Kuratowski: operators satisfying $c(\varnothing) = \varnothing$, $A \subseteq c(A)$, $c(c(A)) = c(A)$ 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) $c(\varnothing) = \varnothing$;
• (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 $c_{\mathcal{T}} : A \mapsto \overline{A}$, 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 $\varnothing$ 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 $c = \mathrm{id}$ and the indiscrete topology to the operator sending $\varnothing$ to $\varnothing$ 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 ↗).