Union of Subset of Ordinals is Ordinal

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Let $A$ be a class of ordinals.

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

Then $\bigcup A$ is an ordinal.


Let $y$ be a set.

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

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


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

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


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

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.