Closure Operator from Closed Elements

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

Let $C \subseteq S$.

Suppose that $C$ is a subset of $S$ with the property that every element of $S$ has a smallest successor in $C$.

Let $\operatorname{cl}: S \to S$ be defined as follows:

For $x \in S$, $\operatorname{cl}(x) = \min \left({ C \cap {\bar\uparrow}x }\right)$, where ${\bar\uparrow}x$ is the upper closure of $x$.

That is, let $\operatorname{cl}(x)$ be the smallest successor of $x$ in $C$.

Then:
 * $\operatorname{cl}$ is a closure operator on $S$.
 * The closed elements of $\operatorname{cl}$ are precisely the elements of $C$.

Inflationary
$x$ is a lower bound of ${\bar\uparrow} x$, so $x$ is a lower bound of $C \cap {\bar\uparrow} x$.

By the definition of smallest element, $x \preceq \operatorname{cl}(x)$.

Order-Preserving
Suppose that $x \preceq y$.

Then $C \cap {\bar\uparrow} y \subseteq C \cap {\bar\uparrow} x$.

By Smallest Element of Subset, $\operatorname{cl}(x) \preceq \operatorname{cl}(y)$.

Idempotent
Let $x \in S$.

For each $x \in S$, $\operatorname{cl}(x) = \min \left({ C \cap {\bar\uparrow}x }\right)$.

Thus $\operatorname{cl}(x) \in \left({ C \cap {\bar\uparrow}x }\right) \subseteq C$.

Thus $\operatorname{cl}(x) = \min \left({ C \cap {\bar\uparrow}\operatorname{cl}(x) }\right) = \operatorname{cl}(\operatorname{cl}(x))$

Elements of $C$ are closed elements with respect to $\operatorname{cl}$ because when $x \in C$, $x$ is the minimum of $C \cap {\bar\uparrow}x$.

Suppose that $x$ is closed with respect to $\operatorname{cl}$. Then $x = \min \left({ C \cap {\bar\uparrow}x }\right)$, so in particular $x \in C$.