Superinductive Class under Strictly Progressing Mapping is Proper Class
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Theorem
Let $A$ be a class.
Let $g: A \to A$ be a strictly progressing mapping on $A$.
Let $A$ be superinductive under $g$.
Then $A$ cannot be a set, and thus is a proper class.
Proof
Aiming for a contradiction, suppose $A$ is a set.
Then from Set which is Superinductive under Progressing Mapping has Fixed Point, $A$ has a fixed point.
However, we have that $g$ is a strictly progressing mapping on $A$.
Hence a fortiori $g$ has no fixed point in $A$.
Hence the result by Proof by Contradiction.
$\blacksquare$
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 $2.9$