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\fancyhead[CE]{M. Etebar, S. Soltanpour} 
\fancyhead[CO]{On rings of real-valued $R _{cl}$-supercontinuous functions}



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{\noindent Journal of Mathematical Extension \\
Vol. XX, No. XX, (2014), pp-pp (Will be inserted by layout editor)}\\
ISSN: 1735-8299\\
URL: http://www.ijmex.com\\
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{\Large \bf 
{On rings of real-valued $R _{cl}$-supercontinuous functions}\\}
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\let\thefootnote\relax\footnote{\scriptsize Received: XXXX; Accepted: XXXX (Will be inserted by editor)}

{\bf M. Etebar^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\
\vspace{2mm} {\small  Department of Mathematics, Shahid Chamran University of Ahvaz, Ahvaz, Iran} \vspace{2mm}

{\bf  S. Soltanpour$
%^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}
\vspace*{-2mm}\\
\vspace{2mm} {\small   Department of Science, Petroleum University of Technology, Ahvaz, Iran} \vspace{2mm}

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{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.} The notion of $\rm{R _{cl}} $-supercontinuity, a strong variant of continuity, is considered.
The ring $\rm{R _{cl}} (X)$ of all real valued $\rm{R _{cl}} $-supercontinuous functions on a topological space $X$ is studied. It is shown that $\rm{R _{cl}} (X) \cong C(Y)$, where $Y$ is an ultra-Hausdorff $\rm{r_{cl}}$-quotient of $X$ and it turns out that whenever $X$ is $\rm{r_{cl}}$-compact, then $Y$ is zero-dimensional. The maximal ideals of $\rm{R _{cl}} (X)$ are specified. The spaces $X$ are determined for which every maximal ideal in $\rm{R _{cl}} (X)$ is fixed. Finally, $\rm{P_{r _{cl}}} $-spaces and almost $\rm{P_{r _{cl}}} $-spaces are defined and characterized both algebraically and topologically.
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification:} {Primary: 54C40 ; Secondary: 54C08. }

\noindent{\bf Keywords and Phrases:} $R _{cl}$-supercontinuous function, $P_{r _{cl}} $-space, almost $P_{r _{cl}} $-space, $r_{cl}$-compact space, $s_{cl}$-completely regular space.
\end{quotation}}

