Supremum Inequality for Ordinals

Theorem
Suppose $A \subseteq \operatorname{On}$ and $B \subseteq \operatorname{On}$.


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

Proof
Therefore, $\bigcup A \subseteq \bigcup B$ and $\bigcup A \le \bigcup B$.

Warning
The converse of this statement does not hold.