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\pagestyle{myheadings} \markboth{\small{ P. Padma and Alias B. Khalaf}} {\small { Pairwise Q*s-(regular and normal) spaces in
bitopological spaces}}
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%----------------------- 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
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%----------------------- Address of Author(s) ------------------
\medskip
\noindent
P.Padma$^1$ and Alias B. 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}
