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\fancyhead[CE]{J. Hashemi} 
\fancyhead[CO]{On the $\cap$-structure spaces}



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\begin{center}

{\Large \bf 
On the $\cap$-structure spaces\\}
%{\bf Do You Have a Subtitle? \\ If so, Write It Here} 





{\bf J. Hashemi}  
\vspace{2mm} 

{\small   Shahid Chamran University of Ahvaz, Ahvaz, Iran} \vspace{2mm}

 \vspace{2mm}

\end{center}

\vspace{4mm}


{\footnotesize
\begin{quotation}
{\noindent \bf Abstract.}
The family $\MM_X\sub\mathcal{P}(X)$ is called an $\cap$-structure, when it is closed under the arbitrary intersection. This concept has been studied and considered in algebra, specially in lattices. Using this concept, we define a quasi topological structure which is called $\cap$-structure space. By studying this space, we attempt to explain some  algebraic concepts through this structure.  
 
 
\end{quotation}
\begin{quotation}
\noindent{\bf AMS Subject Classification 2010:} Primary: 06-XX; Secondary:  54-XX, 13-XX.

\noindent{\bf Keywords and Phrases:} Algebraic Structure, connected, compact, Lattice, $\cap$-structure space,subspace.
\end{quotation}}

\section{Introduction}
% It is advised to give each section and subsection a unique label.

A lattice $L$ is called a complete lattice if $\ve A$ exists for every $A\sub L$; or equivalently, $\we A$ exists for every $A\sub L$, and also is called a distributive lattice if 
$a\ve(b\we c)=(a\ve b)\we(a\ve c)$, for every $a,b,c\in L$.
 Supposing $X$ is an ordered set, a function $f:X\to X$ is said to be a closure operator (interior operator) if it has the following properties:\\
(i) $f$ is an increasing function; i.e., if $a\leq b$, then $f(a)\leq f(b)$ for every $a,b\in X$.\\
(ii) $f$ is idempotent; i.e., $f(f(a))=f(a)$ for every $a\in X$.\\
(iii)  $f$ is extensive (contractive); i.e., $a\leq f(a)$ ($f(a)\leq a$) for every $a\in X$.\\
A nonempty subset $S$ of an ordered set is said to be directed if every pair of $S$ has an upper bound in $S$. A nonempty family $D$ of subsets of  a set $X$ is said to be closed under directed unions if 
$\cup_{i\in I}A_i\in D$ for any directed family $\{A_i\}_{i\in I}$
in $D$.

For any set $X$, an $\cap$-structure on $X$ is a  family  $\MM_X$  of subsets of $X$ which is  closed under arbitrary intersection. We say $(X,\MM_X)$, briefly $X$, is an $\cap$-structure space. Clearly, if $(X,\MM_X)$ is an $\cap$-structure space, then $\MM_X$ is a complete lattice in which  for every nonempty family $\{A_i\}_{i\in I}$, we have
 $$\we _{i\in I}A_i=\cap_{i\in I}A_i~~,~~\ve _{i\in I}A_i=\cap\{B\in \MM_X: \cup_{i\in I}A_i\sub B\}.$$
 If $\MM_X$ is a distributive lattice, we say $\TX$ is a distributive $\cap$-structure space.
Obviously, $X$ is the top element of $\MM_X$. The least element of this complete lattice is denoted by
$\0$ and we  call it zero.
 
It is clear that if X is any algebraic structure (for example, module, ring, group, vector space) and $\MM_X$ is the collection of all substructure of $X$ (resp., submodule, ideal, subgroup, subspace), then $\TX$ is an $\cap$-structure space. Hence this concept is a suitable  model for studying and generalizing algebraic structures.
Throughout this article $R$ is a  commutative ring  with $1\neq \circ$. We use the notations ${\rm Id}(R)$, ${\rm Spec}(R)$, ${\rm Max}(R)$  for the set of all ideals , the set of all prime and the set of all maximal ideals of the ring $R$, respectively. We denote the annihilator of a subset $S\sub R$ by $\An(S)$, i.e., $\An(S)=\{r\in R: rS=\circ\}$. 
\section{general properties of $\cap$-structure spaces}
\defi{Let $\TX$ be an $\cap$-structure space  and $A\sub X$. The element $\lb A\re$ of $\MM_X$ generated (or spanned) by $A$ is the  intersection of all elements $u\in \MM_X$ that  contain $A$. In the case that $A$ is the finite set $\{a_1,a_2,\dots,a_n\}$, $\lb A\re$ is written as $\lb a_1,a_2,\dots,a_n\re$, and is referred as the element generated by $a_1,a_2,\dots,a_n$. If for an element $x$ of $X$, we have $\lb x\re=X$, then we call $x$ is an invertible element of $X$. The set of all invertible elements of $X$ is denoted by ${\rm U}(X)$. }

