Union of Subset of Ordinals is Ordinal

From ProofWiki
Jump to navigation Jump to search

Theorem


Let $A$ be a class of ordinals. That is, $A \subseteq \operatorname{On}$, where $\operatorname{On}$ denotes the ordinal class.

Then $\bigcup A$ is an ordinal.


Corollary

Let $y$ be a set.

Let $\operatorname{On}$ be the class of all ordinals.

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


Then:

$\displaystyle \bigcup F \left({y}\right) \in \operatorname{On}$

where $F \left({y}\right)$ is the image of $y$ under $F$.


Proof

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


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

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

$\displaystyle \bigcup A \subseteq A \subseteq \operatorname{On}$

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