## The easy direction

May 10, 2011 at 15:44 | Posted in Uncategorized | Leave a commentIn this post, we’ll prove a more nuanced version of the result we discussed in our January 31 post. This is the “easy direction” of the theorem we’re aiming to prove, and the argument goes through without any special assumptions on the cofinality of the singular cardinal .

Theorem 1Suppose , , and are cardinals satisfying . Then

*Proof:* Suppose is a progressive set of regular cardinals such that and . Further assume that is a -complete ideal on extending the bounded ideal such that has true cofinality, and let

We must show

Suppose this fails, and let witness . By our choice of , there is a sequence such that

- each is in ,
- , and
- if , then for some .

Since , it follows that for each the range of is covered by a union of fewer than sets from . Since is -complete, we conclude that for each there is a set such that

Now is regular and , so we can assume that there is a single set with the property that

for each .

Now define a function as follows:

Since , we know that is bounded in for all sufficiently large . Since contains the bounded subsets of , it follows that for each ,

But this is a contradiction, as there must exist an such that .

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