## Conclusion 3.2A (part 3)

June 17, 2011 at 13:07 | Posted in Uncategorized | Leave a commentThis post is an immediate continuation of the previous one, so all notation etc. carries over from there. I noticed last night that Conclusion 3.2A is really two separate results which have been smashed together. The main part actually looks like it holds without the assumption that is not a fixed point, so I’m going to work in a slightly more general context today. If things work out, I’ll formulate “the real theorem” underlying things.

Lemma 1Let be a singular cardinal of cofinality . Then there is a such that

*Proof:* First, note that if we restrict ourselves to with , then any relevant to computing the left-hand side of the above equation is necessarily progressive, and hence the cofinality of will just be . Thus, we need only prove that there is a in the interval for which

Suppose this is not the case, so for any there is a of cardinality with . The following proposition takes this hypothesis and moves us closer to a contradiction:

Proposition 2Under the above assumptions, for every there is a such that

- is a progressive subset of ,
- ,
- , and
- is a bounded subset of .

*Proof:* Without loss of generality, we assume and so we can find (progressive) of cardinality such that . If is bounded below then we are done, so assume is unbounded.

Let be an ultrafilter on such that

The above equation, taken together with the fact that is unbounded in of cardinality , tells us (by way of the definition of ) that cannot be disjoint to the bounded subsets of , i.e,. that there is a such that

Now this has all of the required properties.

Armed with the above proposition, we can construct sequences and such that

- is a strictly increasing sequence of cardinals cofinal in
- (hence is progressive)
- for each .

Now define

By construction, is a progressive set of regular cardinals cofinal in of cardinality , and so the definition of tells us that if is an ultrafilter on disjoint to the bounded ideal, then

Let , and consider the ideal on generated by and the bounded subsets of . It is easy to show that must be a proper ideal on : if not, then there would be a such that

but this is impossible as

by our construction of . If is an ultrafilter on disjoint to , then

and we have a contradiction of (6).

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