\section{Introduction}
\label{intro} % It is advised to give each section and subsection a unique label.
In 2007, Singh introduced the concept of a $cl$-\textit{open} set in the study of clopen continuous maps. A set $A$ in a topological space $X$ is called $cl$-\textit{open} if $A$ is a union of clopen sets. The complement of a $cl$-open set is called $cl$\textit{-closed}. In 2013, Tyagi et al. introduced the concept of an $r_{cl}$-\textit{open} set as follows. An open set $U$ in a topological space $X$ is said to be $r _{cl} $-\textit{open}  if $U$ is a union of $cl$-closed sets. The complement of an $r_{cl}$-open set is called an $r _{cl} $-\textit{closed} set. For a set $A$ in a topological space $X$, the set of all $x \in A$ such that $A$ contains an $\rm{r_{cl}}$-open set containing $x$ is called $r_{cl}$-\textit{interior} of $A$ and it is denoted by $\mbox{int}_{r_{\rm{cl}}} A$. Clearly, a subset $G$ of $X$ is $\rm{r_{cl}}$-open if and only if $B = \mbox{int}_{r_{\rm{cl}}} B$. The $r_{cl}$-\textit{closure} of a set $A$ in a topological space $X$, denoted by $\mbox{cl}_{r_{\rm{cl}}} A$, is the set of all $x \in X$ such that each $\rm{r_{cl}}$-open set containing $x$ intersects $A$. Clearly, a subset $B$ of $X$ is $\rm{r_{cl}}$-closed if and only if $B = \mbox{cl}_{r_{\rm{cl}}} B$. According to \cite{13}, a topological space $X$ is called an $R _{cl} $-\textit{space} if each open set in $X$ is $r_{cl}$-open. If $X$ and $Y$ are topological spaces, then a function $f:X \rightarrow Y$ is said to be $R_{cl}$-\textit{supercontinuous} if for each $x \in X$ and each open set $V$ in $Y$ containing $f(x)$, there exists an $r_{cl}$-\textit{open} set $U$ in $X$ containing $x$ such that $f(U) \subseteq V$. A bijection $\sigma : X \rightarrow Y$ is said to be an $R_{cl}$-\textit{homeomorphism} if both $\sigma$ and $\sigma ^{-1}$ are $R_{cl}$-supercontinuous. In this case, $X$ and $Y$ are said to be $R_{cl}$-\textit{homeomorphic} and it is written as $X \cong _{r_{cl}} Y$.\\
\indent
Let $R_{cl} (X)$ be the set of all real-valued $R_{cl}$-supercontinuous functions on $X$. It is easily seen that $R_{cl} (X)$ is a subring and sublattice of $C(X)$ where $C(X)$ is  the ring of all real-valued continuous functions on a space $X$.\\
Throughout this paper, for $f\in C(X)$, the set $Z(f)=\left\{x\in X:f(x)=0\right\}$ is the zero-set of $f$. The set-theoretic complement of $Z(f)$ is denoted by $\coz(f)$ and is called the cozero-set of $f$.We denote by $Z(X)$ the set of all zero-sets in $X$ and $\rm{Z_{r_{cl}}} (X)$ denotes the set of all zero-sets  $Z(f)$ in $X$, where $f \in \rm{R_{cl}} (X)$. We refer the reader to \cite{6} for undefined terms and notations.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{\bf {\large $R_{cl}(X)$ is a $C(Y)$}}
In this section for any topological space $X$, an ultra-Hausdorff space $Y$ is established such that  $R_{cl} (X)$ and $C(Y)$ are isomorphic. First, let us recall some definitions and facts. A $T_1$-space is called \textit{zero-dimensional} if it has a base consisting of clopen sets. According to \cite{11}, a topological space $X$ is called \textit{ultra-Hausdorff} if every pair of distinct points in $X$ are contained in disjoint clopen sets.
\begin{remark}
The following implications hold.
$$\text{zero-dimensional space} \Rightarrow  \text{ultra-Hausdorff space}  \Rightarrow  \text{$R_{cl}$-space}.$$
However, none of the above implications is reversible. For example, the space of strong ultrafilter topology \cite[Example 113]{12} is an ultra-Hausdorff space which is not zero-dimensional. A nondegenerate indiscrete space is an $R_{cl}$-space, but it is not ultra-Hausdorff, see \cite[p. 28]{9}.
\end{remark}
Now, let us make the following observation.
\begin{lemma} \label{2-1}
If $X$ is an  $R_{cl}$-space, then $C(X)= R_{cl} (X)$ and whenever $X$ is completely regular, the converse is also true.
\end{lemma}
\begin{proof}
Using Theorem $8.2$ in \cite{13}, the first implication is immediate. Now, suppose that $X$ is a completely regular space and $C(X)=R_{cl}(X)$. Then by Theorem $3.2$ in \cite{6}, we infer that the collection $\beta = \left\{coz(f):f \in C(X)\right\}$ is a base for open subsets of $X$. Since $C(X)=R_{cl}(X)$, for each $f \in C(X)$ and every $x \in coz (f)$, there is an $r_{cl}$-open set $U$ in $X$ containing $x$ such that $U \subseteq coz(f)$. This shows that $X$ is an $R_{cl}$-space.
\end{proof}
\\

In the following, we give some properties of $R_{cl}$-supercontinuous functions. Before stating our results, let us say a few words. For a point $x$ in a topological space $X$, the maximal connected subset of $X$ containing $x$, denoted by $C_x$, is called the \textit{component of} $x$. A space is called \textit{totally disconnected} if the only nonempty components are one-point sets. For a point $x$ in a topological space $X$, the intersection of all clopen subsets of $X$ containing $x$, denoted by $Q_x$, is called the \textit{quasi-component of} $x$. The collection of all  components (resp. quasi-components) of a topological space $X$ constitute a decomposition of $X$ into pairwise disjoint closed sets. By Theorem $6.1.22$ in \cite{5}, $C_x \subseteq Q_x$ for each $x \in X$ and the inclusion may be proper as it is shown by \cite[Example 6.1.24]{5}. It is easily seen that an open (resp. closed) set in a topological space $X$ is $\rm{r_{cl}}$-open (resp. $\rm{r_{cl}}$-closed) if and only if it is a union of quasi-components of $X$. By Theorem $4 . 1$ in \cite{13}, every $Z \in \rm{Z_{r_{cl}}} (X)$ is $\rm{r_{cl}}$-closed. Hence, each $Z \in \rm{Z_{r_{cl}}} (X)$ is a union of quasi-components in $X$. However, an $\rm{r_{cl}}$-closed set need not be a zero-set in $\rm{Z_{r_{cl}}} (X)$, see problem $4 . N$ in \cite{6}.
\begin{proposition} \label{2-2}
For an $R_{cl}$-supercontinuous function $f:X \rightarrow Y$, the following statements are true.
\begin{enumerate}
\item $f[Q_x] \subseteq Q_{f(x)}$ for each $x \in X$.
\item If $Y$ is a $T_1$-space, then $f[Q_x]=\{f(x)\}$ for each $x \in X$.
\item If $f$ is injective and $Y$ is a $T_1$-space, then $X$ is totally disconnected.
\item If $f$ is an $R_{cl}$-homeomorphism, then $f[Q_x]=Q_{f(x)}$ for each $x \in X$. Furthermore, if $X$ or $Y$ is a  $T_1$-space, then both $X$ and $Y$ are totally disconnected.
\end{enumerate}
\end{proposition}
\begin{proof}
$(1)$ Let $x \in X$, $p \in Q_x$ and $f(p) \notin Q_{f(x)}$. Then there is a clopen set $V$ in $Y$ containing $f(p)$ such that $f(x) \notin V$. Since $f$ is  $R_{cl}$-supercontinuous, $f[Q_x] \subseteq  V$. So $f(x) \in V$ which is a contradiction.\\
$(2)$ Let $x \in X$, $p \in Q_x$ and $f(p) \neq f(x)$. Then there is an open set $V$ in $Y$ containing $f(x)$ such that $f(p) \notin V$. Since $f$ is $R_{cl}$-supercontinuous, $f[Q_x] \subseteq  V$ which yields that $f(p) \in V$, a contradiction.\\
$(3)$ By part $(2)$, $f[Q_x]=\{f(x)\}$ for every $x \in X$. This implies that $Q_x=\{x\}$, since $f$ is injective and this shows that $X$ is totally disconnected.\\
$(4)$ It is an immediate consequence of parts $(1)$ and $(3)$.
\end{proof}
\begin{corollary} \label{2-3}
Let $X$ be a topological space. If $f \in R_{cl}(X)$, then $f[Q_x] = \{f(x)\}$ for each $x \in X$.
\end{corollary}
\begin{definition} Let $X$ be a space and $Y$ be a set and let $p :X \rightarrow Y$ be a surjection. The collection $\tau _p = \{ U \subseteq Y\, : \, p^{-1} (U) \,{\text is} \, r_{cl}\text{-open  in} \,  X\}$ of subsets of $Y$ is called the $r_{cl}$-\textit{quotient topology} \cite{13} on $Y$ induced by $p$. Moreover, $(Y, \tau_p)$ is called an $r_{cl}$-\textit{quotient space of} $X$.
\end{definition}
Next, we state the main result of this section.
\begin{theorem} \label{2-4}
For any topological space $X$, there is an ultra-Hausdorff space $X_u$ which is an $r_{cl}$-quotient space of $X$ and $R_{cl}(X) \cong C(X_u)$.
\end{theorem}
\begin{proof}
Let $X_u=\{ Q_x \, : \, x \in X \}$ and define $p : X \longrightarrow X_u$ by $p(x)=Q_x$, for all $x \in X$. Let $\tau_p$ be the $r_{cl}$-quotient topology on $X_u$ induced by $p$. Then $(X_u, \tau_p)$ is ultra-Hausdorff. In fact, if $Q_x$ and $Q_y$ are distinct points in $X_u$, then $x \notin Q_y$ and hence there is a clopen set $V$ in $X$ such that $x \in V$ and $y \notin V$. Now, let $H=\{ Q_z \, : \, z \in V \}$. Then $H$ is a clopen set in $X_u$, for $V=p^{-1}(H)=\bigcup_{z \in V} Q_z$ is a clopen set in $X$. Clearly, $Q_x \in H$ and $Q_y \notin H$ which shows that $X_u$ is ultra-Hausdorff. \\
To complete the proof, we show that $R_{cl} (X) \cong C(X_u)$. To this end, define
$\theta :R_{cl}(X) \rightarrow C(X_u)$ by $\theta(f)=f_u$, for each $f \in R_{cl} (X)$, where $f_u : X_u \rightarrow \mathbb{R}$ be defined as $f_u (Q_x)=f(x)$, for all $x \in X$. By Corollary \ref{2-3}, $f_u$ and $\theta$ are well-defined. To show that $f_u \in C(X_u)$, let $x \in X$, $Q_x \in X_u$ and let $f_u (Q_x) =f(x) =a$. Then for each $\epsilon >0$, there is an $r_{cl}$-open set $U$ in $X$ containing $x$ such that $f(U) \subseteq (a-\epsilon , a+\epsilon)$. Now $G=\{Q_z \, : \, Q_z \subseteq U \}$ is an open set in $X_u$ containing $Q_x $ such that $f_u(G) \subseteq (a-\epsilon , a+\epsilon)$. It is easily seen that $\theta$ is a one to one homomorphism. Finally, we show that $\theta$ is onto. To this end, let $g \in C(X_u)$ and define $f: X \rightarrow \mathbb{R}$ by $f(x)=g(Q_x)$, for all $x \in X$. To see that $f \in R_{cl}(X)$, let $x \in X$ and let $f(x)=a$. Since $g \in C(X_u)$, there is an open set $G$ in $X_u$ containing $Q_x$ such that $g(G) \subseteq (a-\epsilon , a+\epsilon)$ for every $\epsilon >0$. So $U=\bigcup _{Q_z \in G} Q_z$ is an $r_{cl}$-open set in $X$ containing $x$ such that $f(U) \subseteq (a-\epsilon , a+\epsilon)$ and this shows that $f$ is $R_{cl}$-supercontinuous. Also we have $\theta(f)=g$.
\end{proof}
\begin{remark}
From now on, for each topological space $X$, we consider $X_u$ and the isomorphism $\theta : R_{cl}(X) \rightarrow C(X_u)$ as defined in the proof of Theorem \ref{2-4}.
\end{remark}
\begin{definition}
A topological space $X$ is said to be $r_{cl}$-\textit{compact} if every $r_{cl}$-open cover of $X$ has a finite subcover.
\end{definition}
Note that every compact space is $r_{cl}$-compact, but not conversely. For instance, $\mathbb{R} $ is an $r_{cl}$-compact space which is not compact. Clearly, a space $X$ is $r_{cl}$-compact if and only if $X_u$ is compact.
\begin{corollary} \label{2-5}
If $X$ is $r_{cl}$-compact, then $R_{cl}(X) \cong C(Y)$ for a zero-dimensional space $Y$.
\end{corollary}
\begin{proof}
By Theorem \ref{2-4}, $X_u$ is ultra-Hausdorff and since $X_u$ is compact, $X_u$ is zero-dimensional. Now, let $Y=X_u$, so by Theorem \ref{2-4}, $R_{cl}(X) \cong C(Y)$.
\end{proof}\\

Accourding to \cite{7}, a topological space $X$ is called \textit{sum connected} if each component in $X$ is open. Obviously, if $X$ is sum connected, then $C_x = Q_x$ for every $x \in X$.
\begin{corollary} \label{2-6}
If $X$ is sum connected, then $R_{cl}(X) \cong C(Y)$ for a discrete space $Y$.
\end{corollary}
\begin{proof}
Since $X$ is sum connected, $Q_x$ is $r_{cl}$-open for every $x \in X$. So every one point set $\{Q_x\}$ is open in $X_u$ which yields that $X_u$ is discrete. Now, let $Y=X_u$, so by Theorem \ref{2-4}, $R_{cl}(X) \cong C(Y)$.
\end{proof}
We conclude this section by the following proposition.
\begin{proposition} \label{2-7}
Let $X$ and $Y$ be two topological spaces. If $X \cong _{r_{cl}} Y$, then $X_u \cong Y_u$.
\end{proposition}
\begin{proof}
Let $\varphi : X \rightarrow Y$ be an $R_{cl}$-homeomorphism. Define $\tau : X_u \longrightarrow Y_u$ by $\tau(Q) =\varphi(Q)$, for all $Q \in X_u$. Clearly, $\tau$ is one to one. To see that $\tau$ is onto, let $y \in Y$ and let $Q_y \in Y_u$. Then there is $x \in X$ such that $\varphi (x)=y$ and hence $\tau (Q_x) = \varphi (Q_x) = Q_y$ by part $(4)$ of Proposition \ref{2-2}. Now, we show that $\tau$ and $\tau ^{-1}$ are continuous. To this end, let $x \in X$, $Q_x \in X_u$ and let $H$ be an open set in $Y_u$ containing $\varphi (Q_x)$. Then $V=\bigcup _{Q_z \in H} Q_z$ is an open set in $Y$ containing $\varphi (x)$. So by $R_{cl}$-supercontinuity of $\varphi$, there is an $r_{cl}$-open set $U$ in $X$ containing $x$ such that $\varphi (U) \subseteq V$. Therefore, $G=\{Q_x |x \in U \}$ is an open set in $X_u$ containing $Q_x$ such that $\tau (G) \subseteq H$. Similarly, $\tau^{-1}$ is continuous.
\end{proof}
We remind the reader that the converse of Proposition \ref{2-7} is not true in general. For instance, let  $X=\{a\}$ and let $Y=\mathbb{R}$. Then $X_u$ is homeomorphic to $Y_u$, but $X$ and $Y$ are not $R_{cl}$-homeomorphic.
\section{\bf Maximal ideals of $R_{cl}(X)$}
In this section, we turn our attention to the maximal ideals in the rings $R_{cl}(X)$. Fisrt, let us recall that an ideal $I$ of $C(X)$ (resp. $R_{cl}(X)$) is called a \textit{fixed ideal} if $\bigcap _{f \in I} Z(f) \neq \emptyset$ (resp. $\bigcap _{f \in I} Z_{cl}(f) \neq \emptyset$), otherwise $I$ is called a \textit{free ideal}. We begin with the following easy lemma, its proof is left to the reader.
\begin{lemma} \label{3-1}
An ideal $I$ in $R_{cl}(X)$ is fixed if and only if $\theta (I)$ is fixed in $C(X_u)$.
\end{lemma}
Now, we completely characterize the fixed maximal ideals of a ring $R_{cl}(X)$.
\begin{theorem} \label{3-2}
For a topological space $X$, any fixed maximal ideal in $R_{cl}(X)$ is in the form of
$$M_{Q_x} =\big\{ f \in R_{cl}(X)  \, : \, Q_x \subseteq Z(f) \big \} \quad x \in X.$$
The ideals $M_{Q_x}$ are distinct for distinct $Q_x$. Furthermore, $\frac{R_{cl}(X)}{M_{Q_x}} \cong \mathbb{R}$ for every $x \in X$.
\end{theorem}
\begin{proof}
By Theorem $4.6$ in \cite{6} and Lemma \ref{3-1}, $M$ is a fixed maximal ideal in $R_{cl}(X)$ if and only if $M= \theta ^{-1} (M_y) $ for some $y \in X_u$. So $M= M_{Q_x}$ for some $x \in X$. Now, suppose that $Q_x \neq Q_y$ for $x , y \in X$. Then $Q_x \cap Q_y= \emptyset$, so there is a clopen set $C_x$ in $X$ containing $x$ such that $Q_x \subseteq C_x$ and $C_x \cap Q_y = \emptyset $. Let $f : X \rightarrow \mathbb{R}$ be defined as $f(x)=0$, if $x \in C_x$ and $f(x)=1$, if $x \notin C_x$. Then $f \in R_{cl}(X)$, $f \in M_{Q_x} \setminus M_{Q_y}$ which shows that $M_{Q_x} \neq M_{Q_y}$. For the last assertion, let $x \in X$ and define $\varphi : R_{cl}(X) \rightarrow \mathbb{R}$ by $\varphi (f)=f(x)$, for all $f \in R_{cl}(X)$. Then $\varphi$ is a homomorphism and $\mbox{Ker} \varphi = M_{Q_x}$. Consequently,  $\frac{R_{cl}(X)}{M_{Q_x}} \cong \mathbb{R}$.
