# Union of Subset of Ordinals is Ordinal

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## Contents

## Theorem

Let $A$ be a class of ordinals.

That is, $A \subseteq \On$, where $\On$ denotes the ordinal class.

Then $\bigcup A$ is an ordinal.

### Corollary

Let $y$ be a set.

Let $\On$ be the class of all ordinals.

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

Then:

- $\displaystyle \bigcup \map F y \in \On$

where $\map F y$ is the image of $y$ under $F$.

## Proof

\(\displaystyle x \in \bigcup A\) | \(\leadsto\) | \(\displaystyle \exists y \in A: x \in y\) | |||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \exists y \subseteq \bigcup A: x \subseteq y\) | |||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle x \subseteq \bigcup A\) |

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

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

- $\displaystyle \bigcup A \subseteq A \subseteq \On$

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.

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 7.19$