On quasi $\kappa$-metrizable spaces

TL;DR

This paper explores quasi κ-metrizable spaces, a class that remains stable under arbitrary products and extends Shchepin's κ-metrizable spaces, providing new insights into their structure and properties.

Contribution

It introduces and studies the class of quasi κ-metrizable spaces, expanding the understanding of metrizable space subclasses and their invariance properties.

Findings

Quasi κ-metrizable spaces are invariant under arbitrary products.

This class properly contains Shchepin's κ-metrizable spaces.

The paper characterizes the relationship between quasi κ-metrizable and κ-metrizable spaces.

Abstract

The aim of this paper is to investigate the class of quasi $\kappa$-metrizable spaces. This class is invariant with respect to arbitrary products and contains Shchepin's $\kappa$-metrizable spaces as a proper subclass.

Equations

$$d(y,\overline{U})=\sup \{\rho (x,f^{-1}(\overline{U})):x\in f^{-1}(y)\}.$$

$$[U_1,..,U_k]=\{H\in \exp \mathrm{X}:\mathrm{H}\subset \bigcup _{\mathrm{i}=1}^{\mathrm{k}}{\mathrm{U}}_{\mathrm{i}}\text{mbox}\mathrm{and}\mathrm{H}\cap {\mathrm{U}}_{\mathrm{i}}\ne \varnothing \text{mbox}\mathrm{forall}\mathrm{i}\}$$

$$\xi (F,[U_1,..,U_k])=\frac{1}{k}\min \{\underset{x\in F}{\inf }\underset{i}{\max }{\xi }_{\rho }(x,U_i),\underset{i}{\min }\underset{x\in F}{\sup }{\xi }_{\rho }(x,U_i)\}.$$

$${\mathrm{e}}_1(U)=\bigcup \{\mathrm{e}(V):V\in {\mathcal{T}}_X\text{mbox}and\overline{V}\subset U\}.$$

$$\rho (x,\overline{\bigcup _{\alpha }{\mathrm{e}}_1(U_{\alpha })})=\underset{\alpha }{\inf }\rho (x,\overline{{\mathrm{e}}_1(U_{\alpha })})=\underset{\alpha }{\inf }d(x,C_{\alpha }).$$

Full text

Research supported in part by NSERC Grant 261914-13

1. Introduction

Recall that a $\kappa$-metric [8] on a space $X$ is a non-negative function $\rho(x,C)$ of two variables, a point $x\in X$ and a regularly closed set $C\subset X$, satisfying the following conditions:

• K1)

$\rho(x,C)=0$ iff $x\in C$;

• K2)

If $C\subset C^{\prime}$, then $\rho(x,C^{\prime})\leq\rho(x,C)$ for every $x\in X$;

• K3)

$\rho(x,C)$ is continuous function of $x$ for every $C$;

• K4)

$\rho(x,\overline{\bigcup C_{\alpha}})=\inf_{\alpha}\rho(x,C_{\alpha})$ for every increasing transfinite family $\{C_{\alpha}\}$ of regularly closed sets in $X$.

A $\kappa$-metric on $X$ is said to be regular if it satisfy also next condition

• K5)

$\rho(x,C)\leq\rho(x,C^{\prime})+\overline{\rho}(C^{\prime},C)$ for any $x\in X$ and any two regularly closed sets $C,C^{\prime}$ in $X$, where $\overline{\rho}(C^{\prime},C)=\sup\{\rho(y,C):y\in C^{\prime}\}$.

We say that a function $\rho(x,C)$ is an quasi $\kappa$-metric (resp., a regular quasi $\kappa$-metric) on $X$ if it satisfies the axioms $K2)-K4)$ (resp., $K2)-K5))$ and the following one:

• K$1)^{*}$

For any $C$ there is a dense open subset $V$ of $X\setminus C$ such that $\rho(x,C)>0$ iff $x\in V$.

Obviously, we can assume that $\rho(x,C)\leq 1$ for all $x$ and $C$, in such a case we say that $\rho$ is a normed quasi $\kappa$-metric.

Quasi $\kappa$-metrizable spaces were introduced in [9]. Our interest of this class was originated by Theorem 1.4 from [9] stating that a compact space is quasi $\kappa$-metrizable if and only if it is skeletally generated. Unfortunately, the presented there proof of the implication that any skeletally generated compactum is quasi $\kappa$-metrizable is not correct. Despite of this incorrectness, the class of quasi $\kappa$-metrizable is very interesting. It is closed with respect to arbitrary products and contains as a proper subclass the $\kappa$-metrizable spaces. The aim of this paper is to investigate the class of quasi $\kappa$-metrizable spaces, and to provide a correct characterization of skeletally generated spaces.

The class of skeletally generated spaces was introduced in [10]. According to [9, Theorem 1.1], a space is skeletally generated iff it is $I$-favorable in the sense of [2]. Recall that a map $f:X\to Y$ is skeletal if $\mathrm{Int}\overline{f(U)}\neq\varnothing$) for every open $U\subset X$. A space $X$ is skeletally generated [10] if there is an inverse system $\displaystyle S=\{X_{\alpha},p^{\beta}_{\alpha},A\}$ of separable metric spaces $X_{\alpha}$ and skeletal surjective bounding maps $p^{\beta}_{\alpha}$ satisfying the following conditions: $(1)$ the index set $A$ is $\sigma$-complete (every countable chain in $A$ has a supremum in $A$); $(2)$ for every countable chain $\{\alpha_{n}\}_{n\geq 1}\subset A$ with $\beta=\sup\{\alpha_{n}\}_{n\geq 1}$ the space $X_{\beta}$ is a (dense) subset of $\displaystyle\lim_{\leftarrow}\{X_{\alpha_{n}},p^{\alpha_{n+1}}_{\alpha_{n}}\}$; $(3)$ $X$ is embedded in $\displaystyle\lim_{\leftarrow}S$ and $p_{\alpha}(X)=X_{\alpha}$ for each $\alpha$, where $p_{\alpha}\colon\displaystyle\lim_{\leftarrow}S\to X_{\alpha}$ is the $\alpha$-th limit projection. If in the above definition all bounding maps $p^{\beta}_{\alpha}$ are open, we say that $X$ is openly generated.

