[proofplan] The proof has two parts. First, we verify that the closure operator $A \mapsto \overline{A}$ in a topological space satisfies the four Kuratowski axioms directly from the definition $\overline{A} = \bigcap\{F : F \text{ closed}, A \subset F\}$. Second, given an abstract map $c$ satisfying the four axioms, we show $\tau := \{U \subset X : c(X \setminus U) = X \setminus U\}$ is a topology and $c$ is its closure operator. The topology verification checks the three axioms (empty set and $X$ are open, closure under arbitrary unions, closure under finite intersections), and the identification of $c$ with the closure operator uses the fact that $c$-fixed points are exactly the closed sets. [/proofplan] [step:Verify normalisation: $\overline{\varnothing} = \varnothing$] The set $\varnothing$ is closed (its complement $X$ is open). The closure $\overline{\varnothing}$ is the intersection of all closed sets containing $\varnothing$. Since $\varnothing$ is itself closed and $\varnothing \subset \varnothing$, the set $\varnothing$ appears in this intersection. Therefore $\overline{\varnothing} \subset \varnothing$. The reverse inclusion $\varnothing \subset \overline{\varnothing}$ holds vacuously. [/step] [step:Verify extensivity: $A \subset \overline{A}$] Every closed set $F$ with $A \subset F$ satisfies $A \subset F$. Taking the intersection over all such $F$, we get $A \subset \bigcap\{F : F \text{ closed}, A \subset F\} = \overline{A}$. [/step] [step:Verify idempotency: $\overline{\overline{A}} = \overline{A}$] The set $\overline{A}$ is closed (it is an intersection of closed sets). Since $\overline{A}$ is a closed set containing $\overline{A}$, it appears in the family defining $\overline{\overline{A}}$, so $\overline{\overline{A}} \subset \overline{A}$. The reverse inclusion follows from extensivity applied with $\overline{A}$ in place of $A$. [/step] [step:Verify preservation of binary unions: $\overline{A \cup B} = \overline{A} \cup \overline{B}$] **Inclusion $\overline{A} \cup \overline{B} \subset \overline{A \cup B}$:** Since $A \subset A \cup B$, monotonicity of closure gives $\overline{A} \subset \overline{A \cup B}$. Similarly $\overline{B} \subset \overline{A \cup B}$, so $\overline{A} \cup \overline{B} \subset \overline{A \cup B}$. **Inclusion $\overline{A \cup B} \subset \overline{A} \cup \overline{B}$:** The set $\overline{A} \cup \overline{B}$ is closed (a finite union of closed sets is closed). Since $A \subset \overline{A}$ and $B \subset \overline{B}$ by extensivity, $A \cup B \subset \overline{A} \cup \overline{B}$. The closure $\overline{A \cup B}$ is contained in every closed set that contains $A \cup B$, so $\overline{A \cup B} \subset \overline{A} \cup \overline{B}$. [guided] We prove both inclusions separately. **Forward ($\supset$):** We use the monotonicity of closure, which itself follows from the definition. If $A \subset B$, then every closed set containing $B$ also contains $A$, so the intersection defining $\overline{A}$ is taken over a larger family, giving $\overline{A} \subset \overline{B}$. Since $A \subset A \cup B$, monotonicity gives $\overline{A} \subset \overline{A \cup B}$. By the same argument applied to $B \subset A \cup B$, we get $\overline{B} \subset \overline{A \cup B}$. Taking the union: $\overline{A} \cup \overline{B} \subset \overline{A \cup B}$. **Backward ($\subset$):** We need a closed set containing $A \cup B$ to bound $\overline{A \cup B}$ from above. The set $\overline{A} \cup \overline{B}$ is a candidate: it is closed because $\overline{A}$ and $\overline{B}$ are closed and a finite union of closed sets is closed (the complement $X \setminus (\overline{A} \cup \overline{B}) = (X \setminus \overline{A}) \cap (X \setminus \overline{B})$ is a finite intersection of open sets, hence open). It contains $A \cup B$ because $A \subset \overline{A}$ and $B \subset \overline{B}$. Since $\overline{A \cup B}$ is the smallest closed set containing $A \cup B$, we conclude $\overline{A \cup B} \subset \overline{A} \cup \overline{B}$. [/guided] [/step] [step:Define $\tau$ from the abstract operator $c$ and verify $\varnothing, X \in \tau$] Given a map $c: \mathcal{P}(X) \to \mathcal{P}(X)$ satisfying the four axioms, define \begin{align*} \tau := \{ U \subset X : c(X \setminus U) = X \setminus U \}. \end{align*} We call a set $F$ *$c$-closed* if $c(F) = F$, so $\tau$ consists of complements of $c$-closed sets. **$\varnothing \in \tau$:** We need $c(X \setminus \varnothing) = c(X) = X$. By extensivity, $X \subset c(X)$. Since $c(X) \subset X$ (as $c$ maps $\mathcal{P}(X)$ to $\mathcal{P}(X)$), equality holds. **$X \in \tau$:** We need $c(X \setminus X) = c(\varnothing) = \varnothing$, which is the normalisation axiom. [/step] [step:Verify $\tau$ is closed under arbitrary unions] Let $\{U_i\}_{i \in I}$ be a family of sets in $\tau$, so $c(X \setminus U_i) = X \setminus U_i$ for each $i \in I$. Set $U := \bigcup_{i \in I} U_i$, so $X \setminus U = \bigcap_{i \in I}(X \setminus U_i)$. We must show $c(X \setminus U) = X \setminus U$. By extensivity, $X \setminus U \subset c(X \setminus U)$. For the reverse, note that $X \setminus U \subset X \setminus U_i$ for each $i$. By monotonicity of $c$ (which we derive below), $c(X \setminus U) \subset c(X \setminus U_i) = X \setminus U_i$. Since this holds for every $i \in I$, $c(X \setminus U) \subset \bigcap_{i \in I}(X \setminus U_i) = X \setminus U$. *Monotonicity of $c$:* If $S \subset T$, then $T = S \cup (T \setminus S)$, so $c(T) = c(S \cup (T \setminus S)) = c(S) \cup c(T \setminus S) \supset c(S)$ by the union axiom and extensivity. [guided] Let $\{U_i\}_{i \in I}$ be a family in $\tau$ and set $U := \bigcup_{i \in I} U_i$. By De Morgan's law, $X \setminus U = \bigcap_{i \in I}(X \setminus U_i)$. We need $c(X \setminus U) = X \setminus U$. Extensivity gives the easy inclusion $X \setminus U \subset c(X \setminus U)$. For the reverse, we need a monotonicity property of $c$: if $S \subset T$, then $c(S) \subset c(T)$. We derive this from the axioms. Write $T = S \cup (T \setminus S)$. By preservation of binary unions, $c(T) = c(S) \cup c(T \setminus S)$, so $c(S) \subset c(T)$. Now, for each $i \in I$, $X \setminus U \subset X \setminus U_i$, so by monotonicity $c(X \setminus U) \subset c(X \setminus U_i)$. Since $U_i \in \tau$, we have $c(X \setminus U_i) = X \setminus U_i$. Therefore $c(X \setminus U) \subset X \setminus U_i$ for every $i$, and taking the intersection: \begin{align*} c(X \setminus U) \subset \bigcap_{i \in I}(X \setminus U_i) = X \setminus U. \end{align*} Combined with extensivity, $c(X \setminus U) = X \setminus U$, so $U \in \tau$. [/guided] [/step] [step:Verify $\tau$ is closed under finite intersections] Let $U_1, U_2 \in \tau$. Then $X \setminus (U_1 \cap U_2) = (X \setminus U_1) \cup (X \setminus U_2)$, and by preservation of binary unions: \begin{align*} c(X \setminus (U_1 \cap U_2)) = c((X \setminus U_1) \cup (X \setminus U_2)) = c(X \setminus U_1) \cup c(X \setminus U_2) = (X \setminus U_1) \cup (X \setminus U_2) = X \setminus (U_1 \cap U_2). \end{align*} Hence $U_1 \cap U_2 \in \tau$. By induction, $\tau$ is closed under all finite intersections. [/step] [step:Show that $c$ is the closure operator of $(X, \tau)$] Denote the closure in $(X, \tau)$ by $\operatorname{cl}$. The closed sets of $\tau$ are exactly the $c$-closed sets (sets $F$ with $c(F) = F$), by construction. We show $c(A) = \operatorname{cl}(A)$ for every $A \subset X$. By idempotency, $c(c(A)) = c(A)$, so $c(A)$ is $c$-closed, hence closed in $\tau$. By extensivity, $A \subset c(A)$. So $c(A)$ is a closed set containing $A$, giving $\operatorname{cl}(A) \subset c(A)$. For the reverse, $\operatorname{cl}(A)$ is closed in $\tau$, so $c(\operatorname{cl}(A)) = \operatorname{cl}(A)$. Since $A \subset \operatorname{cl}(A)$, monotonicity gives $c(A) \subset c(\operatorname{cl}(A)) = \operatorname{cl}(A)$. [guided] Let $\operatorname{cl}(A) := \bigcap\{F \subset X : F \text{ is } \tau\text{-closed and } A \subset F\}$ denote the closure in the topology $\tau$. We show $c(A) = \operatorname{cl}(A)$ for all $A \subset X$. **$\operatorname{cl}(A) \subset c(A)$:** We show $c(A)$ is a $\tau$-closed set containing $A$, so it is one of the sets in the intersection defining $\operatorname{cl}(A)$. By extensivity, $A \subset c(A)$. By idempotency, $c(c(A)) = c(A)$, so $c(A)$ is $c$-fixed, meaning $c(A)$ is a $c$-closed set. By the definition of $\tau$, the complement $X \setminus c(A)$ belongs to $\tau$, so $c(A)$ is $\tau$-closed. Hence $\operatorname{cl}(A) \subset c(A)$. **$c(A) \subset \operatorname{cl}(A)$:** The set $\operatorname{cl}(A)$ is $\tau$-closed, so its complement is in $\tau$, meaning $c(\operatorname{cl}(A)) = \operatorname{cl}(A)$. Since $A \subset \operatorname{cl}(A)$ and $c$ is monotone, $c(A) \subset c(\operatorname{cl}(A)) = \operatorname{cl}(A)$. [/guided] [/step]