\documentclass[two side,14pt]{article} \usepackage{amsmath,latexsym,amssymb,graphicx} \input{xypic} \usepackage{epsfig} \usepackage{graphicx} \usepackage{fancyhdr} %----------------------- Page Layout ---------------------------- \headsep0.5cm \topmargin 1cm \textwidth 12.5truecm \textheight 19truecm \oddsidemargin 2.5truecm \evensidemargin 3.0cm %----------------------- Font Setting --------------------------- \font\chuto=cmbx10 at 16pt \font\chudam=cmbxsl10 %----------------------- Page Style ----------------------------- \pagestyle{myheadings} \markboth{\small{ P. Padma and Alias B. Khalaf}} {\small { Pairwise Q*s-(regular and normal) spaces in bitopological spaces}} \date{} \pagenumbering{arabic} \setcounter{page}{1} %----------------------- Command Setting ------------------------ \newtheorem{Theorem}{Theorem}[section] \newtheorem{Proposition}[Theorem]{Proposition} \newtheorem{Corollary}[Theorem]{Corollary} \newtheorem{Lemma}[Theorem]{Lemma} \newtheorem{Example}[Theorem]{Example} \newtheorem{Definition}[Theorem]{Definition} \newtheorem{Remark}[Theorem]{Remark} \numberwithin{equation}{section} \newcommand{\ds}{\displaystyle} \newenvironment{proof} {\begin{trivlist}\item[\hskip% \labelsep{\bf Proof.}]}% {$\hfill\square$\end{trivlist}} \thispagestyle{empty} \begin{document} %----------------------- Title of Article ---------------------- \vskip1.0cm \centerline {\bf \chuto Pairwise Q*s-(regular and normal) spaces in } \vskip0.2cm \centerline {\bf \chuto bitopological spaces} %----------------------- The Author(s) ---------------------- \vskip0.8cm \centerline {P. Padma and Alias B. Khalaf} %----------------------- Abstract ------------------------------ \vskip0.8cm \noindent {\small{\bf Abstract :} The notion of $\tau_1 \tau_2 - Q^*$ - open sets in a bitopological spaces was introduced by K.Kannan and K.Chandrasekhararao. We introduce the notion of pairwise $Q^*s$ - regular, pairwise $Q^*s$ - normal, pairwise $s^*Q^*$ - normal and obtain some characterizations of pairwise $Q^*s$ - regularity and pairwise $Q^*s$ - normality, pairwise $s^*Q^*$ - normal. %----------------------- Keywords ------------------------------ \vskip0.3cm \noindent {\bf Keywords :} pairwise $Q^*s$ - regular ; pairwise $Q^*s$ - normal ; pairwise $s^*Q^*$ - normal; pairwise $Q^*$ - normal. %----------------------- 2000 MSC ------------------------------ \noindent {\bf 2010 Mathematics Subject Classification : } 54E55. %----------------------- Details ------------------------------- \section{Introduction} Separation axioms are properties by which the topology on a space $X$ separates points from points, points from closed sets and closed sets from each other. The various separation axioms give rise to a sequence of successively stronger requirements, which are put upon the topology of a space to separate varying types of subsets. These axioms are also found useful to characterize continuous mappings. In 1963, Levine introduced the concept of semi-open sets. Maheshwari and Prasad have introduced pairwise semi $T_i$ - spaces, $i\in \{ 0, 1, 2\}$. Using the notion of semi - open sets , Maheshwari , Prasad and Bhamini have defined and studied the notions of pairwise s-normal (resp. pairwise irresolutely normal), if for any pair of disjoint $\tau_i$ - closed set A and a $\tau_j$ - closed set B ( $\tau_i$ - semi closed set A and a $\tau_j$ - semi-closed set B ), there exists a $\tau_j$ - semi open set U and a $\tau_i$ -semi open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j =1, 2. The notion of $Q^*$-open sets in a topological space was introduced by Murugalingam and Lalitha \cite{ML1, ML2}. Mean while J . C . Kelly introduced bitopological space in 1963 . There after , several authors studied