All topological spaces are Tychonoff and the single-valued maps are continuous. The paper is organized as follows: Section 2 contains the proof that any product of quasi $\kappa$-metrizable spaces is also quasi $\kappa$-metrizable. In Section 3 we provide some additional properties of quasi $\kappa$-metrizable spaces. For example, it is shown that this property is preserved by open and perfect surjections, and that the Čech-Stone compactification of any pseudocompact quasi $\kappa$-metrizable space is quasi $\kappa$-metrizable. The results from Section 3 imply that there exist quasi $\kappa$-metrizable spaces which are not $\kappa$-metrizable. In Section 4 we introduce a similar wider class of spaces, the weakly $\kappa$-metrizable spaces, and proved that a compact space is skeletally generated if and only if it is weakly $\kappa$-metrizable. Hence, every skeletally generated space is also weakly $\kappa$-metrizable. The converse implication is interesting for spaces with a countable cellularity only, but it is still unknown, see Questions 4.3 - 4.4.

2. Products of quasi $\kappa$-metrizable spaces

Let $\mathcal{B}$ be a base for a space $X$ consisting of regularly open sets. A real-valued function $\xi:X\times\mathcal{B}\to[0,1]$ will be called a $\pi$-capacity if it satisfies the following conditions:

• E1)

$\xi(x,U)=0$ for $x\not\in U$, and $0\leq\xi(x,U)\leq 1$ for $x\in U$.

• E2)

For any $U\in\mathcal{B}$ the set $\{x\in U:\xi(x,U)>0\}$ is dense in $U$.

• E3)

For any $U$ the function $\xi(x,U)$ is lower semi-continuous, i.e if $\xi(x_{0},U)>a$ for some $x_{0}\in X$ and $a\in\mathbb{R}$, then there is a neighborhood $O_{x_{0}}$ with $\xi(x,U)>a$ for all $x\in O_{x_{0}}$.

• E4)

For any two mappings $U:T\to\mathcal{B}$ and $\lambda:T\to X$, where $T$ is a set with an ultrafilter $\mathcal{F}$, such that the limit $\displaystyle\widetilde{\lambda}=\lim_{\mathcal{F}}\lambda(t)$ exists and $\displaystyle\lim_{\mathcal{F}}\xi(\lambda(t),U(t))>a>0$, then there exists $\widetilde{U}\in\mathcal{B}$ such that $\widetilde{U}\subset\overline{\lim}_{\mathcal{F}}U(t)$ and $\xi(\widetilde{\lambda},\widetilde{U})>a$. Here, $\overline{\lim}_{\mathcal{F}}U(t)=\bigcap_{F\in\mathcal{F}}\overline{\bigcup_{t\in F}U(t)}$.

A capacity is called regular if it satisfies also the following condition:

• E5)

If $\xi(x,U)>a>0$, there exists $U_{a}\in\mathcal{B}$ such that $\xi(x,U_{a})\geq a$ and $\xi(y,U)\geq\xi(x,U)-a$ for all $y\in U_{a}$.

Our definition of a $\pi$-capacity is almost the same as the Shchepin’s definition [7] of capacity, the only difference is that Shchepin requires $\xi(x,U)>0$ for all $x\in U$.

Lemma 2.1**.**

Suppose $\xi:X\times\mathcal{B}\to[0,1]$ is a (regular) $\pi$-capacity on $X$. Then the function $\rho_{\xi}(x,C)$, $\rho_{\xi}(x,C)=0$ if $x\in C$ and $\rho_{\xi}(x,C)=\sup\{\xi(x,U):U\cap C=\varnothing\}$ otherwise, is a (regular) quasi $\kappa$-metric on $X$.

Proof.

Suppose $\xi$ is a $\pi$-capacity on $X$. Clearly, $\rho_{\xi}$ satisfies condition $K2)$. According to the proof of [8, Proposition 6, chapter 3] $\rho_{\xi}$ also satisfies conditions $K3)-K4)$. To check condition $K1)^{*}$, let $C$ be a proper regularly closed subset of $X$. Then there is a subfamily $\{U_{\alpha}:\alpha\in A\}$ of $\mathcal{B}$ covering $X\setminus C$. For every $\alpha$ the set $V_{\alpha}=\{x\in U_{\alpha}:\xi(x,U_{\alpha})>0\}$ is dense in $U_{\alpha}$. So $V=\bigcup_{\alpha\in A}V_{\alpha}$ is dense in $X\setminus C$ and $\rho_{\xi}(x,C)>0$ for all $x\in V$.

Let show that if $\xi$ is a regular $\pi$-capacity, then $\rho_{\xi}$ satisfies condition $K5)$. Suppose $C,C^{\prime}$ are two regularly closed subsets of $X$ and $x\in X$. Obviously, $\rho_{\xi}(x,C)\leq\rho_{\xi}(x,C^{\prime})$ implies $\rho_{\xi}(x,C)\leq\rho_{\xi}(x,C^{\prime})+\overline{\rho_{\xi}}(C^{\prime},C)$. So, let $\rho_{\xi}(x,C)>\rho_{\xi}(x,C^{\prime})$, and choose an integer $m$ such that $\rho_{\xi}(x,C)>\rho_{\xi}(x,C^{\prime})+1/n$ for all $n\geq m$. Hence, there is $U\in\mathcal{B}$ such that $U\cap C=\varnothing$ and $\xi(x,U)>a_{n}=\rho_{\xi}(x,C^{\prime})+1/n$. So, according to condition $E5)$, there is $U_{a_{n}}\in\mathcal{B}$ such that $\xi(x,U_{a_{n}})\geq a_{n}$ and $\xi(x,U)\leq\xi(y,U)+a_{n}$ for all $y\in U_{a_{n}}$. Since $\xi(x,U_{a_{n}})\geq a_{n}$, $U_{a_{n}}\cap C^{\prime}\neq\varnothing$ (otherwise we would have $\rho_{\xi}(x,C^{\prime})\geq\rho_{\xi}(x,C^{\prime})+1/n$). Hence, $\xi(x,U)\leq\xi(z,U)+a_{n}$ for every $z\in U_{a_{n}}\cap C^{\prime}$, which yields $\xi(x,U)\leq\overline{\rho_{\xi}}(C^{\prime},C)+\rho_{\xi}(x,C^{\prime})+1/n$ for all $n\geq m$ and $U\in\mathcal{B}$ with $U\cap C=\varnothing$. Therefore, $\rho_{\xi}(x,C)\leq\rho_{\xi}(x,C^{\prime})+\overline{\rho_{\xi}}(C^{\prime},C)$. ∎

Lemma 2.2**.**

Let $\rho$ be a (regular) normed quasi $\kappa$-metric on $X$ and $\mathcal{B}$ be the family of all regularly open subsets of $X$. Then the formula $\xi(x,U)=\sup\{\rho(x,C):C\cup U=X\}$ defines a (regular) $\pi$-capacity on $X$.

