## Claim 3.2 page 377

June 15, 2011 at 11:56 | Posted in Uncategorized | Leave a commentLast time, we left off with the statement of the following theorem:

Theorem 1Suppose is a progressive set of regular cardinals, andThen

The above is just Claim 3.2 on page 377 of The Book with the notation changed a little. Today I want to give a proof of this theorem. If you’ve already read through our June 10th post, then today will be simple because it’s essentially the same proof.

*Proof:* Start by letting be -cofinal in for any of cardinality , and suppose . We will show that , thereby establishing the “” part of the theorem.

Assume by way of contradiction that . Let and witness that is in , i.e.,

- is a -complete ideal on ,
- has true cofinality ,
- , and
- .

Since , there is a function such that

Since , we can find of cardinality such that

Given , we define

Then (4) tells us

A couple of things now come to our rescue: since and is -complete, there must be an such that is not in , and this tells us

This is absurd, of course, for was an upper bound of in . Thus, and the “” part of the theorem has been established.

For the other direction, we once again mirror our June 10th post. Given , let and be the associated ideal and generating set. Basic pcf theory tells us

and we let serve to witness this.

Now define

Clearly , so we will be done if we can show is -cofinal in for every of cardinality .

Suppose this is not the case. Then we can find a set of cardinality and a function such that the restriction of to cannot be covered by fewer than functions in .

Let us now define a collection of subsets of by putting into if and only if there is a family of cardinality such that

It’s easy to check that is an ideal, and for trivial reasons. Our choice of guarantees that is a -complete proper ideal as well.

Now let be minimal with , and define

The ideal is essentially restricted to ; it is easy to see that is a proper (by choice of ) -complete (as is) ideal on . Clearly as well, since this set is in .

The minimality of ensures

which immediately implies

Thus, the functions witness

Putting all of the above together, we see

by way of our ideal .

Our goal now is to reach a contradiction exactly as in the June 10th post by demonstrating that , and .

Part of this is trivial, as was selected so that . For the other part, we know that given our choice of , there is an such that

But since , the definition of tells us

and now the conjunction of (17), (18), and (19) together with the fact that is an ideal shows us that . Contradiction.

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