the above mentioned concepts in bitopological settings. %%%%%%%%%%%%%%%%%%%%%%%%%%%%% The notion of pairwise semi-$T_0$, pairwise semi-$T_1$, pairwise semi-$T_2$, pairwise $s$-regular and pairwise $s$-normal spaces, $s$-normal (resp. semi normal) spaces were introduced and studied by Maheshwari and Prasad \cite{MP-1, MP0, MP1, MP3, MP4, MP5}. In this paper, the notion of pairwise $Q^*s$-regular spaces and pairwise $Q^*s$-normal spaces are introduced and their basic properties in bitopological spaces are discussed. \section{Preliminaries}\rm Let $(X, \tau_1, \tau_2)$ or simply $X$ denote a bitopological space. For any subset $A\subseteq X$, $\tau_i - int(A)$ and $\tau_i - cl(A)$ denote the interior and closure of a set $A$ with respect to the topology $\tau_i$, respectively. $A^C$ denotes the complement of $A$ in $X$ unless explicitly stated. We give the following definitions in bitopological spaces. \begin{Definition}\rm A function $f : ( X, \tau_1, \tau_2 ) \longrightarrow ( Y,\sigma_1, \sigma_2 )$ is said to be pairwise homeomorphism if the induced functions $f : ( X, \tau_1 ) \longrightarrow (Y, \sigma_2 )$ and $f : ( X, \tau_2 ) \longrightarrow (Y, \sigma_1 )$ are homeomorphism. \end{Definition} \begin{Definition}\rm A function $f : ( X, \tau_1, \tau_2 ) \longrightarrow ( Y,\sigma_1, \sigma_2 )$ is said to be pairwise semi - homeomorphism if the induced functions $f : ( X, \tau_1 ) \longrightarrow (Y, \sigma_2 )$ and $f : ( X, \tau_2 ) \longrightarrow (Y, \sigma_1 )$ are semi-homeomorphism, i.e., the induced function are pre-semi open, irresolute and bijective. \end{Definition} \begin{Lemma}\rm\cite{ML1} Let $X$ be a topological space. Then the family of all $Q^*$-open sets in $X$ with $\phi$ is a topology. It is denoted by ${\tau }_{Q^*}={\sigma }^*$. \end{Lemma} \begin{Lemma}\rm\cite{ML1} Let $X$ be a topological space . The set of all $Q^*$ - closed sets with $X$ is a topology. It is denoted by ${\tau }_{Q^*}={\mu }^*$. \end{Lemma} \begin{Definition}\rm \cite{Y} A space $X$ is said to pairwise $\widehat{g}s$-regular if for any point $x$ and a $\tau_i-gs$- closed set $F$ not containing $x$, there exist a $\tau_i$-semi open set $U$ and a $\tau_j$-semi open set $V$ such that $x \in U$ and $F \subseteq V$ and $U \cap V = \phi$, $i \neq j$, and $i, j \in \{ 1, 2\}$. \end{Definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin {Definition}\rm\cite{PU} A bitopological space $X$ is said to be pairwise $Q^*$ - normal if for each $ \tau_i -Q^*$ - closed set A and $ \tau_j -Q^*$ - closed set B with $A \cap B = \phi$ , there exists a $ \tau_i -Q^*$-open set $V \supseteq B$ and there exists a $\tau_i - Q^*$ - open set $U \supseteq A$ such that $U \cap V = \phi$ , where $i, j = 1, 2$ and $i \neq j$. \end {Definition} \begin {Definition}\rm\cite{MP1} A space $X$ is said to be $semi T_2$ if for each pair of distinct points x and y in X, there exist disjoint semi open sets U and V in X such that $x \in U$ and $y \in V$. \end {Definition} \begin {Definition}\rm\cite{MP4} A space $X$ is said to be pairwise semi $T_2$ if for each pair of distinct points x and y in X, there exist disjoint $\tau_1$- semi open set U and $\tau_2$- semi open set V in X such that $x \in U$ and $y \in V$ such that $U \cap V = \phi$ , where $i, j = 1, 2$ and $i \neq j$. \end {Definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{ Pairwise $Q^*$s - regular spaces } In this section, we introduce the concept of pairwise $Q^*$s - regular spaces and we establish some its properties. \begin{Definition}\rm A space $X$ is said to pairwise $Q^*s$ - regular if for any point x and a $\tau_i - Q^*$ - closed set F not containing x, there exist a $\tau_i$ - semi open set U and a $\tau_j$ - semi open set V