Proof.

It is easy to show that $\xi$ satisfies conditions $E1)$ and $E3)$. Condition $E4)$ was established in [7, Lemma 3] in the case $\rho$ is a $\kappa$-metric, but the same proof works for quasi $\kappa$-metrics as well. Let show condition $E2)$. We fix $U\in\mathcal{B}$ and consider the family $\mathcal{B}_{U}=\{G\in\mathcal{B}:G\varsubsetneq U\}$. For every $G\in\mathcal{B}_{U}$ the set $V_{G}=\{x\in G:\rho(x,C_{G})>0\}$ is open and non-empty, where $C_{G}=X\setminus G$. Hence, $V=\bigcup_{G\in\mathcal{B}_{U}}V_{G}$ is dense in $U$. Moreover, $\xi(x,U)\geq\rho(x,C_{G})$ for every $G\in\mathcal{B}_{U}$ and $x\in V_{G}$ because $U\cup C_{G}=X$. So, $\xi(x,U)>0$ for all $x\in V$.

It remains to show that $\xi$ is regular, i.e. it satisfies $E5)$, provided $\rho$ is regular. Let $U\in\mathcal{B}$ and $\xi(x,U)>a>0$ for some $x\in U$. Since $C_{U}=X\setminus U$ is the smallest regularly closed subset $C$ of $X$ with $U\cup C=X$, we have $\xi(z,U)=\rho(z,C_{U})$ for all $z\in X$. The set $W=\{y\in X:\rho(y,C_{U})>\rho(x,C_{U})-a\}$ is open and non-empty because $\rho(\cdot,C)$ is continuous and $x\in W$. Let $U_{a}\in\mathcal{B}$ be the set $U_{a}=\rm{Int}\overline{W}$ and $C_{U_{a}}=X\setminus U_{a}$. Observe that $W\subset U$, which implies $U_{a}\subset U$ and $C_{U}\subset C_{U_{a}}$. Then, by $K5)$, $\rho(x,C_{U})\leq\rho(x,C_{U_{a}})+\overline{\rho}(C_{U_{a}},C_{U})$. Consequently, $\rho(x,C_{U_{a}})\geq\rho(x,C_{U})-\overline{\rho}(C_{U_{a}},C_{U})$. On the other hand, $\overline{\rho}(C_{U_{a}},C_{U})=\sup_{y\in C_{U_{a}}}\rho(y,C_{U})=\sup_{y\in C_{U_{a}}\setminus C_{U}}\rho(y,C_{U})$ because $\rho(y,C_{U})=0$ for all $y\in C_{U}$. Observe also that $\rho(y,C_{U})\leq\rho(x,C_{U})-a$ for all $y\in C_{U_{a}}\setminus C_{U}$. So, $\overline{\rho}(C_{U_{a}},C_{U})\leq\rho(x,C_{U})-a$, and thus $\xi(x,U_{a})=\rho(x,C_{U_{a}})\geq a$. Finally, $\xi(y,U)=\rho(y,C_{U})\geq\rho(x,C_{U})-a=\xi(x,U)-a$ for all $y\in U_{a}$ because $U_{a}\subset\overline{W}$. Hence, $\xi$ satisfies $E5)$. ∎

Let consider the following condition, where $\rho(x,C)$ is a non-negative function with $C$ being a regularly closed subset of $X$:

• K$1)^{**}$

For any regularly closed $C\subsetneqq X$ there is $y\not\in C$ with $\rho(y,C)>0$ and $\rho(x,C)=0$ for all $x\in C$.

Remark 2.3**.**

Observe that in the previous lemma we actually proved the following more general statement: Suppose $\rho$ satisfies conditions $K1)^{**}$ and $K2)-K4)$, and $\rho(x,C)\leq 1$ for all $x\in X$ and all regularly closed sets $C\subset X$. Then $\xi(x,U)=\sup\{\rho(x,C):C\cup U=X\}$ defines a $\pi$-capacity on $X$. Moreover, $\xi$ is regular if $\rho$ satisfies also $K5)$.

Corollary 2.4**.**

Suppose there is a function $\rho$ on $X$ satisfying conditions $K1)^{**}$ and $K2)-K4)$. Then there is a quasi $\kappa$-metric $d$ on $X$. Moreover, $d$ is regular if $\rho$ satisfies also condition $K5)$.

Proof.

We can suppose that $\rho$ is normed. Then, by Lemma 2.2, there is a $\pi$-capacity $\xi$ on $X$. Finally, Lemma 2.1, implies the existence of a quasi $\kappa$-metric $d$ on $X$. Moreover, if $\rho$ satisfies condition $K5)$, then $\xi$ is regular, so is $d$. ∎

Theorem 2.5**.**

Any product of (regularly) quasi $\kappa$-metrizable spaces is (regularly) quasi $\kappa$-metrizable.

Proof.

Suppose $X=\prod_{\alpha\in A}X_{\alpha}$ and for every $\alpha$ there is a normed (regular) quasi $\kappa$-metric $\rho_{\alpha}$ on $X$. Following the proof of [7, Theorem 2], for every $\alpha$ let $\mathcal{T}_{\alpha}$ be the family of all regularly open subsets of $X_{\alpha}$ and let $\mathcal{B}$ be the standard base for $X$ consisting of sets of the form $U=\bigcap_{i=1}^{n}\pi_{\alpha_{i}}^{-1}(U_{i})$ with $U_{i}\in\mathcal{T}_{\alpha_{i}}$ and $U_{i}\neq X_{\alpha_{i}}$, where $\pi_{\alpha}:X\to X_{\alpha}$ is the projection. Denote by $v(U)$ the collection $\{\alpha_{1},...,\alpha_{n}\}$. According to Lemma 2.2, for every $\alpha$ there exists a (regular) $\pi$-capacity $\xi_{\alpha}$ on $X_{\alpha}$. Consider the function $\xi:X\times\mathcal{B}\to\mathbb{R}$ defined by $\displaystyle\xi(x,U)=\inf_{\alpha\in v(U)}\xi_{\alpha}(\pi_{\alpha}(x),\pi_{\alpha}(U))/|v(U)|$. Obviously, condition $E1)$ is satisfied. Moreover, since for each $\alpha_{i}$ the set $W_{i}=\{z\in X_{\alpha_{i}}:\xi_{i}(z,U_{i})>0\}$ is open and dense in $U_{i}$, the set $W=\bigcap_{i=1}^{n}\pi_{\alpha_{i}}^{-1}(W_{i})$ is open and dense in $U$. So, $\xi$ satisfies condition $E2)$. Shchepin has shown that the function $\xi$ is a (regular) capacity provide each $\xi_{\alpha}$ is so, see the proof of [8, Theorem 15] and [7, Theorem 2]. The same arguments show that $\xi$ also satisfies conditions $E3)-E4)$, and condition $E5)$ in case each $\xi_{\alpha}$ is regular. Therefore, $\xi$ is a (regular) $\pi$-capacity. Finally, by Lemma 2.1, there exists a (regular) quasi $\kappa$-metric on $X$. ∎

3. Some more properties of quasi $\kappa$-metrizable spaces

Proposition 3.1**.**

Let $X$ be a quasi $\kappa$-metrizable space and $Y\subset X$. The $Y$ is also quasi $\kappa$-metrizable in each of the following cases: (i) $Y$ is dense in $X$; (ii) $Y$ is regularly closed in $X$; (iii) $Y$ is open in $X$.

Proof.

If $\rho$ is a quasi $\kappa$-metric on $X$ and $Y\subset X$ is dense, the equality $d(y,\overline{U}^{Y})=\rho(y,\overline{U}^{X})$, where $U\subset Y$ is open, defines a quasi $\kappa$-metric on $Y$. The second case follows from the observation that every regularly closed subset of $Y$ is also regularly closed in $X$. The third case is a consequence of the first two because every open subset of $X$ is dense in its closure. ∎

Let consider the following condition.

• K$4)^{*}$

$\rho(x,\overline{\bigcup C_{n}})=\inf_{n}\rho(x,C_{n})$ for every increasing sequence $\{C_{n}\}$ of regularly closed sets in $X$.

Lemma 3.2**.**

Suppose $X$ is a space admitting a non-negative function $\rho(x,C)$ satisfying conditions $K1)^{*}$, $K2)$, $K3)$ and $K4)^{*}$. Then $X$ is quasi $\kappa$-metrizable provided $X$ has countable cellularity. In particular, every compact space admitting such a function $\rho$ is quasi $\kappa$-metrizable.

Proof.

It suffices to show that $\rho$ satisfies condition $K4)$ in case $X$ is of countable cellularity, and this follows from [3, Proposition 2.1]. For reader’s convenience we provide a proof. Let $\{C_{\alpha}\}_{\alpha}$ be an increasing transfinite family of regularly closed sets in $X$. Then $\overline{\bigcup_{\alpha}C_{\alpha}}=\overline{\bigcup_{\alpha}U_{\alpha}}$ and $\{U_{\alpha}\}_{\alpha}$ is also increasing, where $U_{\alpha}$ is the interior of $C_{\alpha}$. Since $X$ has countable cellularity, there are countably many $\alpha_{i}$ such that $\bigcup_{i\geq 1}U_{\alpha_{i}}$ is dense in $\bigcup_{\alpha}U_{\alpha}$. We can assume that the sequence $\{\alpha_{i}\}$ is increasing, so is the sequence $\{U_{\alpha_{i}}\}$. Because $\rho$ satisfies condition $K4)^{*}$, we have $\rho(x,\overline{\bigcup C_{\alpha_{i}}})=\inf_{i}\rho(x,C_{\alpha_{i}})$. This implies that $\rho(x,\overline{\bigcup C_{\alpha}})=\inf_{\alpha}\rho(x,C_{\alpha})$. Indeed, since $\overline{\bigcup C_{\alpha_{i}}}=\overline{\bigcup C_{\alpha}}$, $\inf_{\alpha}\rho(x,C_{\alpha})<\inf_{i}\rho(x,C_{\alpha_{i}})$ for some $x\in X$ would implies the existence of $\alpha_{0}$ with $\rho(x,C_{\alpha_{0}})<\inf_{i}\rho(x,C_{\alpha_{i}})$ for all $i$. Because any two elements of the family $\{C_{\alpha}\}_{\alpha}$ are comparable with respect to inclusion, the last inequality means that $C_{\alpha_{0}}$ contains all $C_{\alpha_{i}}$. Hence, $C_{\alpha_{0}}=\overline{\bigcup_{\alpha}C_{\alpha}}$ and $\rho(x,C_{\alpha_{0}})$ would be equal to $\inf_{i}\rho(x,C_{\alpha_{i}})$, a contradiction.

It was shown in [9, Theorem 1.4] that every compact space $X$ admitting a non-negative function $\rho(x,C)$ satisfying conditions $K1)^{*}$, $K2)$, $K3)$ and $K4)^{*}$ is skeletally generated, and hence $X$ has countable cellularity. Therefore, any such compactum is quasi $\kappa$-metrizable. ∎

It was shown by Chigogidze [1] that the Čech-Stone compactification of every pseudocompact $\kappa$-metrizable space is $\kappa$-metrizable. We have a similar result for quasi $\kappa$-metrizable spaces.

Theorem 3.3**.**

If $X$ is a pseudocompact (regularly) quasi $\kappa$-metrizable space, then $\beta X$ is (regularly) quasi $\kappa$-metrizable.

Proof.

Suppose $\rho(x,C)$ is a quasi $\kappa$-metric on $X$. We can assume that $\rho(x,\overline{U}^{X})\leq 1$ for all $x\in X$ and all open $U\subset X$ ($\overline{U}^{X}$ denotes the closure of $U$ in $X$). For every open $W\subset\beta X$ consider the function $f_{W}$ on $X$ defined by $f_{W}(x)=\rho(x,\overline{W\cap X}^{X})$. Let $\widetilde{f}_{W}:\beta X\to\mathbb{R}$ be the continuous extension of $f_{W}$, and define $d(y,\overline{W})=\widetilde{f}_{W}(y)$, $y\in\beta X$. Obviously, $d(y,\overline{W})=0$ if $y\in W\cap X$. Since $W\cap X$ is dense in $\overline{W}$, $d(y,\overline{W})=0$ for all $y\in\overline{W}$. Moreover, if $\overline{W}\neq\beta X$, then $\overline{W\cap X}^{X}\neq X$. So, there is an open dense subset $V$ of $X\setminus\overline{W}$ with $\rho(x,\overline{W\cap X}^{X})>0$ for all $x\in V$. Since $f_{W}$ is continuous, the set $\widetilde{V}=\{y\in\beta X:f_{W}(y)>0\}$ is open in $\beta X$ and disjoint from $\overline{W}$. Finally, because $V\subset\widetilde{V}$ and $V$ is dense in $X\setminus\overline{W}$, $\widetilde{V}$ is dense in $\beta X\setminus\overline{W}$. So, $d$ satisfies condition $K1)^{*}$. Conditions $K2)$ and $K3)$ also hold. Hence, by Lemma 3.2, it suffices to show that $d$ satisfies $K4)^{*}$. To this end, let $\{\overline{W}_{n}\}$ be an increasing sequence of regularly closed subsets of $\beta X$ and $W=\bigcup_{n\geq 1}W_{n}$. We have $d(y,\overline{W})\leq\inf_{n}d(y,\overline{W}_{n})$ for all $y\in\beta X$. Moreover, since $\rho$ satisfies $K4)$, $d(y,\overline{W})=\inf_{n}d(y,\overline{W}_{n})$ if $y\in X$. Suppose there is $y_{0}\in\beta X\setminus X$ with $d(y_{0},\overline{W})<\inf_{n}d(y_{0},\overline{W}_{n})$. Consequently, for every $n$ there exists a neighborhood $V_{n}$ of $y_{0}$ in $\beta X$ such that $\delta<d(y,\overline{W}_{n})$ for all $y\in V_{n}$, where $d(y_{0},\overline{W})<\delta<\inf_{n}d(y_{0},\overline{W}_{n})$. We also choose a neighborhood $V_{0}$ of $y_{0}$ with $d(y,\overline{W})<\delta$ for all $y\in V_{0}$. This implies that $d(y,\overline{W})<\delta\leq\inf_{n}d(y,\overline{W}_{n})$ provided $y\in V=\bigcap_{n\geq 1}V_{0}\cap V_{n}$. But $V\cap X\neq\varnothing$ because $X$ is pseudocompact. Thus, $d(y,\overline{W})<\inf_{n}d(y,\overline{W}_{n})$ for any $y\in V\cap X$, a contradiction.

It follows from the definition of $d$ that it satisfies condition $K5)$ provided $\rho$ is regular. ∎

Corollary 3.4**.**

Every pseudocompact quasi $\kappa$-metrizable space $X$ is skeletally generated.

Proof.

We already noted that every quasi $\kappa$-metrizable compactum is skeletally generated, see [9]. So, by Theorem 3.3, $\beta X$ is skeletally generated. Finally, by [2] and [9], every dense subset of a skeletally generated space is also skeletally generated. ∎

Proposition 3.5**.**

Suppose $f:X\to Y$ is a perfect open surjection and $X$ is (regularly) quasi $\kappa$-metrizable. Then $Y$ is also (regularly) quasi $\kappa$-metrizable.

Proof.

Let $\rho$ be a quasi $\kappa$-metric on $X$. Since $f$ is open, $f^{-1}(\overline{U})=\overline{f^{-1}(U)}$ for any open $U\subset Y$. So, $f^{-1}(\overline{U})$ is regularly closed set in $X$ and we define

[TABLE]

One can check that $d$ satisfies conditions $K2)$ and $K4)$, and condition $K5)$ in case $\rho$ is regular. Moreover, $\overline{U}\neq Y$ implies $f^{-1}(\overline{U})\neq X$. So, there is a dense open subset $V\subset X\setminus f^{-1}(\overline{U})$ such that $\rho(x,f^{-1}(\overline{U}))>0$ iff $x\in V$. Then $f(V)$ is a dense and open subset of $Y\setminus\overline{U}$ such that $f^{-1}(y)\cap V\neq\varnothing$ for all $y\in f(V)$. Hence, $d(y,\overline{U})>0$ if $y\in f(V)$. If $y\not\in f(V)$, then $f^{-1}(y)\cap V=\varnothing$. Thus, $d(y,\overline{U})>0$ iff $y\in f(V)$. Finally, let check continuity of the functions $d(\cdot,\overline{U})$. Suppose $d(y_{0},\overline{U})<\varepsilon$ for some $y_{0}$ and $U$. Then $\rho(x,f^{-1}(\overline{U}))<\varepsilon$ for all $x\in f^{-1}(y_{0})$. Consequently, there is a neighborhood $W$ of $f^{-1}(y_{0})$ with $\rho(x,f^{-1}(\overline{U}))<\varepsilon$ for all $x\in W$. Since, $f$ is closed, $y_{0}$ has a neighborhood $G$ such that $f^{-1}(G)\subset W$. This implies that $d(y,\overline{U})<\varepsilon$ for all $y\in G$. Now, let $d(y_{0},\overline{U})>\delta$ for some $\delta\in\mathbb{R}$. So, there exists $x_{0}\in f^{-1}(y_{0})$ with $\rho(x_{0},f^{-1}(\overline{U}))>\delta$. Choose a neighborhood $O$ of $x_{0}$ such that $\rho(x,f^{-1}(\overline{U}))>\delta$ for all $x\in O$. Then, $f(O)$ is a neighborhood of $y_{0}$ and $d(y,\overline{U})>\delta$ for any $y\in f(O)$. Therefore, each $d(.,\overline{U})$ is continuous. ∎

Recall that a surjective map $f:X\to Y$ is irreducible provided there is no a proper closed subset $F$ of $X$ with $f(F)=Y$.

