# Principle of Recursive Definition

## Theorem

Let $\N$ be the natural numbers.

Let $T$ be a class (which may be a set).

Let $a \in T$.

Let $g: T \to T$ be a mapping.

Then there exists exactly one mapping $f: \N \to T$ such that:

- $\forall x \in \N: \map f x = \begin{cases} a & : x = 0 \\ \map g {\map f n} & : x = n + 1 \end{cases}$

### General Result

Let $p \in \N$.

Let $p^\ge$ be the upper closure of $p$ in $\N$:

- $p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$

Then there exists exactly one mapping $f: p^\ge \to T$ such that:

- $\forall x \in p^\ge: \map f x = \begin{cases} a & : x = p \\ \map g {\map f n} & : x = n + 1 \end{cases}$

### Strong Version

Let $\omega$ denote the natural numbers as defined by the von Neumann construction.

Let $A$ be a class.

Let $c \in A$.

Let $g: \omega \times A \to A$ be a mapping.

Then there exists exactly one mapping $f: \omega \to A$ such that:

- $\forall x \in \omega: \map f x = \begin{cases} c & : x = \O \\ \map g {n, \map f n} & : x = n^+ \end{cases}$

## Proof 1

Consider $\N$ defined as a Peano structure $\struct {P, 0, s}$.

The result follows from Principle of Recursive Definition for Peano Structure.

$\blacksquare$

## Proof 2

Consider $\N$ defined as the von Neumann construction of the natural numbers.

The result follows from Principle of Recursive Definition for Minimally Inductive Set.

$\blacksquare$

## Proof 3

Recall the general result:

Let $p \in \N$.

Let $p^\ge$ be the upper closure of $p$ in $\N$:

- $p^\ge := \set {x \in \N: x \ge p} = \set {p, p + 1, p + 2, \ldots}$

Then there exists exactly one mapping $f: p^\ge \to T$ such that:

- $\forall x \in p^\ge: \map f x = \begin{cases} a & : x = p \\ \map g {\map f n} & : x = n + 1 \end{cases}$

The result follows from setting $p = 0$.

$\blacksquare$

## Proof 4

From the Principle of Recursive Definition: Strong Version:

Let $\omega$ denote the natural numbers as defined by the von Neumann construction.

Let $A$ be a class.

Let $c \in A$.

Let $g: \omega \times A \to A$ be a mapping.

Then there exists exactly one mapping $f: \omega \to A$ such that:

- $\forall x \in \omega: \map f x = \begin{cases} c & : x = \O \\ \map g {n, \map f n} & : x = n^+ \end{cases}$

Let $h: A \to A$ be defined as:

- $\forall x \in A: \map h x := \map g {a, x}$ for arbitrary $a \in \omega$

That is:

- $\forall y \in \omega: \map g {y, x} = \map h x$

Then *a priori* there exists exactly one mapping $f: \omega \to A$ such that:

- $\forall x \in \omega: \map f x = \begin {cases} c & : x = \O \\ \map h {\map f n} & : x = n^+ \end {cases}$

The result follows on identifying $\omega$ with $\N$, $c$ with $a$, $A$ with $T$, $\O$ with $0$ and $n^+$ with $n + 1$.

$\blacksquare$

## Also presented as

The Principle of Recursive Definition can also be presented as:

- For any mapping $f: T \to T$ and any $a \in T$, there exists an infinite sequence $a_0, a_1, \ldots, a_n, a_{n + 1}, \ldots$ such that:
- $a_0 = a$
- $a_{n + 1} = \map g {a_n}$

## Fallacious Proof

The proofs given (hidden behind the links) are necessarily long, precise and detailed.

There is a temptation to take short cuts and gloss over the important details.

The following argument, for example, though considerably shorter, is incorrect.

Consider $\N$, defined as a naturally ordered semigroup $\struct {S, \circ, \preceq}$.

Let the mapping $f$ be defined as:

- $\map f x = \begin{cases} a & : x = 0 \\ \map s {\map f n} & : x = n \circ 1 \end{cases}$

if $\map f n$ is defined.

Let $S' = \set {n \in S: \map f n \text{ is defined} }$.

Then:

- $0 \in S'$

and:

- $n + 1 \in S'$

so by Principle of Mathematical Induction for Naturally Ordered Semigroup:

- $S' = S$

Thus the domain of $f$ is $S$.

Consequently, $f$ is a mapping from $S$ into $T$ which satisfies:

- $\map f 0 = a$

and:

- $\map f {n \circ 1} = \map s {\map f n}$

for all $n \in S$.

$\blacksquare$

### Objections

$(1):$

In the above argument, $S'$ is not precisely defined.

In order for a set to be defined, that definition should be able to be expressed solely in terms of logic and definitions from set theory.

In this case, the expression "is defined" does not meet that criterion.

$(2):$

The mapping $f$ is not defined properly.

A mapping needs to be defined as a set of ordered pairs which needs to satisfy a condition.

In the above, it is indeed specified that

- $\tuple {0, a} \in f$

and:

- $\tuple {n \circ 1, \map s x} \in f$ whenever $\tuple {n, x} \in f$

Thus it appears either that:

- $f$ itself is used to define $f$

or else that:

- $f$ itself changes during the process in which it is being defined.

Neither of these possibilities can be accepted.

$(3):$

The only property of $\struct {S, \circ, \preceq}$ used in the argument is that it satisfies Principle of Mathematical Induction for Naturally Ordered Semigroup.

However, consider the commutative semigroup $\struct {D, +}$ which has elements $0$ and $1$, such that:

- $D$ is the only subset of $D$:
- containing $0$

- and
- containing $x + 1$ whenever it contains $x$.

Then:

there exists a mapping $f: D \to T$ such that:

- $\map f y = \begin{cases} a & : y = 0 \\ \map s {\map f x} & : y = x + 1 \end{cases}$

But consider the additive group of integers modulo $2$:

- $\struct {\Z_2, +_2}$

which is indeed a commutative semigroup containing $0$ and $1$ which satisfies the hypothesis.

But if $g: \N \to \N$ is the mapping defined as:

- $\map g n = n + 1$

there is no mapping $f: \Z_2 \to \N$ which satisfies:

- $\map f y = \begin{cases} 0 & : y = 0 \\ \map s {\map f x} & : y = x +_2 1 \end{cases}$

for all $x \in \Z_2$.

Hence the argument is invalid.

## Also known as

This result is often referred to as the **Recursion Theorem**.

Some sources only cover the general result.

## Also see

## Sources

- 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Theorem $8.1$