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