such that $x \in U$ and $F \subseteq V$ and $U \cap V = \phi$, $i \neq j$, i, j = 1, 2. \newline Clearly, every pairwise $Q^*s$ - regular space is pairwise $\widehat{g}s$ - regular but converse is not true . \end{Definition} \begin{Example}\rm \label{3.1} Let $X = \{ a, b, c \}$, $\tau_1$ = $\{\phi, X, \{ a \}, \{ b\} , \{ a, b \}\}$ and $\tau_2$ = $\{\phi, X,\{ c \}, \{ b \}, \{ b, c \} \}$. $SO ( X, \tau_1 ) = \{ \phi, X, \{ a \}, \{ b \}, \{ b, c \},\{ a, c \}, \{ a, b \} \}$, $SO ( X, \tau_2 ) = \{\phi, X, \{ c \}, \{ b \}, \{ b, c \},\{ a, c \}, \{ a, b\} \}$ and $\sigma_1^* = \{ \phi, X, \{ a, b \} \}$, $\sigma_2^* = \{\phi, X, \{ b, c \} \}$. Then the space $X$ is pairwise $\widehat{g}s$ - regular but not pairwise $Q^*s$ - regular space. \end{Example} \begin{Theorem}\rm For a space $X$, the following are equivalent : \begin{itemize}\rm \item[(a)] $( X, \tau_1, \tau_2 )$ is pairwise $Q^*s$ - regular. \item[(b)] For each $x \in X$ and every $\tau_i - Q^*$ - open set U containing x, there exists a $\tau_i$ - semi open set H such that $x \in H \subseteq \tau_j - scl ( H )\subseteq U$ ; $i\neq j$, i, j = 1, 2. \item[(c)]For every $\tau_i - Q^*$ - closed set F, the intersection of all $\tau_i$ - semi closed, $\tau_j$ - semi neighborhoods of F is exactly F; $i\neq j$, i, j = 1, 2. \item[(d)]For every set A and a $\tau_i - Q^*$ - open set B such that $A\cap B = \phi$, there exists a $\tau_i$ - semi open set W such that $A \cap W \neq \phi$ and $\tau_j - scl ( W ) \subseteq B$; $i \neq j$, i, j = 1, 2. \item[(e)]For every non empty set A and any $\tau_i - Q^*$ - closed set B satisfying $A \cap B = \phi$, there exists a $\tau_i$ - semi open set U and a $\tau_j$ - semi open set V such that $A \cap U \neq \phi$, and $B \subseteq V$ and $U \cap V = \phi$, $i\neq j$, i, j = 1, 2. \end{itemize} \end{Theorem} \begin{proof} $(a)\longrightarrow(b)$ Let $x \in U$ and U is $\tau_i - Q^*$ - open in X. Therefore, $x \notin X - U$ and X - U is $\tau_i - Q^*$ - closed in X . Since X is pairwise $Q^*S$ - regular, there exists a $\tau_i$ - semi open set V and a $\tau_j$ - semi open set W such that $x \in V$ and $X - U \subseteq W$ and $V \cap W = \phi$. Obviously, $V\subseteq X - W$ and hence $\tau_j - scl ( V ) \subseteq X - W$. Hence $x \in V \subseteq \tau_j - scl ( V ) \subseteq U$. \newline $(b)\longrightarrow(c)$ Let F be a $\tau_i - Q^*$ - closed subset of X and $x \notin F$. Then X - F is a $\tau_i - Q^*$ - open set containing x. Therefore, by ( b ) there exists a $\tau_i$ - semi open set O such that $x \in O \subseteq scl ( O )\subseteq X - F$, which implies that $F \subseteq ( X - \tau_j - scl ( O ) ) \subseteq X - O$. Also $X - O$ is $\tau_i$ - semi closed, $\tau_j$ - semi neighborhood of F which does not contain x . Hence, the intersection of all $\tau_i$ - semi closed, $\tau_j$ - semi neighborhoods of F is exactly F. \newline $(c)\longrightarrow(d)$ Let A be a non empty subset of X and B be a $\tau_i - Q^*$ - open set such that $A \cap B \neq \phi$. Let $x \in A \cap B$. Then $X - B$ is a $\tau_i - Q^*$ - closed such that $x \notin X - B$. Therefore, by (c), the intersection of all $\tau_i$ - semi closed, $\tau_j$ - semi neighborhood of X - B is exactly X - B, ie., there exists a $\tau_i$ - semi closed set, $\tau_j$ - semi - neighborhood of X - B, say V such that $x \notin V$. Thus, there is a $\tau_j$ - semi open set U such that $X - B \subseteq U \subseteq V$. Take $W = X - V$. Then $W$ is a $\tau_i$ - semi open set containing x as $x \notin X - B$ therefore $x \notin V$. Hence $x \in A$ and $x \in W$ which implies that $A \cap W \neq \phi$. Since $X - V \subseteq X - U \subseteq B$, therefore, $\tau_j - scl ( X - V )\subseteq X - U \subseteq B$. Hence $\tau_j - scl ( W )\subseteq B$.