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\begin{document}

\title{On characterizations of new separation axioms and topological properties}


% Please mark \corrauth after the name of the corresponding author.
\author[S. Pious Missier, M. J. Jeyanthi, Adem K{\i}l{\i}\c cman]{S. Pious Missier\affil{1}, M. J. Jeyanthi\affil{2}, Adem K{\i}l{\i}\c cman\affil{3}\comma\corrauth}

\address{\affilnum{1}\ Department of Mathematics, V. O. Chidambaram College, Thoothukudi-628 008 (T.N.), India\\
\affilnum{2}\  Aditanar College of Arts and Science, Tiruchendur(T.N.), India\\
\affilnum{3}\ Institute of Mathematical Research
(INSPEM) and Department of Mathematics, University Putra Malaysia, 43400 UPM, Serdang, Selangor, Malaysia}

\emails{{\tt spmissier@yahoo.com}\ (S. Pious Missier), {\tt jeyanthimanickaraj@gmail.com}\ (M. J. Jeyanthi), {\tt akilicman@putra.upm.edu.my}\ (Adem K{\i}l{\i}\c cman)}

\begin{abstract}
In this paper, we introduce new separations axioms $\Lambda_{r}$-$R_{0}$, $\Lambda_{r}$-$R_{1}$ and $\Lambda_{r}$-$D_{k}$,
and study their properties.
\end{abstract}

\keywords{$\Lambda_{r}$-$R_{0}$, $\Lambda_{r}$-$R_{1}$ and
$\Lambda_{r}$-$D_{k}$, $k = 0,1,2$}
\ams{AMS 54D10, 54D15}

\maketitle

\section{Introduction}
\noindent Caldas and Jafari \cite{[2]} introduced the notions of $\Lambda_{\delta}$-$R_{0}$ and
$\Lambda_{\delta}$-$R_{1}$ topological spaces. In this paper, we define $\Lambda_{r}$-open sets, that is, if (X,$\tau$) is a topological space and A$\subset$X. Then $\Lambda_{r}$-kernel of A is defined by
$ \Lambda_{r}$-ker(A) = $\cap$ $\lbrace$ G / G $\in$ $\Lambda_{r}$O(X,$\tau$) and A$\subset$G$\rbrace$.
Then we introduce some
$\Lambda_{r}$-separation axioms, we call these axioms as $\Lambda_{r}$-$R_{0}, \Lambda_{r}-R_{1}$ and study the properties of these axioms. We also define $\Lambda_{r}$-difference sets and utilize them
to define the $\Lambda_{r}$-$D_{k}$, k = 0,1,2 axioms.\\

\noindent Throughout the paper (X,$\tau$) (or simply X) will always denote a topological
space. Let (X,$\tau$) be a topological space and S $\subset$ X. Then S is called regularly-open
if S = Int(cl S). The complement $S^{c}$ (=X-S) of a regularly-open set S is called the
regularly-closed set. The family of all regularly-open sets(resp. regularly-closed sets) will
be denoted by RO(X,$\tau$) (resp.RC(X,$\tau$)). A subset S of a topological space
(X,$\tau$) is called $\Lambda_{r}$-set if S = $\Lambda_{r}$(S) where $\Lambda_{r}$(S)
= $\cap\lbrace$G / G $\in$ RO(X,$\tau$) and S $\subseteq$ G$\rbrace$. The collection of
all $\Lambda_{r}$-sets is denoted by $\Lambda_{r}$(X,$\tau$).\\

\noindent Throughout this paper, we let A be a subset of a space (X,$\tau$). Then A is called a $\Lambda_{r}$-closed
set if A = T$\cap$C where T is a $\Lambda_{r}$-set and C is a closed set.
The complement of a $\Lambda_{r}$-closed set is called $\Lambda_{r}$-open.
The collection of all $\Lambda_{r}$-open sets is denoted by $\Lambda_{r}$O(X,$\tau$).
The collection of all $\Lambda_{r}$-closed sets is denoted by $\Lambda_{r}$C(X,$\tau$).
A point x $\in$X is called a $\Lambda_{r}$-cluster point of A if for every
$\Lambda_{r}$-open set U containing x, A$\cap$U$\neq\emptyset$.
The set of all $\Lambda_{r}$-cluster points of A is called the $\Lambda_{r}$-closure of A
and is denoted by $\Lambda_{r}$-cl(A).

\section{$\Lambda_{r}$-$R_{0}$ Spaces}

\begin{definition}\label{3.1}
 The topological space (X,$\tau$) is said to be $\Lambda_{r}$-$R_{0}$ if for each $\Lambda_{r}$-open set G, x $\in$ G $\Rightarrow$
 $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ G.
\end{definition}

\begin{theorem}\label{3.3}
For a topological space (X,$\tau$), the following statements are equivalent:
\begin{itemize}
\item[(1)] (X,$\tau$) is $\Lambda_{r}$-$R_{0}$
\item[(2)] For any $\Lambda_{r}$-closed set F and a point x $\notin$ F,
$\exists$ U $\in$ $\Lambda_{r}$O(X,$\tau$) such that x $\notin$ U and F $\subseteq$ U
\item[(3)] For any $\Lambda_{r}$-closed set F and a point x $\notin$ F,
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ F = $\emptyset$
\end{itemize}

