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.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 4$ Well ordering and choice: Theorem $4.6$