\end{proof}
\begin{lemma} \label{3-3}
If $X$ is $r_{cl}$-compact, then each ideal in $R_{cl}(X)$ is fixed.
\end{lemma}
\begin{proof}
If $X$ is $r_{cl}$-compact, then $X_u$ is zero-dimensional by Corollary \ref{2-5}. So in view of Theorem $4.11$ in \cite{6}, every ideal in $C(X_u)$ is fixed. Consequently, each ideal in $R_{cl}(X)$ is fixed by Lemma \ref{3-1}.
\end{proof}
\begin{definition}
A topological space $X$ is called $s_{cl}$-\textit{completely regular} if for each $r_{cl}$-closed set $A$ and each $x \notin A$, there exists an $R_{cl}$-supercontinuous function $f : X \rightarrow \mathbb{R}$ such that $f[A] =\{0\}$ and $f[Q_x]=\{1\}$.
\end{definition}
Clearly, a space $X$ is $s_{cl}$-completely regular if and only if $X_u$ is completely regular.
\begin{theorem} \label{3-4}
For a $s_{cl}$-completely regular space $X$, the following statements are equivalent.
\begin{enumerate}
\item  $X$ is $r_{cl}$-compact.
\item $X_u$ is compact.
\item Every ideal in $R_{cl}(X)$ is fixed.
\item Every maximal ideal in $R_{cl}(X)$ is fixed.
\end{enumerate}
\end{theorem}
\begin{proof}
Clearly, parts $(1)$ and $(2)$ are equivalent. \\
$(2)\Rightarrow(3)$. It is immediate by Lemma \ref{3-3}.\\
$(3)\Rightarrow (4)$. It is immediate.\\
$(4)\Rightarrow (2)$. Since $X$ is $s_{cl}$-completely regular, $X_u$ is completely regular and by Lemma \ref{3-1}, each maximal ideal in $C(X_u)$ is fixed. These follow by Theorem $4.11$ in \cite{6}, that $X_u$ is compact.
\end{proof}
\begin{theorem} \label{3-5}
Let $X$ and $Y$ be two topological spaces. If $X_u \cong Y_u$, then  $R_{cl}(X) \cong R_{cl}(Y)$ and whenever $X$ and $Y$ are $r_{cl}$-compact, then the converse is also true.
\end{theorem}
\begin{proof}
If $X_u \cong Y_u$, then $C(X_u) \cong C(Y_u)$. So $R_{cl}(X) \cong R_{cl}(Y)$ by Theorem \ref{2-4}. Now, suppose that $X$ and $Y$ are $r_{cl}$-compact. Then $X_u$ and $Y_u$ are compact zero-dimensional spaces by Theorem \ref{3-4} and Corollary \ref{2-5}. If $R_{cl}(X) \cong R_{cl}(Y)$, then $C(X_u) \cong C(Y_u)$ by Theorem \ref{2-4}. This implies by Theorem $4.9$ in \cite{6}, that $X_u \cong Y_u$.
\end{proof}
\begin{corollary}
For two topological spaces $X$ and $Y$, if $X \cong _{r_{cl}} Y$, then  $R_{cl}(X) \cong R_{cl}(Y)$.
\end{corollary}
\begin{proof}
It follows immediately from Proposition \ref{2-7} and Theorem \ref{3-5}.
\end{proof}
\begin{remark}
For two $r_{cl}$-compact spaces $X$ and $Y$, the rings $R_{cl}(X)$ and $R_{cl}(Y) $ may be isomorphic, while $X$ and $Y$ may not be homeomorphic. To see that, we utilize the example of Remark $4.7$ in \cite{1}. Let $X=\{ \frac{1}{n}: n \in \mathbb{N} \} \cup \{0\}$ and $Y=\bigcup_{n=1} ^{\infty} (\frac{1}{n+1},\frac{1}{n} ) \cup \{0\}$ as subspaces of $\mathbb{R}$. Since $X$ and $Y$ are compact, $X$ and $Y$ are $r_{cl}$-compact. Using a proof similar to Remark $4.7$ in \cite{1}, we can show that $R_{cl}(X) \cong R_{cl}(Y)$, but $X$ and $Y$ are not homeomorphic.
\end{remark}
\begin{theorem} \label{3-6}
For a $s_{cl}$-completely regular space $X$, the maximal ideals of $R_{cl}(X)$ are precisely the sets