$ \newline (d)\longrightarrow(e)$ Let $A \cap B = \phi$, where A is non empty and B is a $\tau_i - Q^*$ - closed, then $A\cap X - B \neq \phi$, where X - B is a $\tau_i - Q^*$ - open set. Therefore by ( d ), there exists a $\tau_i$ - semi open set G such that $A \cap G \neq \phi$, and $\tau_j - scl ( G ) \subseteq X - B$. Now, put $M = X - \tau_i - scl ( G )$. Then $B \subseteq M$ and G and M are $\tau_j$ - semi open sets such that $G \cap M = \phi$. \newline $(e)\longrightarrow(a)$ Let F be a $\tau_i - Q^*$ - closed subset of X and $x \notin F$. Then \{ x \} and F are disjoint . Therefore by ( e ), there exists a $\tau_i$ - semi open set U and a $\tau_j$ - semi open set V such that $\{ x \} \cap U \neq \phi$, $F \subseteq M$ and $U \cap V = \phi$, ie., $x \in U$. Hence X is pairwise $Q^*s$ - regular. \end{proof} \begin{Definition}\rm A space X is said to bi-$Q^*$-symmetric if every singleton \{ x \} is $\tau_i - Q^*$ - closed, i = 1, 2. \end{Definition} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{Remark}\rm Every bi - $Q^*$ - symmetric is bi – symmetric but the converse need not be true in general. The following example supports our claim. \end{Remark} \begin {Example}\rm Let $X = \{a, b\}$, $\tau_1 = \tau_2 = \{ \phi, X, \{a\},\{b\}\}$. Then $X$ is bi - symmetric but not bi - $Q^*$ - symmetric. \end{Example} \begin{Theorem}\rm Every pairwise $Q^*s$ - regular, bi - $Q^*$ - symmetric space is pairwise semi-$T_2$. \end{Theorem} \begin{proof} Let X be a pairwise $Q^*s$ - regular and bi - $Q^*$ - symmetric space. Let x, y be any two distinct points of X. Since X is bi - $Q^*$ - symmetric implies \{ x \} is $\tau_i - Q^*$ - closed for i = 1, 2. Also $y \notin \{ x \}$. Since X is pairwise $Q^*s$ - regular, there exists a $\tau_i$ - semi open set U and $\tau_j$ - semi open set V such that $\{ x \} \in V$, $y \in U$ and $U \cap V = \phi$; $i\neq j$, i, j = 1,2. Hence X is pairwise semi-$T_2$. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%% \begin {Example}\rm In Example \ref{3.1}, the space $X$ is pairwise $Q^*s$-normal but not bi-$Q^*$-symmetric and pairwise $Q^*s$-regular. \end{Example} \begin{Theorem}\rm Let $f : X \longrightarrow Y$ is a pairwise homeomorphism. Then X is pairwise $Q^*s$ - regular if and only if Y is pairwise $Q^*s$ - regular. \end{Theorem} \begin{proof} Let $f : ( X, \tau_1, \tau_2 ) \longrightarrow ( Y,\sigma_1, \sigma_2 )$ be a pairwise homeomorphism. Let X is pairwise $Q^*s$ - regularity. Let F be a $\sigma_i - Q^*$ - closed subset in Y such that $y \notin F$. Then $x\notin f^{-1}(F)$, where $y = f ( x )$ and $f^{-1}(F)$ is a $\tau_i - Q^*$- closed since f is pairwise homeomorphism . By pairwise $Q^*s$ - regularity of X, there exists a $\tau_i$ - semi open set U and a $\tau_j$ - semi - open set V such that $x \in U$, $f^{-1}(F)\subseteq V$ and $U \cap V = \phi$. Hence $y \in f(U)$, $F\subseteq f(V)$ and $f(U)\cap f(V) = \phi$. Since f is pairwise homeomorphism implies f is pairwise semi homeomorphism implies f is pairwise pre semi open. Therefore, f(U) and f(V) are $\sigma_i$ - semi open and $\sigma_j$ - semi open sets respectively. Hence Y is pairwise $Q^*s$ - regular. Conversely, Let Y be pairwise $Q^*s$ - regular and let G be any $\tau_i - Q^*$ - closed set in X such that $x \notin G$. Then $y \notin f(G)$ a $\sigma_i - Q^*$ - closed set in Y since f is pairwise homeomorphism. By pairwise $Q^*s$ - regularity of Y, there exists a $\sigma_i$ - semi open set U and a $\sigma_j$ - semi open set V in Y such that $y \in U$ and $f(G) \subseteq V$. Hence $x \in f^{-1}(U)$ and $G \subseteq f^{-1}(V)$ with $f^{-1}(U) \cap f^{-1}(V)= \phi$. Since f is pairwise homeomorphism implies f is pairwise semi homeomorphism implies f is pairwise irresolute. Therefore, $f^{-1}(U)$ and $f^{-1}(V)$ are $\tau_i$ - semi open and $\tau_j$ - semi open sets respectively in X. Hence, X is pairwise $Q^*s$ - regular. \end{proof} \section{ Pairwise $Q^*$s-normal spaces} In this section, we introduce the concept of pairwise $Q^*$s-normal spaces and we establish some properties of this concept. \begin{Definition}\rm A bitopological space $( X, \tau_1, \tau_2 )$ is said to be pairwise $Q^*s$ - normal if for every pair of disjoint $\tau_i - Q^*$ - closed set A and $\tau_j - Q^*$ - closed set B, there exists a $\tau_j$ - semi open set U and a $\tau_i$ - semi open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j = 1, 2. \end{Definition} \begin{Example}\rm In Example \ref{3.1}, shows that the space $( X, \tau_1, \tau_2 )$ is pairwise $Q^*s$-normal but not pairwise $Q^*$-normal. \end{Example} \begin{Definition}\rm A bitopological space $( X, \tau_1, \tau_2 )$ is said to be pairwise $s^*Q^*$-normal if for every pair of disjoint $\tau_i$ - semi closed set A and $\tau_j$ - semi closed set B in X , there exists disjoint $\tau_j - Q^*$ - open set U and a $\tau_i - Q^*$ - open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j = 1, 2. \end{Definition} \begin{Definition}\rm \cite{KNH} A space X is said to be pairwise $s^*$ - normal if for any two disjoint $\tau_i$ - semi closed set A and $\tau_j$ - semi closed set B, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j = 1, 2. \end{Definition} \begin{Definition}\rm \cite{Y} A space X is said to be pairwise gs - normal if for any two disjoint $\tau_i$ - g - closed set A and $\tau_j$ - g - closed set B, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j = 1, 2. \end{Definition} \begin{Definition}\rm \cite{KNH} A space X is said to be pairwise $s$ - normal if for any two disjoint $\tau_i$ - closed set A and $\tau_j$ - closed set B, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$, $B \subseteq V$ and $U \cap V = \phi$ ; $i \neq j$, i, j = 1, 2. \end{Definition} \begin{Theorem}\rm For a space $X$, the following are equivalent : \begin{itemize}\rm \item[(a)] $( X, \tau_1, \tau_2 )$ is pairwise $Q^*s$ - normal. \item[(b)]For each $\tau_i - Q^*$ - closed set F and a $\tau_j - Q^*$ - open set K containing F, there exists a $ \tau_j$ - semi open set U such that $F \subseteq U \subseteq \tau_i - scl(U)\subseteq K$. \item[(c)] For every $\tau_i - Q^*$ - closed set A and a $\tau_j - Q^*$ - closed set B disjoint from A, there exists a $\tau_i$-semi open set U containing A such that $ \tau_j - scl(U) \cap B = \phi$. \end{itemize} \end{Theorem} \begin{proof} $(a)\longrightarrow(b)$ Let X be pairwise $Q^*s$ - normal and let K be a $\tau_j - Q^*$ - open set containing a $\tau_i - Q^*$ - closed set F. Then F and X - K are disjoint $\tau_i - Q^*$ - closed and $\tau_j - Q^*$ - closed sets respectively. So by (a), there exists a $\tau_j$ - semi open set U and a $\tau_i$ - semi open set V such that $F \subseteq U$, $X - K \subseteq V$ and $U \cap V = \phi$. Thus $U \subseteq X - V$, which implies that $\tau_i - scl ( U ) \subseteq X - V$. Hence, $F \subseteq U \subseteq \tau_i - scl(U)\subseteq K$. \newline $(b)\longrightarrow(c)$ Let A and B are respectively $\tau_i - Q^*$ - closed and $\tau_j - Q^*$ - closed subsets of X such that $A \cap B = \phi$, which implies $A\subseteq X - B$ , a $\tau_j - Q^*$ - open set. So by (b), there exists a $\tau_j$ - semi open set U such that $A \subseteq U \subseteq \tau_i - scl(U)\subseteq X - B$. Hence, $\tau_i - scl(U) \cap B = \phi$. \newline $(c)\longrightarrow(a)$ Let A be a $\tau_i - Q^*$ - closed set and B be a $\tau_j - Q^*$ - closed set disjoint from A. Then, by (c) there is a $\tau_j$ - semi open set U such that $A \subseteq U$ and $\tau_i - scl(U)\cap B = \phi$. Now $\tau_i - scl(U)$ is semi closed. Hence, $B \subseteq X - \tau_i - scl(U)$, let $V = X - \tau_i - scl(U)$. Then V is a $\tau_i$ - semi open set such that $B \subseteq V$ and $U \cap V = \phi$. Hence, X is pairwise $Q^*s$ - normal. \end{proof} \begin{Theorem}\rm For a space $X$, the following are equivalent : \begin{itemize}\rm \item[(a)] $( X, \tau_1, \tau_2 )$ is pairwise $s^*Q^*$ - normal. \item[(b)] For each $\tau_i$ - semi closed set F and a $\tau_j$ - semi open set K containing F, there exists a $\tau_j - Q^*$ open set U such that $F \subseteq U \subseteq \mu_i^* - cl(U)\subseteq K$. \item[(c)]For every $\tau_i$ - semi closed set A and a $\tau_j$ - semi closed set B disjoint from A, there exists a $\tau_i - Q^*$- open set U containing A such that $\mu_j^* - cl(U)\cap B =\phi$. \end{itemize} \end{Theorem} \begin{proof} $(a)\longrightarrow(b)$ Let X be pairwise $S^*Q^*$ - normal and let K be a $\tau_i$ - semi open set containing a $\tau_j$ - semi closed set F. Then F and X - K are disjoint $\tau_i$ - semi closed set and $\tau_j$ - semi closed sets respectively. So by (a), there exists a $\tau_j - Q^*$ open set U and a $\tau_i - Q^*$ - open set V such that $F \subseteq U$, $X - K \subseteq V$ and $U \cap V = \phi$. Thus $U \subseteq X - V$, which implies that $\mu_i^* - cl(U)\subseteq X - V$. Hence, $F \subseteq U \subseteq \mu_i^* - cl(U)\subseteq K$. $(b)\longrightarrow(c)$ Let A and B are $\tau_i$ - semi closed set and $\tau_j$ - semi closed subsets of X such that $A \cap B = \phi$, which implies $A \subseteq X - B$, a $\tau_j$ - semi open set. So by (b), there exists a $\tau_j - Q^*$ open set U such that $A \subseteq U \subseteq \mu_j^* - cl(U)\subseteq X - B$. Hence, $\mu_j^* - cl(U)\cap B = \phi$. $(c)\longrightarrow(a)$ Let A be $\tau_i$ - semi closed and B be a $\tau_j$ - semi closed set disjoint from A. Then, by (c), there is a $\tau_j - Q^*$ - open set U such that $A \subseteq U$ and $\mu_i^* - cl(U)\cap B = \phi$. Now $\mu_i^*- cl(U)$ is $Q^*$ - closed. Hence, $B \subseteq X - \mu_i^* - cl(U)$, let $V = X - \mu_i^* - cl(U)$. Then V is a $\tau_i - Q^*$ - open set such that $B \subseteq V$ and $U \cap V = \phi$. Hence, X is pairwise $S^*Q^*$ - normal. \end{proof} \begin{Theorem}\rm Every pairwise $Q^*s$-normal and $Q^*$-symmetric space X is $Q^*s$-regular. \end{Theorem} \begin{proof} Let F be a $\tau_j - Q^*$ closed subset of X with $x \notin F$. Since X is bi - $Q^*$ - symmetric so \{ x \}is $\tau_i - Q^*$ - closed; $i \neq j$ and i, j = 1, 2. So \{ x \} and F are disjoint $\tau_i - Q^*$ closed and $\tau_j - Q^*$ closed sets respectively in X. Since X is pairwise $Q^*s$ - normal, there exist disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $\{ x \} \subseteq U$, $F \subseteq V$. Hence $X$ is pairwise $Q^*s$ - regular. \end{proof} \begin{Theorem}\rm Every pairwise $Q^*$ - normal and bi-$Q^*$-symmetric space X is pairwise $Q^*$-regular. \end{Theorem} \begin{proof} Let F be a $\tau_j - Q^*$ - closed subset of X with $x \notin F$. Since X is bi - $Q^*$ - symmetric so \{ x \} is $\tau_i - Q^*$ - closed. So \{ x \} and F are disjoint $\tau_i - Q^*$ - closed and $\tau_j - Q^*$ - closed sets respectively in X. Since X is pairwise $Q^*$ - normal, there exists a disjoint $\tau_j - Q^*$ - open set U and $\tau_j - Q^*$ - open set V such that $\{ x \} \subseteq U$, $F \subseteq V$. Hence $X$ is pairwise $Q^*$- regular. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin {Example}\rm Let $X = \{a, b, c\}$, $\tau_1 = \{ \phi, X, \{b, c\}\}$ and $\tau_2 = \{ \phi, X, \{a, c\}\}$. Therefore, the space $X$ is pairwise $Q^*$-normal but not bi-$Q^*$-symmetric and pairwise $Q^*$-regular. \end{Example} \begin{Theorem}\rm Let $f : X \longrightarrow Y$ is a pairwise homeomorphism. Then $X$ is pairwise $Q^*s$-normal if and only if $Y$ is pairwise $Q^*s$-normal. \end{Theorem} \begin{proof} Let Y be pairwise $Q^*s$ - normal. Let A and B be two disjoint $\tau_i - Q^*$ - closed set and $\tau_j - Q^*$ - closed sets in X. Then f (A) and f (B) are $\sigma_i - Q^*$ - closed set and $\sigma_ j - Q^*$ - closed sets in Y. Since Y is pairwise $Q^*s$ - normal, there exist disjoint $\sigma_i$ - semi open set U and $\sigma_j$ - semi open set V in Y such that $f(A)\sqsubseteq U$, $f(B)\subseteq V$. Hence, $A \subseteq f^{-1}(U)$, $B \subseteq f^{-1}(V)$ , and $f^{-1}(U)\cap f^{-1}(V) = \phi$ as $U \cap V = \phi$. Moreover, $f^{-1}(U)$ and $f^{-1}(V)$ are $\tau_j$ - semi open and $\tau_i$ - semi open sets; since f is pairwise irresolute . Hence X is pairwise $Q^*s$ - normal. Conversely, Let X is pairwise $Q^*s$ - normal. Let A and B be two disjoint $\sigma_i - Q^*$ closed set and $\sigma_j - Q^*$ closed sets in Y . Then $f^{-1}(A)$ and $f^{-1}(B)$ are $\tau_i - Q^*$ - closed and $\tau_j - Q^*$ - closed sets in X. Since X is pairwise $Q^*s$ - normal, there exists a disjoint $\tau_j$ - semi open set U and $\tau_j$ - semi open set V in X such that $f^{-1}(A)\subseteq U$, $f^{-1}(B)\subseteq V$. Hence $A \subseteq f(U)$, $B \subseteq f(V)$, and $f(U)\cap f(V) = \phi$ as $U \cap V = \phi$. Since f is pairwise homeomorphism implies f is pairwise semi homeomorphism implies f is pairwise pre - semi open. Therefore, f(U) and f(V) are $\sigma_i$ - semi open and $\sigma_i$ - semi open in Y respectively. Hence, Y is pairwise $Q^*s$ - normal. \end{proof} \section{ Comparision } \begin{Remark}\rm We summarize the relationship between various special types of normal spaces in the following diagram. None of the implications is reversible. %\newpage %\begin{figure}[ht] %\centering %\begin{align*} %\noindent \includegraphics[width=80ex]{5.jpg} \noindent \textbf{} %\end{align*} %\end{figure} \end{Remark} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \diagram & \text{}& \text{Pairwise S*Q*-normal}\dlto \drto \ddrto& \text{}\\ & \text{Pairwise S*-normal} & \text{Pairwise Q*-normal}\uto \lto \drto \ddto \ddlto & \text{Pairwise S-normal}\\ & \text{}& \text{}& \text{Pairwise semi-normal}\uto\\ & \text{Pairwise Q*s-normal}\uuto \rto& \text{Pairwise normal}\uulto \uurto \urto \enddiagram \begin{Theorem}\rm Every pairwise $Q^*S$-normal space is pairwise $S$-normal. \end{Theorem} \begin{proof} Let X be a pairwise $Q^*s$-normal space . To show that X is pairwise $S$ - normal. Let A be $\tau_i - Q^*$- closed and B be $\tau_j-Q^*$ - closed. Since X is pairwise $Q^*s$-normal, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$ and $B \subseteq V$. Since every $Q^*$-closed set is closed we have A is $\tau_i$ - closed and B is $\tau_j$ - closed. Hence X is pairwise $s$ - normal. \end{proof} \begin{Remark}\rm Converse of the above theorem need not be true in general. \end{Remark} \begin{Example}\rm In Example \ref{3.1}, X is pairwise $S$-normal but not pairwise $Q^*s$-normal. Here \{ b, c \} $\tau_i$ - closed but not $\tau_i - Q^*$ - closed. \end{Example} \begin{Theorem}\rm Every pairwise $Q^*S$ - normal space is pairwise semi - normal. \end{Theorem} \begin{proof} Let X be a pairwise $Q^*s$ - normal space. To show that X is pairwise semi - normal. Let A be $\tau_i - Q^*$ - closed and B be $\tau_j - Q^*$ - closed. Since $X$ is pairwise $Q^*s$ - normal, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$ and $B \subseteq V$. Since every $Q^*$ - closed set is semi - closed we have A is $\tau_i$ - semi closed and B is $\tau_j$ - semi closed. Hence X is pairwise semi - normal. \end{proof} \begin{Remark}\rm But the converse of the above theorem need