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\title{ Nano topology induced by Lattices}
\author{*M. Lellis Thivagar and **V.Sutha Devi}
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\date{}
\maketitle
\begin{center}
*School of Mathematics, Madurai Kamaraj University,\\ %%@
Madurai -625021, Tamilnadu, India.\\
e-mail:mlthivagar@yahoo.co.in
\end{center}
\begin{center}
**School of Mathematics, Madurai Kamaraj University, \\%%@
Madurai -625021, Tamilnadu, India.\\
e-mail:vsdsutha@yahoo.co.in
\end{center}

\textbf{Abstract}\rm: 
 Lattice is a partially ordered set in which all finite subsets have a least upper bound and greatest lower bound. Dedekind worked on lattice theory in the 19th century.  Nano topology explored by Lellis Thivagar et.al. can be described as a collection of nano approximations, a non-empty finite universe and empty set for which equivalence classes are buliding blocks. This is named as Nano topology, because of its size and  what ever may be the size of universe it has atmost five elements in it. The elements of Nano topology are called the Nano open sets. This paper is to study the nano topology within the context of lattices.  In lattice, there is a special class of join-congruence relation which is defined with respect to an ideal. We have defined the nano approximations of a set with respect to an ideal of a lattice. Also some properties of the approximations of a set in a lattice with respect to ideals are studied.   On the other hand, the lower and upper approximations have also been studied within the context various algebraic structures.\\ 
\textbf{Keywords and phrases}: Nano topology, lattices, ideal of a lattice \\\\
\textbf{2010 AMS SUBJECT CLASSIFICATION: 54B05,54CO8}\\
\begin{center}
\section{INTRODUCTION}
\end{center}
\indent\indent Nano topology[5] explored by Thivagar et.al  can be described as a collection of nano approximations, a non-empty finite universe and empty set for which equivalence classes are buliding blocks.  On the other hand, the lower and upper approximations have also been studied within the context various algebraic structures. Lattice[1] is a partially ordered set in which all finite subsets have a least upper bound and greatest lower bound. Dedekind worked on lattice theory in the 19th century. The motivation of this paper is to discuss the algebraic properties of nano topology induced by ideals in lattices. In a lattice, there is a special class of join-congruence relation which is defined with respect to an ideal. \newpage
\section{Preliminaries}
The following recalls requisite ideas and preliminaries necessitated in the sequel of our work.
\begin{definition}\rm[3]: Let  $\mathcal{U}$  be a non-empty finite set of objects called the universe $\mathcal{R}$ be an equivalence relation on   $\mathcal{U}$ named as the indiscerniblity relation. Elements belonging to the same equivalence class are said to be indiscernible with one another. The pair $(\mathcal{U},R)$ is said to be the approximation space. Let $X\subseteq 
\mathcal{U}$.\\
(i) The Lower appproximation of X with respect to R is the set of all objects, which can be for certain classified as X with respect to R and it is denoted by $L_{R}(X)$. That is, $L_{R}(X)= \bigcup \limits_{x\in \mathcal{U}}\{R(x):R(x)\subseteq X\}$, where R(x) denotes the equivalence class\\ determined by x.\\
(ii)  The Upper approximation of X with respect to R is the set of all objects, which can be possibly classified as X with respect to R and it is denoted by\\ $U_{R}(X) = \bigcup\limits_{X\in\mathcal{U}}\{R(x):R(x)\cap X
\neq \emptyset\}$\\
(iii)  The Boundary region of X with respect to R is the set of all objects
which can be classified neither as X nor as not -X with respect to R and it is denoted by\\ $B_{R}(X)=U_{R}(X)-L_{R}(X)$ \\
\end{definition}
\begin{definition}\rm:[3] Let $\mathcal{U}$ be the universe,R be an equivalence relation on $\mathcal{U}$ and \\$\tau_{R}(X)=\left\{\mathcal{U},\emptyset,L_{R}(X),U_{R}(X),B_{R}(X) \right\}$
where $X\subseteq\mathcal{U}$ and $\tau_{R}(X)$satisfies the following axioms.\\
(i) $\mathcal{U}$  and  $\emptyset\in \tau_{R}(X)$\\
(ii) The union of elements of any subcollection $\tau_{R}(X)$  is in $\tau_{R}(X)$.\\
(iii) The intersection of the elements of any finite subcollection of $\tau_{R}(X)$ is in $\tau_{R}(X)$.That is $\tau_{R}(X)$ forms a topology $\mathcal{U}$ called as the nano topology on $\mathcal{U}$ with respect to X. $(\mathcal{U},\tau_{R}(X))$ as the nano topological space. The elements of $\tau_{R}(X)$ are called as nano open sets. A set A is said to be nano closed if its complement is nano open.\\
\end{definition}
\begin{definition}\rm:[1] A relation defined on a set which is reflexive, anti-symmetric and transitive is called a Partially ordering on S. A set S with a partial ordering on $\rho$ defined on it is called a poset denoted by (S,$\rho$)
\end{definition}
\begin{definition}\rm:[1] A Lattice is a poset in which any two elements have greatest lower bound and least upper bound.