\defi{If $\TX$ is  an $\cap$-structure space and $E\sub X$, the closure of $E$, i.e., $\bar{E}$ is defined by
$$\bar{E}=\{x\in X: x\in E~\text{or}~ E\cap u\nsub \0~ \mbox{for all}~ u\in \MM_x\},$$
 where $\MM_x$ is the set of all elements $u$ of $\MM_X$ containing $x$.
If we want to emphasize the set $X$, we  use the notation $cl_XE$ for the closure of  $E$.}

\lem{The mapping  $E\to \bar{E}$ in an $\cap$-structure space $\TX$ is a closure operator on $X$   and moreover has the following properties:\\
(a) For any collection $\{A_i\}_{i\in I}$ of subsets of $X$, $\overline{\cup_{i\in I}A_i}=\cup_{i\in I}\bar {A_i}$.\\
(b) For all $u\in \MM$:\\
i) $\overline{u\set\0}=u\set\0$.\\
ii) $\overline{X\set u}=X\set u$.\\
iii) $\overline{X\set(u\set\0)}=X\set(u\set\0)$.}
\proof{
The proof is straightforward.}
\vspace{3mm}

\exa{Let $R$ be a ring. In the $\cap$-structure space $(R,{\rm Id}(R))$ , for any nonzero ideal $I$ , we have $\bar I=R$ if and only if $I$ is an essential ideal of $R$.}

\defi{If $\TX$ is  an $\cap$-structure space and $A\sub X$, the interior of $A$, i.e., $A^{\0}$  is the set
$$A^{\0}=\{x\in X: u\sub A ~\mbox{for some}~ u\in \MM_x\}.$$
It is easy to see that $A^{\0}=\{x\in X:~ \lb x\re\sub A\}.$}

\lem {The operation $A\to A^{\0}$ in an $\cap$-structure space $\TX$ is an interior operation   and moreover has the following properties:\\
(a) For any collection $\{A_i\}_{i\in I}$ of subsets of $X$, $({\cap_{i\in I}A_i})^{\0}=\cap_{i\in I}A_i^{\0}$.\\
(b) For all $u\in \MM_X$, $u^{\0}=u$.}
\proof{
The proof is straightforward.}
\vspace{3mm}

The previous  lemmas show that the closure and interior have the same properties as in the topological spaces. We denote the  topologies induced by closure and interior maps, by
$\tau$ and  $\tau_{\0}$, respectively. By the definition and previous lemma, it is clear that   $\MM_X$ is a base for the topology  $\tau_{\0}$.  
%The topology $\tau$ and the topology $\tau_{\0}$ are the same as our goal, because their difference is that the zero set in $\tau$   is clopen and it has discrete  topology as subspace, but    in the $\tau_{\0}$ has a trivial  topology.\\ 

The next lemma give us more information regarding 
$\tau$ and $\tau_{\0}$.

\lem {Let $\TX$ be an $\cap$-structure space and $A\sub X$. Then
the folloeing statements hold.\\
(a) $\beta_{\0}=\{\lb x\re:~x\in X\}$ is a base (in fact the smallest base) for the topology $\tau_{\0}$.\\
(b) The set  $\beta=\{\lb x\re\set\0:~ x\in X\set\0\}\cup\0$ is  the smallest base for the topology $\tau$. Hence if $\0\neq\varnothing$, then $\tau_{\0}\subsetneq\tau$.\\
(c) $x\in int_{\tau}A$ if and only if $x\in A$ and $\lb x\re\set\0\sub A$.\\
(d) If $\0\sub A$,   then  
$int_{\tau_{\0}}A=int_{\tau}A$.\\
(e) If $A\cap \0=\varnothing$, then $cl_{\tau}A=cl_{\tau_{\0}}A$. clearly if $\0=\varnothing$, then these two toplogies coincide.\\
(f) $cl_{\tau}A=cl_{\tau_{\0}}(A\setminus\0)\cup A$ and $int_{\tau}A=int_{\tau_{\0}}(A\cup\0)\set (\0\set A)$.}

