Union of Ordinals is Ordinal

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.

$\blacksquare$