$$ M^p = \left\{ f \in R_{cl}(X) :  p \in cl_{\beta X_u} Z(f_u)\right\} \quad p \in \beta X_u.$$
\end{theorem}
\begin{proof}
Using Theorem \ref{2-6}, and in view of Theorem $7.3$ in \cite{6}, it is immediate.
\end{proof}
\begin{remark}
Let $X$ be a space and let $M$ be a maximal ideal of $R_{cl}(X)$. Similar to $C(X)$, we define $$O_{M} =\left\{ f \in R_{cl}(X) : fg=0\, \,\text{for some}\, g \notin M_{Q_x}\right\},$$ see the discussion preceding Theorem 2.12 in \cite{2}  .For each $x \in X$, let $O_{Q_x} =\big\{ f \in R_{cl}(X) \, : \, Q_x \subseteq int_{r_{cl}} Z(f) \big\}$. If $X$ is $s_{cl}$-completely regular and $M$ is a fixed maximal  ideal of $R_{cl}(X)$, then
$O_{M}=O_{Q_x}$ for some $x \in X$. In fact, if $M$ is a fixed maximal ideal in $ R_{cl}(X)$, then by Theorem \ref{3-2}, $M=M_{Q_x}$ for some $x \in X$. Now we show that $O_{M}=O_{Q_x}$. To this end, let $f \in O_{M}$. Then $fg=0$ for some $ g \notin M_{Q_x}$ and hence $Q_x \subseteq X \setminus Z(g) \subseteq Z(f)$. Since $X \setminus Z(g)$ is $r_{cl}$-open, then $Q_x \subseteq int_{r_{cl}} Z(f)$ which follows that $f \in O_{Q_x}$. Now, let  $f \in O_{Q_x}$. Then $Q_x \subseteq int_{r_{cl}} Z(f)$, so there is an $r_{cl}$-open set $U$ in $X$ such that  $Q_x \subseteq U  \subseteq  Z(f)$. Since $X$ is $s_{cl}$-completely regular, there exists $ g \in R_{cl}(X)$ such that $g[X \setminus U] =\{0\}$ and $g[Q_x]=\{1\}$. Therefore $ g \notin M_{Q_x}$ and $fg=0$ which shows that $f \in O_{M}$.
\end{remark}
\begin{theorem} \label{3-7}
Let $X$ be an $r_{cl}$-compact space. Then for every $x \in X$, the ideal $O_{Q_x}$ in $R_{cl}(X)$ is generated by a set of idempotents.
\end{theorem}
\begin{proof}
By Corollary \ref{2-5}, $X_u$ is zero-dimensional. In view of Theorem $2.4$ in \cite{4}, a space $X$ is zero-dimensional if and only if for each $x \in X$, the ideal $O_x$ in $C(X)$ is generated by a set of idempotents. Using this fact, for each $x \in X$, the ideal $O_{Q_x}$ in $C(X_u)$ is generated by a set of idempotents. This follows by Theorem \ref{2-4}, that the ideal $O_{Q_x}$ in $R_{cl}(X)$ is generated by a set of idempotents.
\end{proof}
\section{\bf $P_{r_{cl}}$-spaces and almost $P_{r_{cl}}$-spaces}
In this section the counterparts of $\rm{P}$-spaces and almost $\rm{P}$-spaces are defined and characterized both algebraically and topologically. We recall that a completely regular Hausdorff space $X$ is called a $P$-\textit{space} if $Z(f)$ is open for each $f\in C(X)$, equivalently, $C(X)$ is a von Neumann regular ring, see \cite[4J]{6}. Recall that a ring $R$ is called {\it von Neumann regular} if for every $a\in R$ there is  $x\in R$ for which $axa=a$. We observe trivially that if $C(X)$ is von Neumann regular, then
so is $R_{cl}(X)$, but the converse is not true in general. For example, the space $R_{cl}(\mathbb{R})$ is von Neumann regular but $C(\mathbb{R})$ is not von Neumann regular. Motivated by this, we offer the following definition.
\begin{definition}
A $s_{cl}$-completely regular space $X$ is called $P_{r_{cl}}$-space if $Z(f)$ is open for each $f\in R_{cl}(X)$.
\end{definition}
\begin{lemma}\label{4-1}
Every $P$-space is a $P_{r_{cl}}$-space.
\end{lemma}
\begin{proof}
Suppose that $X$ is a $P$-space. Then  $X$ is zero-dimensional and hence $X$ is $s_{cl}$-completely regular. Also each  $Z \in Z_{r_{cl}}(X)$ is open, for $Z_{r_{cl}}(X) \subseteq Z(X)$.
\end{proof}