\proof{(a). Clearly $\beta_\0$ is a base for the topology $\tau_\0$. Now, suppose that $\beta$ is a base for $\tau_\0$ and $\lb x\re\in\beta_\0$. Thus, $B\in\beta$ exists such that $x\in B\sub\lb x\re$ and consequently $\lb x\re\sub B\sub\lb x\re$. Therefore, $\lb x\re=B\in\beta$.\\
(b). It is similar to (a). \\
(c). It is evident by part (b) and the fact that every point of $\0$ is isolated with respect to the topology $\tau$.\\
(d). Since $\tau_\0\sub\tau$, clearly $int_{\tau_\0}A\sub int_\tau A$. Assume that $x\in int_\tau A$. Then by (c) we have $\lb x\re\set\0\sub A$ and so $\lb x\re=(\lb x\re\set\0)\cup\0\sub A$. Therefore, $x\in int_{\tau_\0}A$.\\
(e). Since $\tau_\0\sub\tau$, clearly $cl_{\tau}A\sub cl_{\tau_\0} A$. Assume that $x\in cl_{\tau_\0} A$. Then, clearly $\tohi\neq\lb x\re\cap A\sub A$ and consequently  $\lb x\re\cap A\nsub\0$. Hence, $x\in cl_\tau A$.\\
(f). By (e) we can write
$$cl_\tau A=cl_\tau((A\set\0)\cup(A\cap\0)=cl_\tau(A\set\0)\cup A=cl_{\tau_\0}(A\set\0)\cup A.$$
Also, by (d) and the fact that $\0\set A$ is clopen with respect to the topology $\tau$, we can write
$$int_\tau A=int_\tau((A\cup\0)\set(\0\set A))=(int_\tau(A\cup\0))\set(\0\set A)=(int_{\tau_\0}(A\cup\0))\set(\0\set A).$$}


%\vspace{-3mm}


\defi{Let $\TX$ be an $\cap$-structure space and $Y\sub X$. We define:
$$\MM_Y=\{u\cap Y:~u\in \MM_X\}.$$
This is easily checked to be  closed under arbitrary intersection. So $\TY$ is an $\cap$-structure space and we call it the subspace of $\TX$. It is clear that if $Y\in \MM_X$ then $\MM_Y=\{u\in \MM_X:~ u\sub Y\}$.} 

\vspace{3mm}
It is clear that $(Y,\MM_Y)$ is a topological subspace of $(X,\MM_X)$ with respect to $\tau_\0$. The following proposition shows that  this  is true so for $\tau$, too.

\pro{If $\TX$ is an $\cap$-structure space and $A\sub Y\sub X$, then $cl_YA=cl_XA\cap Y$.}
\proof{It is easy to see that $\lb y\re\cap A\nsub\0_Y$ if and only $\lb y\re\cap A\nsub\0_X$. By this fact, one can easily see that $cl_YA=cl_XA\cap Y$.}

\rem{Let $R$ be a ring, in the $\cap$-structure space $(R,{\rm Id}(R))$,  for a nonzero subset $A$ of $R$, it is clear that $A\cup U(R)\sub\bar{A}$, where $U(R)$ is the set of all invertible elements of $R$. The natural question is, when  does the equality
	$A\cup U(R)=\bar{A}$ hold. The next proposition is a partial answer to this question.}

