New results regarding the lattice of uniform topologies on $C(X)$

TL;DR

This paper investigates the structural, categorical, and cardinal properties of the lattice of uniform topologies on the space of continuous functions over a Tychonoff space, extending previous work in the area.

Contribution

It advances understanding of the lattice $al U_X$ by exploring its structural, categorical, and cardinal characteristics and their relationships to properties of the underlying space.

Findings

Analyzed the structural properties of $al U_X$.

Established connections between lattice cardinal characteristics and space cardinal functions.

Provided new insights into the categorical aspects of the lattice of uniform topologies.

Abstract

For a Tychonoff space $X$, the lattice $\mathscr U_X$ was introduced in L.A. P\'erez-Morales, G. Delgadillo-Pi\~n\'on, and R. Pichardo-Mendoza, "The lattice of uniform topologies on $C(X)$", Questions and Answers in General Topology, 39 (2021), 65-71. In the present paper we continue the study of $\mathscr U_X$. To be specific, the present paper deals, in its first half, with structural and categorical properties of $\mathscr U_X$, while in its second part focuses on cardinal characteristics of the lattice and how these relate to some cardinal functions of the space $X$.

Equations

$$V(f,A,\epsilon ):=\{g\in C(X):\forall x\in A(|f(x)-g(x)|<\epsilon )\}\subseteq U.$$

$$\alpha \vee \beta :=\{A\cup B:A\in \alpha ,B\in \beta \}\text{and}\alpha \wedge \beta :=\{\overline{A}\cap \overline{B}:A\in \alpha ,B\in \beta \}$$

$$\tilde{e}:=\bigcap \{\overline{e(\delta )}:\delta \in \mathcal{S}\}.$$

$$f^{\ast }\alpha :=\{f[A]:A\in \alpha \}$$

$$\phi =\{(C_{\alpha }(X),C_{f^{\ast }\alpha }(Y)):\alpha \in {\mathcal{D}}_X\}\subseteq {\mathcal{U}}_X\times {\mathcal{U}}_Y.$$

$$\forall \alpha ,\beta \in {\mathcal{D}}_X(C_{\alpha }\le C_{\beta }\to C_{f^{\ast }\alpha }\le C_{f^{\ast }\beta }).$$

$$C_{f[{\mathrm{cl}}_{X}A]}=\phi (C_{{\mathrm{cl}}_{X}A})\le \phi (C_A)=C_{f[A]},$$

$$\forall \alpha ,\beta \in {\mathcal{D}}_X(C_{f^{\ast }\alpha }\le C_{f^{\ast }\beta }\to C_{\alpha }\le C_{\beta })$$

$$\text{for each }x\in X,\phi (C_x(X))=C_{f(x)}(Y).$$

$$\phi (C_x)=C_{f^{\ast }\{\{x\}\}}=C_{f(x)}\text{and}\phi (C_x)=C_{g(x)},$$

$$C_{\alpha }=\bigvee \{C_A:A\in \alpha \}\text{and}{C}_{f^{\ast }\alpha }=\bigvee \{C_{f[A]}:A\in \alpha \};$$

$$\phi (C_{\alpha })=\bigvee \{\phi (C_A):A\in \alpha \}=\bigvee \{C_{f[A]}:A\in \alpha \}=C_{f^{\ast }\alpha },$$

$$\text{for all }A\in \mathcal{A}\text{ and }B\in \beta ,B\subseteq f[A].$$

$$C_{\beta }=\phi (C_G)\le \phi (C_A)=C_{f[A]};$$

$$B\subseteq \bigcap \{f[A]:A\in \mathcal{A}\}=f\left[\bigcap \mathcal{A}\right]=f[G].$$

$$\Sigma (X)=\{\{\varnothing ,X\}\}\text{and}{\mathcal{D}}_X=\{\varnothing ,\{\varnothing \},\{X\},\{\varnothing ,X\}\}.$$

$$p^{\sim }:=p^{-1}\{0\}\cup \{-x:x\in p^{-1}\{1\}\},$$

$$\alpha (d):=\left\{\bigvee p^{\sim }:p\in [d{]}^{<\omega }\right\}$$

$$U\subseteq \bigvee q^{\sim }.$$

$$\varnothing \ne \bigwedge r^{\sim }=U\wedge \left(-\bigvee q^{\sim }\right),$$

$$\text{for any space }X,|\mathrm{RO}(X)|\le 2^{d(X)}.$$

$$w(\beta D(\kappa ))\le 2^{d(\beta D(\kappa ))}=2^{\kappa }.$$

$$|{\mathcal{U}}_{\beta D(\kappa )}|\le {\beth }_2\left(nw\left(\beta D(\kappa )\right)\right)={\beth }_2\left(w\left(\beta D(\kappa )\right)\right)\le {\beth }_3(\kappa ).$$

$${\beth }_3(\kappa )=2^{|\beta D(\kappa )|}\le |{\mathcal{U}}_{\beta D(\kappa )}|.$$

$$|{\mathcal{U}}_{D(2{)}^{\kappa }}|\le {\beth }_2\left(nw\left(D(2{)}^{\kappa }\right)\right)={\beth }_2\left(w\left(D(2{)}^{\kappa }\right)\right)={\beth }_2(\kappa ).$$

$${\beth }_2(\kappa )=2^{|D(2{)}^{\kappa }|}\le |{\mathcal{U}}_{D(2{)}^{\kappa }}|.$$

$$\max \{2^{|{\mathbb{A}}^2|},2^{|\mathrm{RO}({\mathbb{A}}^2)|}\}=2^{\mathfrak{c}}<{\beth }_2(\mathfrak{c})=|{\mathcal{U}}_{{\mathbb{A}}^2}|.$$

$$|{\mathcal{U}}_X|\le {\beth }_2(|X|)<{\beth }_3(|X|)={\beth }_2(|C(X)|).$$

$$|{\mathcal{U}}_X|=2^{\mathfrak{c}}<{\beth }_2(\mathfrak{c})\le {\beth }_2(|C(X)|).$$

$$|{\mathcal{U}}_X|=2^{o(X)}$$

Full text

In the present paper we continue the study of $\mathscr{U}_{X}$. To be specific, the present paper deals, in its first half, with structural and categorical properties of $\mathscr{U}_{X}$, while in its second part focuses on cardinal characteristics of the lattice and how these relate to some cardinal functions of the space $X$.

The research of the second author was supported by CONACyT grant no. 814282.

1. Introduction

In [9] the authors define, given a completely regular Hausdorff space $X$, a partially ordered set $(\mathscr{U}_{X},\subseteq)$ (see Section 2 for details and the corresponding definitions) which turns out to be a bounded lattice (the lattice of uniform topologies on $C(X)$). Here we expand some of the results obtained in that paper and explore new directions. For example, Section 3 is mainly about finding connections between order-isomorphisms and homeomorphisms, while the last two sections deal heavily on finding relations between some cardinal characteristics of $\mathscr{U}_{X}$ and highly common cardinal functions of $X$.

2. Preliminaries

All topological notions and all set-theoretic notions whose definition is not included here should be understood as in [1] and [7], respectively. With respect to lattices, we will follow [8] for notation and results. The same goes for Boolean algebras and [6].

The symbol $\omega$ denotes both, the set of all non-negative integers and the first infinite cardinal. Also, $\mathbb{R}$ is the real line endowed with the Euclidean topology.

Given a set $S$, $[S]^{<\omega}$ denotes the collection of all finite subsets of $S$. For a set $A$, the symbol ${}^{A}S$ is used to represent the collection of all functions from $A$ to $S$. In particular, for $f\in{}^{A}S$, $E\subseteq A$, and $H\subseteq S$ we define $f[E]:=\{f(x):x\in E\}$ and $f^{-1}[H]:=\{x\in A:f(x)\in H\}$. Moreover, if $y\in S$, $f^{-1}\{y\}:=f^{-1}[\{y\}]$.

A nonempty family of sets, $\alpha$, is called directed if for any $A,B\in\alpha$ there is $E\in\alpha$ with $A\cup B\subseteq E$. For example, $[S]^{<\omega}$ is directed, for any set $S$.

Assume $X$ is a set. Hence, $\mathscr{P}(X)$ and $\mathscr{D}_{X}$ represent its power set and the collection of all directed subsets of $\mathscr{P}(X)$, respectively. In [10] the term base for an ideal on $X$ was used to refer to members of $\mathscr{D}_{X}$.

Unless otherwise stated, the word space means Hausdorff completely regular space (i.e., Tychonoff space).

