Past exam of the logic and set theory course of the University of Oxford / 2011 / B / 4 / a / i / Solution

An approximation is a function $s:n+1\to X$ with $s(0)=x_0$ and $s(k+1)=g(s(k))$ for $k<n$. By induction on $n$, there is such a function: start with $\{(0,x_0)\}$ and append the uniquely specified value $g(s(n))$.

Any two approximations agree on their common domain, by induction from their common initial value and the same transition function $g$. The collection $\mathcal{S}$ of all approximations is a set, being a definable subset of $\mathcal{P}(\omega \times X)$. Therefore $f=\bigcup \mathcal{S}$ is a function. Existence of approximations of every finite length gives $\mathrm{dom}(f)=\omega$. Each defining equation appears in a sufficiently long approximation, soThis proves the natural-number recursion theorem without the axiom of choice; no arbitrary sequence of choices was made.