Pairwise Semiregular Properties on Generalized Pairwise Lindelof Spaces

TL;DR

This paper investigates the concept of pairwise semiregular properties in bitopological spaces, establishing which Lindel"{o}f-related properties are preserved under pairwise semiregularization.

Contribution

It introduces the notion of pairwise semiregular properties and characterizes which Lindel"{o}f-type properties are preserved under this process.

Findings

(i,j)-almost Lindel"{o}f, pairwise almost Lindel"{o}f, (i,j)-weakly Lindel"{o}f, and pairwise weakly Lindel"{o}f are pairwise semiregular properties.

Certain types of pairwise Lindel"{o}f spaces are not pairwise semiregular properties.

Abstract

Let $\left( X,\tau _{1},\tau _{2}\right) $ be a bitopological space and $% \left( X,\tau _{\left( 1,2\right) }^{s},\tau _{\left( 2,1\right) }^{s}\right) $ its pairwise semiregularization. Then a bitopological property $\mathcal{P}$\ is called pairwise semiregular provided that $\left( X,\tau _{1},\tau _{2}\right) $\ has the property $\mathcal{P}$\ if and only if $\left( X,\tau _{\left( 1,2\right) }^{s},\tau _{\left( 2,1\right) }^{s}\right) $\ has the same property. In this work we study pairwise semiregular property of $\left( i,j\right) $-nearly Lindel\"{o}f, pairwise nearly Lindel\"{o}f, $\left( i,j\right) $-almost Lindel\"{o}f, pairwise almost Lindel\"{o}f, $\left( i,j\right) $-weakly Lindel\"{o}f and pairwise weakly Lindel\"{o}f spaces. We prove that $\left( i,j\right) $-almost Lindel% \"{o}f, pairwise almost Lindel\"{o}f, $\left( i,j\right) $-weakly Lindel\"{o}% f and pairwise weakly Lindel\"{o}f are pairwise semiregular properties, on the contrary of each type of pairwise Lindel\"{o}f space which are not pairwise semiregular properties.

Equations

$${\tau }_{\left(j,i\right)}^s\text{-}\mathrm{cl}\left({\bigcup }_{n\in \mathbb{N}}{\tau }_i\text{-}\mathrm{int}\left({\tau }_j\text{-}\mathrm{cl}\left(V_{{\alpha }_n}\right)\right)\right)$$

Full text

1. Introduction

Semiregular properties in topological spaces have been studied by many topologist. Some of them related to this research studied by Mrsevic et al. [14, 15] and Fawakhreh and Kılıçman [3]. The purpose of this paper is to study pairwise semiregular properties on generalized pairwise Lindelöf spaces, that we have studied in [16, 9, 17, 13], namely, $\left(i,j\right)$-nearly Lindelöf, pairwise nearly Lindelöf, $\left(i,j\right)$-almost Lindelöf, pairwise almost Lindelöf, $\left(i,j\right)$-weakly Lindelöf and pairwise weakly Lindelöf spaces.

The main results is that the Lindelöf, $B$-Lindelöf, $s$-Lindelöf and $p$-Lindelöf spaces are not pairwise semiregular properties. While $\left(i,j\right)$-almost Lindelöf, pairwise almost Lindelöf, $\left(i,j\right)$-weakly Lindelöf and pairwise weakly Lindelöf spaces are pairwise semiregular properties. We also show that $\left(i,j\right)$-nearly Lindelöf and pairwise nearly Lindelöf spaces are satisfying pairwise semiregular invariant properties.

2. Preliminaries

Throughout this paper, all spaces $\left(X,\tau\right)$ and $\left(X,\tau_{1},\tau_{2}\right)$ $($or simply $X)$ are always mean topological spaces and bitopological spaces, respectively unless explicitly stated. If $\mathcal{P}$ is a topological property, then $\left(\tau_{i},\tau_{j}\right)$-$\mathcal{P}$ denotes an analogue of this property for $\tau_{i}$ has property $\mathcal{P}$ with respect to $\tau_{j}$, and $p$-$\mathcal{P}$ denotes the conjunction $\left(\tau_{1},\tau_{2}\right)$-$\mathcal{P}\wedge\left(\tau_{2},\tau_{1}\right)$-$\mathcal{P}$, i.e., $p$-$\mathcal{P}$ denotes an absolute bitopological analogue of $\mathcal{P}$. As we shall see below, sometimes $\left(\tau_{1},\tau_{2}\right)$-$\mathcal{P}\Longleftrightarrow\left(\tau_{2},\tau_{1}\right)$-$\mathcal{P}$ (and thus $\Leftrightarrow p$-$\mathcal{P}$) so that it suffices to consider one of these three bitopological analogue. Also sometimes $\tau_{1}$-$\mathcal{P}\Longleftrightarrow\tau_{2}$-$\mathcal{P}$ and thus $\mathcal{P}\Longleftrightarrow\tau_{1}$-$\mathcal{P}\wedge\tau_{2}$-$\mathcal{P}$, i.e., $\left(X,\tau_{i}\right)$ has property $\mathcal{P}$ for each $i=1,2$.

