Union of Ordinals is Least Upper Bound

Theorem
Let $A \subset \operatorname{On}$. That is, let $A$ be a class of ordinals (every member of $A$ is an ordinal).

Then, $\bigcup A$ is the least upper bound of $A$.

Proof
First, we must show that $\bigcup A$ is an upper bound. Take any member $a \in A$. Then, $a \subseteq \bigcup A$. By Ordering on an Ordinal is Subset Relation, $a \preceq A$. By generalizing for all $a \in A$,


 * $\displaystyle \forall x \in A: x \preceq \bigcup A$

Similarly, suppose now that $x$ is an upper bound of $A$. We shall denote $\lt$ for ordering on the ordinal numbers, and it is the same as both $\in$ and $\subsetneq$ by Ordering on an Ordinal is Subset Relation and Ordinal Proper Subset Membership


 * $\displaystyle

\begin{align*} z \in \bigcup A &\implies \exists y: ( z \in y \land y \in A ) \\ &\implies \exists y: ( z \prec y \land y \prec x ) \\ &\implies z \prec x \\ &\implies z \in x \end{align*}$

But this means that every member of $\bigcup A$ is a member of $x$, so $\bigcup A \subseteq x$. Therefore,


 * $\displaystyle \forall y \in A: y \preceq x \implies \bigcup A \preceq x$