G-Tower is Nest/Lemma 2

Lemma for $g$-Tower is Nest

Let $M$ be a class.

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

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

Then:

$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

Proof

First a lemma:

Lemma 1

Let $M$ be a class.

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

Let $\RR$ be a relation defined as:

$\forall x, y \in M: \tuple {x, y} \in \RR \iff \map g x \subseteq y \lor y \subseteq x$

where $\lor$ denotes disjunction (inclusive "or").

Then $\RR$ satisfies the $3$ conditions $\text D_1$, $\text D_2$ and $\text D_3$ of the Double Superinduction Principle.

That is:

\((\text D_1)\)	$:$			\(\ds \forall x \in M:\)	\(\ds \map \RR {x, \O} \)
\((\text D_2)\)	$:$			\(\ds \forall x, y \in M:\)	\(\ds \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y} \)
\((\text D_3)\)	$:$			\(\ds \forall x \in M: \forall C: \forall y \in C:\)	\(\ds \map \RR {x, y} \implies \map \RR {x, \bigcup C} \)				where $C$ is a chain of elements of $M$

$\Box$

By definition of $g$-tower:

$M$ is minimally superinductive under $g$.

Hence by the Principle of Superinduction:

$\forall x, y \in M: \tuple {x, y} \in \RR$

That is:

$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

$\blacksquare$

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: $\S 2$ Superinduction and double superinduction: Theorem $2.3$