Class of All Ordinals is Proper Class/Proof 1
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Theorem
Let $\On$ denote the class of all ordinals.
Then $\On$ is a proper class.
That is, $\On$ is not a set.
Proof
We have that Successor Mapping on Ordinals is Strictly Progressing.
The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.
$\blacksquare$
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: Theorem $1.10$