On ADS (Part 2)February 7, 2012 at 11:11 | Posted in Uncategorized | Leave a comment
So our project is to present a proof of the following result of Shelah:
Theorem 1 Suppose is singular and . If , then holds.
We first prove a lemma that is intended to make our task a bit easier:
Lemma 2 Suppose is a singular cardinal of countable cofinality, and suppose and are such that
- each is a countable subset of , and
- for each , there are finite sets such that is pairwise disjoint.
Proof: This is not completely trivial because requires sets that are unbounded in , but it is still an elementary argument. Let be a bijection, and for each let . Given , if we let , then it is certainly the case that
Now we are almost to ; all we are missing is that the need to be unbounded in . This probably won’t be true in general, but there is an so that is unbounded in whenever . To see why, suppose this fails. Then we can find so that is unbounded in .
Now choose so that . Now the collection gives us a collection of disjoint subsets of , and clearly this is absurd.
Thus, if we throw away the first sets from our family, the remainder will witness .
I guess the moral of the above is that in the case of singular , we can safely drop the requirement that each is unbounded in because this will be true from some point on anyway.
Returning now to the proof of the theorem, our plan is to proceed as follows:
We will define a certain family of distinct countable subsets of . We then set to be some bijection, and define
Since the sets are distinct, the collection will form an almost disjoint collection of countable subsets of , and our aim is to show that this collection satisfies the demands of the preceding lemma. Of course, this proof will depend on choosing the sets quite carefully.