Also note that $\left(X,\tau_{i}\right)$ has a property $\mathcal{P}\Longleftrightarrow\left(X,\tau_{1},\tau_{2}\right)$ has a property $\tau_{i}$-$\mathcal{P}$. Sometimes the prefixes $\left(\tau_{i},\tau_{j}\right)$- or $\tau_{i}$- will be replaced by $\left(i,j\right)$- or $i$- respectively, if there is no chance for confusion. By $i$-open cover of $X$, we mean that the cover of $X$ by $i$-open sets in $X$; similar for the $\left(i,j\right)$-regular open cover of $X$ etc. By $i$-$\mathop{\mathrm{i}nt}\left(A\right)$ and $i$-$\mathop{\mathrm{c}l}\left(A\right)$, we shall mean the interior and the closure of a subset $A$ of $X$ with respect to topology $\tau_{i}$, respectively. In this paper always $i,j\in\left\{1,2\right\}$ and $i\neq j$. The reader may consult [2] for the detail notations.

The following are some basic concepts.

Definition 2.1**.**

[6, 19]** A subset $S$ of a bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is said to be $\left(i,j\right)$-regular open $($resp. $\left(i,j\right)$-regular closed$)$ if $i$-$\mathop{\mathrm{i}nt}\left(j\text{-}\mathop{\mathrm{c}l}\left(S\right)\right)=S$ $($resp. $i$-$\mathop{\mathrm{c}l}\left(j\text{-}\mathop{\mathrm{i}nt}\left(S\right)\right)=S)$, where $i,j\in\left\{1,2\right\},i\neq j$. $S$ is called pairwise regular open $($resp. pairwise regular closed$)$ if it is both $\left(1,2\right)$-regular open and $\left(2,1\right)$-regular open $($resp. $\left(1,2\right)$-regular closed and $\left(2,1\right)$-regular closed$)$.**

Definition 2.2**.**

[6, 20]** A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is said to be $\left(i,j\right)$-almost regular if for each point $x\in X$ and for each $\left(i,j\right)$-regular open set $V$ containing $x$, there exists an $\left(i,j\right)$-regular open set $U$ such that $x\in U\subseteq j$-$\mathop{\mathrm{c}l}\left(U\right)\subseteq V$. $X$ is called pairwise almost regular if it is both $\left(1,2\right)$-almost regular and $\left(2,1\right)$-almost regular.**

In any bitopological space $\left(X,\tau_{1},\tau_{2}\right)$, the family of all $\left(i,j\right)$-regular open sets is closed under finite intersections. Thus the family of $\left(i,j\right)$-regular open sets in any bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ forms a base for a coarser topology called $\left(i,j\right)$-semiregularization of $\left(X,\tau_{1},\tau_{2}\right)$, which is defined as follows.

Definition 2.3**.**

[16]** The topology generated by the $\left(i,j\right)$-regular open subsets of $\left(X,\tau_{1},\tau_{2}\right)$ is denoted by $\tau_{\left(i,j\right)}^{s}$ and it is called $\left(i,j\right)$-semiregularization of $X$. The topologies is pairwise semiregularization of $X$ if the first topology is $\left(1,2\right)$-semiregularization of $X$ and the second topology is $\left(2,1\right)$-semiregularization of $X$. If $\tau_{i}\equiv\tau_{\left(i,j\right)}^{s}$, then $X$ is said to be $\left(i,j\right)$-semiregular. $\left(X,\tau_{1},\tau_{2}\right)$ is called pairwise semiregular if it is both $\left(1,2\right)$-semiregular and $\left(2,1\right)$-semiregular, that is, whenever $\tau_{i}\equiv\tau_{\left(i,j\right)}^{s}$ for each $i,j\in\left\{1,2\right\}$ and $i\neq j$. In other words, $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(i,j\right)$-semiregular if the family of $\left(i,j\right)$-regular open sets form a base for the topology $\tau_{i}$.

