http://eteisworth.blogspot.com/

because I can use mathjax more easily over there.

]]>Let us assume that is as usual, with . If we want to produce a generator for , we saw last time that what we need is a universal sequence for that in addition possesses an exact upper bound modulo .

I don’t want to make an excursion into the theory of exact upper bounds, as there have been many high quality write-ups of this material already: in addition to Section 2.1 of the Abraham/Magidor article, there are also some materials due to Kojman that do an excellent job of exposing this material. The point is that one can use results of Shelah to modify the universal sequence so that it ends up with an exact upper bound, and then we get generators by the previous post.

Instead of re-hashing such arguments, I offer the following direct route from universal sequences to generators. The proof takes advantage of the ideal , and we will see similar arguments shortly when we talk about transitivity of generators. The argument is based on the exposition of generators given in the Burke-Magidor pcf paper. It turns out that framing things in terms of simplifies things considerably.

**From universal sequences to generators**

Theorem 1Suppose is a progressive set of regular cardinals and . Then there is a generator for .

*Proof:* We can assume that . (If either or are in , then they have generators consisting of singletons.)

Let be a universal sequence for , that is, a sequence such that

- , and
- is cofinal in whenever is an ultrafilter on satisfying .

Such sequences exist by Theorem 4.2 of the Abraham/Magidor article, or see Lemma 2.1 on page 327 of *Cardinal Arithmetic*.

Let . Since we have assumed , we know there is a stationary set lying in the ideal . (See the first section of [Sh:420], or Theorem 3.18 in my own Handbook article.)

This means that there is a family and a club such that

- is a closed (possibly bounded) subset of
- if then , and
- if then is club in of order-type .

For each with , we define

If , we set .

Since is -directed, we can find a single function bounding the collection modulo .

Now we let be an elementary submodel of for some sufficiently large regular cardinal such that

- , , , , , , and are all in
- is an ordinal

This is possible because is stationary in .

We define

We prove that is a generator for . Part of this is simple, as by choice of we know . To finish, we must establish the following:

Proposition 2If is an ultrafilter on with , then .

The proof is not difficult, but the following lemma is critical.

Lemma 3.

*Proof:*

This is where the assumption pays dividends. Note that as , and therefore we know that is club in with order-type .

Given in , we know that is an initial segment of , and both are initial segments of . Thus

Given and , we know there is a with . Since is cofinal in , we know is an ordinal in and . Thus , and we see

In particular, if , then there is an such that

Since is club in with order-type , we can find a single such that for all ,

In particular,

The set on the right is definable from parameters available in (both and are there), and therefore .

Let us return now to the proof of Proposition 2.

*Proof:* By way of contradiction, suppose is an ultrafilter on forming a counterexample. Since is in , we can assume that is in as well (this is the key point).

This sequence is universal for and therefore the collection is cofinal in . Since , we know furthermore that the sequence is increasing.

Thus, there is an such that whenever . Again, we may assume that is in (and hence less than ).

Choose greater than and let . Then and . In particular, and

So we have

One the other hand, since we know

But now we have obtained a contradiction.

Proposition 2 taken together with the fact that easily establishes that is a generator for , so we are done.

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** Finding Generators**

Theorem 1Suppose

- is a progressive sequence of regular cardinals
- is a universal sequence for
- is an exact upper bound for modulo
Then for any ultrafilter on ,

What does this have to do with generators? Just note the following:

Corollary 2Under the assumptions of the theorem, we have , hence is a generator for .

*Proof:* If is an ultrafilter on containing , then either meets or it does not. In the first case, the cofinality of is less than by definition of , and in the second case the cofinality is exactly by the conclusion of the theorem. In either case, the cofinality is at most and so is in and we have

For the other inclusion, suppose is in but not in . Let be an ultrafilter on containing but disjoint to . Since the cofinality of is , we have contradicted the conclusion of the theorem.

**Digression on exact upper bounds**

Before proving Theorem~1, we need to say a few words about exact upper bounds because different authors treat them in slightly different ways. Let us assume that is an exact upper bound for mod just as in the statement of the theorem. It is easy to see

and so our is equal mod to a function satisfying

If we define , then and are equal modulo the ideal , and for any ultrafilter on disjoint to , we have if and only if .

