of weight $\aleph _1$ which is not weakly selectively pseudocompact.

Proof

Clearly, $w(X)=\omega _1$. To show that X is not weakly selectively pseudocompact, for every $n\in \omega$, let $U_n:=\{n\}\times S=\big (\{n\}\times [0,1]^{\omega _1}\big )\cap X$ be an open subset in X. Take an arbitrary sequence $\{M_n\}_{n\in \omega }$ of finite subsets in X such that $M_n\subseteq U_n$ for every $n\in \omega$. Clearly, $\bigcup _n M_n$ is discrete in X. Any cluster point y of $\bigcup _n M_n$ in $(\omega +1)\times [0,1]^{\omega _1}$ has a form $y=\{w\}\times z$ with $z\in [0,1]^{\omega _1}$. The countability of $\bigcup _n M_n$ implies that $z\in S$. Therefore $y\not \in X$, and hence $\bigcup _n M_n$ is discrete in X. Thus X is not weakly selectively pseudocompact.

To show that X is pseudocompact, we remark first that each open nonempty set U contains an open set of the form $\big (\{n\}\times [0,1]^{\omega _1}\big )\cap X= U_n$. Now, let $(W_n)_n$ be a sequence of open nonempty sets in X. By the remark we can assume that $W_n=\{m_n\}\times V_n$, where $V_n$ is an open subset of S. If the sequence $(m_n)_n$ is bounded by N, then the sequence $(W_n)_n$ lies in the countably compact space $\{0,1, \ldots ,N\}\times S$ and, therefore, it has a limit point. If $(m_n)_n$ is unbounded, then we can assume that $(m_n)_n$ is an increasing sequence of integers.

We show that the sequence $(V_n)_n$ accumulates at some point $x\in [0,1]^{\omega _1}\setminus S$. For $n\in \omega$, there exists an open set $W_n=\prod _{\alpha <\omega _1} W_{n,\alpha }$ in $[0,1]^{\omega _1}$ such that $W_n\cap X\subseteq V_n$, where $W_{n,\alpha }$ is an open subset of [0, 1] and the set $A_n=\{\alpha <\omega _1: W_{n,\alpha }\ne [0,1]\}$ is finite. Then the set $A=\bigcup _{n<\omega } A_n$ is at most countable. Let Z be the set of accumulation points for the sequence of sets $(W_n)_n$ in $[0,1]^{\omega _1}$. Then $Z=F\times [0,1]^{\omega _1\setminus A}$, where F is some closed non-empty subset of $[0,1]^A$. Therefore, the set $Z\setminus S$ is not empty. Let $x\in Z\setminus S$. Then the sequence $(V_n)_n$ accumulates at $x\in [0,1]^{\omega _1}\setminus S$.

Then $(W_n)_n$ accumulates at the point $(\omega ,x)\in X$. Thus X is pseudocompact. $\square$

Proposition 8.14

There exists an abelian pseudocompact $k_\mathbb {R}$-group G of weight $w(G)= \aleph _1$ which is not a k-space.

Proof

Let X be a space of weight $\aleph _1$ that is not a k-space. (For example, X is one point Lindelöfication of a discrete space of cardinality $\aleph _1$.) According to Theorem 5.4, the space X is closed embeddable into an abelian pseudocompact $k_\mathbb {R}$-group G such that $w(G)=\aleph _1$. The group G is not a k-space because, by construction, it contains a closed subspace X which is not a k-space. $\square$

Under $\textrm{CH}$, there is a countably compact group G such that $w(G)=\aleph _1$ and G is not an Ascoli space (see Proposition 8.9). On the other hand, under $\omega _1<\mathfrak {p}$, Corollary 8.12 implies that any pseudocompact group with $w(G) =\aleph _1$ is an Ascoli space.

Problem 8.15

Does there exist in $\textrm{ZFC}$ a pseudocompact (countably compact) group with $w(G)=\aleph _1$ that is not a $k_\mathbb {R}$-space?

Proposition 8.16

Under $\textrm{CH}$, there exists an abelian countably compact $k_\mathbb {R}$-group G of weight $w(G)=\aleph _1$ which is not a k-space.

Proof

Let H be a countably compact abelian group such that $w(H)=\aleph _1$ and H is not a k-space (see Proposition 8.9). According to Theorem 5.1, the group H is topologically isomorphic to a closed subgroup of an abelian countably compact $k_\mathbb {R}$-group G such that $w(G)=\aleph _1$. The group G is not a k-space because it contains a closed subspace H which is not a k-space. $\square$

Note that an $\omega$-bounded group G of weight $w(G)=\aleph _1$ is a k-space (Corollary 8.2) and there exists an abelian $\omega$-bounded $k_\mathbb {R}$-group G of weight $w(G)=\aleph _2$ which is not a k-space (Proposition 6.2).

Problem 8.17

Does there exist in $\textrm{ZFC}$ a countably compact $k_\mathbb {R}$-group of weight $w(G)=\aleph _1$ that is not a k-space?