\noindent \begin{proof}
\noindent (1)$\rightarrow$ (2) Let F be a $\Lambda_{r}$-closed set and x $\notin$ F. Then $F^{c}$ is $\Lambda_{r}$-open
and x $\in$ $F^{c}$. Since X is $\Lambda_{r}$-$R_{0}$, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$
$F^{c}$ and hence F $\subseteq$ X-($\Lambda_{r}$-cl($\lbrace$x$\rbrace$)). Thus X-($\Lambda_{r}$-cl($\lbrace$x$\rbrace$))
is a $\Lambda_{r}$-open set containing F and x $\notin$ X-($\Lambda_{r}$-cl($\lbrace$x$\rbrace$)).\\

\noindent (2)$\rightarrow$ (3) Let F be a $\Lambda_{r}$-closed set and x $\notin$ F. Then $\exists$
U $\in$ $\Lambda_{r}$O(X,$\tau$) such that x $\notin$ U and F $\subseteq$ U. Claim:
U $\cap$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) = $\emptyset$. For, if U $\cap$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\neq$ $\emptyset$, then $\exists$ a point y in X such that y $\in$ U and
y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$). That implies
y is a $\Lambda_{r}$-cluster point of $\lbrace$x$\rbrace$. That implies for every $\Lambda_{r}$-open set G
containing y, G $\cap$ $\lbrace$x$\rbrace$ $\neq$ $\emptyset$. That is, x $\in$ G. Here U is a $\Lambda_{r}$-open
set containing y. Hence x $\in$ U, which is a contradiction. Therefore U $\cap$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
= $\emptyset$ and hence F $\cap$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) = $\emptyset$.\\

\noindent (3)$\rightarrow$ (1) Let G be a $\Lambda_{r}$-open set and x $\in$ G. Then $G^{c}$ is $\Lambda_{r}$-closed
and \\x $\notin$ $G^{c}$. By (3), $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $G^{c}$
 = $\emptyset$ and hence $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ G. Therefore (X,$\tau$) is
 $\Lambda_{r}$-$R_{0}$.
\end{proof}
\end{theorem}

\begin{theorem}\label{3.4}
 A space (X,$\tau$) is $\Lambda_{r}$-$R_{0}$ iff for each pair x,y of distinct points in X,
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) = $\emptyset$
or $\lbrace$x,y$\rbrace$ $\subseteq$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$). \\

\noindent \begin{proof}
Let (X,$\tau$) be a $\Lambda_{r}$-$R_{0}$ space. Let x,y $\in$ X such that x $\neq$ y. Then we have \\

\noindent Case(i): Suppose $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$)
$\neq$ $\emptyset$.If $\lbrace$x,y$\rbrace$ is not a subset of $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) and x $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$),
then x $\in$ X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$)) and X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$))
is $\Lambda_{r}$-open. Since (X,$\tau$) is $\Lambda_{r}$-$R_{0}$, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$)). Therefore $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) = $\emptyset$. This is a contradiction. Hence
$\lbrace$x,y$\rbrace$ $\subseteq$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$).\\

\noindent Case(ii): Suppose $\lbrace$x,y$\rbrace$ is not a subset of
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) and let
x $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Then x $\in$ X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$))
and X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$)) is $\Lambda_{r}$-open. Since (X,$\tau$) is $\Lambda_{r}$-$R_{0}$,
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ X-($\Lambda_{r}$-cl($\lbrace$y$\rbrace$)) and hence
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) = $\emptyset$.
\\

\noindent Conversely, let U be a $\Lambda_{r}$-open set and x $\in$ U. Suppose $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
is not a subset of U. Then $\exists$ a point y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) such that
y $\notin$ U. That implies y $\in$ X-U and X-U is $\Lambda_{r}$-closed. Since $\Lambda_{r}$-cl($\lbrace$y$\rbrace$)
is the smallest $\Lambda_{r}$-closed set containing y, $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$
X-U and hence $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\cap$ U = $\emptyset$. Since x $\in$ U,
x $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) and hence $\lbrace$x,y$\rbrace$ is not a subset of
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Also y $\in$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). That implies
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\neq$ $\emptyset$.
So by this contradiction, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ U and hence
(X,$\tau$) is $\Lambda_{r}$-$R_{0}$.
\end{proof}
\end{theorem}

\begin{theorem}\label{3.5}
For any points x and y in a topological space (X,$\tau$), the following are equivalent:
\begin{itemize}
\item[(1)] $\Lambda_{r}$-ker($\lbrace$x$\rbrace$) $\neq$ $\Lambda_{r}$-ker($\lbrace$y$\rbrace$)
\item[(2)] $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$)
\end{itemize}
\begin{proof}

\noindent (1)$\rightarrow$(2) Suppose $\Lambda_{r}$-ker($\lbrace$x$\rbrace$) $\neq$ $\Lambda_{r}$-ker($\lbrace$y$\rbrace$).
Then $\exists$ a point z in X such that z $\in$ $\Lambda_{r}$-ker($\lbrace$x$\rbrace$) and
z $\notin$ $\Lambda_{r}$-ker($\lbrace$y$\rbrace$) $\Rightarrow$ $\Lambda_{r}$-cl($\lbrace$z$\rbrace$)
$\cap$ $\lbrace$x$\rbrace$ $\neq$ $\emptyset$ and $\Lambda_{r}$-cl($\lbrace$z$\rbrace$) $\cap$
$\lbrace$y$\rbrace$ = $\emptyset$ $\Rightarrow$ x $\in$ $\Lambda_{r}$-cl($\lbrace$z$\rbrace$) and
y $\notin$ $\Lambda_{r}$-cl($\lbrace$z$\rbrace$) $\Rightarrow$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ $\Lambda_{r}$-cl($\lbrace$z$\rbrace$) and y $\notin$ $\Lambda_{r}$-cl($\lbrace$z$\rbrace$)
$\Rightarrow$ y $\notin$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\Rightarrow$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$).\\

\noindent (2)$\rightarrow$ (1) Suppose $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Then $\exists$ a point z in X such that z $\in$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) and z $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). That implies
$\exists$ a $\Lambda_{r}$-open set V containing z such that x $\in$ V and y $\notin$ V. That is, V is a
$\Lambda_{r}$-open set containing x but not y. If y $\in$ $\Lambda_{r}$-ker($\lbrace$x$\rbrace$), then
x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). That implies for every $\Lambda_{r}$-open set G containing x,
G $\cap$ $\lbrace$y$\rbrace$ $\neq$ $\emptyset$. That is, y $\in$ G. By this contradiction, y $\notin$
$\Lambda_{r}$-ker($\lbrace$x$\rbrace$) and hence $\Lambda_{r}$-ker($\lbrace$x$\rbrace$)
$\neq$ $\Lambda_{r}$-ker($\lbrace$y$\rbrace$).
\end{proof}
\end{theorem}

\begin{theorem}\label{3.6}
For a topological space (X,$\tau$), the following are equivalent:
\begin{itemize}
\item[(1)] (X,$\tau$) is $\Lambda_{r}$-$R_{0}$
\item[(2)] For any non-empty set A and G $\in$ $\Lambda_{r}$O(X,$\tau$) such that A $\cap$ G $\neq$
$\emptyset$,\\ $\exists$ F $\in$ $\Lambda_{r}$C(X,$\tau$) such that A $\cap$ F $\neq$ $\emptyset$ and F $\subseteq$ G
\item[(3)] For any G $\in$ $\Lambda_{r}$O(X,$\tau$), G = $\cup\lbrace$F/F $\in$ $\Lambda_{r}$C(X,$\tau$)
and F $\subseteq$ G$\rbrace$
\item[(4)] For any F $\in$ $\Lambda_{r}$C(X,$\tau$), F = $\cap\lbrace$G/G $\in$ $\Lambda_{r}$O(X,$\tau$)
and F $\subseteq$ G$\rbrace$
\item[(5)] For any x $\in$ X, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ $\Lambda_{r}$-ker($\lbrace$x$\rbrace$)
\item[(6)] For any x,y $\in$ X, y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\Leftrightarrow$
x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$)
\end{itemize}
\begin{proof}

\noindent (1)$\rightarrow$(2) Let A be any nonempty subset of X and G be a $\Lambda_{r}$-open set such that
A $\cap$ G $\neq$ $\emptyset$. Let x $\in$ A $\cap$ G. Since (X,$\tau$) is $\Lambda_{r}$-$R_{0}$,
x $\in$ G $\Rightarrow$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ G. Since x $\in$ A,
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ A $\neq$ $\emptyset$. Thus $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
is a $\Lambda_{r}$-closed set contained in G and A $\cap$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$
$\emptyset$.\\

\noindent (2)$\rightarrow$(3) Let G $\in$ $\Lambda_{r}$O(X,$\tau$) and x $\in$ G. Then by (2), $\exists$
F $\in$ $\Lambda_{r}$C(X,$\tau$) such that $\lbrace$x$\rbrace$ $\cap$ F $\neq$ $\emptyset$
and F $\subseteq$ G. That implies x $\in$ F where F $\in$ $\Lambda_{r}$C(X,$\tau$) and F $\subseteq$ G
and hence x $\in$ $\cup$ $\lbrace$F/F $\in$ $\Lambda_{r}$C(X,$\tau$) and F $\subseteq$ G$\rbrace$.
Therefore G $\subseteq$ $\cup$ $\lbrace$F/F $\in$ $\Lambda_{r}$C(X,$\tau$) and F $\subseteq$ G$\rbrace$.
Also $\cup$ $\lbrace$F/F $\in$ $\Lambda_{r}$C(X,$\tau$) and F $\subseteq$ G$\rbrace$ $\subseteq$ G.
Hence G = $\cup$ $\lbrace$F/F $\in$ $\Lambda_{r}$C(X,$\tau$) and F $\subseteq$ G$\rbrace$.\\

