You are here
Homeclosure space
Primary tabs
closure space
Call a set $X$ with a closure operator defined on it a 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:
Proposition 1.
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 $X$.
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. ∎
$\mathcal{T}$ so defined is called the closure topology of $X$ with respect to the closure operator $c$.
Remarks.
1. A closure space can be more generally defined as a set $X$ together with an operator $\operatorname{cl}:P(X)\to P(X)$ such that $\operatorname{cl}$ 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.
2. Even more generally, a closure space can be defined as a set $X$ and an operator $\operatorname{cl}$ on $P(X)$ such that

$A\subseteq\operatorname{cl}(A)$,

$\operatorname{cl}(\operatorname{cl}(A))\subseteq\operatorname{cl}(A)$, and

$\operatorname{cl}$ is orderpreserving, i.e., if $A\subseteq B$, then $\operatorname{cl}(A)\subseteq\operatorname{cl}(B)$.
It can be easily deduced that $\operatorname{cl}(A)\cup\operatorname{cl}(B)\subseteq\operatorname{cl}(A\cup B)$. In general however, the equality fails. The three axioms above can be shown to be equivalent to a single axiom:
$A\subseteq\operatorname{cl}(B)\quad\mbox{ iff }\quad\operatorname{cl}(A)% \subseteq\operatorname{cl}(B).$ 
3. In a closure space $X$, a subset $A$ of $X$ is said to be closed if $\operatorname{cl}(A)=A$. 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 $\operatorname{cl}$ “distributes over” $\cup$, that is, we have the equality $\operatorname{cl}(A)\cup\operatorname{cl}(B)=\operatorname{cl}(A\cup B)$.
4. Also, $\operatorname{cl}(\varnothing)$ 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 $\operatorname{cl}(\varnothing)=\varnothing$. This is equivalent to saying that $\varnothing$ is closed whenever there exist $A,B\subseteq X$ such that $\operatorname{cl}(A)\cap\operatorname{cl}(B)=\varnothing$.
5. Since the distributivity of $\operatorname{cl}$ over $\cup$ does not hold in general, and there is no guarantee that $\operatorname{cl}(\varnothing)=\varnothing$, a closure space under these generalized versions is a more general system than a topological space.
References
 1 N. M. Martin, S. Pollard: Closure Spaces and Logic, Springer, (1996).
Mathematics Subject Classification
54A05 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections