Class of All Ordinals is Minimally Superinductive over Successor Mapping
We need to show that:
We recall immediately that Successor Mapping is Progressing.
By definition of ordinal:
- $S \in \On$
Thus by definition of class intersection:
- $\On$ is a subclass of $\cap S$.
We have by definition of superinductive class:
- $\O \in M$
Thus as $M$ is arbitrary:
- $\O \in \cap S$
- $\O \in \On$
Now let $\alpha \in \On$.
Then by definition of ordinal:
Hence by definition of superinductive class:
where $\alpha^+$ denotes the successor set of $\alpha$.
Hence by definition:
- $\alpha^+ \in \On$
Let $C$ be a chain in $\On$.
By definition of superinductive class:
where $\cup C$ denotes the union of $C$.
- $\cup C \in \On$
demonstrating that $\On$ is closed under chain unions.
Thus $\On$ is a superinductive class.
- $\exists x \in \On: x \notin A$
Hence $\On \not \subseteq A$
It remains to prove uniqueness.
Then we have:
- $\On' \subseteq \On$
but also we have:
- $\On \subseteq \On'$
- $\On' = \On$
- 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