\noindent (3)$\rightarrow$(4) Let F $\in$ $\Lambda_{r}$C(X,$\tau$). Then $F^{c}$ $\in$ $\Lambda_{r}$O(X,$\tau$).
By (3), $F^{c}$ =  $\cup$ $\lbrace$ $G^{c}$ / $G^{c}$ $\in$ $\Lambda_{r}$C(X,$\tau$) and $G^{c}$
$\subseteq$ $F^{c}$ $\rbrace$. That implies F =  $\cap$ $\lbrace$ G / G $\in$ $\Lambda_{r}$O(X,$\tau$)
 and F $\subseteq$ G $\rbrace$.\\

\noindent (4)$\rightarrow$(5) Let y $\notin$ $\Lambda_{r}$-ker($\lbrace$x$\rbrace$). Then
x $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). That implies $\exists$ a $\Lambda_{r}$-open set V
containing x such that V $\cap$ $\lbrace$y$\rbrace$ = $\emptyset$ and hence $\Lambda_{r}$-cl($\lbrace$y$\rbrace$)
$\cap$ V = $\emptyset$. By (4), $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) =
  $\cap$ $\lbrace$ G / G $\in$ $\Lambda_{r}$O(X,$\tau$)
 and $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ G$\rbrace$. Since x $\in$ V, x $\notin$
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$) and hence $\exists$ G $\in$ $\Lambda_{r}$O(X,$\tau$)
such that $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ G and x $\notin$ G. Therefore
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ G = $\emptyset$ and hence y $\notin$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$). Therefore $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
 $\subseteq$ $\Lambda_{r}$-ker($\lbrace$x$\rbrace$).\\

\noindent (5)$\rightarrow$(6) If y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$), then y $\in$
$\Lambda_{r}$-ker($\lbrace$x$\rbrace$) by (5). That implies x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$).
 Similarly, if x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$), then by (5), x $\in$
 $\Lambda_{r}$-ker($\lbrace$y$\rbrace$) and hence y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$).
Thus x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\Leftrightarrow$ y $\in$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$).\\

\noindent (6) $\rightarrow$ (1) Let G $\in$ $\Lambda_{r}$O(X,$\tau$) and x $\in$ G. If y $\notin$ G, then
y $\in$ X-G and hence $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ X-G since
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$) is the smallest $\Lambda_{r}$-closed set containing y.
Therefore $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\cap$ G = $\emptyset$ and hence
x $\notin$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). By (6), y $\notin$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$). $\therefore$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ G and hence (X,$\tau$) is $\Lambda_{r}$-$R_{0}$.
\end{proof}
\end{theorem}

\begin{corollary}\label{3.7}
For a topological space (X,$\tau$), the following properties are equivalent:
\begin{itemize}
\item[(1)] (X,$\tau$) is $\Lambda_{r}$-$R_{0}$
\item[(2)] For any x $\in$ X, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) =
$\Lambda_{r}$-ker($\lbrace$x$\rbrace$)
\end{itemize}
\end{corollary}

\begin{theorem}\label{3.8}
For a topological space (X,$\tau$), the following properties are equivalent:
\begin{itemize}
\item[(1)] (X,$\tau$) is $\Lambda_{r}$-$R_{0}$
\item[(2)] If F is $\Lambda_{r}$-closed, then F = $\Lambda_{r}$-ker(F)
\item[(3)] If F is $\Lambda_{r}$-closed and x $\in$ F, then $\Lambda_{r}$-ker($\lbrace$x$\rbrace$)
$\subseteq$ F
\item[(4)] If x $\in$ X, then $\Lambda_{r}$-ker($\lbrace$x$\rbrace$) $\subseteq$
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
\end{itemize}
\begin{proof}
\noindent (1) $\rightarrow$ (2) Let F be a $\Lambda_{r}$-closed set and x $\notin$ F. Then X-F is
$\Lambda_{r}$-open and x $\in$ X-F. By (1), $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ X-F and hence $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\cap$ F = $\emptyset$.
That implies x $\notin$ $\Lambda_{r}$-ker(F). Therefore $\Lambda_{r}$-ker(F)
$\subseteq$ F. Also F $\subset$ $\Lambda_{r}$-ker(F). Hence F = $\Lambda_{r}$-ker(F).\\

\noindent (2) $\rightarrow$ (3) Let F be a $\Lambda_{r}$-closed set and x $\in$ F. Then
$\Lambda_{r}$-ker($\lbrace$x$\rbrace$) $\subseteq$ $\Lambda_{r}$-ker(F) = F.\\

\noindent (3) $\rightarrow$ (4) Since x $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) and
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) is $\Lambda_{r}$-closed, by (3) \\$\Lambda_{r}$-ker($\lbrace$x$\rbrace$)
$\subseteq$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$).\\

\noindent (4) $\rightarrow$ (1) Since x $\in$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\Leftrightarrow$
y $\in$ $\Lambda_{r}$-cl($\lbrace$x$\rbrace$), (X,$\tau$) is $\Lambda_{r}$-$R_{0}$.
\end{proof}
\end{theorem}

