Extending Operation is a Strictly Progressing Mapping
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Theorem
Let $S$ denote the class of all ordinal sequences.
Let $E: S \to S$ be an extending operation on $S$.
Then $E$ is a strictly progressing mapping.
Proof
Let $\theta \in S$ be an $\alpha$-sequence.
By definition of extending operation:
- $\map E \theta = \theta \cup \tuple {\alpha, x}$
where $x$ is arbitrary.
From Extending Operation is a Slowly Progressing Mapping we have a fortiori that $E$ is a progressing mapping.
By definition of $\alpha$-sequence:
- $\alpha \notin \Dom \theta$
and so:
- $\tuple {\alpha, x} \notin \theta$
Hence:
- $\theta \subsetneqq \map E \theta$
demonstrating that $E$ is a strictly progressing mapping.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems