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 $\bigcup A$, the union of $A$, is the least upper bound of $A$:


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

Proof
First, we must show that $\displaystyle \bigcup A$ is an upper bound.

Take any member $a \in A$.

Then by Subset of Union:
 * $\displaystyle a \subseteq \bigcup A$

By Ordering on Ordinal is Subset Relation:
 * $a \le A$

By generalizing for all $a \in A$:


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

Similarly, suppose now that $x$ is an upper bound of $A$.

We shall denote $<$ for ordering on the ordinal numbers.

By Ordering on Ordinal is Subset Relation and Ordinal Proper Subset Membership, $<$ is the same as both $\in$ and $\subsetneq$.

Then:

Thus, by definition of subset:
 * $\displaystyle \bigcup A \subseteq x$

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