Class of All Ordinals is Only Proper Class of Ordinals

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 $2.2 \ (2)$