# Transfinite Recursion Theorem

I have the following homework assignment:

Let $\textbf{V} \models ZFC$ and let $\mathbb{P} = (P, \leq)$ be a forcing partial order in $\textbf{V}$. Define a class function $F: \textbf{V} \to \textbf{V}$ such that using the transfinite recursion theorem there exists a function $G: \Omega \to \textbf{V}$ such that for all $\alpha \in \Omega$: $$G(\alpha) = V_\alpha^{\mathbb{P}}$$

What follows is my answer, can you please tell me if I have it right:

(i) For $\alpha = \emptyset$ we want $G(\alpha) = V_\emptyset^\mathbb{P} = 0 =: F(0)$

(ii) For $\alpha = 1$ we want $G(1) = V_1^\mathbb{P} = \mathcal{P}( V_\emptyset^\mathbb{P}\times \mathbb{P}) = \mathcal{P}( \emptyset \times \mathbb{P}) \cong \mathcal{P}(\emptyset) = \{\emptyset\} = 1 =: F(G|_1)$

For $\alpha = 2$ we want $G(2) = V_2^\mathbb{P} = \mathcal{P}(V_1^\mathbb{P} \times \mathbb{P}) = \mathcal{P}(\{\emptyset\} \times \mathbb{P}) =: F(G|_2)$

So for a successor ordinal $\alpha \in \Omega$ we define $F(G|_{\alpha + 1}) := \mathcal{P}(F(G|_\alpha) \times \mathbb{P})$

(iii) For a limit ordinal $\alpha \in \Omega$, we have $V_\alpha^\mathbb{P} = \cup_{\beta \in \alpha} V_\beta^\mathbb{P}$ and so we want to define $G(\alpha) = \cup_{\beta \in \alpha} V_\beta^\mathbb{P} = \cup_{\beta \in \alpha} F(G|_\beta) =: F(G|_\alpha)$

Is this all I need to do? Or do I need to show that $F$ is well-defined or anything else? Many thanks for your help!

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Note that $V^\mathbb P_0 = \varnothing$, so $V^\mathbb P_1 = \{\varnothing\}$. Only from the $2$ you get actual content. – Asaf Karagila Dec 1 '11 at 16:06
@AsafKaragila: How so? The definition in the notes says $$V_0^\mathbb{P}=\emptyset$$ $$V_\alpha^\mathbb{P} = \cup_{\beta \in \alpha}V_\beta^\mathbb{P}$$ for a limit ordinal $\alpha$ and $$V_{\alpha + 1}^\mathbb{P} = \mathcal{P}(V_\alpha^\mathbb{P} \times \mathbb{P})$$ So with $\alpha = 1 = \{ \emptyset \}$ I should get $\mathcal{P}(V_{\emptyset}^\mathbb{P} \times \mathbb{P})$ – Rudy the Reindeer Dec 1 '11 at 16:31
Which is $\mathcal P(\varnothing\times\mathbb P) = \mathcal P(\varnothing)$. – Asaf Karagila Dec 1 '11 at 16:37
It depends on the precise statement of the Rec.Thrm. It appears that you are following something close to Kunen, wherein it is is stated as follows. If $F:V\to V$, then there is a unique $G:\mathrm{On}\to V$ such that $(\forall\alpha\in\mathrm{On})(G(\alpha)=F(G\restriction\alpha))$. So first of all, you want $F$ to be defined on all sets (though you can make trivial choices in cases where the value at an argument is not important). You should be aware that the interesting case is when the argument to $F$ is an $\alpha$-sequence (since that is what $G\restriction\alpha$ is). – arjafi Dec 1 '11 at 18:43
@ArthurFischer: Thanks! Do you mean I just need to define $F(X) := \emptyset$ for any $X$ not an ordinal? I don't understand what you mean by $\alpha$-sequence. As for the theorem: you guessed right, we are using what you stated above. – Rudy the Reindeer Dec 1 '11 at 21:26

To answer a small question, by an $\alpha$-sequence we simply mean any function with domain $\alpha$ (where $\alpha$ is an ordinal). And we will follow standard notation that by $\langle x_\xi \rangle_{\xi<\alpha}$ we denote the $\alpha$-sequence $g$ where $g (\xi) = x_\xi$ for all $\xi < \alpha$.
I'm not certain how pedantic you are supposed to be in your own solution, but since they are even asking you to do this, you might have to be fairly thorough, so let's look at the Recursion Theorem closely. It says that given $F : V \to V$ there is a (unique) $G : \mathrm{On} \to V$ such that for all $\alpha$ $$G ( \alpha ) = F ( G \restriction \alpha ).$$ Looking at the right-hand-side, we have "the value of $F$ at the argument $G \restriction \alpha$." Clearly, $G \restriction \alpha$ is an $\alpha$-sequence, so we should focus our main attention to defining $F$ on $\alpha$-sequences. As we intend to define the "$\mathrm{On}$-sequence" $G = \langle V_\alpha^P \rangle_{\alpha \in \mathrm{On}}$, we note for $\alpha \in \mathrm{On}$ that $G \restriction \alpha = \langle V_\xi^P \rangle_{\xi < \alpha}$. So the case that we are really, really interested in is when we have an $\alpha$-sequence of the form $\langle V_\xi^P \rangle_{\xi < \alpha}$. Now how to you move from this to $V_\alpha^P$? Clearly this will depend on whether $\alpha$ is a successor or a limit ordinal, but I am sure that by looking at these cases closely for a few seconds you will see how to do this -- and we have thus begun our definition of $F$. Whatever you do will likely be easily extended to arbitrary $\alpha$-sequences, where the values of the sequence are not necessarily of the form desired, so you might as well define $F$ in this similar way (it doesn't hurt), and then take $F (x) = \emptyset$ whenever $x$ is not an $\alpha$-sequence for any $\alpha$. In this way we will get $F : V \to V$.