# Principle of Recursive Definition for Minimal Infinite Successor Set

## Theorem

Let $\omega$ be the minimal infinite successor set.

Let $T$ be a set.

Let $a \in T$.

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

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

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

where $n^+$ is the successor set of $n$.

## Proof

Take the function $F$ generated in the second principle of transfinite recursion.

Set $f = F {\restriction_\omega}$.

 $\displaystyle \map f \O$ $=$ $\displaystyle \map F \O$ $\O \in \omega$ $\displaystyle \map f {n^+}$ $=$ $\displaystyle \map F {n^+}$ $n^+ \in \omega$ $\displaystyle$ $=$ $\displaystyle \map g {\map F n}$ Second Principle of Transfinite Recursion $\displaystyle$ $=$ $\displaystyle \map g {\map f n}$ $n \in \omega$ and the definition of $f$

Therefore, such a function exists.

Now, suppose there are two functions $f$ and $f'$ that satisfy this:

$\map f \O = \map {f'} \O$

Then:

 $\displaystyle \map f {n^+}$ $=$ $\displaystyle \map g {\map f n}$ By Hypothesis $\displaystyle$ $=$ $\displaystyle \map g {\map {f'} n}$ Inductive Hypothesis $\displaystyle$ $=$ $\displaystyle \map {f'} {n^+}$ By Hypothesis

This completes the proof.

$\blacksquare$