It is very clear that $\tau_{\left(i,j\right)}^{s}\subseteq\tau_{i}$, but it is not necessary $\tau_{i}\subseteq\tau_{\left(i,j\right)}^{s}$. Thus with every given bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ there is associated another bitopological space $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ in the manner described above (see [19]). We provide the following example in order to understand the concept of pairwise semiregular spaces clearly.

Example 2.1**.**

For the set of all real numbers $\mathbb{R}$, let $\tau_{u}$ denotes the usual topology and $\tau_{s}$ denote the Sorgenfrey topology, i.e., topology generated by right half-open intervals $($see [21]$)$. Then $\left(\mathbb{R},\tau_{u},\tau_{s}\right)$ is $\left(\tau_{u},\tau_{s}\right)$-semiregular since $\tau_{u}=\tau_{\left(\tau_{u},\tau_{s}\right)}^{s}$, i.e., $\tau_{u}$ generated by $\left(\tau_{u},\tau_{s}\right)$-regular open subsets of $\mathbb{R}$. $\left(\mathbb{R},\tau_{u},\tau_{s}\right)$ is also $\left(\tau_{s},\tau_{u}\right)$-semiregular since $\tau_{s}=\tau_{\left(\tau_{s},\tau_{u}\right)}^{s}$ because any set $E\in\tau_{s}$ is the union of a collection of $\left(\tau_{s},\tau_{u}\right)$-regular open sets in $\mathbb{R}$. Thus $\left(\mathbb{R},\tau_{u},\tau_{s}\right)$ is pairwise semiregular.

Khedr and Alshibani [6] defined the equivalent definition of $\left(i,j\right)$-semiregular spaces as follows.

Definition 2.4**.**

A bitopological space $X$ is said to be $\left(i,j\right)$-semiregular if for each $x\in X$ and for each $i$-open subset $V$ of $X$ containing $x$, there is an $i$-open set $U$ such that $x\in U\subseteq i$-$\mathop{\mathrm{i}nt}\left(j\text{-}\mathop{\mathrm{c}l}\left(U\right)\right)\subseteq V$. $X$ is called pairwise semiregular if it is both $\left(1,2\right)$-semiregular and $\left(2,1\right)$-semiregular.

Definition 2.5**.**

[1]** A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is said to be $\left(i,j\right)$-extremally disconnected if the $i$-closure of every $j$-open set is $j$-open. $X$ is called pairwise extremally disconnected if it is both $\left(1,2\right)$-extremally disconnected and $\left(2,1\right)$-extremally disconnected.

Recall that a property $\mathcal{P}$ will be called bitopological property (resp. $p$-topological property) if whenever $\left(X,\tau_{1},\tau_{2}\right)$ has property $\mathcal{P}$, then every space homeomorphic (resp. $p$-homeomorphic) to $\left(X,\tau_{1},\tau_{2}\right)$ also has property $\mathcal{P}$ (see [8]). If a bitopological space $X$ has bitopological (or $p$-topological) property $\mathcal{P}$, one may ask, does the pairwise semiregularization of $X$ satisfies the property $\mathcal{P}$ also? Now we arrive to the concept of pairwise semiregular property.

Definition 2.6**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a bitopological space and let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ its pairwise semiregularization. A bitopological property $\mathcal{P}$ is called pairwise semiregular provided that $\left(X,\tau_{1},\tau_{2}\right)$ has the property $\mathcal{P}$ if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ has the property $\mathcal{P}$.

Lemma 2.1**.**

[16]** Let $\left(X,\tau_{1},\tau_{2}\right)$ be a bitopological space and let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ its pairwise semiregularization. Then

$\left(a\right)$ $\tau_{i}$-$\mathop{\mathrm{i}nt}\left(C\right)=\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(C\right)$ for every $\tau_{j}$-closed set $C;$

$\left(b\right)$ $\tau_{i}$-$\mathop{\mathrm{c}l}\left(A\right)=\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{c}l}\left(A\right)$ for every $A\in\tau_{j};$

$\left(c\right)$ the family of $\left(\tau_{i},\tau_{j}\right)$-regular open sets of $\left(X,\tau_{1},\tau_{2}\right)$ are the same as the family of $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-regular open sets of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right);$

$\left(d\right)$ the family of $\left(\tau_{i},\tau_{j}\right)$-regular closed sets of $\left(X,\tau_{1},\tau_{2}\right)$ are the same as the family of $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-regular closed sets of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right);$

$\left(e\right)$ $\left(\tau_{\left(i,j\right)}^{s}\right)_{\left(i,j\right)}^{s}=\tau_{\left(i,j\right)}^{s}$.

