# a question on transfinite recursion on $\mathbf{ON}$

I want to know what's purpose for Kunen to build the theorem I.9.3 on transfinite recursion on $\mathbf{ON}$. Any help will be appreciated.

Theorem I.9.3 says: if $F: V \to V$, then there is a unique $G:\mathbf{ON}\to V$ such that for any $\alpha,$ $$G(\alpha)=F(G|_\alpha).$$

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If you add pointers to what the theorem actually states, it might be of help to people who don't have immediate access to the book in question like me! – user21436 Jan 12 '12 at 8:41
It seems that you avoid trying to understand on your own. You have asked so many questions which seem as though you don't really try on your own. If Kunen is hard for you, try Jech's Set Theory instead. Either way learning mathematics requires sometimes to believe things will be useful later on. – Asaf Karagila Jan 12 '12 at 8:45
It turns out to be useful to define things recursively. – tomcuchta Jan 12 '12 at 8:49
@AsafKaragila I think you're being too judgmental. – Quinn Culver Jan 12 '12 at 9:42
@Quinn: Of course I am being judgmental, this is a personal opinion. However this is an opinion based on a similar experience that I had when I studied these things for the first time. You ask a lot of people and gain intuition. However unless you sit and deconstruct the definitions by hand you cannot really see the details of the picture for yourself. This is fine if you just want to get a general idea about what's going on. However rarely one can understand the difference between $\omega$ and $\omega+1$ without being able to deconstruct the definitions on their own, and vice versa. – Asaf Karagila Jan 12 '12 at 10:17

You need this to do induction on $ON$. The induction you know from analysis is over $\omega \in ON$. The reason why you want induction over $ON$ is because you might want to prove something like this:

Let $X$ be a set, let $\mathcal{C}$ be a family of subsets of $X$ that is closed under unions of subfamilies that are linearly ordered by inclusion and let $f : \mathcal{C} \to \mathcal{C}$ be a function with $f(C) \supseteq C$ for all $C \in \mathcal{C}$. Then there exists a $C$ in $\mathcal{C}$ such that $f(C) = C$.

Here's an outline of the proof: You pick an ordinal $\alpha$ that cannot be mapped into $\mathcal{C}$ injectively. Then you define a function $F: \alpha \to \mathcal{P}(X)$ involving $f$ such that certain properties hold. The definition of $F$ is where the transfinite recursion comes in because $F$ needs to be defined for all $\alpha$ in $ON$. The definition looks something like this:

Assume $\beta \in \alpha$ and $F(\gamma)$ has been defined for all $\gamma < \beta$. Then

$$F(\beta) := \begin{cases} f(\bigcup\{ F(\gamma) : \gamma < \beta \}) ) & \text{if } \bigcup\{ F(\gamma): \gamma < \beta\} \in \mathcal{C} \\ X & \text{otherwise} \end{cases}$$

You can find the full proof of this theorem on page 163 of "Discovering Modern Set Theory" by Just and Weese. If you don't like Kunen have a look at that. I have both and I think Kunen is too painful to read.

For another example of transfinite recursion see here. I'd recommend that you first look at an example of transfinite induction, like for example the proof of $\alpha + \beta = \alpha \cup \{ \alpha + \gamma : \gamma < \beta \}$ for ordinals $\alpha, \beta$. Then you'll get the idea of doing the induction in two parts, once for limit ordinals and once of "normal" ordinals. And from there it's easy to understand what transfinite recursion is -- they're almost the same thing.

Hope this helps.

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Here's p.163 - in case it is not accessible from Google Books preview. I hope such a short excerpt classifies as a fair use. – Martin Sleziak Jan 24 '12 at 21:28
@MartinSleziak Nice, thank you! – Rudy the Reindeer Jan 24 '12 at 21:29