Transfinite Recursion Theorem
Theorem
Formulation 1
Let $M$ be a class.
Let $g$ be a strictly progressing mapping on $M$.
Let $M$ be minimally superinductive under $g$.
For an arbitrary ordinal $\alpha$, let $M_\alpha$ be the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$.
Then:
\((1)\) | $:$ | Zeroth Ordinal: | \(\ds M_0 \) | \(\ds = \) | \(\ds \O \) | ||||
\((2)\) | $:$ | Successor Ordinal: | \(\ds \forall \alpha \in \On:\) | \(\ds M_{\alpha^+} \) | \(\ds = \) | \(\ds \map g {M_\alpha} \) | |||
\((3)\) | $:$ | Limit Ordinal: | \(\ds \forall \lambda \in K_{II}:\) | \(\ds M_\lambda \) | \(\ds = \) | \(\ds \bigcup_{\alpha \mathop < \lambda} M_\alpha \) |
where:
- $\On$ denotes the class of all ordinals
- $K_{II}$ denotes the class of all limit ordinals.
Formulation 2
Let $\On$ denote the class of all ordinals.
Let $S$ denote the class of all ordinal sequences.
Let $g$ be a mapping such that $S \subseteq \Dom g$.
Then there exists a unique mapping $F$ on $\On$ such that:
- $\forall \alpha \in \On: \map F \alpha = \map g {F \restriction \alpha}$
where $F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$.
Formulation 3
Let $\On$ denote the class of all ordinals.
Let $h$ and $f$ be mappings defined for all sets.
Let $c$ be a set.
Then there exists a unique mapping $F$ on $\On$ such that:
\((1)\) | $:$ | \(\ds \map F 0 \) | \(\ds = \) | \(\ds c \) | |||||
\((2)\) | $:$ | \(\ds \forall \alpha \in \On:\) | \(\ds \map F {\alpha^+} \) | \(\ds = \) | \(\ds \map h {\map F \alpha} \) | ||||
\((3)\) | $:$ | \(\ds \forall \lambda \in K_{II}:\) | \(\ds \map F \lambda \) | \(\ds = \) | \(\ds \map f {F \sqbrk \lambda} \) |
where $K_{II}$ denotes the class of all limit ordinals.
Formulation 4
Let $\On$ denote the class of all ordinals.
Let $g$ be a mapping defined for all sets.
Let $c$ be a set.
Then there exists a unique $\On$-sequence $S_0, S_1, \dots, S_\alpha, \dots$ such that:
\((1)\) | $:$ | \(\ds S_0 \) | \(\ds = \) | \(\ds c \) | |||||
\((2)\) | $:$ | \(\ds \forall \alpha \in \On:\) | \(\ds S_{\alpha + 1} \) | \(\ds = \) | \(\ds \map g {S_\alpha} \) | ||||
\((3)\) | $:$ | \(\ds \forall \lambda \in K_{II}:\) | \(\ds S_\lambda \) | \(\ds = \) | \(\ds \bigcup_{\alpha \mathop < \lambda} S_\alpha \) |
where $K_{II}$ denotes the class of all limit ordinals.
Formulation 5
Let $\On$ denote the class of all ordinals.
Let $h$ be a mapping.
Then there exists a unique mapping $F$ such that:
- $\forall \alpha \in \On: \map F \alpha = \map h {F \sqbrk \alpha}$
Also presented as
There are variants in how this result is presented, as the topic can be approached from a number of different directions.
Here are examples:
First Principle of Transfinite Recursion
Let $G$ be a (class) mapping from $\On^{\On}$ to $\On$.
Let $K$ be a class of mappings $f$ that satisfy:
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 $A = F$.
Second Principle of Transfinite Recursion
Let $\Dom x$ denote the domain of $x$.
Let $\Img x$ denote the image of the mapping $x$.
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Let $G$ be a class of ordered pairs $\tuple {x, y}$ satisfying at least one of the following conditions:
- $(1): \quad x = \O$ and $y = a$
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- $(2): \quad \exists \beta: \Dom x = \beta^+$ and $y = \map H {\map x {\bigcup \Dom x} }$
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- $(3): \quad \Dom x$ is a limit ordinal and $y = \bigcup \Rng x$.
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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 $A = F$.
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.1$