3. **Pairwise Semiregularization of Pairwise Lindelöf Spaces

**

Definition 3.1**.**

[5, 7]**. A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is said to be $i$-Lindelöf if the topological space $\left(X,\tau_{i}\right)$ is Lindelöf. $X$ is called Lindelöf if it is $i$-Lindelöf for each $i=1,2$. In other words, $\left(X,\tau_{1},\tau_{2}\right)$ is called Lindelöf if the topological space $\left(X,\tau_{1}\right)$ and $\left(X,\tau_{2}\right)$ are both Lindelöf.

Note that $i$-Lindelöf property as well as Lindelöf property is not a pairwise semiregular property by the following example.

Example 3.1**.**

Let $X$ be a set with cardinality $2^{c}$, where $c=\mathop{\mathrm{c}ard}\left(\mathbb{R}\right)$. Let $\tau_{1}$ be a co-$c$ topology on $X$ consisting of $\emptyset$ and all subsets of $X$ whose complements have cardinality at most $c$ and let $\tau_{2}$ be a cofinite topology on $X$. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\tau_{2}$-Lindelöf but not $\tau_{1}$-Lindelöf and hence not Lindelöf. Observe that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(1,2\right)}^{s}$-Lindelöf and $\tau_{\left(2,1\right)}^{s}$-Lindelöf since $\tau_{\left(1,2\right)}^{s}$ and $\tau_{\left(2,1\right)}^{s}$ are indiscrete topologies. Hence $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

Definition 3.2**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is called $\left(i,j\right)$-Lindelöf [5, 7] if for every $i$-open cover of $X$ there is a countable $j$-open subcover. $X$ is called $B$-Lindelöf [5] or $p_{1}$-Lindelöf [7] if it is both $\left(1,2\right)$-Lindelöf and $\left(2,1\right)$-Lindelöf.

An $\left(i,j\right)$-Lindelöf property as well as $B$-Lindelöf property is not pairwise semiregular property by the following example.

Example 3.2**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a bitopological space as in Example 3.1. Then $\left(X,\tau_{1},\tau_{2}\right)$ is not $\left(\tau_{1},\tau_{2}\right)$-Lindelöf but it is $\left(\tau_{2},\tau_{1}\right)$-Lindelöf and hence not $B$-Lindelöf. Observe that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$-Lindelöf and $\left(\tau_{\left(2,1\right)}^{s},\tau_{\left(1,2\right)}^{s}\right)$-Lindelöf since $\tau_{\left(1,2\right)}^{s}$ and $\tau_{\left(2,1\right)}^{s}$ are indiscrete topologies. Hence $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $B$-Lindelöf.

Definition 3.3**.**

A cover U\of a bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is called $\tau_{1}\tau_{2}$-open [22] if $U\subseteq\tau_{1}\cup\tau_{2}$. If, in addition, $U$ contains at least one nonempty member of $\tau_{1}$ and at least one nonempty member of $\tau_{2}$, it is called $p$-open [4].

Definition 3.4**.**

[5]** A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is called $s$-Lindelöf  $($resp. $p$-Lindelöf$)$ if every $\tau_{1}\tau_{2}$-open $($resp. $p$-open$)$ cover of $X$ has a countable subcover.**

A $p$-Lindelöf property is not pairwise semiregular property by the following example. Thus the $s$-Lindelöf property is also not pairwise semiregular property.

Example 3.3**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a bitopological space as in Example 3.1. Then $\left(X,\tau_{1},\tau_{2}\right)$ is not $p$-Lindelöf and hence not $s$-Lindelöf. Observe that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $p$-Lindelöf and $s$-Lindelöf since $\tau_{\left(1,2\right)}^{s}$ and $\tau_{\left(2,1\right)}^{s}$ are indiscrete topologies.