Assume $X$ is a space. Then, $\tau_{X}$ and $\tau_{X}^{*}$ stand, respectively, for the families of all open and closed subsets of $X$. Moreover, whenever $x\in X$, $\tau_{X}(x)$ will be the set $\{U\in\tau_{X}:x\in U\}$. Now, given $A\subseteq X$, the symbol $\mathop{\rm cl}\nolimits_{X}A$ (or $\overline{A}$ when the space $X$ is clear from the context) represents the closure of $A$ in $X$; similarly, $\mathop{\rm int}_{X}A$ and $\mathop{\rm int}A$ will be used to denote the interior of $A$ in $X$.

$C(X)$ is, as usual, the subset of ${}^{X}\mathbb{R}$ consisting of all continuous functions. Now, given $\alpha\in\mathscr{D}_{X}$ we generate a topology on $C(X)$ as follows: a set $U\subseteq C(X)$ is open if and only if for each $f\in U$ there are $A\in\alpha$ and a real number $\varepsilon>0$ with

[TABLE]

The resulting topological space is denoted by $C_{\alpha}(X)$. As it is explained in [10], $C_{\alpha}(X)$ is a uniformizable topological space which may not be Hausdorff. In fact, one has the following result (whose proof can be found in [10, Proposition 3.1, p. 559]).

Lemma 2.1**.**

For any space $X$ and $\alpha\in\mathscr{D}_{X}$, $C_{\alpha}(X)$ is Hausdorff if and only if $\alpha$ has dense union, i.e., $\overline{\bigcup\alpha}=X$.

Given a space $X$, set $\mathscr{U}_{X}:=\{\tau_{C_{\gamma}(X)}:\gamma\in\mathscr{D}_{X}\}$. In order to simplify our writing, for each $\alpha\in\mathscr{D}_{X}$ we identify the space $C_{\alpha}(X)$ with its topology. Thus, expressions of the form $C_{\alpha}(X)\in\mathscr{U}_{X}$ will be common in this paper. Also, in those occasions where the ground space is clear from the context, we will suppress it from our notation, i.e., we will use $C_{\alpha}$ instead of $C_{\alpha}(X)$. Finally, for any $\alpha,\beta\in\mathscr{D}_{X}$, both, $C_{\alpha}(X)\leq C_{\beta}(X)$ and $C_{\alpha}\leq C_{\beta}$, are abbreviations of the relation $\tau_{C_{\alpha}(X)}\subseteq\tau_{C_{\beta}(X)}$.

It is shown in [9, Proposition 3.2, p. 67] that the poset $(\mathscr{U}_{X},\subseteq)$ is a bounded distributive lattice; to be precise, given $\alpha,\beta\in\mathscr{D}_{X}$, the collections

[TABLE]

are directed and, moreover, $C_{\alpha\vee\beta}$ and $C_{\alpha\wedge\beta}$ are, respectively, the supremum and infimum of $\{C_{\alpha},C_{\beta}\}$ in $\mathscr{U}_{X}$.

The topologies generated on $C(X)$ by the directed sets $\{\emptyset\}$, $[X]^{<\omega}$, and $\{X\}$ are denoted by $C_{\emptyset}(X)$, $C_{p}(X)$, and $C_{u}(X)$, respectively. Let us note that $C_{\emptyset}$ is the indiscrete topology on $C(X)$, while $C_{p}$ and $C_{u}$ are the topologies of pointwise and uniform convergence on $C(X)$, respectively.

The result below (see [10, Theorem 3.4, p. 560] for a proof) will be used several times in what follows.

Proposition 2.2**.**

If $X$ is a space and $\alpha,\beta\in\mathscr{D}_{X}$, then $C_{\alpha}\leq C_{\beta}$ if and only if for each $A\in\alpha$ there is $B\in\beta$ with $A\subseteq\overline{B}$.

We finish this section by mentioning that our notation for topological cardinal functions follows [3]; in particular, all of them are, by definition, infinite.

3. Some structural and categorical results

We begin by improving the result presented in [9, Proposition 3.2, p. 67].

Proposition 3.1**.**

For any space $X$, $\mathscr{U}_{X}$ is a complete lattice.

Proof.

Given an arbitrary set $\mathscr{S}\subseteq\mathscr{D}_{X}$, define $\mathscr{A}:=\{C_{\delta}:\delta\in\mathscr{S}\}$.

By letting $\alpha$ be the family of all sets of the form $\bigcup\mathscr{E}$, where $\mathscr{E}\subseteq\bigcup\mathscr{S}$ is finite, we obtain $\alpha\in\mathscr{D}_{X}$. Also, the fact that $\delta\subseteq\alpha$, whenever $\delta\in\mathscr{S}$, implies (see Proposition 2.2) that $C_{\alpha}$ is an upper bound for $\mathscr{A}$.

Now, assume that $\gamma\in\mathscr{D}_{X}$ is such that $C_{\gamma}$ is an upper bound for $\mathscr{A}$. In order to show that $C_{\alpha}\leq C_{\gamma}$, fix $A\in\alpha$. There is a finite set $\mathscr{E}\subseteq\bigcup\mathscr{S}$ satisfying $A=\bigcup\mathscr{E}$. According to Proposition 2.2, for each $E\in\mathscr{E}$ there exists $E^{*}\in\gamma$ with $E\subseteq\overline{E^{*}}$. Since $\gamma$ is directed, $\bigcup\{E^{*}:E\in\mathscr{E}\}\subseteq G$ for some $G\in\gamma$ and, consequently, $A\subseteq\overline{G}$. In other words, $C_{\alpha}\leq C_{\gamma}$.

From the previous paragraphs we conclude that any subset of $\mathscr{U}_{X}$ has a supremum in $\mathscr{U}_{X}$. Now, regarding infima, let us observe that the infimum of $\emptyset$ in $\mathscr{U}_{X}$ is $C_{u}$. Thus, we will suppose that $\mathscr{S}$ is non-empty.

Denote by $\mathscr{E}$ the set of all choice functions of $\mathscr{S}$, i.e., $e\in\mathscr{E}$ if and only if $e:\mathscr{S}\to\bigcup\mathscr{S}$ and $e(\delta)\in\delta$, for all $\delta\in\mathscr{S}$. Now, for each $e\in\mathscr{E}$, set

[TABLE]

We claim that if $\beta:=\{\widetilde{e}:e\in\mathscr{E}\}$, then $C_{\beta}$ is the infimum of $\mathscr{A}$.

To show that $\beta$ is directed, consider $d,e\in\mathscr{E}$. Since, for any $\delta\in\mathscr{S}$, $\delta$ is directed, we deduce that there is a set $f(\delta)\in\delta$ with $d(\delta)\cup e(\delta)\subseteq f(\delta)$. This produces $f$, a choice function of $\mathscr{S}$, in such a way that $\widetilde{d}\cup\widetilde{e}\subseteq\widetilde{f}$.

The fact that $C_{\beta}$ is a lower bound for $\mathscr{A}$ follows from the observation that for each $e\in\mathscr{E}$ and $\delta\in\mathscr{S}$, $\widetilde{e}\subseteq\overline{e(\delta)}$.

Finally, let $\gamma\in\mathscr{D}_{X}$ be such that $C_{\gamma}$ is a lower bound for $\mathscr{A}$. Fix $G\in\gamma$. Then, for any $\delta\in\mathscr{S}$ there is $e(\delta)\in\delta$ with $G\subseteq\overline{e(\delta)}$. As a consequence, we obtain $e$, a choice function of $\mathscr{S}$, with $G\subseteq\widetilde{e}$. ∎

As in [8], we will use the symbol $\Sigma(E)$ to the represent the collection of all topologies on a fixed set $E$. It is well-known that when we order $\Sigma(E)$ by direct inclusion, the resulting structure is a complete lattice. In particular, the supremum of $\mathscr{A}\subseteq\Sigma(E)$ is the topology on $E$ generated by $\bigcup\mathscr{A}$ (i.e., it has the collection $\bigcup\mathscr{A}$ as a subbase).

Clearly, $\mathscr{U}_{X}$ is a suborder of $\Sigma(C(X))$. Thus, a natural question is, given a family $\mathscr{A}\subseteq\mathscr{U}_{X}$, is the supremum (respectively, infimum) of $\mathscr{A}$ as calculated in $\mathscr{U}_{X}$ the same as the supremum (respectively, infimum) of $\mathscr{A}$ as obtained in $\Sigma(C(X))$? We have a positive answer for suprema.

