# Union of Subset of Ordinals is Ordinal

## Theorem

Let $A$ be a class of ordinals. That is, $A \subseteq \operatorname{On}$, where $\operatorname{On}$ denotes the ordinal class.

Then $\bigcup A$ is an ordinal.

### Corollary

Let $y$ be a set.

Let $\operatorname{On}$ be the class of all ordinals.

Let $F: y \to \operatorname{On}$ be a mapping.

Then:

$\displaystyle \bigcup F \left({y}\right) \in \operatorname{On}$

where $F \left({y}\right)$ is the image of $y$ under $F$.

## Proof

 $\displaystyle x \in \bigcup A$ $\implies$ $\displaystyle \exists y \in A: x \in y$ $\displaystyle$ $\implies$ $\displaystyle \exists y \subseteq \bigcup A: x \subseteq y$ $\displaystyle$ $\implies$ $\displaystyle x \subseteq \bigcup A$

From this, we conclude that $\displaystyle \bigcup A$ is a transitive class.

From Class is Transitive iff Union is Subset, it follows that:

$\displaystyle \bigcup A \subseteq A \subseteq \operatorname{On}$

By Subset of Well-Ordered Set is Well-Ordered, $A$ is also well-ordered by $\Epsilon$.

Thus by Alternative Definition of Ordinal, $\bigcup A$ is an ordinal.