Transfinite Recursion/Corollary

Theorem
Let $x$ be an ordinal.

Let $G$ be a mapping

There exists a unique mapping $f$ that satisfies the following properties:


 * The domain of $f$ is $x$
 * $\forall y \in x: \map f y = \map G {f \restriction y}$

Proof
Construct $K$ and $F$ as in the First Principle of Transfinite Recursion.

Set $f = \paren {F \restriction x}$.

Then since $x \subseteq \Dom F$, the domain of $f$ is $x$.

Thus such a mapping $f$ exists.

Suppose there are two mappings $f$ and $g$ that satisfy these conditions.

We will use the first principle of transfinite induction to show that $f = g$.

Take $y \in x$.

Suppose $\forall z \in y: \map f z = \map g z$.

Then:
 * $\paren {f \restriction y} = \paren {g \restriction y}$

and so:

So $\map f y = \map g y$ for all $y \in x$ by transfinite induction.

Since $x$ is the domain of $f$ and $g$, it follows that $f = g$.

Thus the mapping $f$ is unique.