Corollary 3.2**.**

If $X$ is a space and $\mathscr{A}\subseteq\mathscr{U}_{X}$, then $\bigvee\mathscr{A}$, the supremum of $\mathscr{A}$ in $\mathscr{U}_{X}$, is the topology on $C(X)$ which has $\bigcup\mathscr{A}$ as a subbase.

Proof.

Fix $\mathscr{S}\subseteq\mathscr{D}_{X}$ in such a way that $\mathscr{A}=\{C_{\beta}:\beta\in\mathscr{S}\}$ and denote by $\sigma$ the topology on $C(X)$ generated by $\bigcup\mathscr{A}$. Since $\bigvee\mathscr{A}$ is an upper bound of $\mathscr{A}$ in $\Sigma(C(X))$, we obtain $\sigma\subseteq\bigvee\mathscr{A}$.

Now, let $f\in U\in\bigvee\mathscr{A}$ be arbitrary. According to the proof of Proposition 3.1, there are $\varepsilon>0$ and $\mathscr{E}$, a finite subset of $\bigcup\mathscr{S}$, with $V(f,A,\varepsilon)\subseteq U$, where $A:=\bigcup\mathscr{E}$. When $\mathscr{E}=\emptyset$, we deduce that $U=C(X)\in\sigma$. Hence, let us assume that $\mathscr{E}\neq\emptyset$.

For each $E\in\mathscr{E}$ let $\beta(E)\in\mathscr{S}$ be such that $E\in\beta(E)$. By setting $\mathscr{W}:=\{\mathop{\rm int}\nolimits_{C_{\beta(E)}}V(f,E,\varepsilon):E\in\mathscr{E}\}$ we produce a finite subset of $\bigcup\mathscr{A}$ which satisfies $f\in\bigcap\mathscr{W}\subseteq V(f,A,\varepsilon)\subseteq U$. In conclusion, $\bigvee\mathscr{A}\subseteq\sigma$. ∎

Recall that if $E$ is a set and $\sigma,\tau\in\Sigma(E)$, the infimum of $\{\sigma,\tau\}$ in $\Sigma(E)$ is $\sigma\cap\tau$; consequently, for any space $X$ and $\alpha,\beta\in\mathscr{D}_{X}$, $C_{\alpha}\wedge C_{\beta}\subseteq C_{\alpha}\cap C_{\beta}$. Now, assume that $X$ is a non-empty space which is resolvable (i.e., it can be written as the union of two disjoint dense subsets of it). In [9, Proposition 4.5, p. 69], it is shown that there are two Hausdorff topologies $\sigma,\tau\in\mathscr{U}_{X}$ with $\sigma\wedge\tau=C_{\emptyset}$. Consequently, $\sigma\cap\tau$ is a $T_{1}$ topology, but $\sigma\wedge\tau$ fails to be $T_{0}$. Hence, the question posed in the paragraph preceding Corollary 3.2 has a negative answer for infima.

Question 3.3*.*

Given a space $X$, find conditions on $\alpha,\beta\in\mathscr{D}_{X}$ in order to obtain $C_{\alpha}\wedge C_{\beta}=C_{\alpha}\cap C_{\beta}$.

As in [9], the symbol $\mathscr{C}_{X}$ represents the collection of all members of $\mathscr{U}_{X}$ which have a complement in $\mathscr{U}_{X}$. Thus, from the fact that $\mathscr{U}_{X}$ is a bounded distributive lattice, we deduce that $\mathscr{U}_{X}$ is a Boolean algebra if and only if $\mathscr{U}_{X}=\mathscr{C}_{X}$. Our next result shows that this condition is attained only in trivial cases.

Proposition 3.4**.**

For any space $X$, $\mathscr{U}_{X}$ is a Boolean algebra if and only if $X$ is finite.

Proof.

Firstly observe that, in virtue of [9, Proposition 3.3, p. 68], we only need to show that $X$ is a finite space if and only if for each $\alpha\in\mathscr{D}_{X}$ there is $E\in\alpha$ with $\overline{E}\in\tau_{X}$ and $\bigcup\alpha\subseteq\overline{E}$. Now, evidently any finite $X$ satisfies the latter condition. For the converse let us assume that $X$ is infinite. Since $X$ is Hausdorff, there is $\{U_{n}:n<\omega\}$, a family of non-empty open subsets of $X$, with $U_{m}\cap U_{n}=\emptyset$, whenever $m<n<\omega$. By setting $\alpha:=\left\{\bigcup_{k=0}^{n}U_{k}:n<\omega\right\}$ we obtain a member of $\mathscr{D}_{X}$ in such a way that, for each $E\in\alpha$, there is $m<\omega$ with $U_{m}\cap E=\emptyset$ and thus, $\bigcup\alpha\not\subseteq\overline{E}$. ∎

For our next results we will need some auxiliary concepts. First of all, assume that $f$ is function from the space $X$ into a space $Y$. One easily verifies that for any $\alpha\in\mathscr{D}_{X}$ the family

[TABLE]

belongs to $\mathscr{D}_{Y}$ and so, we have the following notion (recall that for any space $Z$ and $\gamma\in\mathscr{D}_{Z}$ we are identifying the space $C_{\gamma}(Z)$ with its topology).

Definition 3.5**.**

If $X$, $Y$, and $f$ are as in the previous paragraph, the phrase $\varphi$ is the $f$-induced relation means that

[TABLE]

With the notation used above, the domain of $\varphi$, ${\mathop{\rm dom}}(\varphi)$, is equal to $\mathscr{U}_{X}$ and its range, ${\mathop{\rm ran}}(\varphi)$, is a subset of $\mathscr{U}_{Y}$.

Proposition 3.6**.**

If $X$ and $Y$ are spaces and $f:X\to Y$, then $f$ is continuous if and only if $\varphi$, the $f$-induced relation, is an order-preserving function.

Proof.

Let us begin by assuming that $f$ is continuous and prove the statement below.

[TABLE]

Given $\alpha,\beta\in\mathscr{D}_{X}$ with $C_{\alpha}\leq C_{\beta}$, fix $A\in f^{*}\alpha$. There is $B\in\alpha$ with $A=f[B]$ and so (see Proposition 2.2), for some $E\in\beta$, $B\subseteq{\mathop{\rm cl}}_{X}E$. Finally, $f$’s continuity produces $A=f[B]\subseteq f[{\mathop{\rm cl}}_{X}E]\subseteq{\mathop{\rm cl}}_{Y}f[E]$ and, clearly, $f[E]\in f^{*}\beta$.

The final step for this implication is to note that the properties required for $\varphi$ are consequences of (3.1).

Suppose that $\varphi$ is an order-preserving function and fix $A\subseteq X$. According to Proposition 2.2, $C_{{\mathop{\rm cl}}_{X}A}\leq C_{A}$ and so,

[TABLE]

i.e., $f[{\mathop{\rm cl}}_{X}A]\subseteq{\mathop{\rm cl}}_{Y}f[A]$. ∎

For the rest of the paper, given a space $X$, a point $x\in X$, and a set $A\subseteq X$, we use the symbols $C_{x}(X)$ and $C_{A}(X)$ to represent the topological spaces $C_{\{\{x\}\}}(X)$ and $C_{\{A\}}(X)$, respectively. As expected, if the space $X$ is clear from the context, we only write $C_{x}$ and $C_{A}$; also, as we have done before, $C_{x}$ and $C_{A}$ are, as well, the topologies of the corresponding spaces.

A function $f$ from the space $X$ into the space $Y$ is called open onto its range if, for any $U\in\tau_{X}$, $f[U]\in\tau_{f[X]}$. Note that if $f$ is one-to-one, then $f$ is open onto its range if and only if $f$ is closed onto its range (i.e., whenever $G$ is a closed subset of $X$, $f[G]$ is a closed subset of the subspace $f[X]$).

Proposition 3.7**.**

Assume $X$ and $Y$ are spaces. For any $f:X\to Y$, the following are equivalent.

1. (1)

$f$* is one-to-one and open onto its range.* 2. (2)

$\varphi^{-1}$, the inverse relation of the $f$-induced relation, is an order-preserving function.

Proof.