\end{definition}
\begin{definition}\rm:[1] Let L be a lattice, a non-empty subset $\mathcal{I}$ of L is  called an ideal if 
\begin{itemize}
\item[(i)] a,b $\in \mathcal{I}$ $\Rightarrow $  $a \vee b \in \mathcal{I}$
\item[(ii)] a$\in L , b \in I$  and a$\leq b$  $\Rightarrow a \in \mathcal{I}$
\end{itemize}
\end{definition}
\begin{definition}\rm:[1] An equivalence relation $\theta$ on L is called a join-congruence and meet congruence if, for all $a,b,c,d\in L$, $a\cong b(mod\theta)$ and $c\cong d(mod\theta)$ imply $a \vee c \cong b \vee d (mod \theta)$ and $a   c \cong b d (mod \theta)$,respectively. $\theta$ is a congruence relation if it is both a join congruence and a meet congruence. 
\end{definition}
\begin{theorem}\rm:[1] Let $\mathcal{I}$ be an ideal of L and define a relation $\theta_{\mathcal{I}}$ on L by for all $a,b,c \in L$, a $\cong b(mod \theta_{\mathcal{I}})\Leftrightarrow $ there exist d $\in \mathcal{I}$ such that $a \vee d = b\vee d$. Then the following statements hold:
\begin{itemize}
\item[(i)]: $\theta_{\mathcal{I}}$ is a join-congruence on L
\item[(ii)]: $\theta_{\mathcal{I}}$ is a congruence on L iff L is distributive.
\end{itemize}
\end{theorem} 
\section{Nano topology in Lattices}
In this section we have framed Nano topology over Lattices by means of a special class of join congruence relation which is defined with respect to an ideal.