Proposition 3.6**.**

Let $f:X\to Y$ be a perfect irreducible surjection, and $Y$ is (regularly) quasi $\kappa$-metrizable. Then $X$ is also (regularly) quasi $\kappa$-metrizable.

Proof.

Suppose $\rho$ is a quasi $\kappa$-metric on $Y$. For every regularly closed $C\subset X$ define $d(x,C)=\rho(f(x),f(C))$. This definition is correct because $f(C)$ is regularly closed in $Y$. Indeed, let $C=\overline{U}$ with $U$ open in $X$. Since $f$ perfect and irreducible, we have $f(C)=\overline{U_{\sharp}}$, where $U_{\sharp}=\{y\in Y:f^{-1}(y)\subset U\}=Y\setminus f(X\setminus U)$ is open in $Y$. It is easily seen that $d$ satisfies conditions $K2)$ and $K3)$. Condition $K4)$ follows from the equality $f(\overline{\bigcup_{\alpha}C_{\alpha}})=\overline{\bigcup_{\alpha}f(C_{\alpha})}$ for any family of regularly closed sets in $X$. To see that $d$ satisfies also condition $K1)^{*}$, we observe that for every regularly closed $C\subsetneqq X$ there is a dense open subset $V\subset Y\setminus f(C)$ such that $\rho(y,f(C))>0$ iff $y\in V$. Then $W=f^{-1}(V)$ is open in $X$ and disjoint from $C$. Moreover, $d(x,C)>0$ iff $x\in W$. It remains to show that $W$ is dense in $X\setminus C$. And that is really true because for every open $O\subset X\setminus\ C$ the set $O_{\sharp}$ is a non-empty open subset of $Y\setminus f(C)$. So, $O_{\sharp}\cap V\neq\varnothing$, which implies $W\cap O\neq\varnothing$.

One can also see that $d$ is regular provided so is $\rho$. ∎

It is well known that for every space $X$ there is a unique extremally disconnected space $\widetilde{X}$ and a perfect irreducible map $f:\widetilde{X}\to X$. The space $\widetilde{X}$ is said to be the absolute of $X$. A space $Y$ is called co-absolute to $X$ if their absolutes are homeomorphic.

Corollary 3.7**.**

The absolute of any (regularly) quasi $\kappa$-metrizable space is (regularly) quasi $\kappa$-metrizable.

Remark 3.8**.**

The last corollary shows that the class of $\kappa$-metrizable spaces is a proper subclass of the quasi $\kappa$-metrizable spaces. Indeed, let $X$ be a $\kappa$-metrizable compact infinite space. Then its absolute $aX$ is quasi $\kappa$-metrizable. On the other hand, $aX$ being extremally disconnected can not be $\kappa$-metrizable (otherwise, it should be discrete by [8, Theorem 11]).

Corollary 3.9**.**

Every compact space co-absolute to a quasi $\kappa$-metrizable space is skeletally generated.

Proof.

Let $X$ and $Y$ be compact spaces having the same absolute $Z$. So, there are perfect irreducible surjections $g:Z\to Y$ and $f:Z\to X$. If $Y$ is quasi $\kappa$-metrizable, then so is $Z$, see Proposition 3.6. Hence, $Z$ is skeletally generated, and by [5, Lemma 1], $X$ is also skeletally generated. ∎

Recall that the hyperspace $\rm{exp}X$ consists of all compact non-empty subsets $F$ of $X$ such that the sets of the form

[TABLE]

form a base $\mathcal{B}_{\rm{exp}}$ for $\rm{exp}X$, where each $U_{i}$ belongs to a base $\mathcal{B}$ for $X$, see [6].

Proposition 3.10**.**

If $X$ is (regularly) quasi $\kappa$-metrizable, so is $\rm{exp}X$.

Proof.

Let $\mathcal{B}$ be a base for $X$ and $\rho$ be a (regular) quasi $\kappa$-metric on $X$. Then $\rho$ generates a (regular) $\pi$-capacity $\xi_{\rho}:X\times\mathcal{B}\to\mathbb{R}$ on $X$. Following the proof of [7, Theorem 3], we define a function $\xi:\rm{exp}X\times\mathcal{B}_{\rm{exp}}\to\mathbb{R}$ by

[TABLE]

It was shown in [7] that $\xi$ satisfies conditions $E1)$, $E3)$ and $E4)$, and that $\xi$ is regular provided $\xi_{\rho}$ is regular. Let show that $\xi$ satisfies condition $E2)$. Since $\xi$ is satisfies $E3)$, it suffices to prove that for every $[U_{1},..,U_{k}]$ there is a dense subset $V_{\rm{exp}}\subset[U_{1},..,U_{k}]$ with $\xi(F,[U_{1},..,U_{k}])>0$ for all $F\in V_{\rm{exp}}$. To this end, for each $i$ fix an open dense subset $V_{i}$ of $U_{i}$ such that $\xi_{\rho}(x,U_{i})>0$ if $x\in V_{i}$. Let $V_{\rm{exp}}$ consists of all finite sets $F\subset X$ such that $F\subset\bigcup_{i=1}^{n}V_{i}$ and $F\cap V_{i}\neq\varnothing$ for all $i$. Then $V_{\rm{exp}}$ is dense in $[U_{1},..,U_{k}]$ and $\xi(F,[U_{1},..,U_{k}])>0$ for all $F\in V_{\rm{exp}}$. Hence, by Lemma 2.1, $\rm{exp}X$ is (regularly) quasi $\kappa$-metrizable. ∎

Shchepin [7, Theorem 3a] has shown that if $\rm{exp}X$ is $\kappa$-metrizable, then so is $X$. We don’t know if a similar result is true for quasi $\kappa$-metrizable spaces.

