Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element/Proof 1
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Theorem
Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Let $x$ be a fixed point of $g$.
Then $x$ is the greatest element of $M$.
Proof
Let $x$ be an element of $M$ such that $\map g x = x$.
From Empty Set is Subset of All Sets, we have that:
- $\O \subseteq x$
Suppose that $y \subseteq x$.
Then by Characteristics of Minimally Inductive Class under Progressing Mapping:
- $\map g y \subseteq \map g x$
But we have that $\map g x = x$.
Thus:
- $\map g y \subseteq x$
That is:
- $\O \subseteq x$
and:
- $y \subseteq x \implies \map g y \subseteq x$
and the result follows by the Principle of General Induction.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.14$