Sandwich Principle for Slowly Progressing Mapping

Theorem

Let $M$ be a class.

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

Let $M$ be a minimally inductive class under $g$.

Then $x \subsetneqq y \subsetneqq \map g x$ can never hold.

Corollary

Let $N$ be a nest which is closed under $g$.

Then:

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

Proof

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Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional: Lemma $S_3 \ (1)$