\begin{definition}\rm: Let the Lattice L be the universe, $\theta$ be a congruence relation with respect to the ideal $\mathcal{I}$ on L and A be a non-empty subset of L. Then the sets,
\begin{itemize}
\item[(i)] $\underline{\theta _\mathcal{I}}(A)$ = $\{ x \in L : [x]_{\theta _\mathcal{I}}\subseteq A\} $
\item[(ii)] $\overline{\theta _\mathcal{I}}(A)$ = $\{ x \in L : [x]_{\theta _\mathcal{I}}\cap A\neq \emptyset\} $
\item[(iii)] $B_{\theta _\mathcal{I}}(A)$ = $\overline{\theta _\mathcal{I}}(A)$ - $\underline{\theta_\mathcal{I}}(A)$
\end{itemize}
\end{definition} 
\begin{definition}\rm: Let the Lattice L be the universe, $\theta$ be a congruence relation with respect to the ideal $\mathcal{I}$ on L \\$\tau_{\theta _\mathcal{I}}(A) = \left\{I,\emptyset,\underline{\theta_\mathcal{I}}(A),\overline{\theta _\mathcal{I}}(A), B_{\theta _\mathcal{I}}(A)\right\}$
where $A\subseteq I$ and $\tau_{\theta _\mathcal{I}}(A)$ satisfies the following axioms.\\
(i) L  and  $\emptyset\in \tau_{\theta _\mathcal{I}}(A)$\\
(ii) The union of elements of any subcollection $\tau_{\theta _\mathcal{I}}(A)$  is in $\tau_{\theta _\mathcal{I}}(A)$.\\
(iii) The intersection of the elements of any finite subcollection of $\tau_{\theta _\mathcal{I}}(A)$ is in $\tau_{\theta _\mathcal{I}}(A)$.That is $\tau_{\theta _\mathcal{I}}(A)$ forms a topology on L called as the nano topology on Lattice. \\
\end{definition}
\begin{example}\rm: Let L be a lattice, the Hasse diagram of L is illustrated by Fig.1. Let $\mathcal{I}$ = $\{c,e\}$ be an ideal of L, \\
\begin{center}
$$\includegraphics[width = 4cm , height = 6cm]{lat1.eps}$$
\end{center}
then the congruence classes with respect to $\mathcal{I}$, 
$\theta _{\mathcal{I}}$ = $\{\{a\},\{b,d\},\{c,e\}\}$ and $A=\{a,b\}\subseteq L$. Thus $\underline{\theta_\mathcal{I}}(A) = \{a\}$, $\overline{\theta _\mathcal{I}}(A) =\{a,b,d\}$, $B_{\theta _\mathcal{I}}(A) = \{b,d\}$ and hence the nano topology on L $\tau_{\theta _\mathcal{I}}(A) = \{\{L, \emptyset, \{a\}, \{a,b,d\}, \{b,d\}\}$.
\end{example}
\begin{theorem}\rm: Let L be a lattice and $\mathcal{I}$ be an ideal of L and for very subset A$\subseteq$L,we have
\begin{itemize}
\item[(i)] $\underline{\theta_\mathcal{I}}(A)\subseteq A\subseteq \overline{\theta _\mathcal{I}}(A)$
\item[(ii)] $\underline{\theta_\mathcal{I}}(\emptyset) = \emptyset = \overline{\theta _\mathcal{I}}(\emptyset)$ 
\item[(iii)] $\underline{\theta_\mathcal{I}}(L) = L =  \overline{\theta _\mathcal{I}}(L)$
\end{itemize}
\end{theorem}
\begin{theorem}\rm: Let L be a lattice and $\mathcal{I}$ be an ideal of L and for very subsets A , B $\subseteq$L, If $A\subseteq B$ then 
\begin{itemize}
\item[(i)]   $\underline{\theta_\mathcal{I}}(A)\subseteq \underline{\theta_\mathcal{I}}(B)$ 
\item[(ii)] $\overline{\theta _\mathcal{I}}(A)\subseteq \overline{\theta _\mathcal{I}}(B)$
\item[(iii)] $B_{\theta _\mathcal{I}}(A)\subseteq B_{\theta _\mathcal{I}}(B)$
\end{itemize}
\end{theorem}
\begin{proof}\rm:(i).  Let $x\in \underline{\theta_\mathcal{I}}(A)$. Then $[x]_{\theta _\mathcal{I}}\subseteq A $ since $A\subseteq B$  implies $[x]_{\theta _\mathcal{I}}\subseteq B$. Hence x$\in \underline{\theta_\mathcal{I}}(B)$. Thus $\underline{\theta_\mathcal{I}}(A)\subseteq \underline{\theta_\mathcal{I}}(B)$.\\
(ii).  Let x$\in \overline{\theta _\mathcal{I}}(A)$,then $[x]_{\theta _\mathcal{I}}\cap A\neq \emptyset$ and hence $[x]_{\theta _\mathcal{I}}\cap B\neq \emptyset$, which implies x$\in \overline{\theta _\mathcal{I}}(A)$. Thus $\overline{\theta _\mathcal{I}}(A)\subseteq \overline{\theta _\mathcal{I}}(B)$.\\
(iii). Subtracting (ii) and (i) on both sides we get $\overline{\theta _\mathcal{I}}(A) - \underline{\theta_\mathcal{I}}(A) \subseteq \overline{\theta _\mathcal{I}}(B) - \underline{\theta_\mathcal{I}}(B)$, which implies $B_{\theta _\mathcal{I}}(A)\subseteq B_{\theta _\mathcal{I}}(B)$. 
\end{proof}
\section{Characterization based on nano approximation}
In this section, we have given some characterisations based on nano approximations on the nano topology induced by lattices. 
\begin{theorem}\rm: Let $\mathcal{I}$ be an ideal of L, then the nano topology on L is $\tau_{\theta _\mathcal{I}}(\mathcal{I}) = \{L, \emptyset, \underline{\theta_\mathcal{I}}(\mathcal{I})\}$.
\end{theorem}
\begin{proof}\rm: It's enough if we prove that, $\underline{\theta_\mathcal{I}}(\mathcal{I})\} = \mathcal{I} = \overline{\theta _\mathcal{I}}(\mathcal{I})\}$. From Proposition 3.4(iii). we have, $\underline{\theta_\mathcal{I}}(\mathcal{I})\subseteq \mathcal{I}\subseteq \overline{\theta _\mathcal{I}}(\mathcal{I})$. On the other hand, let x$\in \overline{\theta _\mathcal{I}}(\mathcal{I})$,then $[x]_{\theta _\mathcal{I}}\cap \mathcal{I}\neq \emptyset$, then there exist $a\in \mathcal{I}\,\, and\,\, d\in \mathcal{I}$ such that x v d = a v d. Since $\mathcal{I}$ is an ideal, we have x v d$\in \mathcal{I}$ and thus $x \in \mathcal{I}$. This means $\overline{\theta _\mathcal{I}}(\mathcal{I})\subseteq \mathcal{I}$.\\
Moreover, let $x \in \mathcal{I}$ and $a\in [x]_{\theta _\mathcal{I}}$. Then there exists $d\in \mathcal{I}$ such that $a\vee d = x \vee d$. Then we have $a \vee d \in \mathcal{I}$, and thus $a\in \mathcal{I}$. So, $[x]_{\theta _\mathcal{I}}\subseteq\mathcal{I}$. Therefore $\mathcal{I}\subseteq \underline{\theta_\mathcal{I}}(\mathcal{I})\}$.\\
From the above, we have $\underline{\theta_\mathcal{I}}(\mathcal{I})\} = \mathcal{I} = \overline{\theta _\mathcal{I}}(\mathcal{I})\}$. Hence $\tau_{\theta _\mathcal{I}}(\mathcal{I}) = \{L, \emptyset, \underline{\theta_\mathcal{I}}(\mathcal{I})\}$.
\end{proof} 
\begin{theorem}\rm: Let L be a lattice and $\mathcal{I}$ and  $\mathcal{J}$be ideals of L, then $\overline{\theta_\mathcal{I}}(\mathcal{I}\cap \mathcal{J}) = \mathcal{I}$.
\end{theorem}
\begin{proof}\rm: Let $x\in \mathcal{I}$, we have x$\wedge y \in \mathcal{I}\cap \mathcal{J}$ for every $y \in \mathcal{J}$. On the other hand, x $\vee x = (x \wedge y)\vee x $ which means that $x\wedge y \equiv x(mod \theta_{\mathcal{I}})$. Hence, $[x]_{\theta_{\mathcal{I}}}\cap (\mathcal{I}\cap \mathcal{J})\neq \emptyset$, so $x\in  \overline{\theta_\mathcal{I}}(\mathcal{I}\cap \mathcal{J}) $. Conversely, by Theorem 3.5.(ii) and Theorem 4.1 we have  $\overline{\theta_\mathcal{I}}(\mathcal{I}\cap \mathcal{J}) \subseteq \overline{\theta _\mathcal{I}}(\mathcal{I}) = \mathcal{I}$.