\noindent The following Examples \ref{ex1} and \ref{ex2} show that $\Lambda_{r}$-$T_{0}$ and $\Lambda_{r}$-$R_{0}$ are independent.
\begin{example} \label{ex1}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace b,c \rbrace\rbrace$. Then \\ $\Lambda_{r} O(X,\tau) = \lbrace X,\emptyset,\lbrace a \rbrace,
\lbrace b,c \rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$R_{0}$ but it is not
$\Lambda_{r}$-$T_{0}$.\end{example}

\begin{example} \label{ex2}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$a,b$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$a,b$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$T_{0}$ but it is not
$\Lambda_{r}$-$R_{0}$.\end{example}



\section{$\Lambda_{r}$-$R_{1}$ Spaces}

\begin{definition}\label{4.1}
A space (X,$\tau$) is $\Lambda_{r}$-$R_{1}$ if for each x,y $\in$ X with \\
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$),
$\exists$ $\Lambda_{r}$-open sets U and V such that $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ U, \\$\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ V and U $\cap$ V = $\emptyset$.
\end{definition}

\begin{proposition}\label{4.2}
If (X,$\tau$) is $\Lambda_{r}$-$R_{1}$, then (X,$\tau$) is $\Lambda_{r}$-$R_{0}$.\\

\noindent \begin{proof}
Let (X,$\tau$) be $\Lambda_{r}$-$R_{1}$. Let U be $\Lambda_{r}$-open in X and x $\in$ U.
For each y $\in$ X-U, $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$).
Then $\exists$ disjoint $\Lambda_{r}$-open sets $U_{y}$ and $V_{y}$ such that $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ $U_{y}$ and $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ $V_{y}$. Take
V = $\cup$ $\lbrace$ $V_{y}$ / y $\in$ X-U $\rbrace$. Then V is $\Lambda_{r}$-open, X-U $\subseteq$ V and
x $\notin$ V. Therefore $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ X-V $\subseteq$ U and hence
(X,$\tau$) is $\Lambda_{r}$-$R_{0}$.
\end{proof}
\end{proposition}

\begin{theorem}\label{4.3}
If (X,$\tau$) is $\Lambda_{r}$-$T_{2}$, then (X,$\tau$) is $\Lambda_{r}$-$R_{1}$.\\

\noindent \begin{proof}
Let x,y $\in$ X such that x $\neq$ y and $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Since (X,$\tau$) is $\Lambda_{r}$-$T_{2}$, $\exists$
$\Lambda_{r}$-open sets U and V such that x $\in$ U, y $\in$ V and U $\cap$ V = $\emptyset$.
That is, $\lbrace$x$\rbrace$ $\subseteq$ U and $\lbrace$y$\rbrace$ $\subseteq$ V. Since
(X,$\tau$) is $\Lambda_{r}$-$T_{2}$, it is $\Lambda_{r}$-$T_{1}$. Therefore for every x $\in$ X,
$\lbrace$x$\rbrace$ = $\Lambda_{r}$-cl($\lbrace$x$\rbrace$). Thus $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\subseteq$ U , $\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ V and U $\cap$ V = $\emptyset$.
Hence (X,$\tau$) is $\Lambda_{r}$-$R_{1}$.
\end{proof}
\end{theorem}

\begin{remark}\label{4.4}
The converse of the above theorem need not be true. For example, let X = $\lbrace$a,b,c,d$\rbrace$
and $\tau$ = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,$\lbrace$b,c$\rbrace$,
$\lbrace$a,b,c$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) =
$\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,$\lbrace$b,c$\rbrace$,
$\lbrace$a,b,c$\rbrace$
$\lbrace$b,c,d$\rbrace$
$\lbrace$a,d$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$R_{1}$ but not
$\Lambda_{r}$-$T_{2}$.
\end{remark}

\noindent Note that  $\Lambda_{r}$-$T_{0}$ and $\Lambda_{r}$-$R_{1}$ are independent as in the following examples \ref{ex3} and \ref{ex4}.

\begin{example} \label{ex3}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace$,
$\lbrace$b,c$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,
$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace$,
$\lbrace$b,c$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$T_{0}$ but it is not
\\$\Lambda_{r}$-$R_{1}$.\end{example}

\begin{example} \label{ex4}
Let X = $\lbrace$a,b,c,d$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$a,d$\rbrace$,$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace$,$\lbrace$b,c,d$\rbrace\rbrace$.
Here (X,$\tau$) is $\Lambda_{r}$-$R_{1}$ but it is not
$\Lambda_{r}$-$T_{0}$.\end{example}

\begin{theorem}\label{4.6}
For a space (X,$\tau$), the following statements are equivalent:
\begin{itemize}
\item[(1)] (X,$\tau$) is $\Lambda_{r}$-$R_{1}$
\item[(2)] If x,y $\in$ X such that $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$), then $\exists$ $\Lambda_{r}$-closed sets $F_{1}$ and
$F_{2}$ such that x $\in$ $F_{1}$, y $\notin$ $F_{1}$, x $\notin$ $F_{2}$, y $\in$ $F_{2}$
and X = $F_{1}$ $\cup$ $F_{2}$.
\end{itemize}

\noindent \begin{proof}
\noindent (1)$\rightarrow$(2) Let x,y $\in$ X such that $\Lambda_{r}$-cl($\lbrace$x$\rbrace$)
$\neq$ $\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Then by (1), $\exists$ disjoint $\Lambda_{r}$-open
sets U and V such that $\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\subseteq$ U and
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$) $\subseteq$ V. Take $F_{1}$ = X-V and $F_{2}$ = X-U.
Then $F_{1}$ and $F_{2}$ are $\Lambda_{r}$-closed sets such that x $\in$ $F_{1}$, y $\notin$ $F_{1}$,
x $\notin$ $F_{2}$, y $\in$ $F_{2}$ and X = $F_{1}$ $\cup$ $F_{2}$.\\

\noindent (2)$\rightarrow$(1) Let x,y $\in$ X such that
$\Lambda_{r}$-cl($\lbrace$x$\rbrace$) $\neq$
$\Lambda_{r}$-cl($\lbrace$y$\rbrace$). Then by (2) $\exists$ $\Lambda_{r}$-closed sets $F_{1}$ and
$F_{2}$ such that x $\in$ $F_{1}$, y $\notin$ $F_{1}$, x $\notin$ $F_{2}$, y $\in$ $F_{2}$
and X = $F_{1}$ $\cup$ $F_{2}$. Take U = X-$F_{2}$ and V = X-$F_{1}$. Then U and V are
$\Lambda_{r}$-open sets, x $\in$ U, y $\in$ V and U $\cap$ V = $\emptyset$. Therefore (X,$\tau$)
is $\Lambda_{r}$-$T_{2}$ and hence (X,$\tau$) is $\Lambda_{r}$-$R_{1}$.
\end{proof}
\end{theorem}

\section{$\Lambda_{r}$-$D_{k}$ Spaces}

\begin{definition}\label{5.1}
Let (X,$\tau$) be a topological space and A be a subset of X. Then A is called $\Lambda_{r}$-difference set
(shortly $\Lambda_{r}$-D set) if $\exists$ U,V $\in$ $\Lambda_{r}$O(X,$\tau$) such that U $\neq$ X and
A = U-V. The collection of all $\Lambda_{r}$-difference sets of (X,$\tau$) is denoted by $\Lambda_{r}$D(X,$\tau$).\end{definition}

\begin{remark}\label{5.2}
Every $\Lambda_{r}$-open set A different from X is a $\Lambda_{r}$-D set if U = A and V = $\emptyset$.
But the converse need not be true. For example, let
X = $\lbrace$a,b,c,d$\rbrace$
and $\tau$ = $\lbrace$X,$\emptyset$,$\lbrace$b,d$\rbrace$,$\lbrace$b,c,d$\rbrace$,
$\lbrace$a,b,d$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) =
$\lbrace$X,$\emptyset$,$\lbrace$b,d$\rbrace$,$\lbrace$b,c,d$\rbrace$,
$\lbrace$a,b,d$\rbrace\rbrace$ and
$\Lambda_{r}$D(X,$\tau$) =
$\lbrace$ $\emptyset$,$\lbrace$b,d$\rbrace$,$\lbrace$b,c,d$\rbrace$,
$\lbrace$a,b,d$\rbrace$,$\lbrace$c$\rbrace$,$\lbrace$a$\rbrace\rbrace$.
Here $\lbrace$a$\rbrace$ is a $\Lambda_{r}$-D set but not $\Lambda_{r}$-open set.
\end{remark}

\begin{definition}\label{5.3}
A space (X,$\tau$) is called
\begin{itemize}
\item[(1)]$\Lambda_{r}$-$D_{0}$ if for x,y $\in$ X, x $\neq$ y, $\exists$ a $\Lambda_{r}$-D set
containing one of x \\and y but not the other
\item[(2)]$\Lambda_{r}$-$D_{1}$ if for x,y $\in$ X, x $\neq$ y, $\exists$  $\Lambda_{r}$-D sets
U and V in X such that \\x $\in$ U, y $\notin$ U and y $\in$ V, x $\notin$ V
\item[(3)]$\Lambda_{r}$-$D_{2}$ if for x,y $\in$ X, x $\neq$ y, $\exists$  $\Lambda_{r}$-D sets
U and V in X such that \\x $\in$ U, y $\in$ V and U$\cap$V = $\emptyset$
\end{itemize}
\end{definition}

\begin{theorem}\label{5.4}
A space (X,$\tau$) is $\Lambda_{r}$-$D_{0}$ iff it is  $\Lambda_{r}$-$T_{0}$.\\
\begin{proof}
Suppose (X,$\tau$) is $\Lambda_{r}$-$D_{0}$. Let x,y $\in$ X such that x $\neq$ y. Then
$\exists$ an $\Lambda_{r}$-D set A containing one of x and y but not the other, say
x $\in$ A but y $\notin$ A. Since A is a $\Lambda_{r}$-D set, A = U-V where U $\neq$ X
and U,V $\in$ $\Lambda_{r}$O(X,$\tau$). Since x $\in$ A, x $\in$ U and x $\notin$ V.
For y $\notin$ A, we have two cases (a) y $\notin$ U \\(b) y $\in$ U and y $\in$ V.
In case (a), x $\in$ U but y $\notin$ U. In case (b), y $\in$ V but x $\notin$ V. Hence
(X,$\tau$) is  $\Lambda_{r}$-$T_{0}$. Conversely, suppose (X,$\tau$) is $\Lambda_{r}$-$T_{0}$.
Let x,y $\in$ X such that x $\neq$ y.
Then $\exists$ an $\Lambda_{r}$-open set U containing one of x and y but not the other, say
x $\in$ U but y $\notin$ U. Then U $\neq$ X and hence U is a $\Lambda_{r}$-D set. Therefore
U is a $\Lambda_{r}$-D set containing x but not y. Hence (X,$\tau$) is $\Lambda_{r}$-$D_{0}$.
\end{proof}
\end{theorem}

