PlanetMath: closure space

(more info)

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closure space

(Derivation)

Call a set 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 be a closure space with the associated closure operator. Define a “closed set” of as a subset of such that , and an “open set” of as the complement of some closed set of . Then the collection $\mathcal{T}$ of all open sets of is a topology on .

Proof. Since $\varnothing^c=\varnothing$ , $\varnothing$ is closed. Also, $X\subseteq X^c$ and $X^c\subseteq X$ imply that

, or

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

are closed. Let $B=\bigcap A_i$ . For each

, $A_i=B\cup A_i$ , 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

, or that $\bigcap A_i$ is closed. $\qedsymbol$

$\mathcal{T}$ so defined is called the closure topology of with respect to the closure operator .

Remarks.

A closure space can be more generally defined as a set 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.
Even more generally, a closure space can be defined as a set and an operator $\operatorname{cl}$ on 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:
$\displaystyle A\subseteq \operatorname{cl}(B)$ iff $\displaystyle \quad\operatorname{cl}(A)\subseteq \operatorname{cl}(B).$
In a closure space , a subset of is said to be closed if $\operatorname{cl}(A)=A$ . Let be the set of all closed sets of . It is not hard to see that if is closed under $\cup$ , then $\operatorname{cl}$ “distributes over” $\cup$ , that is, we have the equality $\operatorname{cl}(A)\cup \operatorname{cl}(B)= \operatorname{cl}(A\cup B)$ .
Also, $\operatorname{cl}(\varnothing)$ is the smallest closed set in ; it is the bottom element in . This means that if there are two disjoint closed sets in , then $\operatorname{cl}(\varnothing)=\varnothing$ . This is equivalent to saying that $\varnothing$ is closed whenever there exist $A,B\subseteq X$ such that $\operatorname{cl}(A)\cap\operatorname{cl}(B)=\varnothing$ .
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.