\end{proof} 

\begin{theorem}\rm: Let L be a lattice and $\mathcal{I}$ be an ideal of L and for very subsets A , B $\subseteq$L, If $A\subseteq B$ then
\begin{itemize}  
\item[(i)] $\underline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A)) = \underline{\theta_\mathcal{I}}(A)$
\item[(ii)] $\overline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A)) = \overline{\theta_\mathcal{I}}(A)$
\item[(iii)] $\overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A)) = \underline{\theta_\mathcal{I}}(A)$
\item[(iv)] $\underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A)) = \overline{\theta_\mathcal{I}}(A)$
\end{itemize}
\end{theorem}
\begin{proof}\rm: (i): By definition, $\underline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$ = $\{ x \in L : [x]_{\theta _\mathcal{I}}\subseteq \underline{\theta _\mathcal{I}}(A)\}$ =$\underline{\theta _\mathcal{I}}(A)$ = $\{ x \in L : [x]_{\theta _\mathcal{I}}\subseteq A\}$ since $\underline{\theta _\mathcal{I}}(A)\subseteq A$ = $\underline{\theta _\mathcal{I}}(A)$.\\
(ii): $\overline{\theta _\mathcal{I}}(A)$. = $\{ x \in L : [x]_{\theta _\mathcal{I}}\cap A\neq \empty\} $ = $\{ x \in L : [x]_{\theta _\mathcal{I}}\cap \overline{\theta _\mathcal{I}}(A)\neq \empty\}$ since $A\subseteq \overline{\theta _\mathcal{I}}(A)$ = $\overline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A))$.\\
(iii): Let $a\in \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$ $\Rightarrow [a]_{\mathcal{I}}\cap \underline{\theta_\mathcal{I}}(A)\neq \emptyset$ $\Rightarrow x\in [a]_{\mathcal{I}} \,\,\,and\,\,\, x\in \underline{\theta_\mathcal{I}}(A)$ $\Rightarrow [x]_{I}\subseteq A$ $\Rightarrow [a]_{\mathcal{I}}\subseteq A$ $\Rightarrow a\in \underline{\theta_\mathcal{I}}(A)$. Thus $\overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A)) \subseteq \underline{\theta_\mathcal{I}}(A))$.\\
Suppose if possible $\underline{\theta_\mathcal{I}}(A)) \nsubseteq \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$. Then there exist  \,\,$a \in  \underline{\theta_\mathcal{I}}(A)) $ such that $a\notin \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$. Now $ a \notin \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$ implies $[a]_{\mathcal{I}}\cap \underline{\theta_\mathcal{I}}(A)) = \emptyset$ and hence $a\notin \underline{\theta_\mathcal{I}}(A))$, which is a contradiction. Hence $\underline{\theta_\mathcal{I}}(A)) \subseteq \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$. Thus $\underline{\theta_\mathcal{I}}(A)) = \overline{\theta_\mathcal{I}}(\underline{\theta_\mathcal{I}}(A))$.\\ 
(iv). Let  $a\in \underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A))$. Then $ [a]_{\mathcal{I}}\subseteq \overline{\theta_\mathcal{I}}(A)$. This implies $a\in \overline{\theta_\mathcal{I}}(A)$. Thus  $\underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A))\subseteq \overline{\theta_\mathcal{I}}(A)$.\\ If possible $\underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A))\nsubseteq \overline{\theta_\mathcal{I}}(A)$. Then there exist  $ x \in [a]_{\mathcal{I}}$ such that $x\notin \overline{\theta_\mathcal{I}}(A)$. This implies $[x]_{\mathcal{I}}\cap A = \emptyset$. Thus $[a]_{\mathcal{I}} \cap A = \emptyset$ which is a contradiction. Hence  $\underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A))\subseteq \overline{\theta_\mathcal{I}}(A)$. Therefore  $\underline{\theta_\mathcal{I}}(\overline{\theta_\mathcal{I}}(A)) = \overline{\theta_\mathcal{I}}(A)$. 
