Transfinite Recursion Theorem/Theorem 1

Statement
Let $G$ be a function.

Let $K$ be a class of mappings $f$ that satisfy:


 * The domain of $f$ is some ordinal $y$.
 * $\forall x \in y: f(x) = G(f \restriction x)$

Let $F = \bigcup K$. Then,


 * $F$ is a function with domain $\operatorname{On}$
 * $\forall x \in \operatorname{On}: F(x) = G(F\restriction x)$
 * $F$ is unique. If another mapping $A$ has the above two properties, then $A = F$.

Proof
First, we must show that $K$ is a chain of functions under the subset relation.

Take any $f,g \in K$. We shall set $B$ equal to the domain of $f$ and set $C$ equal to the domain of $g$. $B \subseteq C \lor C \subseteq B$ (see Relation between Two Ordinals). By Lemma 1, we have that $\forall x \in B \cap C: f(x) = g(x)$. So $f \subseteq g \lor g \subseteq f$ (depending on whether $B \subseteq C \lor C \subseteq B$). Thus, $\subseteq$ creates a total ordering on $K$.

$K$ is a chain of functions. Since the union of a chain of functions is a function, $F$ is a function. Moreover, $\forall x \in B f(x) = F(x)$, because $F$ is an extension of the mapping $f$.

The domain of $F$ is a subset of the ordinals because it is the union of a collection of functions whose domains are themselves subsets of ordinals.

From this we can conclude that $\operatorname{Dom} F$ is an ordinal, since it is a transitive subset of $\operatorname{On}$.

Assume the domain of $F$ is a set. Since $F$ is a function, the range of $F$ is a set and $F$ is a set. Then $\operatorname{Dom} F = \gamma$ for some ordinal $\gamma$. Define $C$ as follows:


 * $\displaystyle C = F \cup \{ ( \gamma, G ( F \restriction \gamma ) ) \}$

Then $C$ is a mapping with domain $\gamma^+$ and satisfies $\forall x < \gamma^+: C(x) = G ( F \restriction \gamma )$, so $C$ is a member of $K$ (note that $F \restriction \gamma = C \restriction \gamma$).


 * $\displaystyle C \in K \implies C \subseteq \bigcup K$

So $C \subseteq F$. But if $F$ is a set, $F \subsetneq C$, a contradiction. Therefore, $\operatorname{Dom} F$ cannot be a set and must therefore be the set of all ordinals $\operatorname{On}$. This proves the first part.


 * $\displaystyle \forall x \in \operatorname{On}: \exists f \in K: f(x) = F(x)$. Since $f \in K$, $f(x) = G(f \restriction x )$.  But $\forall y < x: f(y) = F(y)$, so $f \restriction x = F \restriction x$.  Therefore,


 * $\displaystyle F(\alpha) = G(F \restriction \alpha)$

Finally, assume there is another function $H$ that satisfies the first two properties. Then, $H$ is equal to $F$ by transfinite induction.

So for all ordinals, $H(y) = F(y)$. Thus, $H = F$.

Source

 * : $\S 7.40$