The point of the above is that we can replace by and not change anything, so we may as well assume that our function satisfies

**Proof of Theorem**

*Proof:* Suppose first that is an ultrafilter on containing but disjoint to . The sequence is -increasing, so if we can show it is cofinal in we will know that the cofinality of is exactly .

Suppose . Then by setting equal to zero outside of we produce a function that is equal to mod and less than everywhere. Our assumptions on then give us an such

and since , we achieve

as required.

For the other direction, suppose by way of contradiction that is an ultrafilter on satisfying with .

Outside of the set , we have and so is -equivalent to a function . Since is universal for , there is an such that and hence

as well.

But is an exact upper bound for mod , and so

Since the cofinality of is we know , and therefore

Putting all of this together yields

and this is a contradiction.

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**Introduction**

I’m not sure exactly where things left off, so I’ll just begin with a series of posts on generators for pcf. I will also try to keep the posts short so that I can ease back into the routine of writing them.

Let us assume that is a progressive set of regular cardinals, and . I will assume we know already some of the basics about the ideal and I will use these facts without much comment.

What we want to look at is the existence of a *generator* for : we will sketch the proof that the ideal is generated over by a single set . And I do mean “sketch”, as the details are worked out nicely in Section 4 of the Abraham/Magidor article in the Handbook of Set Theory.

**Universal Sequences**

We’ll start with the following definition:

Definition 1Suppose . A sequence of functions in is auniversal sequence forif

- is -increasing, and
- is cofinal in whenever is an ultrafilter on with .

Universal sequences are tightly related to the existence of generators for pcf, as we shall see. I want to point out the following result:

Theorem 2Suppose is a progressive set of regular cardinals and Then the following statements are equivalent:

- There is a universal sequence for .
- There is a family such that for any ultrafilter , if has cofinality then remains unbounded in .
- There is a family of subsets of such that

- modulo , and
- is the ideal generated by together with the sets .

The above is basically Fact 2.2 on page 13 of *Cardinal Arithmetic*, and it follows quite easily from the work done in the first section of the book. (The 3rd statement says, in the notation of the book, that is *semi-normal*.) Abraham and Magidor develop basic pcf theory in a slightly different order, and deriving the above result from the material they present prior to defining universal sequences is a bit more difficult.

Of course, the main point is the following result:

Theorem 3If is a progressive set of regular cardinals, then every has a universal sequence.

Again, we will not prove the above as this is Theorem 4.2 of the Abraham-Magidor article, and their proof is quite clear.

What have we learned from the above? We have outlined the first steps towards proving that generators exist. Putting the two theorems presented here together, we see that if , then is pretty simply generated over .

In our next post, we’ll see how “tuning up” a universal sequence leads to the existence of generators.

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What this means in practical terms is that I’m going to be doing some translation of Section 1 of Chapter VIII into the language of the Handbook and see how well the proofs go through.

Tedious, but probably worthwhile in the interest of making the material accessible!

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In this post, I’m going to write down some assumptions [(a)-(e) of Claim 1.2] that will be used in the next few posts.

**The Assumptions**

We assume the following

- is a set of regular cardinals satisfying
- For every , we let satisfy
- ,
- is strictly increasing modulo ,
- if and , then for each we have
- for every and , there is a such that (everwhere)

- is a sufficiently large regular cardinal, is a well-ordering of
- is an increasing continuous sequence of elementary submodels of such that
- and .

I’m going to do some “dictionary work” to get the official names for such objects. I just want to check which terms have become standard, and I’ll use the Abraham/Magidor Handbook article as the final word.

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~~harder~~ more interesting than I thought! I’ll start posting pieces of what I know starting next week after our Spring Break is over…

As noted earlier, Shelah’s second proof of the “cov vs. pp Theorem” ([Sh:400] 3.5) contains an error (acknowledged in [Sh:513]), and although it is possible to effect a repair of most of it, the full version is still unproven. Now if we look ahead to [Sh:410] (one of the first papers continuing The Book) we find that a great many proofs of theorems end with a sentence saying roughly “now repeat the proof of [Sh:400, 3.5]” (I count at least four instances of this in Section 2 of the paper).

So on the face of it, it’s not clear how many of these results are actually valid. It may be that the weaker version of [Sh:400, 3.5] is strong enough to push through the arguments, but since there are very few details it’s hard to tell.

This looks like a nice task for me to tackle in the blog, so that’s probably what I will do.

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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.

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