\end{proof}
\begin{theorem}\rm: Let $\mathcal{I}$ and  $\mathcal{J}$ be ideals of L and $\mathcal{I}\subseteq\mathcal{J}$. If A is a non-empty subset of L, then 
$\underline{\theta_\mathcal{J}}(A)\subseteq \underline{\theta_\mathcal{I}}(A)$ 
 and $\overline{\theta_\mathcal{I}}(A)\subseteq \overline{\theta_\mathcal{I}}(A)$ 
\end{theorem}
\begin{remark}\rm: In the above theorem, if we choose A as an ideal of L then the  $\overline{\theta_\mathcal{I}}(\mathcal{J})$, $\underline{\theta_\mathcal{I}}(\mathcal{J})$ will be an ideal of L which can be shown by the following theorem.
\end{remark}
\begin{theorem}\rm: Let $\mathcal{I}$ and  $\mathcal{J}$ be ideals of L and $\mathcal{I}\subseteq\mathcal{J}$ then the following statements are equivalent: 
\begin{itemize}
\item[(i)] $\mathcal{I}\subseteq \mathcal{J}$
\item[(i)]  $\mathcal{J} = \overline{\theta_\mathcal{I}}(\mathcal{J})$     
\item[(ii)]  $\mathcal{J} = \underline{\theta_\mathcal{I}}(\mathcal{J})$
\end{itemize}
\end{theorem}
\begin{proof}\rm: $(i)\Leftrightarrow (ii)$: If $\mathcal{I}\subseteq\mathcal{J}$, let $x \in \overline{\theta_\mathcal{I}}(\mathcal{J})$, then there exist $y\in \mathcal{J}$ and $d\in \mathcal{I}\subseteq \mathcal{J}$ such that x V d = y V d. Since $\mathcal{J}$ is an ideal we have x V d = y V d $\in \mathcal{J}$, then $x\in \mathcal{J}$ which implies $\overline{\theta_\mathcal{I}}(\mathcal{J})\subseteq \mathcal{J}$, and since $\mathcal{J} \subseteq \overline{\theta_\mathcal{I}}(\mathcal{J})$ thus we obtain $\mathcal{J} = \overline{\theta_\mathcal{I}}(\mathcal{J})$.\\
 $(ii)\Leftrightarrow (i)$ If $\mathcal{J} =\overline{\theta_\mathcal{I}}(\mathcal{J})$. By Theorem 3.5 (ii)and Theorem 4.2,  $\mathcal{I} = \overline{\theta_\mathcal{I\cap \mathcal{J}}} \subseteq \overline{\theta_\mathcal{I}}(\mathcal{J}) = \mathcal{J}$.\\
$(ii)\Leftrightarrow (iii)$: If $\mathcal{J} = \overline{\theta_\mathcal{I}}(\mathcal{J})$, let $x \in \mathcal{J}$ and $y\equiv x(mod \theta_{\mathcal{I}})$. Assume that $y\notin \overline{\theta_\mathcal{I}}(\mathcal{J})$. Hence, $[x]_ {\theta_{\mathcal{I}}}\cap \mathcal{J} = [y]_ {\theta_{\mathcal{I}}}\cap \mathcal{J} = \emptyset$ which implies that $x\notin \overline{\theta_\mathcal{I}}(\mathcal{J}) = \mathcal{J}$. It contradicts with $x\in \mathcal{J}$, so $y\in \mathcal{J}$. Thus $[x]_ {\theta_{\mathcal{I}}}\subseteq \mathcal{J}$ this means that $x \in \underline{\theta_\mathcal{I}}(\mathcal{J})$ and by Theorem 3.5(i) we have $\mathcal{J} = \underline{\theta_\mathcal{I}}(\mathcal{J})$.\\
$(iii)\Leftrightarrow (ii)$:  Now suppose that $\mathcal{J} = \underline{\theta_\mathcal{I}}(\mathcal{J})$. Let $x\in \overline{\theta_\mathcal{I}}(J)$, there exist $y\in\mathcal{J}$ such that $ y \equiv x(mod \theta_{\mathcal{I}})$. Since  $\mathcal{J} = \underline{\theta_\mathcal{I}}(\mathcal{J})$. we have $[x]_ {\theta_{\mathcal{I}}} = [y]_ {\theta_{\mathcal{I}}} = \mathcal{J}$ which means $x\in \underline{\theta_\mathcal{I}}(\mathcal{J}) = \mathcal{J}$. Thus by Theorem 2.5(i) we have $\overline{\theta_\mathcal{I}}(\mathcal{J}) = \mathcal{J}$. 
