Closed Elements Uniquely Determine Closure Operator

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $f, g: S \to S$ be closure operators on $S$.

Suppose that $f$ and $g$ have the same closed elements.

Then $f = g$.

Proof
Let $C = f(S)$. Then $C$ is the set of closed elements of $S$ with respect to either $f$ or $g$.

Let $x \in S$.

By Closure is Smallest Closed Successor, $f(x)$ and $g(x)$ are smallest closed successors of $x$.

By Smallest Element is Unique, $f(x) = g(x)$.

Since this holds for all $x \in S$, $f = g$ by Equality of Mappings.