Let us recall that is the minimum cardinality of a set such that for any (equivalently ), there is a such that .

Proposition 1If is singular of cofinality , then

*Proof:* Suppose by way of contradiction that

By the definition of , we can find a (not necessarily increasing) sequence of regular cardinals and an ultrafilter on such that

and

Let be a family of cardinality standing as witness to , and let be a -increasing and cofinal sequence of functions in . For each , the range of is a subset of of cardinality at most , and so we can find such that

Since is a regular cardinal greater than , there is a single such that

Thus, (by passing to a subsequence of ) we may as well assume that the range of each is a subset of .

But , and so

by our choice of . Let us now define a function by setting if , and

whenever is in . Our assumptions imply that is in .

If , then for each we have

as .

It follows that is a function in such that for all . This is absurd, given our choice of , and the proof is complete.

[Updated 2-7-11]

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Fix a sufficiently large regular cardinal ; we will be working with elementary submodels of the structure where is some appropriate well-ordering of used to give us definable Skolem functions.

Let be a -approximating sequence, that is, is a -increasing and continuous sequence of elementary submodels of such that for each ,

- ,
- has cardinality ,
- is an initial segment of (so ), and
- .

Let denote . Then is an elementary submodel of of cardinality containing as both an element and subset. Define

Clearly and , so we need only verify that for any , there is a with .

In broad terms, this will be done via an “ argument” with , but we’ll fill in the details in further posts.

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Theorem 1If , then .

I am pretty sure that this isn’t what Shelah meant to write here — the assumption that has uncountable cofinality means that the result is a trivial consequence of a deeper theorem in Cardinal Arithmetic [Update: Perhaps not! See 2nd update below]. I think that the proof he gives works even for the countable cofinality case, and this gives us something of interest because could very well be a fixed point. So, here’s what I conjecture he meant to say (after changing notation and translating the “cov” statement into more standard form):

Theorem 2Let be a singular cardinal. Then if and only if there is a family of cardinality such that every member of is a subset of some element of .

We’ll work through his proof as best we can, and see if my conjecture is correct.

UPDATE 1: I don’t think the conjecture is correct. It looks to me like the proof was originally written for singular of countable cofinality, but a mistake was discovered and the statement was corrected but a lot of the old proof didn’t get revised properly. The annotated content was revised, but the abstract was not. Anyway, I’ll ask Saharon if I can’t figure out what’s going on.

UPDATE 2: I think the theorem doesn’t follow from the Cardinal Arithmetic stuff. The problem I run into is that when Shelah says informally that “cov = pp when the cardinal has uncountable cofinality”, there are lots of disclaimers hidden in the background. But this underscores why I think what I’m doing here is important — I want to pin down what exactly is known and what exactly is still open in this area.

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My first project will be to develop a treatment of the first section of [Sh:430] “Further Cardinal Arithmetic”; my intermediate goal is to concentrate on resuts related to the “cov vs. pp” problem.

It is safe to say that most of what I will be doing is presenting results of Shelah; if something is unattributed then consider it a result of Shelah (although I tend to try and reformulate things to make them more understandable to myself).

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