Class of All Cardinals is Subclass of Class of All Ordinals

Theorem
Let $\mathcal N$ denote the class of all cardinal numbers.

Let $\operatorname{On}$ denote the class of all ordinals.

Then:


 * $\mathcal N \subseteq \operatorname{On}$

Proof
By definition of the cardinal class:


 * $\mathcal N = \left\{ x \in \operatorname{On} : \exists y: x = \left|{ y }\right| \right\}$

Every element of $\mathcal N$ is thus an element of $\operatorname{On}$.