# Mapping from Set to Ordinal Class is Bounded Above

## Theorem

Let $x$ be a set.

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

Let $f: x \to \operatorname{On}$ be a mapping.

Then $f$ has an upper bound.

### Corollary

Let $\left\langle{x_n}\right\rangle$ be any ordinal-valued sequence.

Then $\left\langle{x_n}\right\rangle$ is bounded above.

## Proof

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Let $I$ be the image of $f$.

By Union of Subset of Ordinals is Ordinal: Corollary, $\bigcup I$ is an ordinal.

But by Union is Smallest Superset, each element of $I$ is a subset of $\bigcup I$.

Thus $\bigcup I$ is an upper bound of $f$.

$\blacksquare$