4. Skeletally generated spaces

In this section we provide a characterization of skeletally generated compact spaces in terms of functions similar to quasi $\kappa$-metrics. We say that a non-negative function $d:X\times\mathcal{C}\to\mathbb{R}$ is a weak $\kappa$-metric, where $\mathcal{C}$ is the family of all regularly closed subsets of $X$, if it satisfies conditions $K1)^{*}$, $K2)-K3)$ and the following one:

• $K4)_{0}$

For every increasing transfinite family $\{C_{\alpha}\}_{\alpha}\subset\mathcal{C}$ the function $f(x)=\inf_{\alpha}d(x,C_{\alpha})$ is continuous.

Theorem 4.1**.**

A compact space is skeletally generated if and only if it is weakly $\kappa$-metrizable.

Proof.

First, let show that every skeletally generated compactum $X$ is weakly $\kappa$-metrizable. We embed $X$ as a subset of $\mathbb{R}^{\tau}$ for some cardinal $\tau$. Then, according to [9, Theorem 1.1], there is a function $\rm e:\mathcal{T}_{X}\to\mathcal{T}_{\mathbb{R}^{\tau}}$ between the topologies of $X$ and $\mathbb{R}^{\tau}$ such that: $(i)$ $\rm e(U)\cap\rm e(V)=\varnothing$ provided $U$ and $V$ are disjoint; $(ii)$ $\rm e(U)\cap X$ is dense in $U$. We define a new function $\mathrm{e}_{1}:\mathcal{T}_{X}\to\mathcal{T}_{\mathbb{R}^{\tau}}$,

[TABLE]

Obviously $\mathrm{e}_{1}$ satisfies conditions $(i)$ and $(ii)$, and it is also monotone, i.e. $U\subset V$ implies $\mathrm{e}_{1}(U)\subset\mathrm{e}_{1}(V)$. Moreover, for every increasing transfinite family $\gamma=\{U_{\alpha}\}$ of open sets in $X$ we have $\mathrm{e}_{1}(\bigcup_{\alpha}U_{\alpha})=\bigcup_{\alpha}\mathrm{e}_{1}(U_{\alpha})$. Indeed, if $z\in\mathrm{e}_{1}(\bigcup_{\alpha}U_{\alpha})$, then there is an open set $V\in\mathcal{T}_{X}$ with $\overline{V}\subset\bigcup_{\alpha}U_{\alpha}$ and $z\in\mathrm{e}(V)$. Since $\overline{V}$ is compact and the family is increasing, $\overline{V}$ is contained in some $U_{\alpha_{0}}$. Hence, $z\in\mathrm{e}(V)\subset\mathrm{e}_{1}(U_{\alpha_{0}})$. Consequently, $\mathrm{e}_{1}(\bigcup_{\alpha}U_{\alpha})\subset\bigcup_{\alpha}\mathrm{e}_{1}(U_{\alpha})$. The other inclusion follows from monotonicity of $\mathrm{e}_{1}$.

Because $\mathbb{R}^{\tau}$ is $\kappa$-metrizable (see [7]), there is a $\kappa$-metric $\rho$ on $\mathbb{R}^{\tau}$. For every regularly closed $C\subset X$ and $x\in X$ we can define the function $d(x,C)=\rho(x,\overline{\mathrm{e}_{1}(\rm{Int}C)})$, where $\overline{\mathrm{e}_{1}(U)}$ is the closure of $\mathrm{e}_{1}(U)$ in $\mathbb{R}^{\tau}$. It is easily seen that $d$ satisfies conditions $K2)-K3)$. Let show that it also satisfies $K4)_{0}$ and $K1^{*})$. Indeed, assume $\{C_{\alpha}\}$ is an increasing transfinite family of regularly closed sets in $X$. We put $U_{\alpha}=\mathrm{Int}C_{\alpha}$ for every $\alpha$ and $U=\bigcup_{\alpha}U_{\alpha}$. Thus, $\mathrm{e}_{1}(U)=\bigcup_{\alpha}\mathrm{e}_{1}(U_{\alpha})$. Since $\{\overline{\mathrm{e}_{1}(U_{\alpha})}\}$ is an increasing transfinite family of regularly closed sets in $\mathbb{R}^{\tau}$, for every $x\in X$ we have

[TABLE]

Hence, the function $f(x)=\inf_{\alpha}d(x,C_{\alpha})$ is continuous on $X$ because so is $\rho(.,\overline{\bigcup_{\alpha}\mathrm{e}_{1}(U_{\alpha})})$. To show that $K1^{*})$ also holds, observe that $d(x,C)=0$ if and only if $x\in X\cap\overline{\mathrm{e}_{1}(\rm{Int}C)}$. Because $\mathrm{e}_{1}(\rm{Int}C)\cap X$ is dense in $C$, $C\subset\overline{\mathrm{e}_{1}(\rm{Int}C)}$. Hence, $V=X\setminus\overline{\mathrm{e}_{1}(\rm{Int}C)}$ is contained in $X\setminus C$ and $d(x,C)>0$ iff $x\in V$. To prove $V$ is dense in $X\setminus C$, let $x\in X\setminus C$ and $W_{x}\subset X\setminus C$ be an open neighborhood of $x$. Then $W\cap\rm{Int}C=\varnothing$, so $\mathrm{e}_{1}(W)\cap\mathrm{e}_{1}(\rm{Int}C)=\varnothing$. This yields $\mathrm{e}_{1}(W)\cap X\subset V$. On the other hand, $\mathrm{e}_{1}(W)\cap X$ is a non-empty subset of $W$, hence $W\cap V\neq\varnothing$. Therefore, $d$ is a weak $\kappa$-metric on $X$.

