Transfinite Recursion Theorem/Theorem 1

Theorem
Let $G$ be a (class) mapping from $\On^{\On}$ to $\On$.

Let $K$ be a class of mappings $f$ that satisfy:
 * the domain of $f$ is some ordinal $y$
 * $\forall x \in y: \map f x = \map G {f {\restriction_x} }$

where $f {\restriction_x}$ denotes the restriction of $f$ to $x$.

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

Then:
 * $(1): \quad F$ is a mapping with domain $\On$
 * $(2): \quad \forall x \in \On: \map F x = \map G {F {\restriction_x} }$
 * $(3): \quad F$ is unique. That is, if another mapping $A$ has the above two properties, then $A = F$.

Proof
First a lemma:

$K$ is a chain
First, it is necessary to 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:
 * $\forall x \in B \cap C: \map f x = \map g x$

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, $\forall x \in B: \map f x = \map F x$, because $F$ is an extension of the mapping $f$.

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

From this it follows that $\Dom F$ is an ordinal, since it is a transitive subset of $\On$.

Domain of $F$ is $\On$
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 $\Dom F = \gamma$ for some ordinal $\gamma$.

Define $C$ as follows:
 * $C = F \cup \set {\tuple {\gamma, \map G {F {\restriction_\gamma} } } }$

Then $C$ is a mapping with domain $\gamma^+$ and satisfies:
 * $\forall x < \gamma^+: \map C x = \map G {F {\restriction_\gamma} }$

Therefore $C$ is a member of $K$ (note that $F {\restriction_\gamma} = C {\restriction_\gamma}$).
 * $\ds C \in K \implies C \subseteq \bigcup K$

So $C \subseteq F$.

But if $F$ is a set, $F \subsetneq C$, a contradiction.

Therefore, $\Dom F$ cannot be a set and must therefore be the set of all ordinals $\On$.

Value of $F$ is $\map G {F {\restriction_\alpha} }$

 * $\forall x \in \On: \exists f \in K: \map f x = \map F x$

Since $f \in K$:
 * $\map f x = \map G {f {\restriction_x} }$

But $\forall y < x: \map f y = \map F y$, so by Equality of Restrictions:
 * $f {\restriction_x} = F {\restriction_x}$

Therefore:
 * $\map F \alpha = \map G {F {\restriction_\alpha} }$

$F$ 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:
 * $\map H y = \map F y$

Thus:
 * $H = F$

Also see

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