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Math Help - Help with Kunen and recursion on wellfounded sets.

  1. #1
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    Help with Kunen and recursion on wellfounded sets.

    Hi everyone,

    I'm a bit stuck on the proof (page 103-4 in Kunen's Set theory and independence book) of the existence of a functional class G for any class A wellfounded relation R on A, and functional class F from A x V to V, which is set-like such that:

    (*) forallx in A(G(x) = F(x, G|pred(x))

    where pred(x) are the predecessors of x under R and in A.

    So Kunen first wants to prove that for every x of A the set {x}Ucl(x) (cl(x) is the closure of x under R) admits of a function g with domain {x}Ucl(x) and such that:

    (**) forall y in {x}Ucl(x)(g(x) = F(x, g|pred(x))

    Now, his way of doing this is by induction. He assumes that for each predecessor of x there is such a function g and then takes the union of these functions h, and adds to that the pair <x, F(x, h)> to get an new function h'.

    The claim is then that h' satisfies (**). One way to prove this is by showing that h is satisfies (**) on cl(x) and then to show that F(x, h) = F(x, h'|pred(x)). This in turn we could do if we could show that h'|pred(x) = h.

    But in general it seems that ~(h = h'|pred(x)). The reason is that h has as a domain the closure of x whereas h'|pred(x) only has x's predecessors as its domain.

    Any help in sorting this out would be really great.
    Thanks
    Sam
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  2. #2
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    I'll try to remember to take a look at the book tonight. Maybe you'll get an answer sooner, though.
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  3. #3
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    I'm pretty sure I found your mistake. You can check me on it.

    Quote Originally Posted by sroberts View Post
    forall y in {x}Ucl(x)(g(x) = F(x, g|pred(x))
    That's where you first went wrong. You need to reletter some of the 'x' to 'y':

    For all y in {x}uCl(x) we have g(y) = F(y g|pred(y))

    That leads to correcting the following:

    Quote Originally Posted by sroberts View Post
    h has as a domain the closure of x
    No, dom(h) is the union of the domains of each of the g's, where for each g, we have dom(g) = {y}uCl(y) for some y less than x. Now with that adjustment, I think that if you work it out, you'll find that dom(h) = dom(h'|pred(x)).
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  4. #4
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    Hey MoeBlee,

    Thanks ever so much for the help.

    Just a quick question. If dom(h) = Udom(g) where dom(g) ={y}Ucl(y) for some y below x, then all of x's predecessors are in dom(h) and the predecessors of their predecessors (because cl(y) is a subset of dom(g)). Isn't that equivalent to cl(x)?

    Unless R is transitive I don't see how in general dom(h) = dom(h'|pred(x)) = pred(x).

    Any (further) help would be great.

    Regards
    Sam
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  5. #5
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    Quote Originally Posted by sroberts View Post
    dom(h) = Udom(g) where dom(g) ={y}Ucl(y) for some y below x
    No, dom(h) = U{dom(g_y) | yRx}, where g_y is the the {y}uCl(y)-approximation function.

    Recall, h = U{g_y | yRx}.

    So dom(h) = U{dom(g_y) | yRx}.
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  6. #6
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    U{{y}uCl(y):y is below x} = {y: y is below x}uU{Cl(y): y is below x} = Cl(x), no?
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  7. #7
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    I think you're right that we need U{{y}uCl(y) | yRx} = Cl(x).

    Let me check your equations when I get a chance, hopefully later today.
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  8. #8
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    Now I am myself stumped.

    Your first equation looks right; but I don't know how you got the second equation.

    Previously you mentioned the lack of transitivity, and now I too don't see how we get the result without transitivity.
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  9. #9
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    Hi MoeBlee,

    My solution, if it is one, was to take h to be the union as normal, but then add to that <x, F(x, h|pred(x))> to get h'. Then it's pretty straight forward to show that h|pred(x) = h'|pred(x) as desired.

    What do you think?

    Regards
    Sam
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  10. #10
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    Whatever may be the case regarding h'|pred(x), what we have to show is:

    dom(h') = {x}uCl(x)
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