Supremum Inequality for Ordinals

Theorem
Let $A \subseteq \On$ and $B \subseteq \On$ be ordinals.

Then:
 * $\ds \forall x \in A: \exists y \in B: x \le y \implies \bigcup A \le \bigcup B$

Proof
Therefore:
 * $\ds \bigcup A \subseteq \bigcup B$

and:
 * $\ds \bigcup A \le \bigcup B$

Warning
The converse of this statement does not hold.