\begin{theorem} \label{4-2}
Let $X$ be a $s_{cl}$-completely regular space $X$. The following statements are equivalent.
\begin{enumerate}
\item $X$ is a $P_{r_{cl}}$-space.
\item $X_u$ is a $P$-space.
\item $M_{Q_x} = O_{Q_x}$ for every $x \in X$.
\item Each countable intersection of $r_{cl}$-open sets is $r_{cl}$-open.
\item $R_{cl}(X)$ is von Neumann regular.
\end{enumerate}
\end{theorem}
\begin{proof}
$(1)\Leftrightarrow(2)$ If $f \in R_{cl}(X)$, then $Z(f)= \bigcup _{Q_x \in Z(f_u)} Q_x$. So $Z(f)$ is open in $X$ if and only if $Z(f_u)$ is open in $X_u$. This implies that $X$ is a $P_{r_{cl}}$-space if and only if $X_u$ is a $P$-space. \\
Using Problem $4J$ in \cite{6}, parts (2), (3) and (4) are equivalent.\\
$(2)\Leftrightarrow(5)$ By Problem $4J$ in \cite{6}, $X_u$ is a $P$-space if and only if $C(X_u)$ is a von Neumann regular ring. This implies by Theorem \ref{2-4}, that $X_u$ is a $P$-space if and only if $R_{cl}(X)$ is a von Neumann regular ring.
\end{proof}
\begin{proposition} \label{4-3}
A Hausdorff space $X$ is a $P$-space if and only if  $X$ is both an $R_{cl}$-space and a $P_{r_{cl}}$-space.
\end{proposition}
\begin{proof}
If $X$ is a $P$-space, then $X$ is zero-dimensional and hence $X$ is an $R_{cl}$-space. Furthermore $X$ is a $P_{r_{cl}}$-space by Lemma \ref{4-1}. Conversely, suppose that $X$ is an $R_{cl}$-space and a $P_{r_{cl}}$-space. Then $X$ is completely regular, since every $s_{cl}$-completely regular  $R_{cl}$-space is completely regular. Now, since $X$ is a $P_{r_{cl}}$-space, $R_{cl}(X)$ is a von Neumann regular ring by Theorem \ref{4-2}. This implies that  $C(X)$ is von Neumann regular, for $R_{cl}(X)=C(X)$ by Lemma \ref{2-1}.
\end{proof}
We recall that a completely regular Hausdorff space $X$ is an {\it almost $P$-space} if every nonempty zero-set in $Z(X)$ has non-empty interior.  Motivated by this, we offer the following definition.
\begin{definition}
A $s_{cl}$-completely regular space $X$ is called an \textit{almost} $P_{r_{cl}}$-\textit{space} if every non-empty zero-set in $Z_{r_{cl}}(X)$ has non-empty $r_{cl}$-interior.
\end{definition}
The following shows that the classes of almost $P$-spaces and almost $P_{r_{cl}}$-spaces are independent of each other.
\begin{example}
The space  $\mathbb{R}$ is an almost $P_{r_{cl}}$-space but it is not an almost $P$-space. Now, let $Y= X \cup \mathbb{N}$, where $X$ is a connected almost $P$-space, see Example 5.3 in \cite{1}. Then $Y$ is an almost $P$-space. To see that $Y$ is not an almost $P_{r_{cl}}$-space, let $f :Y \rightarrow \mathbb{R}$ be defined as $f(n)= \frac{1}{n}$, if $n \in \mathbb{N}$ and $f(x)=0$, if $x \in X$. Then $f \in R_{cl}(Y)$ and $Z(f)=X$, but $int _{r_{cl}} Z(f)= \emptyset$, for if  $x \in int _{r_{cl}} Z(f)$, then there is an open set $U$ in $Y$ such that $X=Q_x\subseteq U \subseteq Z(f) =X$ which shows that $X$ is open in $Y$, a contradiction.
\end{example}
We conclude this section by the following theorem which characterizes almost $P_{r_{cl}}$-space. We call a set $A$ in a space $X$ is  $r_{cl}$-{\it dense} in $X$ if every $r_{cl}$-open set in $X$ intersects $A$ nontrivially.
\begin{theorem}
The following statements are equivalent for a $s_{cl}$-completely regular space $X$.
\begin{enumerate}
\item $X$ is an almost $P_{r_{cl}}$-space.
\item $X_u$ is an almost $P$-space.
\item Every non-unit element in $R_{cl}(X)$ is a zero-divisor.
\item $\bigcup _{x \in X} O_{Q_x} = \bigcup _{x \in X} M_{Q_x}$.
\item If $G= \bigcap _{i \in \mathbb{N}} G_i$, where each $G_i$ is  $r_{cl}$-open, then $int _{r_{cl}} G$ is $r_{cl}$-dense in $G$.
\item
Every non-empty zero-set in $Z_{r_{cl}}(X)$ has non-empty $r_{cl}$-interior.
\end{enumerate}
\end{theorem}
\begin{proof}
Using Theorem 2.2 in \cite{3} and by Theorem \ref{2-4}, the proof is easy.
\end{proof}

%\section*{Acknowledgements}
%\vspace{4mm}\noindent{\bf Acknowledgements}\\
%\noindent If you'd like to thank anyone, place your comments here.


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{\small

\noindent{\bf M. Etebar}

\noindent Department of Mathematics

\noindent Assistant Professor of Mathematics

\noindent Shahid Chamran University of Ahvaz

\noindent Ahvaz, Iran

\noindent E-mail: m.etebar@scu.ac.ir}\\

{\small
\noindent{\bf  S. Soltanpour  }

\noindent  Department of Science

\noindent Associate Professor of Mathematics

\noindent Petroleum University of Technology


\noindent Ahvaz, Iran

\noindent E-mail: s.soltanpour@put.ac.ir}\\



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