What Does the Recursion Theorem in Set Theory State?

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The discussion centers on the Recursion Theorem in set theory, which asserts that for any well-ordered set X and any set Y, there exists a function f that can be defined recursively using another function G. Specifically, f is defined such that f(x) equals G applied to the initial segment of f up to x. This theorem is related to Cantor's theorem and extends inductive proof to recursive construction. The participants clarify that the theorem allows for the unique definition of functions based on previously defined values, illustrating this with examples from natural numbers. Overall, the theorem provides a foundational framework for understanding recursion in mathematical functions.
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There are probably a million theorems called "the recursion theorem" and I'm not sure if this is actually one of them, but there's a remark saying that it defines recursion.
It's proven by piecing together 'attempts' (functions defined on subsets of a domain) and states:

For X a well-ordered set, Y any set:
For any G:P(XxY)->Y,
there exists f:X->Y
such that f(x)=G(f|Ix), for all x in X.

where:
Ix is the initial subset {y in X iff y<x}
f|A = {(x,f(x) : x in A} - the restriction of f to A.
P denotes power set.

I just have no idea what it's saying.

Googling suggests that it's "Cantor's theorem, originally stated for ordinals, which extends inductive proof to recursive construction." but I can't find anything on it, possibly because the lecturer has mixed a few proofs together so it's a weaker form of a better-known theorem, but much easier to prove.

Thanks :\

[edit]
Wait...is this just saying that, for any G:{functions}->{numbers}, there's a function, f, that satisfies f(5)=G[f(4)]? And so on.
 
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If X is natural numbers, then f(5) = G(f(0), f(1), f(2), f(3), f(4)).

I've taken the liberty of writing the set {(0, f(0)), (1, f(1)), (2, f(2)), (3, f(3)), (4, f(4))} -- the graph of f restricted to {0,1,2,3,4} -- as an ordered sequence[/color]
 
Wait...is this just saying that, for any G:{functions}->{numbers}, there's a function, f, that satisfies f(5)=G[f(4)]? And so on.

Not exactly: as Hurkyl already said, f(5) is more complicated; what it states precisely is that, given G as you defined it above (where the "functions" are the ones from the well-ordered set X to any set Y) you can define an unique function f:X\rightarrow Y, given by f(x)=G(f|Ix), where Ix is the initial x-segment of X. For example, if X=Y= N, this means that:

<br /> f\left(0\right)=G\left( \emptyset \right)<br />
<br /> f\left(n\right)=G\left(In\right)=G\left( f \left( 0 \right),f \left( 1 \right),\cdots,f \left( n-1 \right) \right)<br />

This form is usually called course-of-values recursion and it's equivalent to the recursive definition that you are familiar with.
 
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