Equivalence of Definitions of Closure Operator

Proof
Let $\left({S, \preceq}\right)$ 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.

It is necessary to show that for all $x, y \in S$:
 * $x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

Sufficient Condition
Suppose that $\operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$.

Definition 2 implies Definition 1
Suppose that:
 * $x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

It is necessary to show that $\operatorname{cl}$ is inflationary, increasing and idempotent.

Inflationary
Let $x \in S$.

That is, $\operatorname{cl}$ is inflationary.

Increasing
It has been demonstrated that $\preceq$ is inflationary.

Let $x, y \in S$ such that $x \preceq y$.

Then:

That is, $\operatorname{cl}$ is increasing.

Idempotent
It has been demonstrated that $\preceq$ is inflationary.

Let $x \in S$.

Then:

Then as $\operatorname{cl}$ is inflationary:
 * $\operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right)$

As $\preceq$ is antisymmetric by dint of being an ordering:
 * $\operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right) = \operatorname{cl} \left({x}\right)$

That is, $\operatorname{cl}$ is idempotent.

Thus $\operatorname{cl}$ has been shown to be inflationary, increasing and idempotent as required.