I'll try to remember to take a look at the book tonight. Maybe you'll get an answer sooner, though.
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
I'm pretty sure I found your mistake. You can check me on it.
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:
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)).
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