Class of All Ordinals is Proper Class

From ProofWiki
Jump to navigation Jump to search


Let $\On$ denote the class of all ordinals.

Then $\On$ is a proper class.

That is, $\On$ is not a set.

Proof 1

We have that Successor Mapping on Ordinals is Strictly Progressing.

The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.


Proof 2

Aiming for a contradiction, suppose $\On$ is a set.

Then from the Burali-Forti Paradox, a contradiction could be deduced.

Hence by Proof by Contradiction, $\On$ cannot be a set.

Thus $\On$ is a class that is not a set.

Hence $\On$ is a proper class.