not be true in general . ie) every pairwise semi - normal space is not pairwise $Q^*s$ - normal. \end{Remark} \begin{Example}\rm Let $X = \{ a, b, c \}$, $\tau_1 = \{ \phi, X, \{ a \}, \{ a, c \}, \{ a, b \} \}$, $\tau_2 = \{\phi, X, \{ a \}, \{ a, c \} \}$ .Then the space X is pairwise semi normal but not pairwise $Q^*s$ - normal. \end{Example} \begin{Theorem}\rm Every pairwise $S^*Q^*$ - normal space is pairwise semi - normal. \end{Theorem} \begin{proof} Let X be a pairwise $S^*Q^*$ - normal space. To show that X is pairwise semi - normal. Let A and B be a two disjoint $\tau_i$ - semi closed set A and $\tau_j$ - semi closed set B in X. Since X is $S^*Q^*$ - normal, there exists a disjoint $\tau_j - Q^*$ - open set U and $\tau_i - Q^*$ - open set V such that $A \subseteq U$ and $B \subseteq V$. Since every $Q^*$ - open set is semi - open, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$ and $B \subseteq V$. Hence X is semi - normal. \end{proof} \begin{Remark}\rm But the converse of the above theorem need not be true in general . ie) every pairwise semi - normal space is not pairwise $S^*Q^*$ - normal. \end{Remark} \begin{Example}\rm In example 5.2, the space X is pairwise semi normal but not pairwise $S^*Q^*$ - normal. \end{Example} \begin{Theorem}\rm Every pairwise $Q^*$ - normal space is pairwise $Q^*s$ - normal. \end{Theorem} \begin{proof} Let X be a pairwise $Q^*$ - normal space. To show that X is pairwise $Q^*s$ - normal. Let A and B be two disjoint $\tau_i - Q^*$ closed set A and $\tau_j - Q^*$ closed set B in X. Since X is pairwise $Q^*$ - normal, there exists a disjoint $\tau_j - Q^*$ open set U and $\tau_i - Q^*$ open set V such that $A \subseteq U$ and $B \subseteq V$. Since every $Q^*$ - open set is semi - open, there exists a disjoint $\tau_j$ - semi open set U and $tau_i$ - semi open set V such that $A \subseteq U$ and $B \subseteq V$. Hence X is pairwise $Q^*s$ - normal. \end{proof} \begin{Remark}\rm But the converse of the above theorem need not be true in general . ie) every pairwise $Q^*s$ - normal space is not pairwise $Q^*$ - normal. \end{Remark} \begin{Theorem}\rm Every pairwise $Q^*s$ - normal space is pairwise $gs$ - normal. \end{Theorem} \begin{proof} Let X be a pairwise $Q^*s$ - normal space. To show that X is pairwise $gs$ - normal. Let A and B be two disjoint $\tau_i - Q^*$ closed set A and $\tau_j - Q^*$ closed set B in X. Since X is pairwise $Q^*s$ - normal, there exists a disjoint $\tau_j$ - semi open set U and $\tau_i$ - semi open set V such that $A \subseteq U$ and $B \subseteq V$. Since every $Q^*$ - closed set is $g$ - closed we have A and B are $\tau_i - g$ closed and $\tau_i - g$ closed sets. Hence X is pairwise $gs$ - normal. \end{proof} \begin{Remark}\rm But the converse of the above theorem need not be true in general . ie) every pairwise $gs$ - normal space is not pairwise $Q^*s$ - normal. \end{Remark} \begin{Example}\rm Let $X = \{ a, b, c \}$, $\tau_1$ = $\{\phi, X, \{ a \}\}$ and $\tau_2$ = $\{\phi, X,\{ a \}, \{ b \}, \{ a, b \} \}$. Then the space $X$ is pairwise $gs$ normal but not pairwise $Q^*s$ - normal. \end{Example} %\newpage \begin{thebibliography}{99} \bibitem{A} {H.M. Abu-Donia }, {New types of generalized closed sets in bitopological spaces }, {\bf BJournal of the Egyptian Mathematical Society}, (2013) 21, 318 - 323. \bibitem{D} { C.Dorcett} , {\bf Kyungpook Math. 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Khalaf$^2$ \\ \noindent $^1$Department of Mathematics, PRIST University , Thanjavur , India \noindent Email :padmaprithivirajan@gmail.com \noindent $^2$Department of Mathematics, College of Science, University of Duhok, Kurdistan Region, Iraq. \noindent Email :aliasbkhalaf@uod.ac \end{document}