Slow g-Tower is Slowly Well-Ordered under Subset Relation

Theorem

Let $M$ be a class.

Let $g: M \to M$ be a slowly progressing mapping on $M$.

Let $M$ be a slow $g$-tower.

Then $M$ is slowly well-ordered under the subset relation.

Proof

Let $M$ be a slow $g$-tower.

By $g$-Tower is Well-Ordered under Subset Relation: Empty Set:

$\O$ is the smallest element of $M$.

This is condition $(\text S_1)$ of the definition of slowly well-ordered class under subset relation.

By $g$-Tower is Well-Ordered under Subset Relation: Union of Limit Elements:

Each limit element $x$ of $M$ is the union of its lower section $\ds \bigcup x^\subset$

This is condition $(\text S_3)$ of the definition of slowly well-ordered class under subset relation.

It remains to establish condition $(\text S_2)$ of the definition of slowly well-ordered class under subset relation.

Let $x \in M$ be arbitrary.

Let $y$ be the immediate successor of $x$.

Then $x$ is not the greatest element of $M$.

Hence by $g$-Tower is Well-Ordered under Subset Relation: Successor of Non-Greatest Element:

$y = \map g x$

Because $g$ is slowly progressing, $y$ contains at most $1$ more element than $x$.

But we also have that $y$ is the immediate successor of $x$.

Thus $x$ is a proper subset of $y$.

So $y$ contains at least $1$ more element than $x$.

Hence $y$ contains exactly $1$ element which is not in $x$.

The result follows.

$\blacksquare$

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 $4.3$