Class of All Ordinals is Only Proper Class of Ordinals
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Theorem
Let $A$ be a transitive proper class of ordinals.
Then $A$ is the class of all ordinals $\On$.
Proof
Let $A$ be a transitive class of ordinals.
Let there exist $\alpha \in \On$ such that $\alpha \notin A$.
Then by Transitive Class of Ordinals is Subset of Ordinal not in it:
- $A \subseteq \alpha$
But that makes $A$ a set.
So if $A$ is a proper class, it must contain all ordinals.
That is:
- $A = \On$
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 2$ Ordinals and transitivity: Theorem $2.2 \ (2)$