\end{proof}
\begin{remark}\rm: The above theorem can also revealed by the following example.
\end{remark}
\begin{example}\rm: Let L = $\{1,2,4,5,10,20,25,50,100\}$, the factors of 100 which forms a lattice under divisibility. Its Hasse Diagram is given below.\\
\begin{center}
$$\includegraphics[width = 5cm , height = 4cm]{lat2.eps}$$
\end{center}
Let $\mathcal{I} = \{1,5,25\}$ and $\mathcal{J} =\{1,2,5,10\}$ be ideals of L, then $\mathcal{I}\cap \mathcal{J} = \{1,5\}$. The congruence classes with respect to $\mathcal{I}$  and $\mathcal{J}$ are $\theta_{\mathcal{I}} = \{\{1,5,25\}, \{2,10,50\}, \{4,20,100\}$ $\theta_{\mathcal{J}} = \{\{1,2,5,10\}, \{4,20\},  \{25,50\},\{100\}\}$. Thus $\overline{\theta_\mathcal{I}}(\mathcal{I \cap J}) =\{1,5,25\} = \mathcal{I}$ and also $\overline{\theta_\mathcal{J}}(\mathcal{I \cap J}) =\{1,2,5,10\} = \mathcal{J}$. Hence the Nano toplogy $\tau_{\theta _\mathcal{I}}(\mathcal{I \cap J}) = \{L, \emptyset, \{1,5,25\}\}$. $\tau_{\theta _\mathcal{J}}(\mathcal{I \cap J}) = \{L, \emptyset, \{1,2,5,10\}\}$.
\end{example} 
\begin{theorem}\rm: Suppose $\mathcal{I}$ and  $\mathcal{J}$ be ideals of L and $\mathcal{I} \subseteq \mathcal{J}$, then the nano topology $\tau_{\theta _\mathcal{I}}(\mathcal{J})$ = $\{L, \emptyset, \mathcal{J}\}$.
\end{theorem}
\begin{proof}\rm: By the above theorem we have if  $\mathcal{I} \subseteq \mathcal{J}$ then $\mathcal{J} = \overline{\theta_\mathcal{I}}(\mathcal{J}) =\underline{\theta_\mathcal{I}}(\mathcal{J})$ which implies that,$B_{\theta _\mathcal{I}}(\mathcal{J})= \emptyset$. Thus $\tau_{\theta _\mathcal{I}}(\mathcal{J}) = \{L, \emptyset, \mathcal{J}\}$.
\end{proof} 
\begin{remark}\rm: The following example reveals the above theorem
\end{remark}
\begin{example}\rm: Consider the lattice L  in Example 4.8 and let $\mathcal{I}= \{1,5\}$ and  $\mathcal{J} = \{1,5,25\}$ be ideals of L such that $\mathcal{I}\subseteq \mathcal{J}$ and the congruence class with respect to the ideal $\mathcal{I}$ relation $\theta_{\mathcal{I}} = \{\{1,5\},\{2,10\},\{4,20\}, \{25\},\{50\},\{100\}$ then the nano topology $\tau_{\theta _\mathcal{I}}(\mathcal{J})$ = $\{L, \emptyset, \{1,5,25\}\}$.
\end{example}    
\begin{conclusion}\rm: In this paper the universe of objects is endowed with a lattice structure and a join congruence relation is defined with respect to an ideal. This universal set can also be  further extended over quotient ideals to give a new structure.
\end{conclusion}


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