ClosureSpace 2007-03-06 01:35:03 2007-05-10 19:27:27 Derivation closure axioms closure topology \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\cl}{\operatorname{cl}} Call a set $X$ with a closure operator defined on it a \emph{closure space}. Every topological space is a closure space, if we define the closure operator of the space as a function that takes any subset to its closure. The converse is also true: \begin{prop} Let $X$ be a closure space with $c$ the associated closure operator. Define a ``closed set'' of $X$ as a subset $A$ of $X$ such that $A^c=A$, and an ``open set'' of $X$ as the complement of some closed set of $X$. Then the collection $\mathcal{T}$ of all open sets of $X$ is a topology on $X$. \end{prop} \begin{proof} Since $\varnothing^c=\varnothing$, $\varnothing$ is closed. Also, $X\subseteq X^c$ and $X^c\subseteq X$ imply that $X^c=X$, or $X$ is closed. If $A,B\subseteq X$ are closed, then $(A\cup B)^c=A^c\cup B^c=A\cup B$ is closed as well. Finally, suppose $A_i$ are closed. Let $B=\bigcap A_i$. For each $i$, $A_i=B\cup A_i$, so $A_i=A_i^c=(B\cup A_i)^c=B^c\cup A_i^c=B^c\cup A_i$. This means $B^c\subseteq A_i$, or $B^c\subseteq \bigcap A_i=B$. But $B\subseteq B^c$ by definition, so $B=B^c$, or that $\bigcap A_i$ is closed. \end{proof} $\mathcal{T}$ so defined is called the \emph{closure topology} of $X$ with respect to the closure operator $c$. \textbf{Remarks}. \begin{enumerate} \item A closure space can be more generally defined as a set $X$ together with an operator $\cl:P(X)\to P(X)$ such that $\cl$ satisfies all of the Kuratowski's closure axioms where the equal sign ``$=$'' is replaced with set inclusion ``$\subseteq$'', and the preservation of $\varnothing$ is no longer assumed. \item Even more generally, a closure space can be defined as a set $X$ and an operator $\cl$ on $P(X)$ such that \begin{itemize} \item $A\subseteq \cl(A)$, \item $\cl(\cl(A))\subseteq \cl(A)$, and \item $\cl$ is order-preserving, i.e., if $A\subseteq B$, then $\cl(A)\subseteq \cl(B)$. \end{itemize} It can be easily deduced that $\cl(A)\cup \cl(B)\subseteq \cl(A\cup B)$. In general however, the equality fails. The three axioms above can be shown to be equivalent to a single axiom: $$A\subseteq \cl(B)\quad\mbox{ iff }\quad\cl(A)\subseteq \cl(B).$$ \item In a closure space $X$, a subset $A$ of $X$ is said to be closed if $\cl(A)=A$. Let $C(X)$ be the set of all closed sets of $X$. It is not hard to see that if $C(X)$ is closed under $\cup$, then $\cl$ ``distributes over'' $\cup$, that is, we have the equality $\cl(A)\cup \cl(B)= \cl(A\cup B)$. \item Also, $\cl(\varnothing)$ is the smallest closed set in $X$; it is the bottom element in $C(X)$. This means that if there are two disjoint closed sets in $X$, then $\cl(\varnothing)=\varnothing$. This is equivalent to saying that $\varnothing$ is closed whenever there exist $A,B\subseteq X$ such that $\cl(A)\cap\cl(B)=\varnothing$. \item Since the distributivity of $\cl$ over $\cup$ does not hold in general, and there is no guarantee that $\cl(\varnothing)=\varnothing$, a closure space under these generalized versions is a more general system than a topological space. \end{enumerate} \begin{thebibliography}{7} \bibitem{mp} N. M. Martin, S. Pollard: {\em Closure Spaces and Logic}, Springer, (1996). \end{thebibliography}