# Class of All Ordinals is Proper Class

## Theorem

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

Then $\On$ is a proper class.

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

## Proof 1

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

$\blacksquare$

## Proof 2

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

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.
$\blacksquare$