Minimally Inductive Class under Progressing Mapping with Fixed Element is Finite

Theorem

Let $M$ be a class which is minimally inductive under a progressing mapping $g$.

Let there exist an element $x \in M$ such that $x = \map g x$.

Then $M$ is a finite class.

Proof

By Set of Subsets of Element of Minimally Inductive Class under Progressing Mapping is Finite, the set:

$\set {y \in M: y \subseteq x}$

is finite.

From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $x$ is the greatest element of $M$.

Thus:

$M = \set {y \in M: y \subseteq x}$

Hence the result.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 6$ Finite Sets: Exercise $6.5$