\pro{Let $R$ be a ring. In the $\cap$-structure space 
$(R,{\rm Id}(R))$, for any proper  nonzero ideal $I$ of $R$, $I\cup U(R)=\bar{I}$ if and only if  one of the following conditions  hold.\\
	(a) ${\rm Max}(R)=\{M_\0\}$ and $I=M_\0$.\\
	(b) ${\rm Spec}(R)=\{M_\0,M_1\}$, where $I=M_\0$ and $M_1=\An(I)$.\\
	(c) ${\rm Max}(R)=\{M_\0\}$, $I\sub\An(I)=M_\0$ and $I\sub P$ for each $P\in {\rm Spec}(R)$. } 
\proof{$\RTA$)
	First we notice that $R\setminus {\rm U}(R)=I\cup\An(I)$. Let $I\sub M_\0\in {\rm Max}(R)$, then $M_\0\sub I\cup \An(I)$, and so $M_\0\sub \An(I)$ or $M_\0\sub I$. Hence, $M_\0=I$  or otherwise $I\subsetneq M_\0=\An(I)$. Now assume that  $I=M_\0$. In this case we show that if (a) does not hold, then (b) hold. Suppose that $M_1$ is an arbitrary maximal ideal different from $M_\0$. Clearly, $M_1\sub R\set {\rm U}(R)=I\cup\An(I)$ and so $M_1=\An(I)$. Now, we show that ${\rm Spec}(R)=\{M_\0, M_1\}$. To see this, suppose that $P\in {\rm Spec}(R)$. Then $M_\0M_1=\0\sub P$ and so $M_\0=P$ or $M_1=P$.\\   
	$\LTA$) If (a) or (c) hold, then it is clear that    $\overline{I}=I\cup {\rm U}(R)$. Otherwise, suppose that     ${\rm Spec}(R)=\{M_\0, M_1\}$, $I=M_\0$ and $M_1=\An(I)$. Now, assume that $x\notin I\cup {\rm U}(R)$, then it is sufficient to show that $\lb x\re\cap I=\0$. Let $y=rx\in I=M_\0$. Clearly $x\in M_1$ and so $xM_\0=\0$. Therefore, $y^2=r^2x^2=yrx\in M_\0M_1=\0\sub M_\0$ and so $r\in M_\0$. Consequently, $y=rx\in M_\0M_1=\0$.}

\vspace{3mm}
The following corollary is immediate.

\coro{Let $R$ be a ring, in the $\cap$-structure space 
$(R,{\rm Id}(R))$, let $I$ be  a proper ideal of $R$ such that $I\subsetneqq \An(I)$ (for example in the reduced ring this condition holds). Then $I\cup {\rm U}(R)=\bar{I}$ if and only if  one of the following conditions  hold.\\
	(a) ${\rm Max}(R)=\{M_1\}$ and $I=M_\0$.\\
	(b) ${\rm Spec}(R)=\{M_\0,M_1\}$, where $I=M_\0$ and $M_1=\An(I)$. }

\pro{Let $X$ be a vector space over a field $F$, $\MM_X$ be the set of all subspaces of $X$ and $\0\in A\sub X$. Then the following statements are equivalent:\\
	(a) $A=\bar{A}$.\\
	(b) $A$ is closed under scaler multiplication.\\
	(c) $A$ is the union of  a family of subspaces of $X$.}
\proof{(a) $\RTA$ (b). Let $a\in A$ and $x\in F$. If $a=0$, then we have nothing to do. Otherwise $a=\frac{1}{x}xa$. Hence, $a\in\lb xa\re\cap A$ and so $xa\in\bar{A}=A$.\\
(b) $\RTA$ (c). It is clear, since in this case $\lb a\re=\{xa:~x\in F\}\sub A$ and so $A=\cup_{a\in A}\lb a\re$.\\
(c) $\RTA$ (a). Suppose that $b\in\bar{A}$, then $\lb b\re\cap(\cup_{a\in A}\lb a\re)=\lb b\re\cap A\nsub\0$ and so there exists $a\in A$ such that $\lb b\re\cap\lb a\re\nsub\0$. Therefore, $b\in\lb a\re\sub A$.}

\vspace{3mm}
The following corollary is immediate.