Observe that for the implication $(1)\to(2)$, it suffices to prove that the statement

[TABLE]

follows from $(1)$. Thus, suppose $(1)$ and fix $\alpha,\beta\in\mathscr{D}_{X}$ with $C_{f^{*}\alpha}\leq C_{f^{*}\beta}$. Given $A\in\alpha$, Proposition 2.2 guarantees the existence of $B\in\beta$ with $f[A]\subseteq{\mathop{\rm cl}}_{Y}f[B]$, i.e., $A\subseteq f^{-1}[{\mathop{\rm cl}}_{Y}f[B]]$. Thus, we only need to show that $f^{-1}[{\mathop{\rm cl}}_{Y}f[B]]\subseteq{\mathop{\rm cl}}_{X}B$. If $x\in f^{-1}[{\mathop{\rm cl}}_{Y}f[B]]$ and $U\in\tau_{X}(x)$ are arbitrary, then $f(x)\in f[X]\cap{\mathop{\rm cl}}_{Y}f[B]={\mathop{\rm cl}}_{f[X]}f[B]$ and $f[U]\in\tau_{f[X]}(f(x))$; consequently, $f[U]\cap f[B]\neq\emptyset$. Since $f$ is one-to-one, $f[U\cap B]\neq\emptyset$ and so, $U\cap B\neq\emptyset$, as required.

For the rest of the argument, assume $(2)$. In order to verify that $f$ is one-to-one, let $x,y\in X$ be such that $f(x)=f(y)$. Hence, $C_{f(x)}=C_{f(y)}$ and, as a consequence, $C_{x}=\varphi^{-1}(C_{f(x)})=\varphi^{-1}(C_{f(y)})=C_{y}$. The use of Proposition 2.2 produces $x=y$.

Given that $f$ is one-to-one, we only need to argue that $f$ is closed onto its range. Suppose $G$ is a closed subset of $X$. By letting $E:={\mathop{\rm cl}}_{Y}f[G]$ and $A:=f^{-1}[E]$, we deduce that $f[A]=E\cap f[X]={\mathop{\rm cl}}_{f[X]}f[G]$. Therefore, $C_{f[A]}\leq C_{E}\leq C_{f[G]}$ and so, $C_{A}=\varphi^{-1}(C_{f[A]})\leq\varphi^{-1}(C_{f[G]})=C_{G}$. Hence, $A\subseteq{\mathop{\rm cl}}_{X}G=G$ and, consequently, ${\mathop{\rm cl}}_{f[X]}f[G]=f[A]\subseteq f[G]$, i.e., $f[G]$ is a closed subset of $f[X]$. ∎

Proposition 3.8**.**

If $X$ and $Y$ are spaces and $f:X\to Y$, then $f$ is onto if and only if ${\mathop{\rm ran}}(\varphi)=\mathscr{U}_{Y}$, where $\varphi$ is the $f$-induced relation.

Proof.

When $f$ is onto and $\alpha\in\mathscr{D}_{Y}$, the collection $\beta:=\{f^{-1}[A]:A\in\alpha\}$ belongs to $\mathscr{D}_{X}$ and $f^{*}\beta=\alpha$. Thus, $(C_{\beta},C_{\alpha})\in\varphi$ and so, $C_{\alpha}\in{\mathop{\rm ran}}(\varphi)$.

For the remaining implication, fix $y\in Y$ and note that $C_{y}\in\mathscr{U}_{Y}={\mathop{\rm ran}}(\varphi)$, i.e., for some $\alpha\in\mathscr{D}_{X}$, $(C_{\alpha},C_{y})\in\varphi$. Now, our definition of $\varphi$ produces $\beta\in\mathscr{D}_{X}$ with $C_{\alpha}=C_{\beta}$ and $C_{y}=C_{f^{*}\beta}$. Since $C_{y}\leq C_{f^{*}\beta}$, there is $B\in\beta$ in such a way that $y\in{\mathop{\rm cl}}_{X}f[B]$ and so, $B\neq\emptyset$. From the relation $C_{f^{*}\beta}\leq C_{y}$ we obtain $f[B]\subseteq{\mathop{\rm cl}}_{Y}\{y\}=\{y\}$ and therefore, $\emptyset\neq B\subseteq f^{-1}\{y\}$. ∎

Since any topological embedding is a continuous one-to-one function that is open onto its range, we obtain the following result.

Corollary 3.9**.**

If $Y$ is a space which can be embedded into a space $X$, then there is an order-embedding from $\mathscr{U}_{Y}$ into $\mathscr{U}_{X}$. In particular, $|\mathscr{U}_{Y}|\leq|\mathscr{U}_{X}|$.

Assume $X$ and $Y$ are spaces for which there is $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$, an (order) isomorphism. According to [9, Proposition 5.1, p. 70], for each $x\in X$, $C_{x}(X)$ is an atom of $\mathscr{U}_{X}$ (i.e., a minimal element of $\mathscr{U}_{X}\setminus\{C_{\emptyset}\}$) and so, $\varphi(C_{x}(X))$ happens to be an atom of $\mathscr{U}_{Y}$; consequently (see [9, Proposition 5.1, p. 70]), there exists a point $y\in Y$ with $\varphi(C_{x}(X))=C_{y}(Y)$. Moreover, as one easily deduces from Proposition 2.2, $y$ is the only member of $Y$ with this property.

Definition 3.10**.**

Let $X$ and $Y$ be a pair of spaces. If $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$ is an isomorphism, we will say that $f:X\to Y$ is the $\varphi$-induced function if

[TABLE]

Observe that if $f$ is a homeomorphism from a space $X$ onto a space $Y$ and $\varphi$ is the $f$-induced relation, the previous results imply that $\varphi$ is an isomorphism. Now, when $g$ is the $\varphi$-induced function, we obtain that, for each $x\in X$,

[TABLE]

i.e., $f(x)=g(x)$. In conclusion, $f=g$. Hence, the following is a natural question.

Question 3.11*.*

Assume $X$ and $Y$ are spaces for which there is an isomorphism $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$. If $f$ is the $\varphi$-induced function and $\psi$ is the $f$-induced relation, do we get $\varphi=\psi$?

With the idea in mind of giving a positive answer to this question for a class of spaces (zero-dimensional spaces), we will present some auxiliary results.

Lemma 3.12**.**

Assume $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$ is an isomorphism, where $X$ and $Y$ are spaces. If $f$ is the $\varphi$-induced function, then the following statements hold.

1. (1)

$f$* is a bijection and $f^{-1}$ is the $\varphi^{-1}$-induced function.* 2. (2)

If $A\subseteq X$ and $\beta\in\mathscr{D}_{Y}$ satisfy $\varphi(C_{A}(X))=C_{\beta}(Y)$, then $f[\mathop{\rm cl}\nolimits_{X}A]\subseteq\bigcup\overline{\beta}$.

Proof.

For (1), let $g$ be the $\varphi^{-1}$-induced function. Given $x\in X$, the relation $\varphi(C_{x})=C_{f(x)}$ implies that $C_{x}=\varphi^{-1}(C_{f(x)})=C_{g(f(x))}$ and so, $g\circ f$ is the identity function on $X$. Similarly, $f\circ g$ is the identity function on $Y$.

Given $x\in\overline{A}$, Proposition 2.2 produces $C_{x}\leq C_{A}$ and so, $C_{f(x)}=\varphi(C_{x})\leq\varphi(C_{A})=C_{\beta}$; hence, $f(x)\in\bigcup\overline{\beta}$. ∎

Proposition 3.13**.**

Let $X$ and $Y$ be spaces in such a way that there is an isomorphism $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$. Denote by $f$ the $\varphi$-induced function and consider the following statements.

1. (1)

$\varphi$* is the $f$-induced relation.* 2. (2)

For any $A\subseteq X$, $\varphi(C_{A}(X))=C_{f[A]}(Y)$. 3. (3)

Whenever $G$ is a closed subset of $X$, $\varphi(C_{G}(X))=C_{f[G]}(Y)$.

Then, (1) is equivalent to (2) and if $f$ is continuous, (2) and (3) are equivalent.

Proof.

The implications (1)$\to$(2) and (2)$\to$(3) are immediate. On the other hand, it follows from the work done in the first paragraphs of the proof of Proposition 3.1 that, for any $\alpha\in\mathscr{D}_{X}$,

[TABLE]

therefore, by assuming (2) we obtain

[TABLE]

i.e., (1) holds.

