Union of Ordinals is Ordinal

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Theorem

Let $y$ be a set.

Let $F \left({z}\right)$ be a mapping such that:

$\displaystyle \forall z \in y: F \left({z}\right) \in \operatorname{On}$


Then:

$\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right) \in \operatorname{On}$


Proof

\(\displaystyle x \in \bigcup_{z \mathop \in y} F \left({z}\right)\) \(\implies\) \(\displaystyle \exists z \in y: x \in F \left({z}\right)\) Definition of Set Union
\(\displaystyle \) \(\implies\) \(\displaystyle x \in \operatorname{On}\) by hypothesis

So:

$\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right) \subseteq \operatorname{On}$
\(\displaystyle x \in \bigcup_{z \mathop \in y} F \left({z}\right)\) \(\implies\) \(\displaystyle \exists z \in y: x \in F \left({z}\right)\) Definition of Set Union
\(\displaystyle \) \(\implies\) \(\displaystyle \exists z \in y: x \subseteq F \left({z}\right)\) Ordinals are Transitive
\(\displaystyle \) \(\implies\) \(\displaystyle x \subseteq \bigcup_{z \mathop \in y} F \left({z}\right)\) Union Preserved Under Subset Relation

So $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is a transitive set.

Therefore $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is an ordinal.

\(\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)\) \(=\) \(\displaystyle \bigcup \operatorname{Im} \left({y}\right)\)

Let $U$ denote the universal class.

\(\displaystyle y \in V\) \(\implies\) \(\displaystyle \operatorname{Im} \left({y}\right) \in V\) Axiom of Replacement Equivalents
\(\displaystyle \) \(\implies\) \(\displaystyle \bigcup \operatorname{Im} \left({y}\right) \in V\) Axiom of Unions Equivalents
\(\displaystyle \) \(\implies\) \(\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right) \in V\) Equality given above

Therefore $\displaystyle \bigcup_{z \mathop \in y} F \left({z}\right)$ is a set, so it is a member of the ordinal class.

$\blacksquare$