G-Tower is Nest
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Theorem
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is a nest.
Proof
We need to show that:
- $\forall x, y \in M: x \subseteq y$ or $y \subseteq x$
First some lemmata:
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 $M$ |
$\Box$
Lemma 2
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then:
- $\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
$\Box$
But as $g: M \to M$ is a progressing mapping:
- $\forall x \in M: x \subseteq \map g x$
The result follows by Subset Relation is Transitive.
$\blacksquare$
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 $2.3$