Class of All Cardinals is Subclass of Class of All Ordinals

Theorem
Let $\NN$ denote the class of all cardinals.

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

Then:


 * $\NN \subseteq \On$

Proof
By definition of the class of all cardinals:


 * $\NN = \set {x \in \On: \exists y: x = \card y}$

Every element of $\NN$ is thus an element of $\On$.