# Supremum Inequality for Ordinals

## Theorem

Let $A \subseteq \operatorname{On}$ and $B \subseteq \operatorname{On}$ be ordinals.

Then:

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

## Proof

 $\displaystyle x < \bigcup A$ $\implies$ $\displaystyle \exists z: \left({x < z \land z < A}\right)$ Definition of Set Union $\displaystyle$ $\implies$ $\displaystyle \exists y \in B: \exists z: \left({x < z \land z \le y}\right)$ By Hypothesis $\displaystyle$ $\implies$ $\displaystyle \exists y \in B: x < y$ $\displaystyle$ $\implies$ $\displaystyle x < \bigcup B$ Union of Ordinals is Least Upper Bound

Therefore:

$\displaystyle \bigcup A \subseteq \bigcup B$

and:

$\displaystyle \bigcup A \le \bigcup B$

$\blacksquare$

## Warning

The converse of this statement does not hold.