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. ∎
$\mathcal{T}$ 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 $\mathrm{cl}:P(X)\to P(X)$ 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

–
$A\subseteq \mathrm{cl}(A)$,

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

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$\mathrm{cl}$ is orderpreserving, 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.
References
 1 N. M. Martin, S. Pollard: Closure Spaces and Logic, Springer, (1996).
Title  closure space 

Canonical name  ClosureSpace 
Date of creation  20130322 16:48:08 
Last modified on  20130322 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 