\noindent The following examples \ref{ex5} and \ref{ex6} show that $\Lambda_{r}$-$R_{0}$ and $\Lambda_{r}$-$D_{0}$ are independent.
\begin{example} \label{ex5}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$a,c$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$a,c$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$R_{0}$ but it is not
$\Lambda_{r}$-$D_{0}$.\end{example}

\begin{example} \label{ex6}
Let X = $\lbrace$a,b$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,
$\lbrace$a$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,
$\lbrace$a$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$D_{0}$ but it is not
$\Lambda_{r}$-$R_{0}$.\end{example}

\begin{remark}\label{5.6}
Examples \ref{ex7} and \ref{ex8} below show that $\Lambda_{r}$-$R_{1}$ and $\Lambda_{r}$-$D_{0}$ are independent.\\
\begin{example} \label{ex7} Let X = $\lbrace$a,b,c,d$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$b,c,d$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$b,c,d$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$R_{1}$ but it is not
$\Lambda_{r}$-$D_{0}$.\end{example}

\begin{example} \label{ex8}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$a,b$\rbrace\rbrace$. Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$a,b$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$D_{0}$ but it is not
$\Lambda_{r}$-$R_{1}$.\end{example}
\end{remark}

\begin{theorem}\label{5.7}
A space (X,$\tau$) is $\Lambda_{r}$-$D_{1}$ iff it is $\Lambda_{r}$-$D_{2}$.\\

\noindent \begin{proof}
Suppose (X,$\tau$) is $\Lambda_{r}$-$D_{1}$. Then for each pair of distinct points x,y $\in$ X,
we have $\Lambda_{r}$-D sets A and B such that x $\in$ A, y $\notin$ A and y $\in$ B, x $\notin$ B.
Let A = $U_{1}$-$V_{1}$ and B = $U_{2}$-$V_{2}$. Then  $U_{1}$,$V_{1}$,$U_{2}$ and $V_{2}$ are
$\Lambda_{r}$-open sets, $U_{1}$ $\neq$ X and $U_{2}$ $\neq$ X. For x $\notin$ B, we have two
cases (i) x $\notin$ $U_{2}$ (ii) x $\in$ $U_{2}$ and x $\in$ $V_{2}$.\\

\noindent Case(i) x $\notin$ $U_{2}$ then since y $\notin$ A, either y $\notin$ $U_{1}$ or ( y $\in$ $U_{1}$ and y $\in$ $V_{1}$ ).\\

\noindent If y $\notin$ $U_{1}$, from y $\in$ B = $U_{2}$-$V_{2}$, it follows that
y $\in$ $U_{2}$-($V_{2}$ $\cup$ $U_{1}$).\\ From x $\in$ A = $U_{1}$-$V_{1}$ and x $\notin$ $U_{2}$,
x $\in$ $U_{1}$-($V_{1}$ $\cup$ $U_{2}$). \\Also ($U_{2}$-($V_{2}$ $\cup$ $U_{1}$)) $\cap$
($U_{1}$-($V_{1}$ $\cup$ $U_{2}$)) = $\emptyset$.\\If y $\in$ $U_{1}$ and y $\in$ $V_{1}$,
then x $\in$ $U_{1}$-$V_{1}$. That implies ($U_{1}$-$V_{1}$) $\cap$ $V_{1}$ = $\emptyset$.\\

\noindent Case(ii) x $\in$ $U_{2}$ and x $\in$ $V_{2}$ then we have y $\in$ B = $U_{2}$-$V_{2}$, x $\in$ $V_{2}$
and ($U_{2}$-$V_{2}$) $\cap$ $V_{2}$ = $\emptyset$. \\Hence (X,$\tau$) is $\Lambda_{r}$-$D_{2}$.\\

