Successor Mapping on Ordinals is Strictly Progressing
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Theorem
Let $\On$ denote the class of all ordinals.
Let $s$ denote the successor mapping on $\On$:
- $\forall \alpha \in \On: \map s \alpha := \alpha \cup \set \alpha$
where $\alpha$ is an ordinal.
Then $s$ is a strictly progressing mapping.
Proof
From Ordinal is Proper Subset of Successor:
- $\alpha \subsetneqq \alpha^+$
where $\alpha^+$ is identified as $\map s \alpha$.
The result follows by definition of strictly progressing mapping.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers