# Supremum Inequality for Ordinals

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## 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.

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 8.6$