# closure space

###### 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. 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. 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 order-preserving, 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. 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. 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. 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).
Title closure space ClosureSpace 2013-03-22 16:48:08 2013-03-22 16:48:08 CWoo (3771) CWoo (3771) 12 CWoo (3771) Derivation msc 54A05 closure topology