\noindent Conversely, suppose (X,$\tau$) is $\Lambda_{r}$-$D_{2}$.
Let x,y $\in$ X such that x $\neq$ y. Then
$\exists$ $\Lambda_{r}$-D sets A and B such that x $\in$ A, y $\in$ B and A $\cap$ B = $\emptyset$.
Therefore x $\in$ A, y $\notin$ A and y $\in$ B, x $\notin$ B. Hence (X,$\tau$) is
$\Lambda_{r}$-$D_{1}$.
\end{proof}
\end{theorem}

\begin{corollary}\label{5.8}
If (X,$\tau$) is $\Lambda_{r}$-$D_{1}$, then it is $\Lambda_{r}$-$T_{0}$.
\end{corollary}

\begin{remark}\label{5.9}
The converse of the above corollary need not be true. For example, let
X = $\lbrace$a,b,c$\rbrace$
and $\tau$ = $\lbrace$X,$\emptyset$,$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) =  $\lbrace$X,$\emptyset$,$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace\rbrace$
and $\Lambda_{r}$D(X,$\tau$) =  $\lbrace\emptyset$,$\lbrace$a$\rbrace$,$\lbrace$c$\rbrace$,
$\lbrace$a,c$\rbrace\rbrace$. Here (X,$\tau$) is $\Lambda_{r}$-$T_{0}$ but not $\Lambda_{r}$-$D_{1}$.
\end{remark}

\begin{remark}\label{5.10}
Examples \ref{ex9} and \ref{ex10} below show that $\Lambda_{r}$-$R_{0}$ and $\Lambda_{r}$-$D_{2}$ are\\independent.\\
\begin{example} \label{ex9}
Let X = $\lbrace$a,b,c,d$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$a,d$\rbrace$,$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace$,$\lbrace$b,c,d$\rbrace\rbrace$.
Here (X,$\tau$) is $\Lambda_{r}$-$R_{0}$ but it is not
$\Lambda_{r}$-$D_{2}$.\end{example}

\begin{example} \label{ex10}
Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$c$\rbrace$,$\lbrace$b,c$\rbrace$,$\lbrace$a,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$b$\rbrace$,
$\lbrace$c$\rbrace$,$\lbrace$b,c$\rbrace$,$\lbrace$a,c$\rbrace\rbrace$.
Here (X,$\tau$) is $\Lambda_{r}$-$D_{2}$ but it is not
$\Lambda_{r}$-$R_{1}$. \end{example}
\end{remark}

\noindent Similar to the previous cases the examples \ref{ex11} and \ref{ex12} show that $\Lambda_{r}$-$R_{1}$ and $\Lambda_{r}$-$D_{1}$ are independent.

\begin{example} \label{ex11}
Let X = $\lbrace$a,b,c,d$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$a,d$\rbrace$,$\lbrace$b,c$\rbrace$,$\lbrace$a,b,c$\rbrace$,$\lbrace$b,c,d$\rbrace\rbrace$.
Here (X,$\tau$) is $\Lambda_{r}$-$R_{1}$ but it is not
$\Lambda_{r}$-$D_{1}$.\end{example}
\begin{example} \label{ex12} Let X = $\lbrace$a,b,c$\rbrace$ and $\tau$ = $\lbrace$ X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace$,$\lbrace$b,c$\rbrace\rbrace$.
Then $\Lambda_{r}$O(X,$\tau$) = $\lbrace$X,$\emptyset$,$\lbrace$a$\rbrace$,
$\lbrace$c$\rbrace$,$\lbrace$a,c$\rbrace$,$\lbrace$b,c$\rbrace\rbrace$.
Here (X,$\tau$) is $\Lambda_{r}$-$D_{1}$ but it is not
$\Lambda_{r}$-$R_{1}$.\end{example}

\section{References}
\bibliography{EJPAM_Sample}
\bibliographystyle{plain}
\begin{itemize}
\bibitem{[1]} S. Balasubramanian and P. Aruna, $\upsilon$D-sets and separation axioms,
Int. Journal of Math. Analysis,Vol.4, 2010,no.19, 909--919.
\bibitem{[2]} M. Caldas and S. Jafari, On some low separation axioms in topological space,
Houston Journal of Math. 29(2003), 93--104.
\bibitem{[3]} K. K. Dube, A note on $R_{0}$ topological spaces, Mat.Vesnik 11(1974), 203--208.
\bibitem{[4]} K. K. Dube, A note on $R_{1}$ topological spaces, Period. Math. Hung. 13(1982), 267--271.
\bibitem{[5]} M. G. Murdeshwar and S. A. Naimpally, $R_{1}$-topological spaces, Canad. Math. Bull.
9(1966),5 21--523.
\bibitem{[6]} S. A. Naimpally, On $R_{0}$-topological spaces, Ann. Univ. Budapest Eotvos sect.Math.
10(1967), 53--54.


\end{itemize}









\end{document} 


\section{References}
The bibliography should be titled "References", with the section heading centered horizontally. It is preferred that bibliographic references be compiled with bibtex, using the "natbib" package with the "plain" style. Inline citations should be numeric references of the form [\#]; they can be done with \verb \cite{}, \verb \citep{}, etc\ldots. Regarding formatting the bibliographic entries, the primary requirement is that they be in alphabetic order by the first author's surname. In the following sample bibliography, we have:
\begin{itemize}
\item \cite{EngleGranger1987, WestLinster2003} are journal articles
\item this is a book: \citep{Holland1975}
\item \cite{Akaike1973} is an article in a conference proceedings
\item an article in a collection: \citep{Bauwens1993}
\item \cite[see also][]{Chen1976} as a technical report
\item \cite{Thomaz2004} is a PhD Thesis
\end{itemize}

\bibliography{EJPAM_Sample}
\bibliographystyle{plain}
\end{document}