\coro{Let $X$ be a vector space over a field $F$, $\MM_X$ be the set of all subspaces of $X$ and $\0\in A\sub X$. If $\tau$ is the topology on $X$ induced by the closure or interior map, then $A$ is closed if and only if $A\in \tau$.}













\section{compactness and connectedness}
In this section we define essential concepts like compactness and connectedness in 
 $\TX$,  and establish their elementary properties.
 \defi{Let $\TX$ be an $\cap$-structure space, $A\sub X$ and $\KK\sub\MM_X$.  $A$ is said to be  $\KK$-compact ($\KK$-join compact) if  each cover (join cover) of elements of $\KK$ for $A$, has a finite subcover (finite join subcover). For simplicity, the $\MM_X$-compact ($\MM_X$-join compact) subset of $X$ is called compact (join compact).} 

The following examples show that  these concepts need not imply each other.\\

\exa{(1). Let $m_1\subsetneq m_2\subsetneq\cdots$ be a strictly ascending chain of sets. Assume that $A=\cup_{i=1}^{\infty}m_i$ and $X=m_\0=A\cup\{x\}$, for some $x\notin A$. If we consider $\MM_X=\{m_i: i=0,1,2,\cdots\}$, then   $\TX$  is an $\cap$-structure space. In this $\cap$-structure space, according to the equality
 $m_\0=\ve_{i=1}^{\infty}m_i$, it is easy to see that $m_\0$ is compact while it is not join compact.\\
(2). In the $\cap$-structure space $(\Z, {\rm Id}(\Z))$, let $A$ be  the set of all prime numbers. Then clearly, $A$ is join compact but it  is not compact.}


\pro{Suppose $\TX$ is an $\cap$-structure space in which $\MM_X$ is closed under directed unions. If $A\sub X$ is compact, then it is join compact. }
\proof{Suppose  $\mathcal{U}\sub\MM_X$ is a join cover of $A$. Set 
$$\mathcal{V}=\{\ve_{_{m\in F}}m:~F~\mbox{is a finite set of}~ \mathcal{U}\}.$$
By assumption $\cup_{_{n\in \mathcal{V}}}n\in \MM_X$ and also we have 
$\cup_{_{n\in \mathcal{V}}}n=\ve_{_{n\in \mathcal{V}}}n=\ve_{_{m\in \mathcal{U}}}m$. Since $A$ is compact, there exists a finite set $F_{\0}\sub\mathcal{V}$ such that $A\sub \cup_{_{n\in F_{\0}}}n$. Clearly, there exists a finite subset $F\sub\mathcal{U}$ such that $A\sub \cup_{_{n\in F_\0}}n\sub\ve_{_{n\in F_\0}}n=\ve_{_{m\in F}}m$.}

\vspace{3mm}
The previous proposition immediately shows that if $X$ is an $R$-module and $\MM_X$ is the set  of all submodules of $X$, then every compact subset of $X$ is join compact.

\pro{Let $\TX$ be an $\cap$-structure space in which $\MM$ is closed under directed unions, and $A\sub X$. The  following statements  are equivalent:\\
(a) $A$ is join compact.\\
(b) $A\sub\ve_{i=1}^n\lb a_i\re$, where $\{a_1,\dots, a_n\}$ is a finite subset of $A$.\\
Furthermore, if $A\in \MM$
then\\
(c) $A$ is finitely generated.}
\proof{ The proof is straightforward.}

\coro{Let $\TX$ be an $\cap$-structure space. The  following statements  are equivalent:\\
(a) $X$ is Noetherian (i.e., satisfies the ascending chain condition on elements of $\MM_X$).\\
(b) Each $u\in\MM_X$ is join compact.\\
(c) Each $u\in \MM_X$ is finitely generated.}

\proof{It is enough to prove (a) $\LRTA$ (c). The proof of this equivalence is similar to what we have seen in algebra.}

\defi{Let $\TX$ be an $\cap$-structure space and $u,v\in\MM_X$. We say that $\{u,v\}$ is a separation of $X$ if u is a complement of $v$. A separation $\{u,v\}$ of $X$
is called trivial if one of them is zero. We say that $X$ is $\MM_X$-connected (briefly, connected) if each separation of $X$ is trivial.}
 
