Peter Krautzberger · on the web

# BLAST 2010 chalk slides

The organizers of BLAST 2010 [Wayback Machine] asked the speakers to upload or give a link to the slides of their talk. Since I gave a chalk talk at the blackboard I had no slides. So instead of slides let me give a short recollection of my talk here.

## The semigroup \mathbb{F}.

Definition Denote the non-empty, finite subsets of by with semigroup operation (or if you prefer, ).

To offer different choices for the operation may seem odd, but as mentioned in this post the real concern is with the restriction to disjoint sets, i.e., where both operations coincide.

FU-set Accordingly, in what follows, every sequence in is supposed to be pairwise disjoint. Then the FU-set (generated by ) is the set of all finite unions, i.e.,

An FU-set is ordered if the generating sequence is, i.e., .

## Union Ultrafilters

One of the most important examples of idempotent ultrafilters are union ultrafilters on .

Union Ultrafilters
a) is called union ultrafilter if it has a base of FU-sets.
b) is an ordered union ultrafilter if it has a base of ordered FU-sets.
c) is a stable union ultrafilter if it is a union ultrafilter and for every choice of countably many () there exists such that for all

Here is the usual almost inclusion, i.e., up to a finite set; in other words, is a pseudointersection of the . Note that is not a pseudointersection — idempotents (in fact, products) can never be P-points.

These notions were introduced by Andreas Blass (MR). Union ultrafilters are idempotent since . To digress.

Question Do union ultrafilters have a base of FU-sets generated by unnested sequences? (where unnested means )

This small question came up while preparing the talk; it has no ulterior motive (that I know of) and I never really thought about it.

## Differences — orderedness.

As Andreas Blass said at the beginning of his BLAST 2010 tutorials, the very first question when it comes to ultrafilters is, how ultrafilters can be different from each other. In this case, the question becomes: are the three notions any different? Some results about ordered union ultrafilters are as follows.

• Blass (MR) If is ordered union, then the images , are non-isomorphic Q-points.
• Blass, Hindman If is union, then the images , are P-points.
• Blass, Hindman There consistently exist union, such that , are not Q-points. (in particular, such is not ordered union).
• There consistenly exists union with , both Q-points, but is not ordered. (hopefully on the arXiv soon…)

The actual results are usually a bit stronger, but that’s not important right now.

So on the one hand, ordered unions are really stronger than unions; on the other it is not enough for a union ultrafilter to map to Q-points to imply that it is ordered union. So it stays difficult to differentiate the two notions.

As a typical example of clever arguments with union ultrafilters, let’s prove something. This is from Blass, Hindman.

Theorem(Blass, Hindman) If is a union ultrafilter, then is a P-point.

Proof.

1. Fix .
2. Consider .
3. If , then is bounded (hence constant) on a set in .
1. Take disjoint from .
2. It’s a nice exercise to show that if then so are all the FU-sets generated by the with (for each ).
3. Fix . For all but finitely many we have (say from index onwards).
4. Hence we calculate
In other words, is bounded by .
4. If , then is finite-to-one on a set in .
1. Take included in A.
2. We calculate .
3. But and only finitely many have .

If you look closely, the last line shows that is a rapid P-point; rapidity is attributed to Pierre Matet. The argument for is a bit more complicated (and longer).

## Stability.

So it is difficult to differentiate ordered from unordered — maybe the third notion helps? For this, Andreas Blass proved the following amazing theorem.

Theorem (Blass, MR) If is a ordered union ultrafilter, then the following are equivalent.

1. Stability
2. Whenever is partitioned into two pieces, there exists such that is homogeneous.
3. The analogue for (and even if one part of the partition is analytic).
4. The ultrapower has exactly 5 constellations generated by , , , and .
5. generates an initial segment of .

This result is just amazing. It connects a P-point-like property to a Ramsey property — quite unlike the classical situation. At the time it was not yet known that union ultrafilters give P-points as and (which follows easily from stability), so it was even more surprising.

Now, you’d think there must be a difference if you drop the orderedness of the union ultrafilter, right? Unfortunately, this is not really the case.

Theorem If is a union ultrafilter, then 1-4 are equivalent and implied by 5. (again, arXiv, hopefully, soon…)

The proof follows Andreas Blass’s proof. He needed orderedness only in ‘1 implies 2’ and ‘5 implies 1’ so this where one has to work a little.

This part 5) is a bit of a rogue since I hadn’t thought about that part until recently and so it is a ‘new’ observation. I do not know if it is strictly stronger and there is evidence that it might not be. But again there seems little hope to differentiate ordered from unordered by means other than the definition.

Regarding the restriction to ordered pairs: this cannot be weakened to disjoint pairs, since any FU-set (ordered or not) will contain ordered and unordered disjoint elements (if ordered, compare to ).

Now, all constructions in the literature yield stable union ultrafilters, hence the big open question for me is:

Question Does there consistently exist an unstable union ultrafilter, i.e. a union ultrafilter that is not stable?

I have worked on this for a while now and even though I hope these exist (say under CH or MA) I think at the moment it is wide open.

## The other world.

My own interest lies more in the world of .

Strongly Summable Ultrafilters An ultrafilter on is called strongly summable if it has a base of FS-sets.

FS-sets stand for the analogue of FU-set, i.e., for ‘finite sums set’ — but here with no extra conditions on the sequence in .

If is a union ultrafilter, , then is a strongly summable ultrafilter.

This is easy since disjoint unions map to correct sums (there is no carrying over after all). There is also a way back, i.e., for every strongly summable there is a similar looking function that maps it to a union ultrafilter, see Blass, Hindman, however

Question If is strongly summable, is a union ultrafilter?

I hope this is not true, since it would simplify too many interesting questions, but it seems possible.

As an example why the other world is interesting, let me give (a special case of a) theorem (going back to a theorem by Neil Hindman and Dona Strauss).

Theorem If is a union ultrafilter, with , then (arXiv, soon…).

What does this mean? The only way to write these kinds of strongly summable ultrafilters as a sum is the trivial way via integers. Ah, I should mention: it is an exercise to show that is a left ideal in (so this actually makes sense…) and then that the integers commute (so this trivial way is always possible).

This is a fascinating property and these (and some more) strongly summables are the only known examples with this property (unlike nearly all other results for idempotents which are in ZFC). I hope to blog more about more special properties sometime soon but this is all I covered in my talk, so enough for today.