# Help with Kunen and recursion on wellfounded sets.

• Nov 2nd 2010, 04:51 AM
sroberts
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
• Nov 2nd 2010, 12:44 PM
MoeBlee
I'll try to remember to take a look at the book tonight. Maybe you'll get an answer sooner, though.
• Nov 3rd 2010, 09:26 AM
MoeBlee
I'm pretty sure I found your mistake. You can check me on it.

Quote:

Originally Posted by sroberts
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
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)).
• Nov 3rd 2010, 01:45 PM
sroberts
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
• Nov 3rd 2010, 02:54 PM
MoeBlee
Quote:

Originally Posted by sroberts
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}.
• Nov 3rd 2010, 03:30 PM
sroberts
U{{y}uCl(y):y is below x} = {y: y is below x}uU{Cl(y): y is below x} = Cl(x), no?
• Nov 4th 2010, 08:38 AM
MoeBlee
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.
• Nov 4th 2010, 01:54 PM
MoeBlee
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.
• Nov 5th 2010, 01:57 AM
sroberts
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
• Nov 5th 2010, 09:47 AM
MoeBlee
Whatever may be the case regarding h'|pred(x), what we have to show is:

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