 For a subspace of an $\cap$-structure space,  we can define different types of connectivity as bellow.
\defi{Suppose $\TX$ is  an $\cap$-structure space and $Y\sub X$. We say that $Y$ is\\
i) $\MM_X$-connected  if whenever  $u,v\in \MM_X$ and $u\we v=\0$ such that $Y\sub u\ve v$, then $Y\sub u$ or $Y\sub v$;\\
ii) weakly $\MM_X$-connected if for  $u,v\in \MM_X$ with $u\we v=\0$ such that $Y\sub u\ve v$, we have $Y\cap u\sub\0$ or $Y\cap v\sub\0$.}

\vspace{3mm}
If $X$ is a ring, $\MM_X$ is the set of all ideals and $I$ is an ideal of $X$, then the previous definition is rewritten as follows:\\
i) $I$ is $\MM_X$-connected , if $I\subseteq J\oplus K$, where $J$ and $K$ are two ideals of $X$, then 
$I\subseteq J$ or $I\subseteq K$.\\
ii)  $I$ is  Weakly $\MM_X$-connected, if $J$ and $K$ are two ideals of $X$ and $I\subseteq J\oplus K$, then $I\cap J=\0$ or $I\cap K=\0$.

\vspace{3mm}
The next proposition shows that the $\MM_X$-connectedness  implies the weakly $\MM_X$-connectedness. 
 
 \pro{Let $\TX$ be an $\cap$-structure space and $Y\sub X$. 
  If $Y$ is $\MM_X$-connected, then it is weakly $\MM_X$-connected.\label{XCXWC}}
\proof{
Assume that
	$u,v\in \MM_X$,
	 $u\we v=\circ$
	  and 
	  $Y\sub u\ve v$.
	  By assumption and without loss of generality, we may assume that 
	   $Y\sub u$.
	   We show that
	     $ Y\cap v\sub \circ$.
	    Let
	    $y\in Y\cap v$,
	    then
	$y\in u\cap v=\circ$.
	 Thus
	 $Y\cap v\sub\circ$ and the proof is complete.}
  
  \vspace{3mm}
   
In the next examples, first we show that  the converse of the 
   previous proposition is not true in general. This examples also show that
  connected and weakly $\MM_X$-connected   are independent from each other, and in addition, weakly $\MM_X$-connected  does not implies $\MM_X$-connected.\\
   \vspace{3mm}
    
 \exa{Let $X$ be  the vector space $\mathbb{R}^2$ over $\mathbb{R}$ and $\MM_X$  be the set of  all subspaces of $X$. Then we can  easily  see that  every one dimensional subspace is weakly $\MM_X$-connected but it is not $\MM_X$-connected. }
 \vspace{3mm}
 
 \exa{ (1). Let $X=\Z$,  $\MM_X={\rm Id}(\Z)$ and let $I$ and $J$ be two incomparable ideals of $\Z$. If $Y=I\Delta J$, then we have 
$(Y\cap I)\cap(Y\cap J)=Y\cap I\cap J=\tohi$ and 
$(Y\cap I)\cap(Y\cap J)=I\Delta J=Y$. Hence, $Y$ is 
not connected whereas it is $\MM_X$-connected , for, $\Z$ is a uniform ring.\\
(2). In the $\cap$-structure space $(\Z_{30}, {\rm Id}(\Z_{30})$,  the ideal $\lb 4\re\in\MM$ is not indecomposable; i.e., $\lb 4\re$ is connected whereas it is not even weakly $\MM_X$-connected.}
\vspace{3mm}

\exa{Let $X$ be any set and  $Y$, A, and $B$ are subsets of $X$, such that $A\cap B=\tohi$. In addition, suppose that $A_1=Y\cap A$ and $B_1=Y\cap B$ are nonempty sets  for which $A_1\cup B_1$ is a proper nonempty set. Set
$\MM_X=\{\tohi,A_1, B_1, A_1\cup B_1,A, B, X\}$ and
$\MM_Y=\{\tohi, A_1, B_1, A_1\cup B_1, Y\}$. Now, we show that $Y$ is $\MM_Y$-connected whereas is not weakly $\MM_X$-connected. 
We have $Y\sub A\ve B$, $A\cap B=\tohi$, $Y\cap A=A_1\neq\tohi$ and  $Y\cap B=B_1\neq\tohi$, then $Y$ is not weakly $\MM_X$-connected. Now, let $Y=D\ve C$ such that $C\cap D=\tohi$. By the assumption and the definition of $\MM_Y$, we see that  $D$ or $C$ must be $Y$. Thus $Y$ is $\MM_Y$-connected.}
 \vspace{3mm}
 
