# Transfinite Recursion Theorem/Theorem 2

## Theorem

Let $\Dom x$ denote the domain of $x$.

Let $\Img x$ denote the image of the mapping $x$.

Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions:

$(1): \quad x = \O$ and $y = a$

$(2): \quad \exists \beta: \Dom x = \beta^+$ and $y = \map H {\map x {\bigcup \Dom x} }$

$(3): \quad \Dom x$ is a limit ordinal and $y = \bigcup \Rng x$.

Let $\map F \alpha = \map G {F \restriction \alpha}$ for all ordinals $\alpha$.

Then:

$F$ is a mapping and the domain of $F$ is the class of all ordinals, $\On$.
$\map F \O = a$
$\map F {\beta^+} = \map H {\map F \beta}$
For limit ordinals $\beta$, $\ds \map F \beta = \bigcup_{\gamma \mathop \in \beta} \map F \gamma$
$F$ is unique.
That is, if there is another function $A$ satisfying the above three properties, then $A = F$.

## Proof

 $\ds \map F \O$ $=$ $\ds \map G {F \restriction \O}$ by hypothesis $\ds$ $=$ $\ds \map G \O$ Restriction of $\O$ $\ds$ $=$ $\ds a$ Definition of $G$

$\Box$

 $\ds \map F {\beta^+}$ $=$ $\ds \map G {F \restriction \beta^+}$ by hypothesis $\ds$ $=$ $\ds \map H {\map {F \restriction \beta^+} {\bigcup \beta^+} }$ Definition of $G$ $\ds$ $=$ $\ds \map H {\map F \beta}$ Union of successor set is the original set by Union of Ordinals is Least Upper Bound

$\Box$

 $\ds \map F \beta$ $=$ $\ds \map G {F \restriction \beta}$ by hypothesis $\ds$ $=$ $\ds \bigcup \Img {F \restriction \beta}$ Definition of $G$ $\ds$ $=$ $\ds \bigcup_{\gamma \mathop \in \beta} \map F \gamma$

$\Box$

We can proceed in the fourth part by Transfinite Induction.

### Basis for the Induction

 $\ds \map F \O$ $=$ $\ds a$ First part $\ds$ $=$ $\ds \map A \O$ by hypothesis

This proves the basis for the induction.

$\Box$

### Induction Step

 $\ds \map F \beta$ $=$ $\ds \map A \beta$ $\ds \leadsto \ \$ $\ds \map H {\map F \beta}$ $=$ $\ds \map H {\map A \beta}$ Substitutivity of equality $\ds \leadsto \ \$ $\ds \map F {\beta^+}$ $=$ $\ds \map A {\beta^+}$ Second part

This proves the induction step.

$\Box$

### Limit Case

 $\ds \forall \gamma \in \beta: \,$ $\ds \map F \gamma$ $=$ $\ds \map A \gamma$ $\ds \leadsto \ \$ $\ds \bigcup_{\gamma \mathop \in \beta} \map F \gamma$ $=$ $\ds \bigcup_{\gamma \mathop \in \beta} \map A \gamma$ Substitutivity of equality $\ds \leadsto \ \$ $\ds \map F \beta$ $=$ $\ds \map A \beta$ Third part

This proves the limit case.

$\blacksquare$