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### Exercises in Kunen's Introduction to Independence Proofs

Posted: Sat Mar 10, 2012 9:29 pm UTC
Let me start by clarifying this is not homework. While I got a math degree, that was quite a few years ago, and these days I work in a completely unrelated area.

However, I've always been fascinated with the big topics of logic and set theory, and so I'm now working my way through Kunen's book, including the exercises.

And I got a little hang-up there now, in the first chapter exercise 5 (yes, I'm rusty).

Here's the objective: Let a be a limit ordinal. Show that the following are equivalent:
(A) F.a. b, c < a: b+c<a
(B) F.a. b<a: b+a=a
(C) F.a. X subset of a: type(X) = a or type(a\X) = a
(D) a = omegadelta for some ordinal delta

The bit I'm having issues with here is (C) - I've gotten the equivalence for the (A), (B) and (D). And while I'm certain it's really just a stupid little thing I'm overlooking, I just don't see it. My initial thought was to work via type(X) + type(a\X), but that sum can (err, note the current forum madness, that's a "c a n") be < a even for limit ordinals (consider X the final omega segment of omega2 + omega) - I believe "=" only happens for indecomposable ordinals - i.e. the ones of this type.

Note that Cantor normal form is the next exercise (to which this one is meant to lead), so that can't be used. Any pointers?

### Re: Exercises in Kunen's Introduction to Independence Proofs

Posted: Sun Mar 11, 2012 10:40 am UTC
One direction is easy: if (A) is false, then for some b,c < a we have b+c >= a. b+c >=a implies a\b can be order injected into c, so type (a\b) <= type (c) < a, and of course type (b) = b < a, so (C) is false.

For the other direction, we prove (D) => (C) by induction on delta.

For a = omega, either X or a\X will be infinite, and thus have type omega.

Now assume (C) is true for delta. Let a = omega^(delta+1), and X be a subset. Divide a into omega intervals of type omega^delta. For each interval I, the either the intersection of X with I or the intersection of a\X with I will have type omega^delta. Since there are infintely many such intervals, for either X or a\X there will be infinitely many intervals such that their intersection is of type omega^delta. Thus either X or a\X is of type >= omega^delta * omega = omega^(delta+1); since it can't be greater, either X or a\X is of type omega^(delta+1).

Assume (C) is true for all delta < alpha, where alpha is a limit ordinal. Let a = omega^alpha, and X be a subset. For each delta < alpha, either the intersection of X with omega^delta or the intersection of a\X with omega^delta will have type omega^delta. Let Y be the set of delta such that the intersection of X with omega^delta has type omega^delta. Either Y or a\Y is unbounded in a. If Y is unbounded in alpha, then type (X) >= omega^delta for each delta in Y, so type (X) = omega^alpha = a. Similarly for a\Y and a\X. So (C) is true for delta = alpha.

There are probably better ways to prove the (D)=>(C) direction, but that is what I came up with.

### Re: Exercises in Kunen's Introduction to Independence Proofs

Posted: Sun Mar 11, 2012 11:00 pm UTC
Deedlit wrote:There are probably better ways to prove the (D)=>(C) direction, but that is what I came up with.

Maybe, but this certainly is clear and concise, so thanks!

### Re: Exercises in Kunen's Introduction to Independence Proofs

Posted: Wed Mar 14, 2012 5:56 pm UTC
And I already have another problem in the next exercise; specifically, this: "Show that if K is an uncountable cardinal, then K is an epsilon number and there are K epsilon numbers below K". For reference, an ordinal alpha is an epsilon number if and only if omegaalpha=alpha. The other (previous) part of the exercise is to prove Cantor normal form, which is easy given the previous exercise.

Showing that K is an epsilon number is easy (Kunen gives the lemma that a union of cardinality K of sets of cardinality at most K has cardinality K, and K is obviously indecomposable, from previous exercise), but I lack an idea on how to prove there are K epsilon numbers below K.

It seems clear that one needs to look at the series of epsilon numbers; for the first uncountable ordinal omega_1 this would mean showing that epsilonomega_1 = omega_1, or generally epsilonK=K.

One thing that comes into mind is cofinality arguments, but it's possible that cf(K)<K, so that won't work.

There also is the fix point lemma for normal class functions, but Kunen doesn't discuss this in that chapter, and while it's simple to prove, it doesn't give a cardinality for the number of fix points, nor do I see how to show this in general.

Yet another approach I thought of was looking at the union (=sup) of all epsilon numbers below K - if this was equal to K, then we'd have the result (edit: or maybe not, singular cardinals mean the conclusion doesn't work as I'd expect even if the other gap was resolved), but I don't see this either. K is a limit ordinal, but why is it guaranteed that epsilon numbers below K are unbounded in K?

I'm not necessarily looking for a full proof, but a pointer in the right direction would be appreciated!

### Re: Exercises in Kunen's Introduction to Independence Proofs

Posted: Sat Mar 17, 2012 12:04 pm UTC
I would use transfinite induction again. It's not hard to show that epsilonalpha+1 has the same cardinality as epsilonalpha. Let K be a successor cardinal. epsilonalpha for alpha a limit ordinal of cofinality J < K will be a limit of J ordinals of cardinality less than K, so the limit will be less than K. So epsilonalpha for all ordinals alpha < K will be less than K, and epsilonK = K. Limit cardinals are easy - for any alpha < K, a < J for some successor cardinal J < K, and so epsilonalpha < J < K.