## On ADS (Part 4)

February 10, 2012 at 14:40 | Posted in Uncategorized | Leave a commentSo what does it mean when we say “almost works”? We’ll start with a lemma:

Lemma 1For a fixed , the set contains at most one element from each set , and hence is at most countable.

*Proof:* Suppose by way of contradiction what and are distinct members of for which

Then

and this contradicts the fact that is a transversal for .

Now given , we are going to define to be those for which has failed to work, that is,

Lemma 2The set is at most countable.

*Proof:* If not, find for which the set is uncountable, and set

Then for each , we have

which contradicts the preceding lemma.

So for a given , the function will “disjointify” from all but countably many with : the function “almost works”.

Notice that if and only if , so that we can define a graph on by connecting and if and only if . By the preceding lemma, every vertex of the graph has at most countably many edges coming out of it. An easy argument tells us that the connected components of our graph are at most countable as well.

Now if and are members of different connected components of then will disjointify and . But each connected component of can be disjointified easily (by induction) as it is at most countable.

Thus, it is straightforward now to “correct” to a function which will work everywhere.

## On ADS (Part 3)

February 9, 2012 at 11:40 | Posted in Uncategorized | Leave a commentContinuing the construction, let us assume that is singular of countable cofinality and . For each , let be a sequence of sets such that

- ,
- ,
- for all , and
- .

By induction on we choose sets such that for no and is a subset of .

Notice that each set of the form is of cardinality less than , and there are sets of this form. Our assumption that is greater than tells us that a suitable can always be found.

Let be a bijection, and let . We will show that the collection will witness .

What this amounts to is that for each , we need a function so that

for all , where

Lemma 1For each , the family has a transversal (i.e., a one-to-one choice function) .

*Proof:* We define by induction on . This is trivial for and limit. If , then is not a subset of so can be defined easily.

Armed with the preceding lemma, we define a function as follows:

Given , we know for some unique . We define to be the unique such that .

In English, answers the question “when does enumerate the value of ?”.

Our goal is to show that the function almost works, and then show how to modify to a function that actually does the job.

## On ADS (Part 2)

February 7, 2012 at 11:11 | Posted in Uncategorized | Leave a commentSo our project is to present a proof of the following result of Shelah:

Theorem 1Suppose is singular and . If , then holds.

We first prove a lemma that is intended to make our task a bit easier:

Lemma 2Suppose 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.

Then holds.

*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.

## On ADS

February 6, 2012 at 11:13 | Posted in Uncategorized | Leave a commentSo, I think I’ve finished my “busy period”, and I have a little time to write. What I want to do today is post about the topic of my second lecture at this years Czech Winter School. The very first post I made on this blog dealt with an argument of Shelah that can be used to prove Solovay’s theorem that the Singular Cardinals Hypothesis holds above a strongly compact cardinal. The argument used a consequence of as a “black box”; I want to take a look at the interior of this black box today.

Definition 1Let a cardinal. We say that holds if there is a sequence such that each is unbounded in , and for each , the collection isessentially disjoint, in the sense that there is a function such that the collection is disjoint.

Some things to note:

- always holds if is regular (any “almost disjoint” family of size has the required property).
- If is singular and holds, then we may assume each is of order-type .
- for singular implies that that there is no countably complete uniform ultrafilter on — this is the content of the very first post I made on this blog.
- follows from the existence of a
*better scale*(and hence it follows from each of weak square and ). The details of this are in my Handbook of Set Theory article.

What I want to do here is derive from the assumption that is a singular cardinal of countable cofinality satisfying . Remember this means that for any family of cardinality , there is a countable not covered by any member of . I’ll just state the theorem, and spread the proof out over the next couple of days in some more blog posts.

Theorem 2(Shelah) Suppose is singular and . If , then holds.

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