Now suppose $f$ is continuous and (3) is true. In order to prove (2), fix $A\subseteq X$ and set $G:=\overline{A}$. According to Proposition 2.2, $C_{A}=C_{G}$ and, consequently, $\varphi(C_{A})=\varphi(C_{G})=C_{f[G]}$. From the relation $f[A]\subseteq f[G]$ we deduce that $C_{f[A]}\leq C_{f[G]}$. The continuity of $f$ produces $f[G]\subseteq\overline{f[A]}$ and so, $C_{f[G]}\leq C_{f[A]}$. In conclusion, $\varphi(C_{A})=C_{f[G]}=C_{f[A]}$, as needed. ∎

Recall that for any space $Z$, $\mathop{\rm CO}(Z)$ is the collection of all subsets of $Z$ which are closed and open in $Z$. Consequently, $Z$ is zero-dimensional when $\mathop{\rm CO}(Z)$ is a base for $Z$.

Lemma 3.14**.**

Assume $X$ and $Y$ are spaces for which there is $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$, an isomorphism. If $f$ is the $\varphi$-induced function, the following statements hold.

1. (1)

For each $A\in\mathop{\rm CO}(X)$, $f[A]\in\mathop{\rm CO}(Y)$ and $\varphi(C_{A}(X))=C_{f[A]}(Y)$. 2. (2)

If $Y$ is zero-dimensional, $f$ is continuous.

Proof.

Given $A\in\mathop{\rm CO}(X)$, the proof of [9, Proposition 3.3, p. 68] shows that $C_{A}$ and $C_{X\setminus A}$ are complements of each other in $\mathscr{U}_{X}$ and so, $\varphi(C_{A})$ and $\varphi(C_{X\setminus A})$ have the same relation in $\mathscr{U}_{Y}$. Then, according to [9, Proposition 5.3, p. 70], there exists $B\in\mathop{\rm CO}(Y)$ with $\varphi(C_{A})=C_{B}$ and $\varphi(C_{X\setminus A})=C_{Y\setminus B}$. From Lemma 3.12(2), $f[\overline{A}]\subseteq\overline{B}$ and $f[\overline{X\setminus A}]\subseteq\overline{Y\setminus B}$, i.e., $f[A]\subseteq B$ and $Y\setminus B\supseteq f[X\setminus A]=Y\setminus f[A]$. Thus, $f[A]=B$.

For the second part, fix $B\in\mathop{\rm CO}(Y)$. According to Lemma 3.12(1), $f^{-1}$ is the $\varphi^{-1}$-induced function and so, we can apply part (1) of this lemma to $f^{-1}$ in order to get $f^{-1}[B]\in\tau_{X}$. Thus, the assumption that $\mathop{\rm CO}(Y)$ is a base for $Y$ gives $f$’s continuity. ∎

Lemma 3.15**.**

Let $X$ and $Y$ be spaces, with $X$ zero-dimensional. If $\varphi$ is an isomorphism from $\mathscr{U}_{X}$ onto $\mathscr{U}_{Y}$ and $f$ is the $\varphi$-induced function, then $\varphi(C_{G})\leq C_{f[G]}$, whenever $G$ is a closed subset of $X$.

Proof.

Given $G$, a closed subset of $X$, there are $\mathscr{A}\subseteq\mathop{\rm CO}(X)$ and $\beta\in\mathscr{D}_{X}$ in such a way that $G=\bigcap\mathscr{A}$ and $\varphi(C_{G})=C_{\beta}$. Let us argue that

[TABLE]

Suppose $A\in\mathscr{A}$ and $B\in\beta$ are arbitrary. Since $G\subseteq A$, we deduce that $C_{G}\leq C_{A}$ and, consequently, the use of Lemma 3.14(1) gives

[TABLE]

in particular, $B\subseteq\overline{f[A]}$. To complete this part, invoke lemmas 3.12(1) and 3.14(2) in order to get the continuity of $f^{-1}$, i.e., the closedness of $f$.

From (3.4) and the fact that $f$ is one-to-one, we obtain that, for any $B\in\beta$,

[TABLE]

In other words, $C_{\beta}\leq C_{f[G]}$, as claimed. ∎

Proposition 3.16**.**

Let $X$, $Y$, $\varphi$, $f$, and $\psi$ be as in Question 3.11. If $X$ and $Y$ are zero-dimensional, then $\varphi=\psi$.

Proof.

First of all, lemmas 3.14(2) and 3.12(1) guarantee that $f$ is a homeomorphism.

With the idea in mind of verifying condition (3) of Proposition 3.13, fix $G$, a closed subset of $X$. According to Lemma 3.15, $\varphi(C_{G})\leq C_{f[G]}$. On the other hand, $f[G]$ is a closed subset of $Y$ and so, by applying Lemma 3.15 to $\varphi^{-1}$ and $f^{-1}$, we obtain $\varphi^{-1}(C_{f[G]})\leq C_{f^{-1}[f[G]]}=C_{G}$, i.e., $C_{f[G]}\leq\varphi(C_{G})$. Thus, $\varphi(C_{G})=C_{f[G]}$.

We conclude that $\varphi$ is the $f$-induced relation or, in other words, $\varphi=\psi$. ∎

Corollary 3.17**.**

Let $X$ and $Y$ be a pair of zero-dimensional spaces. For any function $\varphi:\mathscr{U}_{X}\to\mathscr{U}_{Y}$, the following statements are equivalent.

1. (1)

$\varphi$* is an isomorphism.* 2. (2)

For some homeomorphism $f:X\to Y$, $\varphi$ is the $f$-induced relation.

Question 3.18*.*

Is the assumption of zero-dimensionality necessary in Corollary 3.17? To be more precise, are there non-homeomorphic spaces $X$ and $Y$ for which the lattices $\mathscr{U}_{X}$ and $\mathscr{U}_{Y}$ are isomorphic?

4. Some cardinal characteristics

Definition 4.1**.**

For a space $X$, set $\mathscr{U}_{X}^{+}:=\mathscr{U}_{X}\setminus\{C_{\emptyset}\}$. Also, given a family $\mathscr{S}\subseteq\mathscr{U}_{X}^{+}$, we say that

1. (1)

$\mathscr{S}$ is an antichain in $\mathscr{U}_{X}$ if for any $\sigma,\tau\in\mathscr{S}$, the condition $\sigma\neq\tau$ implies that $\sigma\wedge\tau=C_{\emptyset}$; 2. (2)

$\mathscr{S}$ is dense in $\mathscr{U}_{X}$ if for each $\sigma\in\mathscr{U}_{X}^{+}$ there is $\tau\in\mathscr{S}$ with $\tau\leq\sigma$.

For a space $X$, the cellularity of $\mathscr{U}_{X}$, $c(\mathscr{U}_{X})$, is the supremum of all cardinals of the form $|\mathscr{W}|$, where $\mathscr{W}$ is an antichain in $\mathscr{U}_{X}$. The density of $\mathscr{U}_{X}$, $\pi(\mathscr{U}_{X})$, is the minimum size of a dense subset of $\mathscr{U}_{X}$.

Proposition 4.2**.**

If $X$ is a space, then $c(\mathscr{U}_{X})=\pi(\mathscr{U}_{X})=|X|$.

Proof.

As one easily verifies, $\mathscr{A}:=\{C_{x}:x\in X\}$ is an antichain in $\mathscr{U}_{X}$. Thus, $|X|\leq c(\mathscr{U}_{X})$. On the other hand, if $\alpha\in\mathscr{D}_{X}$ satisfies $C_{\alpha}\in\mathscr{U}_{X}^{+}$, then $C_{\alpha}\not\leq C_{\emptyset}$, i.e., there are $A\in\alpha$ and $z\in A$. Therefore, $C_{z}\leq C_{\alpha}$ and, consequently, $\mathscr{A}$ is a dense subset of $\mathscr{U}_{X}$. Hence, $\pi(\mathscr{U}_{X})\leq|X|$.

In order to prove that $c(\mathscr{U}_{X})\leq\pi(\mathscr{U}_{X})$, let us fix $\mathscr{W}$, an antichain in $\mathscr{U}_{X}$, and $\mathscr{S}$, a dense subset of $\mathscr{U}_{X}$. Then, there is $e:\mathscr{W}\to\mathscr{S}$ such that $e(\tau)\leq\tau$, whenever $\tau\in\mathscr{W}$. Given $\sigma,\tau\in\mathscr{W}$ with $\sigma\neq\tau$, one gets $e(\sigma)\wedge e(\tau)\leq\sigma\wedge\tau=C_{\emptyset}$ and so, $e(\sigma)\neq e(\tau)$; in other words, $e$ is one-to-one and, as a consequence, $|\mathscr{W}|\leq|\mathscr{S}|$. ∎

Now we turn our attention to $|\mathscr{U}_{X}|$ and $|\mathscr{D}_{X}|$, for an arbitrary space $X$. With this in mind, given a cardinal $\kappa$, let us recursively define $\beth_{0}(\kappa):=\kappa$ and, for each integer $n$, $\beth_{n+1}(\kappa):=2^{\beth_{n}(\kappa)}$.

Proposition 4.3**.**

The following statements hold for any finite space $X$.

1. (1)

When $|X|=1$, $|\Sigma(X)|<2^{|X|}<|\mathscr{D}_{X}|=\beth_{2}(|X|)$. 2. (2)

If $X$ has at least two points, then $2^{|X|}\leq|\Sigma(X)|<|\mathscr{D}_{X}|<\beth_{2}(|X|)$. 3. (3)

$|\mathscr{U}_{X}|=2^{|X|}$.

Proof.

If $X$ has exactly one element, then

[TABLE]

With respect to (2), since the function $\eta:\mathscr{P}(X)\setminus\{\emptyset\}\to\Sigma(X)$ given by $\eta(A):=\{\emptyset,A,X\}$ is one-to-one, we deduce that $2^{|X|}-1=|\mathop{\rm ran}\nolimits(\eta)|\leq|\Sigma(X)|$. Let us fix $p,q\in X$ with $p\neq q$. From the fact that $\{\emptyset,\{p\},\{q\},\{p,q\},X\}$ is a member of $\Sigma(X)\setminus\mathop{\rm ran}\nolimits(\eta)$, it follows that $2^{|X|}\leq|\Sigma(X)|$.

The relations $\Sigma(X)\subseteq\mathscr{D}_{X}$ and $\{X\}\in\mathscr{D}_{X}\setminus\Sigma(X)$ clearly imply that $|\Sigma(X)|<|\mathscr{D}_{X}|$. Lastly, the inequality $|\mathscr{D}_{X}|<\beth_{2}(|X|)$ follows from the facts $\mathscr{D}_{X}\subseteq\mathscr{P}(\mathscr{P}(X))$ and $C_{p}\vee C_{q}\in\mathscr{P}(\mathscr{P}(X))\setminus\mathscr{D}_{X}$.

In order to prove (3), start by noticing that from $|X|<\omega$ one gets $C_{p}=C_{u}$. Thus, [9, Proposition 5.2, p. 70] implies that $\mathscr{P}(X)$, ordered by direct inclusion, and the closed interval $[C_{\emptyset},C_{u}]$, equipped with the order it inherits from $\mathscr{U}_{X}$, are order-isomorphic. Finally, (1) in [9, Proposition 3.2, p. 67] guarantees that $\mathscr{U}_{X}=[C_{\emptyset},C_{u}]$. ∎

Given a space $X$, let us denote by $\mathop{\rm RO}\nolimits(X)$ the collection of all regular open subsets of $X$. According to [6, Theorem 1.37, p. 26], when we order $\mathop{\rm RO}(X)$ by direct inclusion, the resulting structure is a complete Boolean algebra.

Proposition 4.4**.**

The following relations hold for any infinite topological space $X$.

1. (1)

$|\mathscr{D}_{X}|=\beth_{2}(|X|)$. 2. (2)

$\max\left\{2^{|X|},2^{|\mathop{\rm RO}(X)|}\right\}\leq|\mathscr{U}_{X}|\leq 2^{o(X)}$, where $o(X):=|\tau_{X}|$.

Proof.

The inequality $|\mathscr{D}_{X}|\leq\beth_{2}(|X|)$ follows from the relation $\mathscr{D}_{X}\subseteq\mathscr{P}(\mathscr{P}(X))$. On the other hand, according to [5, Theorem 7.6, p. 75], there are $\beth_{2}(|X|)$ filters on the set $X$ and, naturally, each one of them is a member of $\mathscr{D}_{X}$. This proves (1).

With respect to (2), recall that $\tau_{X}^{*}$ is the collection of all closed subsets of $X$. Clearly, $|\tau_{X}^{*}|=o(X)$. An immediate consequence of Proposition 2.2 is that for each $\alpha\in\mathscr{D}_{X}$ the family $\overline{\alpha}:=\{\overline{A}:A\in\alpha\}$ is a directed set and $C_{\alpha}=C_{\overline{\alpha}}$. Therefore, $\mathscr{U}_{X}$ is equal to $\{C_{\beta}:\beta\in\mathscr{D}_{X}\ \wedge\ \beta\subseteq\tau_{X}^{*}\}$, which, in turn, implies that $|\mathscr{U}_{X}|\leq|\mathscr{P}(\tau_{X}^{*})|=2^{o(X)}$.

Now, [9, Proposition 5.2, p. 70] guarantees the existence of a one-to-one map from $\mathscr{P}(X)$ into $\mathscr{U}_{X}$ and so, $2^{|X|}\leq|\mathscr{U}_{X}|$.

For the remaining inequality we need some notation. First, given a finite function $p\subseteq\mathop{\rm RO}(X)\times 2$, set

[TABLE]

where $-x$ is the Boolean complement of $x\in\mathop{\rm RO}(X)$. Hence, a set $\mathscr{A}\subseteq\mathop{\rm RO}(X)$ is called independent if for any finite function $p\subseteq\mathscr{A}\times 2$ one has $\bigwedge p^{\sim}\neq\emptyset$.

The fact that $X$ is an infinite Tychonoff space implies that $\mathop{\rm RO}(X)$ is infinite as well and so, by Balcar-Franěk’s Theorem (see [6, Theorem 13.6, p. 196]), there is an independent set $\mathscr{A}\subseteq\mathop{\rm RO}(X)$ with $|\mathscr{A}|=|\mathop{\rm RO}(X)|$.

Let us argue that, for each $d:\mathscr{A}\to 2$, the collection

[TABLE]

is a member of $\mathscr{D}_{X}$. Indeed, if $p,q\in[d]^{<\omega}$, then $r:=p\cup q$ is a finite subset of $d$ with $\bigvee r^{\sim}=\left(\bigvee p^{\sim}\right)\vee\bigvee q^{\sim}$ and since $\mathop{\rm RO}(X)$ is ordered by direct inclusion, we conclude that $\bigvee r^{\sim}$ is an element of $\alpha(d)$ which is a superset of $\bigvee p^{\sim}$ and $\bigvee q^{\sim}$.

Claim. If $d,e\in{}^{\mathscr{A}}2$ and $U\in\mathscr{A}$ satisfy $d(U)=0$ and $e(U)=1$, then, for any $V\in\alpha(e)$, $U\not\subseteq V$.

Before we present the proof of our Claim, let’s assume it holds and fix $d,e\in{}^{\mathscr{A}}2$ with $d\neq e$. Without loss of generality, we may assume that, for some $U\in\mathscr{A}$, $d(U)=0$ and $e(U)=1$. Thus, $U\in\alpha(d)$ and if $V$ were a member of $\alpha(e)$ with $U\subseteq\overline{V}$, we would get $U=\mathop{\rm int}U\subseteq\mathop{\rm int}\overline{V}=V$, a contradiction to the Claim. As a consequence of this argument, we obtain that the function from ${}^{\mathscr{A}}2$ into $\mathscr{U}_{X}$ given by $d\mapsto C_{\alpha(d)}$ is one-to-one and so, $2^{|\mathop{\rm RO}(X)|}=2^{|\mathscr{A}|}\leq|\mathscr{U}_{X}|$.

Suppose $d$, $e$, and $U$ are as in the Claim. Seeking a contradiction, let us assume that $U\subseteq\bigvee p^{\sim}$, for some $p\in[e]^{<\omega}$. We affirm that if $q:=p\restriction(\mathop{\rm dom}\nolimits(p)\setminus\{U\})$ (the restriction of the function $p$ to the given set), then

[TABLE]

Indeed, when $U\notin\mathop{\rm dom}\nolimits(p)$, $p=q$. On the other hand, if $U\in\mathop{\rm dom}\nolimits(p)$, the relation $p\subseteq e$ gives $p(U)=1$ and so, $\bigvee p^{\sim}=(-U)\vee\bigvee q^{\sim}$ which, clearly, implies (4.1).

Let us define $r:\mathop{\rm dom}\nolimits(q)\cup\{U\}\to 2$ by $r(V)=1-q(V)$, whenever $V\in\mathop{\rm dom}\nolimits(q)$, and $r(U)=0$. Obviously, $r\subseteq\mathscr{A}\times 2$ is a finite function and thus, the independence of $\mathscr{A}$ and the De Morgan’s laws produce

[TABLE]

a contradiction to (4.1). ∎

Let us recall that a $T_{6}$-space (equivalently, perfectly normal space) is a Hausdorff normal space in which all open sets are of type $F_{\sigma}$.

