Set with Slowly Progressing Mapping on Power Set with Self as Fixed Point is Well-Orderable

Theorem
Let $S$ be a set of sets.

Let there exist a slowly progressing mapping $g: \powerset S \to \powerset S$, where $\powerset S$ denotes the power set of $S$.

Let $g$ have exactly one fixed point $S$.

Then $S$ is well-orderable.

Proof
Let the hypothesis be assumed.

We have that:
 * $\O \in \powerset S$
 * $\powerset S$ is closed under $g$
 * $\powerset S$ is closed under chain unions.

Hence $\powerset S$ is superinductive under $g$.

Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.

From:
 * Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
 * Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions

it follows by definition that $M$ is minimally superinductive under $g$.

Hence by definition $M$ is a slow $g$-tower.

By Union of $g$-Tower is Greatest Element and Unique Fixed Point:
 * $\ds \bigcup M \in M$ and $\ds \bigcup M$ is a fixed point of $g$.

But $S$ is the only fixed point of $g$.

Hence:
 * $\ds \bigcup M = S$

From Union of Slow $g$-Tower is Well-Orderable, it follows that $S$ is well-orderable.