# Uniqueness in transfinite recursion.

Theorem of transfinite recursion:

Given a well-ordered set $A$ let $\varphi(g,y)$ be a ZF formula such that for every $a \in A$ and every function $g$ with domain $I_a$ (where $I_a$ is the initial segment of $a$) exists a unique $y$ that satisfies $\varphi(g,y)$. Then there exists a unique function $f$ with domain $A$ such that $\varphi(f\restriction_{I_a}, f(a))$ for every $a$.

In my lecture notes there is written that "An obvious induction shows that $f$, if it exists, is unique".

I've proven the uniqueness in this way: Let us suppose that there exists two functions $f'$ and $f''$, with domain $A$ such that both formulas $\varphi(f'\restriction_{I_a}, f'(a))$ and $\varphi(f''\restriction_{I_a}, f''(a))$ are true. Let $A' \subseteq A$ be the set where $f'(a) \neq f''(a)$. Being $A$ well-ordered $A'$ has a minimal element $m$. If $f'\restriction_{I_m}=f''\restriction_{I_m}$ then by hypotesis $f'(m) = f''(m)$, so $A'$ is empty.

But this proof is not by induction.

What is the "obvious induction" my professor is speaking of?

I think what your professor had in mind is the following: one can prove "by induction" that $f'(a)=f''(a)$ for all $a\in A$, just by noting that if the statement holds for all $b<a$, then it holds as well for $a$, by definition.
What you did is in fact more: you justified that such a reasoning is indeed correct, by considering the smallest $a$ such that the property can possibly fail. This is how the "induction principle" is usually proved.