\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{J. Hashemi} \fancyhead[CO]{On the $\cap$-structure spaces} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} %\newtheorem{definition, notation}[theorem]{Definition, Notation} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \def\exa{\noindent{\bf Example. }} \newtheorem{ttheo}{Theorem}[section] \newtheorem{ccoro}[ttheo]{Corollary} \newtheorem{llem}[ttheo]{Lemma} \newtheorem{rrem}[ttheo]{Remark} \newtheorem{ppro}[ttheo]{Proposition} \newtheorem{ddefi}[ttheo]{Definition} \newtheorem{ddefin}[ttheo]{Definition and Notation} \newcommand{\MM}{\mathcal{M}} \newcommand{\KK}{\mathcal{K}} \newcommand{\ve}{\vee} \newcommand{\we}{\wedge} \newcommand {\T}{{(X,\mathcal{M})}} \newcommand {\TX}{{(X,\mathcal{M}_X)}} \newcommand {\TY}{{(Y,\mathcal{M}_Y)}} \newcommand{\TOO}{\longrightarrow} \newcommand{\LRTA}{\Leftrightarrow} \newcommand{\ex}{\exists} \newcommand{\set}{\setminus} \newcommand{\An}{\rm{Ann}} \newcommand{\tohi}{\varnothing} \newcommand{\0}{\circ} \newcommand{\la}{\lambda} \newcommand{\La}{\Lambda} \newcommand{\ga}{\gamma} \newcommand{\al}{\alpha} \newcommand{\lb}{\langle} \newcommand{\re}{\rangle} \def\defi#1{\begin{ddefi}#1\end{ddefi}} \def\defin#1{\begin{ddefin}#1\end{ddefin}} \def\theo#1{\begin{ttheo}#1\end{ttheo}} \def\pro#1{\begin{ppro}#1\end{ppro}} \def\coro#1{\begin{ccoro}#1\end{ccoro}} \def\lem#1{\begin{llem}#1\end{llem}} \def\rem#1{\begin{rrem}#1\end{rrem}} \newenvironment{pproof}{\noindent{\bf Proof.}}{} \def\proof#1{\begin{pproof}#1\end{pproof}} \renewenvironment{proof}{{\bfseries \noindent Proof.~}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter \def\iff {if and only if\ \ } \def\sub {\subseteq} \def\set {\setminus } \def\nsub {\nsubseteq} \def\set {\setminus } \def\RTA {\Rightarrow } \def\LTA {\Leftarrow } \def\Z {{\Bbb Z}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm %\setcounter{page}{} \thispagestyle{plain} \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.} % BibTeX users please use one of %\bibliographystyle{spbasic} % basic style, author-year citations %\bibliographystyle{spmpsci} % mat %hematics and physical sciences %\bibliographystyle{spphys} % APS-like style for physics %\bibliography{} % name your BibTeX data base % Non-BibTeX users please use \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. \bibitem{DAV} B. A. Davey, H.A. Priestly, {\em Introduction to Lattices and Order}, Cambridge University Press, 2002. \bibitem{Lam} T. Y. Lam, {\em A First Course in Noncommutative Rings}, Springer-Verlag, 1991. \bibitem{SHARP} R. Y. Sharp, {\em Steps in Commutative Algebra}, Cambridge University Press, 2000. \bibitem{WIL} S. Willard, {\em General Topology}, Addison Wesly, Reading, Mass., 1970. \end{thebibliography} \end{center} \end{document}