# Recursion on ordinals

I have some kind of "homework" question. I have the following theorem:

Theorem (Transfinite Recursion over the class of of ordinals $\mathbf{ON}$: Let $\mathbf{V}$ be the class of all sets. If $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ then there exists a unique $\mathbf{G} : \mathbf{ON} \to \mathbf{V}$ such that for all $\alpha$ $\mathbf{G}(\alpha) = \mathbf{F}(\mathbf{G}|_{\alpha}))$.

Now I want to show that if $\hat{x} = \{y : y < x\}$ and if $\mathbf{F} : \mathbf{V} \to \mathbf{V}$ and $(X, <)$ is a well-ordering then there exists a function $f$ with domain $X$ such that $f(x) = \mathbf{F}(f|_{\hat{x}}$.

So I let $0 = \text{min}(X)$, and then I note that I want $f(0) = \mathbf{F(\emptyset})$. I continue like this and I note that $f(1) = \mathbf{F}(f(0))$ and so on. Is this correct? If so, can someone give me a hint how I apply the above theorem to say that this construction works?

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Ordinals are not that elementary... – Asaf Karagila Dec 7 '10 at 4:25

If you can use the transfinite recursion theorem you stated, why not simply apply it to $F$, and then use an isomorphism between $X$ and some ordinal to transfer the function $G$ from $ON$ to $X$? It seems like you are trying to reprove the transfinite recursion theorem rather than just applying it.