Corollary 4.5**.**

If $X$ is a $T_{6}$-space, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

We only need to mention that, according to [3, Theorem 10.5, p. 40], $|\mathop{\rm RO}(X)|=o(X)$. ∎

Our next result is a direct consequence of corollaries 4.5 and 3.9 (recall that any infinite Tychonoff space contains a copy of the discrete space of size $\omega$).

Corollary 4.6**.**

If $Y$ is an infinite discrete subspace of a space $X$, $\beth_{2}(|Y|)\leq|\mathscr{U}_{X}|$. In particular, when $X$ is infinite, $2^{\mathfrak{c}}\leq|\mathscr{U}_{X}|$.

Standard arguments show that if $X$ is an arbitrary space and $D$ is a dense subspace of it, then the function from $\mathop{\rm RO}(X)$ into $\mathscr{P}(D)$ given by $U\mapsto U\cap D$ is one-to-one. Therefore (recall that $d(X)$ is the density of $X$),

[TABLE]

Regarding the accuracy of the bounds presented in Proposition 4.4(2), we have the result below.

Proposition 4.7**.**

The following statements are true.

1. (1)

If $X$ is the Moore-Niemytzki plane (see **[1, Example 1.2.4, p. 21]**), then $|X|=|\mathop{\rm RO}(X)|=\mathfrak{c}$ and $o(X)=2^{\mathfrak{c}}$. 2. (2)

When $X$ is the Stone-Čech compactification of the integers, $|\mathop{\rm RO}(X)|=\mathfrak{c}$ and $|X|=o(X)=2^{\mathfrak{c}}$. 3. (3)

If $X$ is the Arens-Fort space, **[1, Example 1.6.19, p. 54]**, then $|X|=\omega$ and $|\mathop{\rm RO}(X)|=o(X)=\mathfrak{c}$.

Proof.

Let us prove (1). Clearly, $|X|=\mathfrak{c}$. The equality $|\mathop{\rm RO}(X)|=\mathfrak{c}$ follows from the facts, (i) property (4.2) (recall that $X$ is separable) and (ii) the canonical base for $X$ consists of $\mathfrak{c}$ many regular open sets. Note that from (ii) we also deduce the relation $o(X)\leq 2^{\mathfrak{c}}$. Finally, since $X\setminus(\mathbb{R}\times\{0\})$ is an open subset of $X$ which is homeomorphic to an open subspace of the Euclidean plane, we conclude that $2^{\mathfrak{c}}\leq o(X)$.

Suppose $X$ is as in (2). From [1, Corollary 3.6.12, p. 175], $|X|=2^{\mathfrak{c}}$. On the other hand, the relation $|\mathop{\rm RO}(X)|=\mathfrak{c}$ is a consequence of (4.2) and the fact that, according to Theorem 3.6.13 and Corollary 3.6.12 of [1, p. 175], $X$ is a space of weight $\mathfrak{c}$ possessing a base of closed-and-open sets. This last statement also implies that $o(X)\leq 2^{\mathfrak{c}}$. Now, [1, Example 3.6.18, p. 175] guarantees that $X$ has a pairwise disjoint family consisting of $\mathfrak{c}$ many non-empty open sets and so, $2^{\mathfrak{c}}\leq o(X)$.

Finally, when $X$ is as in (3), one clearly gets $|X|=\omega$ and, therefore, $o(X)\leq\mathfrak{c}$. On the other hand, by definition, $X$ has a base consisting of $\mathfrak{c}$ many closed-and-open sets; hence, $\mathfrak{c}\leq|\mathop{\rm RO}(X)|\leq o(X)$. ∎

In the next section we focus on the problem of calculating $|\mathscr{U}_{X}|$, for some spaces $X$.

5. The size of $\mathscr{U}_{X}$

Unless otherwise stated, all spaces considered from now on are infinite. Also, recall that [1] is our reference for topological cardinal functions.

In Corollary 4.5 we were able to calculate the precise value of $|\mathscr{U}_{X}|$ in terms of the cardinal function $o(X)$, when $X$ belongs to the class of $T_{6}$-spaces. Here, we present some other classes of topological spaces in which the cardinality of the lattice $\mathscr{U}_{X}$ can be determined in a similar fashion.

Proposition 5.1**.**

Given a space $X$, if any of the following statements holds, then $|\mathscr{U}_{X}|=2^{\mathfrak{c}}$.

1. (1)

$X$* is hereditarily Lindelöf and first countable.* 2. (2)

$X$* admits a countable network.* 3. (3)

$X$* is hereditarily separable and has countable pseudocharacter.*

Proof.

From Proposition 4.4 and Figure 1 we deduce that $|\mathscr{U}_{X}|\leq 2^{\mathfrak{c}}$. The reverse inequality is a consequence of Corollary 4.6. ∎

In what follows, given a space $X$, we will employ the inequalities presented in Figure 1 together with Proposition 4.4(2) in order to get bounds for $|\mathscr{U}_{X}|$.

Now, regarding compact spaces we have the following results.

Lemma 5.2**.**

For any compact space $X$, $|\mathscr{U}_{X}|\leq\beth_{2}(hL(X))$.

Proof.

Given the hypotheses on $X$, we obtain $\chi(X)=\psi(X)\leq hL(X)$ and thus, the inequality needed follows from Figure 1 and Proposition 4.4. ∎

Proposition 5.3**.**

If $X$ is a compact space in which every open subset of it is an $F_{\sigma}$-set, then $|\mathscr{U}_{X}|=2^{\mathfrak{c}}$. In particular, every compact metrizable space satisfies the previous equality.

Proof.

It is sufficient to notice that our assumptions on $X$ imply $hL(X)=\omega$. Thus, Corollary 4.6 and Lemma 5.2 give the desired result. ∎

Given an infinite cardinal $\kappa$, let us denote by $D(\kappa)$ and $\beta D(\kappa)$ the discrete space of size $\kappa$ and its Stone-Čech compactification, respectively. The regularity of $\beta D(\kappa)$ implies that (see [3, Theorem 3.3, p. 11])

[TABLE]

Therefore, from Figure 1 and the compactness of $\beta D(\kappa)$ we deduce that

[TABLE]

On the other hand, since $\left|\beta D(\kappa)\right|=\beth_{2}(\kappa)$, Proposition 4.4(2) gives

[TABLE]

In conclusion, for any infinite cardinal $\kappa$, $|\mathscr{U}_{\beta D(\kappa)}|=\beth_{3}(\kappa)$.

Once again, let $\kappa\geq\omega$ be a cardinal. If $D(2)$ is the discrete space of size $2$, then $D(2)^{\kappa}$ is the Cantor cube of weight $\kappa$. Clearly (see Figure 1),

[TABLE]

Also, Proposition 4.4(2) produces

[TABLE]

Hence, for any infinite cardinal $\kappa$, $|\mathscr{U}_{D(2)^{\kappa}}|=\beth_{2}(\kappa)$.

Let $\mathbb{L}$ be the lexicographic square (i.e., $\mathbb{L}$ is the cartesian product $[0,1]^{2}$ endowed with the topology generated by the lexicographical ordering). By setting $Y:=[0,1]\times\{\tfrac{1}{2}\}$ one gets a discrete subspace of $\mathbb{L}$ and so, according to Corollaries 4.5 and 3.9, $\beth_{2}(\mathfrak{c})=|\mathscr{U}_{Y}|\leq|\mathscr{U}_{\mathbb{L}}|$. Finally, our definition of $\mathbb{L}$ gives $o(\mathbb{L})\leq 2^{\mathfrak{c}}$ and, as a consequence, $|\mathscr{U}_{\mathbb{L}}|\leq\beth_{2}(\mathfrak{c})$. In other words, $|\mathscr{U}_{\mathbb{L}}|=\beth_{2}(\mathfrak{c})$.

The subspace $[0,1]\times\{0,1\}$ of $\mathbb{L}$ is called the double arrow space and we will denote it by $\mathbb{A}$. Since the subspace $(0,1)\times\{0\}$ of $\mathbb{A}$ is homeomorphic to Sorgenfrey’s line, the space $\mathbb{A}^{2}$ contains a discrete subspace of size $\mathfrak{c}$. Therefore, as we did for $\mathbb{L}$, $|\mathscr{U}_{\mathbb{A}^{2}}|\geq\beth_{2}(\mathfrak{c})$. For the reverse inequality note that $o(\mathbb{A}^{2})\leq o(\mathbb{L}^{2})\leq 2^{\mathfrak{c}}$ and so, $|\mathscr{U}_{\mathbb{A}^{2}}|=\beth_{2}(\mathfrak{c})$.