 In the following two  propositions we  give some 
 conditions such that the weakly  $\MM_X$-connectedness   implies  the   $\MM_X$-connectedness. 
\pro{Let $\TX$ be an $\cap$-structure space and $Y\sub X$. The following statements are hold.\\
(a) If $Y$ is weakly $\MM_X$-connected and $(X\set Y)\cup\0$ includes no any nonzero element of $\MM_X$, then $Y$ is  $\MM_X$-connected.\\
(b) If $Y\in\MM_X$  is weakly $\MM_X$-connected, then it is  $\MM_X$-connected.\\
(c) If  $Y\in\MM_X$ is $\MM_X$-connected, then $Y$ is connected.\label{3.7}}
\proof{
(a). Assume that
$u,v\in\MM_X$,
 $(u\cap Y)\cap(v\cap Y)=\circ_Y$
 and
  $Y=(u\cap Y)\ve(v\cap Y)$.
  In this case it is clear that 
$Y\sub u\ve v$
 and
  $u\cap v\sub (X\set Y)\cup\circ$.
 Hence
   $Y\sub u\ve v$
   and
    $u\cap v\sub \circ$.
   Therefore,  by assumption
     $Y\cap u\sub \circ$ or $Y\cap v\sub \circ$.\\
(b).
Assume that 
$u,v\sub Y$,
 $u\we v=\circ$
and
  $Y=u\ve v$. 
 Then by assumption
   $Y\cap u=u\sub \circ$ or $Y\cap v=v\sub \circ$.\\
   (c). Let $Y=u\ve v$, where $u, v\in \MM_X$ are two subsets of $Y$ and $u\we v=\circ$. Since $Y$ is $\MM_X$-connected, we have $Y\sub u$
or $Y\sub v$. Therefore, $Y=u$ or $Y=v$ and consequently $Y$ is connected.}
   \vspace{3mm}
   
%The next proposition shows that in a 
%distributive $\cap$-structure space, the weakly $\MM_X$-connectedness and the $\MM_X$-connectedness for  a subspace $Y$, where $Y\in \MM_X$, are equivalent.
 \pro{Let  $\TX$ be a distributive $\cap$-structure space and let $Y\in \MM_X$. If $Y$ is a connected, then $Y$ is $\MM_X$-connected subspace.}
\proof{Suppose that  
  $Y\sub m\ve n$
  and
  $m\we n=m\cap n=\circ$.
  In this case by the distributive assumption we have: 
  $Y\cap(m\ve n)=(Y\cap m)\ve(Y\cap n)=Y$.
  Since $Y$ is 
 connected, we must have
   $Y\cap m=Y$
 or
  $Y\cap n=n$.
  Hence
  $Y\sub m$
 or 
  $Y\sub n$.
Thus, $Y$ is a $\MM_X$-connected subspace.\\
% Now assume that $Y$ is a weakly $\MM_X$-connected space and 
 %           $Y\sub m\ve n$
  % where
    %$m\we n=\circ$.
   %In this case again  by the distributive assumption we have: 
    %$Y\cap(m\ve n)=(Y\cap m)\ve(Y\cap n)=Y$.
     %Since $Y$ is a weakly $\MM_X$-connected space, we must have 
      %     $Y\cap m\sub\circ$
      %or
      %$Y\cap n\sub\circ$. 
     %If
      %      $Y\cap m\sub\circ$, then
        %               $(Y\cap m)\ve(Y\cap n)=Y$, therefore
       %                $Y\cap n=Y$, and consequently
         %           $Y\sub n$.
          %   The proof  of the case 
%$Y\cap n\sub \circ$,
 %          is similar.}
 \pro{Let $\TX$ be an $\cap$-structure space and $Y\sub X$. Then the following statements are hold.\\
(a) If $Y$ is weakly $\MM_X$-connected  and $Y\sub B\sub \bar{Y}$, then  $B$ is weakly $\MM_X$-connected.\\
(b) If for each $\lambda\in\Lambda$, $Y_{\lambda}\sub X$ is  $\MM_X$-connected and $\cap_{\lambda\in\Lambda}Y_{\lambda}\nsubseteq\0$, then $\cup_{\lambda\in\Lambda}Y_{\lambda}$ is $\MM_X$-connected.\\
(c) If $Y\sub X$ is $\MM_X$-connected, then $\lb Y\re$ is $\MM_X$-connected.\\
(d) If for each $\lambda\in\Lambda$, $Y_{\lambda}\sub X$ is $\MM_X$-connected and $\cap_{\lambda\in\Lambda}Y_{\lambda}\nsubseteq\0$, then $\ve_{\lambda\in\Lambda}Y_{\lambda}$ is $\MM_X$-connected.\label{3.8}}

\proof{(a). Suppose that $B\sub m\ve n$, where $m,n\in\MM_X$ and $m\ve n=\circ$.  By assumption, $Y\cap m\sub\0$ or $Y\cap n\sub\0$. This is equivalent to the fact that $B\cap m\sub\0$ or $B\cap n\sub\0$.\\
(b). Let $\cup_{\la\in\La}Y_\la\sub m\ve n$ such that $m,n\in\MM_X$ and  $m\ve n=\circ$.  By hypothesis, for every $\la\in\La$ we have $Y_\la\sub m$ or $Y_\la\sub n$. If, on the contrary, there exist $\la_1,\la_2\in\La$ such that $Y_{\la_1}\sub m$ and $Y_{\la_2}\sub n$, then $\0\neq Y_{\la_1}\cap Y_{\la_2}\sub m\cap n=\0$ and this is a contradiction. Therefore, $Y_\la\sub m$ for every $\la\in\La$ or $Y_\la\sub n$ for every $\la\in\La$ and consequently $\cup_{\la\in\La}Y_\la\sub m$  or $\cup_{\la\in\La}Y_\la\sub n$.\\
(c). Suppose that 
$\lb Y\re\sub m\ve n$ such that $m,n\in\MM_X$ and $m\ve n=\circ$. Hence, by hypothesis, $Y\sub m$ or $Y\sub n$ and consequently $\lb Y\re\sub m$ or $\lb Y\re\sub n$.}

\pro{
In the $\cap$-structure space 
$(R,{\rm Id}(R))$, let  $I$ be an ideal  such that  any element $a\in I$  has a root; i.e., there exists a natural number $n>1$ and $b\in I$ such that $b^n=a$. Then the three kinds of connectedness for $I$ are equivalent.}
\proof{By part (b), (c) of Proposition \ref{3.7} and by  proposition \ref{XCXWC}, it is enough to show that if $I$ is $\MM_X$-connected, then $I$ is connected. To see this, suppose that $J,H$ are two  ideals where $J\cap H=\circ$ and $I\sub J\ve H=J+H$. It suffices to show that $I=I\cap J+I\cap H$. Clearly, we have $I\cap J+I\cap H\sub I$. Now, assume that $a\in I$. Then, there exist $b\in I$ and $n>1$ such that $b^n=a$. By hypothesis, there exist $c\in J$ and $d\in H$ for which $b=c+d$. Hence, $a=bb^{n-1}=cb^{n-1}+db^{n-1}\in I\cap J+I\cap H$.\label{3.9}}

\pro{Let $X$ be a ring, $\MM_X$ be the set of all semiprime  ideals of $X$ and $I\in\MM_X$. Then the three kinds of connectedness for $I$ are equivalent.}
\proof{Similar to previus proposition, it is enough to show that for every  $J,H\in\MM_X$ with $J\cap H=\circ$, if $I\sub J+H$, then $I\sub I\cap J+I\cap H$. Assuming $a\in I$, there exist $b\in J$ and $c\in H$ for which $a=b+c$. It suffices to show that $b\in I\cap J$ and $c\in I\cap H$. Clearly, $ab=b^2+bc=b^2$ and so $b^2\in I\cap J$, hence $b\in I\cap J$. The proof of  $c\in I\cap H$ is similar. Therefore 
$a\in I\cap J+I\cap H$, and the proof is complete.}




 

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\begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format.
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\end{thebibliography}
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