Union of Set of Ordinals is Ordinal

Theorem

Let $A$ be a set of ordinals.

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:

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

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

Proof 1

By Class of All Ordinals is Well-Ordered by Subset Relation, a set of ordinals forms a chain.

The result then follows from Union of Chain of Ordinals is Ordinal.

Proof 2

\(\ds x\)

\(\in\)

\(\ds \bigcup A\)

\(\ds \leadsto \ \ \)

\(\ds \exists y \in A: \, \)

\(\ds x\)

\(\in\)

\(\ds y\)

\(\ds \leadsto \ \ \)

\(\ds \exists y \subseteq \bigcup A: \, \)

\(\ds x\)

\(\subseteq\)

\(\ds y\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\subseteq\)

\(\ds \bigcup A\)

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

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

$\bigcup A \subseteq A \subseteq \On$

Note that the above applies as well to sets as it does to classes.

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.

$\blacksquare$