A final note regarding $\mathbb{A}$ is pertinent. From (4.2) and the fact that $\mathbb{A}$ is separable, we deduce that $|\mathop{\rm RO}(\mathbb{A}^{2})|\leq\mathfrak{c}$ and hence,

[TABLE]

This shows that the lower bounds for $|\mathscr{U}_{X}|$ presented in Proposition 4.4(2) need to be improved.

Proposition 5.4**.**

If $X$ is hereditarily Lindelöf, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

With Corollary 4.5 in mind, we only need to show that all open subsets of $X$ are $F_{\sigma}$. Let $U\in\tau_{X}$ be arbitrary. For each $x\in U$ there is $U_{x}\in\tau_{X}$ such that $x\in U_{x}\subseteq\overline{U_{x}}\subseteq U$. Since $U$ is Lindelöf, for some $F\in[U]^{\leq\omega}$ we obtain $U=\bigcup\left\{\overline{U_{x}}:x\in F\right\}$. ∎

We present now our findings regarding the following question.

Question 5.5*.*

Given a space $X$, what conditions on $X$ imply that $|\mathscr{U}_{X}|=2^{o(X)}$?

Lemma 5.6**.**

If $X$ is a space with $|X|^{hd(X)}=|X|$, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

It follows from Figure 1 and our hypotheses that $o(X)\leq|X|$. On the other hand, the fact that $X$ is Tychonoff clearly implies the relation $|X|\leq o(X)$. Hence, the equality we need is a consequence of Proposition 4.4(2). ∎

Proposition 5.7**.**

If $X$ is a space for which there is a cardinal $\kappa$ with $|X|=2^{\kappa}$ and $\kappa\geq hd(X)$, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

Our choice for $\kappa$ gives $|X|^{hd(X)}=|X|$ and so, the hypotheses of Lemma 5.6 are satisfied. ∎

As usual, the acronym GCH stands for the Generalized Continuum Hypothesis and $\mathop{\rm cf}(\alpha)$ denotes the cofinality of an ordinal $\alpha$.

Proposition 5.8**.**

Assuming GCH, if $X$ is a space satisfying $\mathop{\rm cf}\nolimits(|X|)>hd(X)$, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

According to [7, Lemma 10.42, p. 34], $|X|^{hd(X)}=|X|$ and therefore we only need to invoke Lemma 5.6. ∎

Proposition 5.9**.**

Given a space $X$, if $|X|$ is a singular strong limit cardinal, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

The hypothesis allows us to use [2, Theorem 3, p. 22] to find a discrete set $D\subseteq X$ such that $|D|=|X|$. Hence, Proposition 4.4(2) and Corollary 4.6 imply that $|\mathscr{U}_{X}|=2^{o(X)}$. ∎

Let us denote by A the statement “GCH holds and there are no inaccessible cardinals.”

Corollary 5.10**.**

Assume A holds. Then, for any space $X$ whose cardinality is a limit cardinal we obtain $|\mathscr{U}_{X}|=2^{o(X)}$.

With the idea in mind of finding the effect that GCH has on $|\mathscr{U}_{X}|$, let us recall that, for a cardinal number $\kappa$, $\kappa^{+}$ represents the successor cardinal of $\kappa$.

Proposition 5.11**.**

If GCH holds, then, for any space $X$, $|\mathscr{U}_{X}|$ is a regular uncountable cardinal.

Proof.

On the one hand, Corollary 4.6 implies that $|\mathscr{U}_{X}|$ is uncountable. On the other hand, since $2^{|X|}\leq|\mathscr{U}_{X}|\leq 2^{o(X)}\leq\beth_{2}\left(|X|\right)=(2^{|X|})^{+}$, we deduce that $|\mathscr{U}_{X}|\in\{|X|^{+},(2^{|X|})^{+}\}$. In either case, $|\mathscr{U}_{X}|$ is regular. ∎

Proposition 5.12**.**

Under the assumptions $\mathfrak{c}=\omega_{1}$ and $2^{\mathfrak{c}}=\omega_{2}$, if $X$ is a hereditarily separable space, then $|\mathscr{U}_{X}|=2^{o(X)}$.

Proof.

According to [3, Theorem 4.12, p. 21], the relation $hd(X)=\omega$ guarantees that $|X|\leq 2^{\mathfrak{c}}$ and consequently, $|X|\in\{\omega,\omega_{1},\omega_{2}\}$.

When $|X|\in\{\omega_{1},\omega_{2}\}$, Proposition 5.7 gives us the desired equality. Finally, if $|X|=\omega$, then $X$ admits a countable network and thus (see Proposition 5.1), $|\mathscr{U}_{X}|=2^{\mathfrak{c}}=2^{o(X)}$. ∎

Suppose $X$ is a space. Since $\mathscr{U}_{X}$ is a subset of $\Sigma(C(X))$, we obtain $|\mathscr{U}_{X}|\leq|\Sigma(C(X))|$. With the idea in mind of showing two examples for which this inequality is strict, let us note first that the fact $|C(X)|\geq\omega$ implies, according to [8, Theorem 1.4, p. 179], that $|\Sigma(C(X))|=\beth_{2}(|C(X)|)$.

When $X$ is an infinite discrete space, we obtain $|C(X)|=2^{|X|}$ and so, by Proposition 4.4(2),

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On the other hand, if $X$ is any infinite countable space, then it follows from Proposition 5.1(2) that

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Our final result of this section establishes some conditions for a family of topological spaces under which the corresponding Tychonoff product $X$ satisfies the equality $|\mathscr{U}_{X}|=|\Sigma(C(X))|$. For this proposition we won’t require for our spaces to be infinite.

Proposition 5.13**.**

Assume that $\kappa$ is an infinite cardinal. Let $X$ be the topological product of a family of spaces $\{X_{\xi}:\xi<2^{\kappa}\}$. If $|X_{\xi}|\geq 2$ and $d(X_{\xi})\leq\kappa$ for each $\xi<\kappa$, then $|\mathscr{U}_{X}|=|\Sigma(C(X))|$.

Proof.

Since we always have the inequality $|\mathscr{U}_{X}|\leq|\Sigma(C(X))|$, we only need to show that $|\mathscr{U}_{X}|\geq\beth_{2}(|C(X)|)$.

According to Proposition 4.4(2), $|\mathscr{U}_{X}|\geq 2^{|X|}$. Now, the fact that each $X_{\xi}$ has at least two points gives $|X|\geq\beth_{2}(\kappa)$ and so, $2^{|X|}\geq\beth_{3}(\kappa)$. On the other hand, the Hewitt-Marczewski-Pondiczery Theorem (see [1, Theorem 2.3.15, p. 81]) implies that $d(X)\leq\kappa$ and therefore, from the well-known relation $2^{d(X)}\geq|C(X)|$ we deduce that $2^{\kappa}\geq|C(X)|$. In conclusion, $|\mathscr{U}_{X}|\geq\beth_{3}(\kappa)\geq\beth_{2}(|C(X)|)$, as required. ∎

For example, if $X$ is a Cantor cube of the form $D(2)^{2^{\kappa}}$, where $\kappa$ is an infinite cardinal, then $|\mathscr{U}_{X}|\geq\beth_{2}(|C(X)|)$.

We close the paper with a list of open questions.

Question 5.14*.*

Does Corollary 4.5 remain true if we replace $T_{6}$ with $T_{5}$ in the hypotheses?

Question 5.15*.*

Regarding Proposition 5.4, is it true that for any compact space $X$, $|\mathscr{U}_{X}|=2^{o(X)}$?

Question 5.16*.*

Can we drop the set-theoretic assumptions $\mathfrak{c}=\omega_{1}$ and $2^{\mathfrak{c}}=\omega_{2}$ in Proposition 5.12?

We conjecture that, under A, the equality

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holds for any space $X$. Even though we did not prove or refute this conjecture, we were able to obtain some partial results (for example, if one assumes A, then (i) for any space $X$, $\beth_{2}(s(X))\leq|\mathscr{U}_{X}|$, and (ii) we possess a short list of classes $\mathscr{S}$ in such a way that $X\in\mathscr{S}$ implies that (5.1) holds). Consequently, we pose the following problem.

Question 5.17*.*

Does it follow from A that (5.1) is true for any space $X$?