Supremum Inequality for Ordinals
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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
| \(\ds x\) | \(<\) | \(\ds \bigcup A\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists z: \, \) | \(\ds x\) | \(<\) | \(\ds z\) | Definition of Set Union | |||||||||
| \(\, \ds \land \, \) | \(\ds z < A\) | \(<\) | \(\ds A\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in B: \exists z: \, \) | \(\ds x\) | \(<\) | \(\ds z\) | by hypothesis | |||||||||
| \(\, \ds \land \, \) | \(\ds z\) | \(\le\) | \(\ds y\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in B: \, \) | \(\ds x\) | \(<\) | \(\ds y\) | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(<\) | \(\ds \bigcup B\) | Union of Ordinals is Least Upper Bound |
Therefore:
- $\ds \bigcup A \subseteq \bigcup B$
and:
- $\ds \bigcup A \le \bigcup B$
$\blacksquare$
Warning
The converse of this statement does not hold.
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Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.6$