Transfinite Recursion Theorem/Theorem 1

Theorem
Let $G$ be a mapping.

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

where $f \restriction x$ denotes the restriction of $f$ to $x$.
 * The domain of $f$ is some ordinal $y$.
 * $\forall x \in y: f \left({x}\right) = G \left({f \restriction x}\right)$

Let $F = \bigcup K$, the union of $K$.

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

$K$ is a chain
First, it is necessary to must show that $K$ is a chain of mappings under the subset relation.

Take any $f, g \in K$.

Set $B$ equal to the domain of $f$ and set $C$ equal to the domain of $g$.

From Relation between Two Ordinals:
 * $B \subseteq C \lor C \subseteq B$

From Lemma 1:
 * $\forall x \in B \cap C: f \left({x}\right) = g \left({x}\right)$

So depending on whether $B \subseteq C$ or $C \subseteq B$:
 * $f \subseteq g \lor g \subseteq f$

From Subset Relation is Ordering, it follows that $\subseteq$ is a total ordering on $K$.

Domain of $F$ is an ordinal
$K$ is a chain of mappings.

Since the union of a chain of mappings is a mapping, $F$ is a mapping.

Moreover, $\forall x \in B: f \left({x}\right) = F \left({x}\right)$, 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 mappings whose domains are themselves subsets of ordinals.

From this it follows that $\operatorname{Dom} \left({F}\right)$ is an ordinal, since it is a transitive subset of $\operatorname{On}$.

Domain of $F$ is $\operatorname{On}$
Assume the domain of $F$ is a set.

Since $F$ is a mapping, 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:


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

Then $C$ is a mapping with domain $\gamma^+$ and satisfies:
 * $\forall x < \gamma^+: C\left({x}\right) = G \left({ F \restriction \gamma }\right)$

Therefore $C$ 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, $\operatorname{Dom} \left({F}\right)$ cannot be a set and must therefore be the set of all ordinals $\operatorname{On}$.

Value of $F$ is $G \left({F \restriction \alpha}\right)$

 * $\forall x \in \operatorname{On}: \exists f \in K: f \left({x}\right) = F \left({x}\right)$

Since $f \in K$:
 * $f\left({x}\right) = G \left({f \restriction x }\right)$

But $\forall y < x: f \left({y}\right) = F \left({y}\right)$, so by Equality of Restrictions:
 * $f \restriction x = F \restriction x$

Therefore,:


 * $F \left({\alpha}\right) = G \left({F \restriction \alpha}\right)$

$F$ is Unique
Finally, assume there is another mapping $H$ that satisfies the first two properties.

Then $H$ is equal to $F$ by transfinite induction:

So for all ordinals:
 * $H \left({y}\right) = F \left({y}\right)$

Thus $H = F$.

Also see

 * Well-Founded Recursion
 * Transfinite Induction
 * Burali-Forti Paradox