The other implication was actually established in the proof of Theorem 1.4] from [9], and we sketch the proof here. Suppose $d$ is a weak $\kappa$-metric on $X$ and embed $X$ in a Tychonoff cube $\mathbb{I}^{A}$ with uncountable $A$, where $\mathbb{I}=[0,1]$. For any countable set $B\subset A$ let $\mathcal{A}_{B}$ be a countable base for $X_{B}=\pi_{B}(X)$ consisting of all open sets in $X_{B}$ of the form $X_{B}\cap\prod_{\alpha\in B}V_{\alpha}$, where each $V_{\alpha}$ is an open subinterval of $\mathbb{I}$ with rational end-points and $V_{\alpha}\neq\mathbb{I}$ for finitely many $\alpha$. Here $\pi_{B}:\mathbb{I}^{A}\to\mathbb{I}^{B}$ denotes the projection, and let $p_{B}=\pi_{B}|X$. For any open $U\subset X$ denote by $f_{U}$ the function $d(\cdot,\overline{U})$. We also write $p_{B}\prec g$, where $g$ is a map defined on $X$, if there is a map $h:p_{B}(X)\to g(X)$ such that $g=h\circ p_{B}$. Since $X$ is compact this is equivalent to the following: if $p_{B}(x_{1})=p_{B}(x_{2})$ for some $x_{1},x_{2}\in X$, then $g(x_{1})=g(x_{2})$. We say that a countable set $B\subset A$ is $d$-admissible if $p_{B}\prec f_{p_{B}^{-1}(V)}$ for every $V\in\mathcal{A}_{B}^{<\omega}$. Here $\mathcal{A}_{B}^{<\omega}$ is the family of all finite unions of elements from $\mathcal{A}_{B}$. Denote by $\mathcal{D}$ the family of all $d$-admissible subsets of $A$. We are going to show that all maps $p_{B}:X\to X_{B}$, $B\in\mathcal{D}$, are skeletal and the inverse system $S=\{X_{B}:p^{B}_{D}:D\subset B,D,B\in\mathcal{D}\}$ is $\sigma$-continuous. Since $\displaystyle X=\lim_{\leftarrow}S$, this would imply that $X$ is skeletally generated, see [9] and [10].

Following the proof of [9, Theorem 1.4], one can show that for any countable set $B\subset A$ there is $D\in\mathcal{D}$ with $B\subset D$, and the union of any increasing sequence of $d$-admissible sets is also $d$-admissible. So, we need to show only that $p_{B}:X\to X_{B}$ is a skeletal map for every $B\in\mathcal{D}$. Suppose there is an open set $U\subset X$ such that the interior in $X_{B}$ of $\overline{p_{B}(U)}$ is empty. Then $W=X_{B}\setminus\overline{p_{B}(U)}$ is dense in $X_{B}$. Let $\{W_{m}\}_{m\geq 1}$ be a countable cover of $W$ with $W_{m}\in\mathcal{A}_{B}$ for all $m$. Since $\mathcal{A}_{B}^{<\omega}$ is finitely additive, we may assume that $W_{m}\subset W_{m+1}$, $m\geq 1$. Because $B$ is $d$-admissible, $p_{B}\prec f_{p_{B}^{-1}(W_{m})}$ for all $m$. Hence, there are continuous functions $h_{m}:X_{B}\to\mathbb{R}$ with $f_{p_{B}^{-1}(W_{m})}=h_{m}\circ p_{B}$, $m\geq 1$. Recall that $f_{p_{B}^{-1}(W_{m})}(x)=d(x,\overline{p_{B}^{-1}(W_{m})})$ and $p_{B}^{-1}(W)=\bigcup_{m\geq 1}p_{B}^{-1}(W_{m})$. Therefore, $f_{p_{B}^{-1}(W)}(x)=d(x,\overline{p_{B}^{-1}(W)})\leq f(x)=\inf_{m}f_{p_{B}^{-1}(W_{m})}(x)$ for all $x\in X$. Moreover, $f$ is continuous and $f_{p_{B}^{-1}(W_{m+1})}(x)\leq f_{p_{B}^{-1}(W_{m})}(x)$ because $W_{m}\subset W_{m+1}$. The last inequalities together with $p_{B}\prec f_{p_{B}^{-1}(W_{m})}$ yields that $p_{B}\prec f$. So, there exists a continuous function $h$ on $X_{B}$ with $f(x)=h(p_{B}(x))$ for all $x\in X$. But $f(x)=0$ for all $x\in p_{B}^{-1}(W)$, so $f(\overline{p_{B}^{-1}(W)})=0$. This implies that $h(\overline{W})=0$. Since $p_{B}(\overline{p_{B}^{-1}(W)})=\overline{W}=X_{B}$, we have that $h$ is the constant function zero. Consequently, $f(x)=0$ for all $x\in X$. Finally, the inequality $d(x,\overline{p_{B}^{-1}(W)})\leq f(x)$ yields that $d(x,\overline{p_{B}^{-1}(W)})=0$ for all $x\in X$. On the other hand, $\overline{p_{B}^{-1}(W)}\cap U=\varnothing$. So, according to $K1^{*})$, there is an open subset $U^{\prime}$ of $U$ with $d(x,\overline{p_{B}^{-1}(W)})>0$ for each $x\in U^{\prime}$, a contradiction. ∎

Because any compactification of a skeletally generated space is skeletally generated (see [9]) and the weakly $\kappa$-metrizability is a hereditary property with respect to dense subsets, we have the following

Corollary 4.2**.**

Every skeletally generated space is weakly $\kappa$-metrizable.

All results in Section 3, except Proposition 3.10, remain valid for weakly $\kappa$-metrizable spaces. Theorem 4.1 and a result of Kucharski-Plewik [5, Theorem 6] imply that Proposition 3.10 is also true for weakly $\kappa$-metrizable compacta. But the following questions are still open.

Question 4.3**.**

Is any product of weakly $\kappa$-metrizable spaces weakly $\kappa$-metrizable?

If there exists a counter example to Question 4.3 which, in addition has a countable cellularity, then next question would have also a negative answer.

Question 4.4**.**

Is any weakly $\kappa$-metrizable space with a countable cellularity skeletally generated?

Acknowledgments. The author would like to express his gratitude to A. Kucharski for his careful reading and suggesting some improvements of the paper.