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 $A^c\mathrm{=}A$, and an “open set” of $X$ as the complement of some closed set of $X$. Then the collection $\mathrm{T}$ of all open sets of $X$ is a topology on $X$.

Proof.

Since ${\mathrm{\varnothing }}^c=\mathrm{\varnothing }$, $\mathrm{\varnothing }$ 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. ∎

$𝒯$ 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 $\mathrm{cl}:P(X)\to P(X)$ such that $\mathrm{cl}$ satisfies all of the Kuratowski’s closure axioms where the equal sign “$=$” is replaced with set inclusion “$\subseteq$”, and the preservation of $\mathrm{\varnothing }$ is no longer assumed.

2. Even more generally, a closure space can be defined as a set $X$ and an operator $\mathrm{cl}$ on $P(X)$ such that

   • $A\subseteq \mathrm{cl}(A)$,

   • $\mathrm{cl}(\mathrm{cl}(A))\subseteq \mathrm{cl}(A)$, and

   • $\mathrm{cl}$ is order-preserving, i.e., if $A\subseteq B$, then $\mathrm{cl}(A)\subseteq \mathrm{cl}(B)$.

   It can be easily deduced that $\mathrm{cl}(A)\cup \mathrm{cl}(B)\subseteq \mathrm{cl}(A\cup B)$. In general however, the equality fails. The three axioms above can be shown to be equivalent to a single axiom:

   $$A\subseteq \mathrm{cl}(B)\mathit{\hspace{1em}}\text{ iff }\mathit{\hspace{1em}}\mathrm{cl}(A)\subseteq \mathrm{cl}(B).$$

3. In a closure space $X$, a subset $A$ of $X$ is said to be closed if $\mathrm{cl}(A)=A$. 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 $\mathrm{cl}$ “distributes over” $\cup$, that is, we have the equality $\mathrm{cl}(A)\cup \mathrm{cl}(B)=\mathrm{cl}(A\cup B)$.

4. Also, $\mathrm{cl}(\mathrm{\varnothing })$ 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 $\mathrm{cl}(\mathrm{\varnothing })=\mathrm{\varnothing }$. This is equivalent to saying that $\mathrm{\varnothing }$ is closed whenever there exist $A,B\subseteq X$ such that $\mathrm{cl}(A)\cap \mathrm{cl}(B)=\mathrm{\varnothing }$.

5. Since the distributivity of $\mathrm{cl}$ over $\cup$ does not hold in general, and there is no guarantee that $\mathrm{cl}(\mathrm{\varnothing })=\mathrm{\varnothing }$, a closure space under these generalized versions is a more general system than a topological space.

Title	closure space
Canonical name	ClosureSpace
Date of creation	2013-03-22 16:48:08
Last modified on	2013-03-22 16:48:08
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	12
Author	CWoo (3771)
Entry type	Derivation
Classification	msc 54A05
Defines	closure topology