No Largest Ordinal

Theorem
Let $a$ be a set which is a subset of the ordinal class, $\operatorname{On}$.


 * $\displaystyle \forall x \in a: x \prec \left({\bigcup a}\right)^+$

Proof
For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably. We are justified in doing this because of Ordering on an Ordinal is Subset Relation and Ordinal Proper Subset Membership.

Remark
This theorem allows us to create an ordinal strictly greater than any ordinal in the set. Thus, this is another means of proving the Burali-Forti Paradox. If the ordinals are a set, then we may construct an ordinal greater than the set of all ordinals, a contradiction.