Mapping from Set to Class of All Ordinals 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.

Proof
Let $I$ be the image of $f$:

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

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

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