# Transfinite Recursion/Theorem 1

## Theorem

Let $G$ be a mapping.

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 $F = \bigcup K$, 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

### $K$ 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 $g$.

$B \subseteq C \lor C \subseteq B$

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 ordinals because it is the union of a collection of mappings whose domains are themselves subsets of ordinals.

 $\ds a \in \Dom F \land x \in a$ $\leadsto$ $\ds \exists f: \exists \beta \in \On: \paren {f: \beta \to \Img f \land a \in \beta}$ Definition of $F$ (above) $\ds$ $\leadsto$ $\ds a \in \On \land x \in \beta$ Transitivity of $\beta$ and $\On$ $\ds$ $\leadsto$ $\ds x \in \Dom F$

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 $\On$.

$\Box$

### 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:

$\map F \alpha = \map G {F {\restriction_\alpha} }$

$\Box$

### $F$ is Unique

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

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

 $\ds \forall x \in y: \map H x = \map F x$ $\leadsto$ $\ds \paren {H \restriction y} = \paren {F \restriction y}$ Equality of Restrictions $\ds$ $\leadsto$ $\ds \map G {H \restriction y} = \map G {F \restriction y}$ by hypothesis $\ds$ $\leadsto$ $\ds \map H y = \map F y$ by hypothesis

So for all ordinals:

$\map H y = \map F y$

Thus $H = F$.

$\blacksquare$