Mapping from Set to Class of All Ordinals is Bounded Above

Theorem

Let $x$ be a set.

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

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

Then $f$ has an upper bound.

Corollary

Let $\sequence {x_n}$ be any ordinal-valued sequence.

Then $\sequence {x_n}$ is bounded above.

Proof

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

By Union of Set 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 $f$.

$\blacksquare$