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6:15 PM
Hi, I'm looking for someone familiar with the transfinite recursion theorem because I don't really understand what it's saying or what's its significance
Quoting from Kunen, theorem 9.3, chapter 1 (in my edition): if $F:\mathbf{V}\to\mathbf{V}$, then there is a unique $G:\mathbf{ON}\to\mathbf{V}$ such that $\forall\alpha[G(\alpha)=F(G\upharpoonright\alpha)]$.
Where $\mathbf{V}$ is the class of all sets, $\mathbf{ON}$ is the class of all ordinals and $G\upharpoonright\alpha$ is the restriction of $G$ to $\alpha$
 
 
3 hours later…
9:41 PM
@AlessandroCodenotti This just formalizes the definition by .
You want to "define" some object $G(\alpha)$ for each ordinal $\alpha$.
And this theorem says that $G(\alpha)$ is uniquely defined, if you have some kind of assignment which gives unique object based on values of $G(\beta)$ for $\beta<\alpha$.
I'd say that the explanation on Wikipedia gives the basic idea, even though it is rather short.
4
Q: What does $\upharpoonright$ in $G(F\upharpoonright\alpha)$ mean?

Transfinite Lover More formally, we can state the Transfinite Recursion Theorem as follows. Given a class function $G\colon V\to V$, there exists a unique transfinite sequence $F\colon\mathrm{Ord}\to V$ (where $\mathrm{Ord}$ is the class of all ordinals) such that $F(\alpha) = G(F\upharpoonright\alpha)$ for al...

4
Q: Intuition and "proof" of transfinite recursion

BlakeI am trying to understand transfinite recursion. So far I have encountered two different definitions of this theorem (not completely sure whether they describe the same Principle of Transfinite Recursion). The first one: If I have a map $I:X^{<\alpha}\rightarrow X$ (for $\alpha$ some ordinal ...

0
Q: Transfinite Recursion

Sahiba AroraI'm trying to understand the concept of transfinite recursion. Can someone provide me examples which clearly illustrates transfinite recursion or provide some references which I can go through?

 
10:36 PM
ohhh, wait, in $G\upharpoonright\alpha$ only the values of $G(\beta)$ for $\beta<\alpha$ are needed, so we're getting $G(\alpha)$ in terms of the previous ones
$\alpha$ as a set contains the ordinals smaller than $\alpha$
derp, I got it now though, thanks!
 
@AlessandroCodenotti Yes, exactly.
Perhaps sometimes this notation is a bit obscure if somebody is not used to it.
Gut $G\upharpoonright\alpha$ is shorter than writing $G\upharpoonright\{\beta\in\mathbf{ON}; \beta<\alpha\}$.
When reading set-theoretical text, it is probably better get used to this. I believe that this convention is often used in papers from general topology (especially set-theoretic topology).
 
yeah I had never seen it before, it makes sense now
 

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