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