Equivalence of Definitions of Closure Operator

Theorem
The following definitions of closure operator are equivalent:

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

Let $\operatorname{cl}: S \to S$ be a mapping.

Definition 1 implies Definition 2
Let $ \operatorname{cl}$ be an inflationary, increasing, and idempotent mapping.

We must show that for all $x, y \in S$:
 * $x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$.

Reverse implication
Suppose that $\operatorname{cl}(x) \preceq \operatorname{cl}(y)$.

Definition 2 implies Definition 1
Suppose that $x \preceq \operatorname{cl}(y) \iff \operatorname{cl}(x) \preceq \operatorname{cl}(y)$.

We must show that $\operatorname{cl}$ is inflationary, increasing, and idempotent.

Inflationary
Let $x \in S$.

Because $\preceq$ is reflexive, $\operatorname{cl}(x) \preceq \operatorname{cl}(x)$.

Thus $x \preceq \operatorname{cl}(x)$ by our supposition.

Increasing
Let $x, y \in S$ with $x \preceq y$.

Because $\operatorname{cl}$ is inflationary:
 * $y \preceq \operatorname{cl}(y)$

Since $\preceq$ is transitive:
 * $x \preceq \operatorname{cl}(y)$.

By our supposition:
 * $\operatorname{cl}(x) \preceq \operatorname{cl}(y)$

Idempotent
Let $x \in S$.

Since $\preceq$ is reflexive:
 * $\operatorname{cl}(x) \preceq \operatorname{cl}(x)$

By the supposition (in the forward direction):
 * $\operatorname{cl}(\operatorname{cl}(x)) \preceq \operatorname{cl}(x)$

Since $\operatorname{cl}$ is inflationary:
 * $\operatorname{cl}(x) \preceq \operatorname{cl}(\operatorname{cl}(x))$

Thus because $\preceq$ is antisymmetric:
 * $\operatorname{cl}(\operatorname{cl}(x)) = \operatorname{cl}(x)$