Kuratowski: operators satisfying $c(\mathrm{\varnothing })=\mathrm{\varnothing }$, $A\subseteq c(A)$, $c(c(A))=c(A)$ and $c(A\cup B)=c(A)\cup c(B)$ correspond bijectively to topologies

Let $X$ be a set. A Kuratowski closure operator on $X$ is a function $c:\mathcal{P}(X)\to \mathcal{P}(X)$ such that, for all $A,B\subseteq X$:

• (K1) $c(\mathrm{\varnothing })=\mathrm{\varnothing }$;
• (K2) $A\subseteq c(A)$;
• (K3) $c(c(A))=c(A)$;
• (K4) $c(A\cup B)=c(A)\cup c(B)$.

Then:

1. For every topology $\mathcal{T}$ on $X$ the operator $c_{\mathcal{T}}:A\mapsto \bar{A}$, the closure taken in $(X,\mathcal{T})$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space), is a Kuratowski closure operator on $X$.
2. For every Kuratowski closure operator $c$ on $X$ the family ${\mathcal{C}}_c:=\{A\subseteq X:c(A)=A\}$ of its fixed points satisfies the closed-set axioms (C1), (C2), (C3) of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, so ${\mathcal{T}}_c:=\{X\setminus A:A\in {\mathcal{C}}_c\}$ is a topology on $X$ whose closed sets are exactly the members of ${\mathcal{C}}_c$; and the closure operator of ${\mathcal{T}}_c$ is $c$ itself.
3. The assignments $\mathcal{T}\mapsto c_{\mathcal{T}}$ and $c\mapsto {\mathcal{T}}_c$ are mutually inverse, hence bijections (Injection, surjection, bijection) between the set of topologies on $X$ and the set of Kuratowski closure operators on $X$.

So a topology may be specified by naming its closure operator, and the four axioms above are exactly the conditions under which such a specification is legitimate. Note that monotonicity is not among the axioms: it is a consequence of (K4), derived in the proof.

• (K3) is what makes $c$ recoverable as the closure operator of its fixed-point topology. Dropping it leaves an operator whose fixed points still satisfy (C1), (C2) and (C3) — steps 1.3, 1.4 and 2.1 do not use it — but the closure operator of the resulting topology is then only the smallest fixed point above $A$, which need not be $c(A)$. It is step 4.1 that spends (K3).

• (K1) is genuinely independent of the others. The operator $c(A):=X$ for all $A$, on a nonempty $X$, satisfies (K2), (K3) and (K4) and fails (K1); its fixed points are $\{X\}$ alone, which is not the family of closed sets of any topology, since $\mathrm{\varnothing }$ is missing.

• The correspondence is order reversing in the natural sense. A finer topology has more closed sets, hence more fixed points, hence a smaller closure operator pointwise; the discrete topology corresponds to $c=\mathrm{i}\mathrm{d}$ and the indiscrete topology to the operator sending $\mathrm{\varnothing }$ to $\mathrm{\varnothing }$ and every nonempty set to $X$.

• How many sets can be produced by closure and complement together is a separate question with a finite answer, fourteen, worked out on the companion page (Closure and complement generate at most fourteen sets from any subset, and $(0,1)\cup (1,2)\cup \{3\}\cup ([4,5]\cap \mathbb{Q})$ attains fourteen ↗).