4. Pairwise Semiregularization of Generalized Pairwise Lindelöf Spaces

Definition 4.1**.**

[16, 9, 13]** A bitopological space $X$ is said to be $\left(i,j\right)$-nearly Lindelöf $($resp. $\left(i,j\right)$-almost Lindelöf, $\left(i,j\right)$-weakly Lindelöf$)$ if for every $i$-open cover $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ of $X$, there exists a countable subset $\left\{\alpha_{n}:n\in\mathbb{N}\right\}$ of $\Delta$ such that $X=\mathop{\displaystyle\bigcup}\limits_{n\in\mathbb{N}}i$-$\mathop{\mathrm{i}nt}\left(j\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)\right)$ $\left(\text{{resp.}}\mathit{\ }X=\mathop{\displaystyle\bigcup}\limits_{n\in\mathbb{N}}j\text{{-}}\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)\text{{, }}X=j\text{{-}}\mathop{\mathrm{c}l}\left(\mathop{\displaystyle\bigcup}\limits_{n\in\mathbb{N}}\left(U_{\alpha_{n}}\right)\right)\right)$. $X$ is called pairwise nearly Lindelöf $($resp. pairwise almost Lindelöf, pairwise weakly Lindelöf$)$ if it is both $\left(1,2\right)$-nearly Lindelöf $($resp. $\left(1,2\right)$-almost Lindelöf, $\left(1,2\right)$-weakly Lindelöf$)$ and $\left(2,1\right)$-nearly Lindelöf $($resp. $\left(2,1\right)$-almost Lindelöf, $\left(2,1\right)$-weakly Lindelöf$)$.**

Our first result is analogue with the result of Mršević et al. [15, Theorem 1].

Theorem 4.1**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a $\left(\tau_{i},\tau_{j}\right)$-nearly Lindelöf and let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$ and since for each $\alpha_{x}\in\Delta,U_{\alpha_{x}}\in$ $\tau_{\left(i,j\right)}^{s}$, there exists a $\left(\tau_{i},\tau_{j}\right)$-regular open set $V_{\alpha_{x}}$ in $\left(X,\tau_{1},\tau_{2}\right)$ such that $x\in V_{\alpha_{x}}\subseteq U_{\alpha_{x}}$. So $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}V_{\alpha_{x}}$ and hence $\left\{V_{\alpha_{x}}:x\in X\right\}$ is a $\left(\tau_{i},\tau_{j}\right)$-regular open cover of $X$. Since $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-nearly Lindelöf, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}V_{\alpha_{x_{n}}}\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Conversely, suppose that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf and let $\left\{V_{\alpha}:\alpha\in\Delta\right\}$ be a $\left(\tau_{i},\tau_{j}\right)$-regular open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $V_{\alpha}\in\tau_{\left(i,j\right)}^{s}$ for each $\alpha\in\Delta,\left\{V_{\alpha}:\alpha\in\Delta\right\}$ is a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf, there exists a countable subcover such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}V_{\alpha_{n}}$. This implies that $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-nearly Lindelöf.

Corollary 4.1**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

Proposition 4.1**.**

A bitopological space $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

The sufficient condition is obvious by the definitions. So we need only to prove necessary condition. Suppose that $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ is a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-semiregular, there exists a $\tau_{\left(i,j\right)}^{s}$-open set $V_{\alpha_{x}}$ in $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ such that $x\in V_{\alpha_{x}}\subseteq\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x}}\right)\right)\subseteq U_{\alpha_{x}}$. Hence $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}V_{\alpha_{x}}$ and thus the family $\left\{V_{\alpha_{x}}:x\in X\right\}$ forms a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-nearly Lindelöf, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x_{n}}}\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Corollary 4.2**.**

A bitopological space $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

From the Definition 2.6, if the property $\mathcal{P}$ is not bitopological property but it satisfies the condition $\left(X,\tau_{1},\tau_{2}\right)$ has the property $\mathcal{P}$ if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ has the property $\mathcal{P}$, then the property $\mathcal{P}$ will be called pairwise semiregular invariant property. The following theorem prove that $\left(i,j\right)$-nearly Lindelöf as well as pairwise nearly Lindelöf property satisfying the pairwise semiregular invariant property since $\left(i,j\right)$-nearly Lindelöf and pairwise nearly Lindelöf are not $i$-topological property [8] and bitopological property, respectively. This is because the $i$-continuity and $\left(i,j\right)$-$\delta$-continuity (resp. continuity and $p$-$\delta$-continuity) are independent notions (see [12]).

Theorem 4.2**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-nearly Lindelöf.

Proof.

It is obvious by Theorem 4.1 and Proposition 4.1.

Corollary 4.3**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise nearly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise nearly Lindelöf.

Theorem 4.3**.**

[19]** If $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise semiregular, then $\left(X,\tau_{1},\tau_{2}\right)=\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$.**

The converse of Theorem 4.3 is also true by the definitions.

Proposition 4.2**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(i,j\right)$-nearly Lindelöf if and only if it is $i$-Lindelöf.

Proof.

By Theorem 4.3, $\left(X,\tau_{1},\tau_{2}\right)=\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. The result follows immediately by Proposition 4.1.

Corollary 4.4**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise nearly Lindelöf if and only if it is Lindelöf.

Unlike all types of pairwise Lindelöf properties, the $\left(i,j\right)$-almost Lindelöf, pairwise almost Lindelöf, $\left(i,j\right)$-weakly Lindelöf and pairwise weakly Lindelöf properties are pairwise semiregular properties as we prove in the following theorems.

Theorem 4.4**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf.

Proof.

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf and let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Since $\tau_{\left(i,j\right)}^{s}\subseteq\tau_{i}$, $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ is a $\tau_{i}$-open cover of the $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf space $\left(X,\tau_{1},\tau_{2}\right)$. Then there is a countable subset $\left\{\alpha_{n}:n\in\mathbb{N}\right\}$ of $\Delta$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{j}$-$\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)$. By Lemma 2.1, we have $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(j,i\right)}^{s}$-$\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)$, which implies $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf.

Conversely suppose that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$* *is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf and let $\left\{V_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{i}$-open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $V_{\alpha}\subseteq\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha}\right)\right)$ and $\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha}\right)\right)\in$ $\tau_{\left(i,j\right)}^{s}$, we have $\left\{\tau_{i}\text{-}\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha}\right)\right):\alpha\in\Delta\right\}$ is a $\tau_{\left(i,j\right)}^{s}$-open cover of the $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf space $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. So there is a countable subset $\left\{\alpha_{n}:n\in\mathbb{N}\right\}$ of $\Delta$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}$ $\tau_{\left(j,i\right)}^{s}$-$\mathop{\mathrm{c}l}\left(\tau_{i}\text{-}\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{n}}\right)\right)\right)$. By Lemma 2.1, we have $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}$ $\tau_{j}$-$\mathop{\mathrm{c}l}\left(\tau_{i}\text{-}\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{n}}\right)\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}$ $\tau_{j}$-$\mathop{\mathrm{c}l}\left(V_{\alpha_{n}}\right)$. This implies that $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf.

Corollary 4.5**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise almost Lindelöf.

Note that, the $\left(i,j\right)$-almost Lindelöf property and the pairwise almost Lindelöf property are both bitopological properties (see [18]). Utilizing this fact, Theorem 4.4 and Corollary 4.5, we easily obtain the following corollary.

Corollary 4.6**.**

The $\left(i,j\right)$-almost Lindelöf property and the pairwise almost Lindelöf property are both pairwise semiregular properties.

Proposition 4.3**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a $\left(\tau_{i},\tau_{j}\right)$-almost regular space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf and let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$ and since $U_{\alpha_{x}}\in$ $\tau_{\left(i,j\right)}^{s}$, there exists a $\left(\tau_{i},\tau_{j}\right)$-regular open set $V_{\alpha_{x}}$ in $\left(X,\tau_{1},\tau_{2}\right)$ such that $x\in V_{\alpha_{x}}\subseteq U_{\alpha_{x}}$. Since $\left(X,\tau_{1},\tau_{2}\right)$* *is $\left(\tau_{i},\tau_{j}\right)$-almost regular, there is a $\left(\tau_{i},\tau_{j}\right)$-regular open set $C_{\alpha_{x}}$ in $\left(X,\tau_{1},\tau_{2}\right)$ such that $x\in C_{\alpha_{x}}\subseteq\tau_{j}$-$\mathop{\mathrm{c}l}\left(C_{\alpha_{x}}\right)\subseteq V_{\alpha_{x}}$. Hence $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}C_{\alpha_{x}}$ and thus the family $\left\{C_{\alpha_{x}}:x\in X\right\}$ forms a $\left(\tau_{i},\tau_{j}\right)$-regular open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{j}$-$\mathop{\mathrm{c}l}\left(C_{\alpha_{x_{n}}}\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}V_{\alpha_{x_{n}}}\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf. Conversely, let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ be a $\tau_{\left(i,j\right)}^{s}$-Lindelöf and let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{i}$-open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $U_{\alpha}\subseteq\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right)$ and $\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right)\in$ $\tau_{\left(i,j\right)}^{s}$, $\left\{\tau_{i}\text{-}\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right):\alpha\in\Delta\right\}$ is $\tau_{\left(i,j\right)}^{s}$-open cover of the $\tau_{\left(i,j\right)}^{s}$-Lindelöf space $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Then there exists a countable subset $\left\{\alpha_{n}:n\in\mathbb{N}\right\}$ of $\Delta$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{j}$-$\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)$. This implies that $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-almost Lindelöf.

Corollary 4.7**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise almost regular space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

Proposition 4.4**.**

Let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ be a $\left(\tau_{\left(j,i\right)}^{s},\tau_{\left(i,j\right)}^{s}\right)$-extremally disconnected space. Then $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

The sufficient condition is obvious by the definitions. So we need only to prove necessary condition. Suppose that $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ is a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-semiregular, there exists a $\tau_{\left(i,j\right)}^{s}$-open set $V_{\alpha_{x}}$ in $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ such that $x\in V_{\alpha_{x}}\subseteq\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x}}\right)\right)\subseteq U_{\alpha_{x}}$. Hence $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}V_{\alpha_{x}}$ and thus the family $\left\{V_{\alpha_{x}}:x\in X\right\}$ forms a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-almost Lindelöf and $\left(\tau_{\left(j,i\right)}^{s},\tau_{\left(i,j\right)}^{s}\right)$-extremally disconnected, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(j,i\right)}^{s}$-$\mathop{\mathrm{c}l}\left(V_{\alpha_{x_{n}}}\right)=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x_{n}}}\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Corollary 4.8**.**

Let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ be a pairwise extremally disconnected space. Then $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise almost Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf

Proposition 4.5**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular and $\left(j,i\right)$-extremally disconnected space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(i,j\right)$-almost Lindelöf if and only if it is $i$-Lindelöf.

Proof.

By Theorem 4.3, $\left(X,\tau_{1},\tau_{2}\right)=\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. The result follows immediately by Proposition 4.4.

Corollary 4.9**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular and pairwise extremally disconnected space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise almost Lindelöf if and only if it is Lindelöf.

Theorem 4.5**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-weakly Lindelöf.

Proof.

The proof is similar to the proof of Theorem 4.4 by using the fact that

[TABLE]

Thus we choose to omit the details.

Corollary 4.10**.**

A bitopological space $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise weakly Lindelöf.

Note that, the $\left(i,j\right)$-weakly Lindelöf property and the pairwise weakly Lindelöf property are both bitopological properties (see [18]). Utilizing this fact, Theorem 4.5 and Corollary 4.10, we easily obtain the following corollary.

Corollary 4.11**.**

The $\left(i,j\right)$-weakly Lindelöf property and the pairwise weakly Lindelöf property are both pairwise semiregular properties.

Recall that, a bitopological space $X$ is called $\left(i,j\right)$-weak $P$-space [13] if for each countable family $\left\{U_{n}:n\in\mathbb{N}\right\}$ of $i$-open sets in $X$, we have $j$-$\mathop{\mathrm{c}l}\left(\mathop{\displaystyle\bigcup}\limits_{n\in\mathbb{N}}U_{n}\right)=\mathop{\displaystyle\bigcup}\limits_{n\in\mathbb{N}}j$-$\mathop{\mathrm{c}l}\left(U_{n}\right)$. $X$ is called pairwise weak $P$-space if it is both $\left(1,2\right)$-weak $P$-space and $\left(2,1\right)$-weak $P$-space.

Proposition 4.6**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a $\left(\tau_{i},\tau_{j}\right)$-almost regular and $\left(\tau_{i},\tau_{j}\right)$-weak $P$-space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

Necessity: Let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$ and since $U_{\alpha_{x}}\in$ $\tau_{\left(i,j\right)}^{s}$ for each $\alpha_{x}\in\Delta$, there exists a $\left(\tau_{i},\tau_{j}\right)$-regular open set $V_{\alpha_{x}}$ in $\left(X,\tau_{1},\tau_{2}\right)$ such that $x\in V_{\alpha_{x}}\subseteq U_{\alpha_{x}}$. Since $\left(X,\tau_{1},\tau_{2}\right)$* *is $\left(\tau_{i},\tau_{j}\right)$-almost regular, there is a $\left(\tau_{i},\tau_{j}\right)$-regular open set $C_{\alpha_{x}}$ in $\left(X,\tau_{1},\tau_{2}\right)$ such that $x\in C_{\alpha_{x}}\subseteq\tau_{j}$-$\mathop{\mathrm{c}l}\left(C_{\alpha_{x}}\right)\subseteq V_{\alpha_{x}}$. Hence $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}C_{\alpha_{x}}$ and thus the family $\left\{C_{\alpha_{x}}:x\in X\right\}$ forms a $\left(\tau_{i},\tau_{j}\right)$-regular open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-weakly Lindelöf and $\left(\tau_{i},\tau_{j}\right)$-weak $P$-space, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\tau_{j}$-$\mathop{\mathrm{c}l}\left(\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}C_{\alpha_{x_{n}}}\right)=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{j}$-$\mathop{\mathrm{c}l}\left(C_{\alpha_{x_{n}}}\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}V_{\alpha_{x_{n}}}\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Sufficiency: Let $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ be a $\tau_{i}$-open cover of $\left(X,\tau_{1},\tau_{2}\right)$. Since $U_{\alpha}\subseteq\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right)$ and $\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right)\in$ $\tau_{\left(i,j\right)}^{s}$, $\left\{\tau_{i}\text{-}\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha}\right)\right):\alpha\in\Delta\right\}$ is $\tau_{\left(i,j\right)}^{s}$-open cover of the $\tau_{\left(i,j\right)}^{s}$-Lindelöf space$\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Then there exists a countable subset $\left\{\alpha_{n}:n\in\mathbb{N}\right\}$ of $\Delta$ such that $X=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{i}$-$\mathop{\mathrm{i}nt}\left(\tau_{j}\text{-}\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{j}$-$\mathop{\mathrm{c}l}\left(U_{\alpha_{n}}\right)=\tau_{j}$-$\mathop{\mathrm{c}l}\left(\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{n}}\right)$. This implies that $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(\tau_{i},\tau_{j}\right)$-weakly Lindelöf.

Corollary 4.12**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise almost regular and pairwise weak $P$-space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

Proposition 4.7**.**

Let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ be a $\left(\tau_{\left(j,i\right)}^{s},\tau_{\left(i,j\right)}^{s}\right)$-extremally disconnected and $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-weak $P$-space. Then $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Proof.

The sufficient condition is obvious by the definitions. So we need only to prove necessary condition. Suppose that $\left\{U_{\alpha}:\alpha\in\Delta\right\}$ is a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. For each $x\in X$, there exists $\alpha_{x}\in\Delta$ such that $x\in U_{\alpha_{x}}$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-semiregular, there exists a $\tau_{\left(i,j\right)}^{s}$-open set $V_{\alpha_{x}}$ in $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ such that $x\in V_{\alpha_{x}}\subseteq\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x}}\right)\right)\subseteq U_{\alpha_{x}}$. Hence $X=\mathop{\displaystyle\bigcup}\nolimits_{x\in X}V_{\alpha_{x}}$ and thus the family $\left\{V_{\alpha_{x}}:x\in X\right\}$ forms a $\tau_{\left(i,j\right)}^{s}$-open cover of $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. Since $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-weakly Lindelöf, $\left(\tau_{\left(j,i\right)}^{s},\tau_{\left(i,j\right)}^{s}\right)$-extremally disconnected and $\left(\tau_{\left(i,j\right)}^{s},\tau_{\left(j,i\right)}^{s}\right)$-weak $P$-space, there exists a countable subset of points $x_{1},\ldots,x_{n},\ldots$ of $X$ such that $X=\tau_{\left(j,i\right)}^{s}$-$\mathop{\mathrm{c}l}\left(\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}V_{\alpha_{x_{n}}}\right)=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(j,i\right)}^{s}$-$\mathop{\mathrm{c}l}\left(V_{\alpha_{x_{n}}}\right)=\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}\tau_{\left(i,j\right)}^{s}$-$\mathop{\mathrm{i}nt}\left(\tau_{\left(j,i\right)}^{s}\text{-}\mathop{\mathrm{c}l}\left(V_{\alpha_{x_{n}}}\right)\right)\subseteq\mathop{\displaystyle\bigcup}\nolimits_{n\in\mathbb{N}}U_{\alpha_{x_{n}}}$. This shows that $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is $\tau_{\left(i,j\right)}^{s}$-Lindelöf.

Corollary 4.13**.**

Let $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ be a pairwise extremally disconnected and pairwise weak $P$-space. Then $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is pairwise weakly Lindelöf if and only if $\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$ is Lindelöf.

Proposition 4.8**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular, $\left(j,i\right)$-extremally disconnected and $\left(i,j\right)$-weak $P$-space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is $\left(i,j\right)$-weakly Lindelöf if and only if it is $i$-Lindelöf.

Proof.

By Theorem 4.3, $\left(X,\tau_{1},\tau_{2}\right)=\left(X,\tau_{\left(1,2\right)}^{s},\tau_{\left(2,1\right)}^{s}\right)$. The result follows immediately by Proposition 4.7.

Corollary 4.14**.**

Let $\left(X,\tau_{1},\tau_{2}\right)$ be a pairwise semiregular, pairwise extremally disconnected and pairwise weak $P$-space. Then $\left(X,\tau_{1},\tau_{2}\right)$ is pairwise weakly Lindelöf if and only if it is Lindelöf.