Superinductive Class under Strictly Progressing Mapping is Proper Class

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 $2.9$