Principle of Recursive Definition/Proof 2

Theorem
Let $\N$ be the natural numbers.

Let $T$ 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: f \left({x}\right) = \begin{cases}

a & : x = 0 \\ g \left({f \left({n}\right)}\right) & : x = n + 1 \end{cases}$

Proof
Consider $\N$ defined as elements of the minimal infinite successor set $\omega$.

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

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

Therefore, such a function exists.

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


 * $f \left({\varnothing}\right) = f' \left({\varnothing}\right)$

This completes the proof.