Peter Krautzberger · on the web

# Rapid idempotent ultrafilters

Welcome back to the second (and final) part on why strongly summable ultrafilters are rapid.

Why the new title? Well, first, I needed another one (I've had too many posts with "Part X" in them, I think). Second, after I proved the results I mentioned last time, I quickly found an additional, somewhat more general observation regarding idempotents.

Here's the problem with this post though. I want to give the argument. But you know how it is in mathematics: standing on the shoulders of the huddled masses giants and all that. All work relies on established results. In the first post, I gave a lot of "unnecessary" proofs to motivate the story behind the result. In this post, I'll get to the actual proof and I'm torn, I'm not sure how much to quote and how much to prove. So bear with me and leave comments whenever I screw up.

### Strongly summables are rapid.

Why could this be true? On the one hand, because we already established partial results last time. On the other hand there's an old result attributed to Pierre Matet which, for now, I can only state in a obfuscated fashion.

Theorem (Matet 87)
If is a strongly summable ultrafilter, then there exists a function "" such that is rapid.

So you see, strongly summable ultrafilters imply the existence of rapid ultrafilters -- via a very simple -function.

Ah yes, I should talk about ...

### \max to the Max.

For FS-sets, there is a natural notion of maximum. If you have a bunch of elements summed up, then it makes sense to call the largest summand the maximum. So intuitively, the maximum should simply map each element in the FS-set to the largest generator involved in producing it. This might not appear very well defined but bear with me.

Consider an FS-set, let's keep calling it . If we're lucky, there is a unique way to write each as a sum of (or should I write ? indices are hard...). For example, the sequence has this property while fails to have this property (quite badly, I suppose).

It turns out that there's an easy property to ensure this: the sequence just has to grow quickly.

Proposition (folklore? can be found in Blass, Hindman 87)
If for all , then

In other words, each element in has a unique representation.

[Edit May 22, 2012: modified attribution]

The proof is an easy induction on , using the growth factor to argue that the maximal element of and must be equal.

• If , then follows immediately from the growth assumption.
• Now assume inductively that we've proved the claim for smaller sets.
• Let be the maximum of .
• Wlog, .
• Then we must also have -- otherwise

a contradiction to the assumed equality of both sums.

• But if both and contain , we also have .
• Using our induction hypothesis, we have , hence .

So if we have unique representations of elements in , we can define a function

### Growth, growth, growth and more growth

Unfortunately, we need to tweak this a little bit more. Remember that FS-sets are all about sums. Very often will have . So assuming growth, each have a unique representation in terms of the .

It's natural to assume that there's some kind of connection between the representations of . For one thing, we know that if , then . The growth as above does not guarantee the reverse, though (just consider ), and the reverse often simplifies things. Fortunately, all we have to do is improve the growth!

Proposition (Hindman, Blass 87)
If , then if and only if and .

The proof is much like the earlier proof.

• Again, we're doing an induction on .
• If , then or must be empty and the growth condition does the rest.
• For the inductive step, let .
• Due to the growth condition, must be in .
• But it must also lie in .
• Else .
• But it can't be in both and .
• Else .
• Wlog
• Th .
• Applying our induction hypothesis to , we get and .
• This in turn gives us (as ) and -- as desired.

Why all this trouble? Well, there are many uses for this. For what's coming below, it simplifies an important calculation. In general, it is extremely important since it allows us to switch from the addition of numbers to the union of disjoint, finite sets. (I don't know about you, but I find the union operation on disjoint sets much easier to comprehend.)

Corollary
If has growth as above, then if we ever have (and we will), then, assuming , we have .
In particular, if , then .

### Strongly summable ultrafitlers are rapid -- the proof.

Anyway, let's get back to where we started. First, we should make a connection to strongly summable ultrafilters.

Lemma (Blass, Hindman 87)
If is strongly summable, then has a base of FS-sets whose sequences satisfy the growth condition (the stronger one with factor , of course).

This is a great lemma (though maybe not a true lemma) and the reason why I spend so much time above talking about growth conditions -- it comes in handy in many situations and really tells us something about strongly summable ultrafilters and the sets they contain. The proof, however, is weird so I'll skip it (unless you insist in the comments).

And now it makes sense to state the initial theorem.

Theorem (Matet, 87 / Blass, Hindman 87)
Let be a strongly summable ultrafilter and with growth (or just unique representations); fix the -function for as above. Then is a rapid P-point.

[Edit on May 21, 2012: I rephrased the theorem to improve clarity -- thanks to the comment-by-email who suggested it!]

You can skip the proof if you like because it's not important to us (and I'm cheating a little on the important part, rapidity). But I find the argument appealing and since I've had to go through all the trouble to introduce the growth condition and so forth, I think I might as well include this, too. It's a typical proof for strongly summable ultrafilters -- just write down a good partition and let it do the work for you.

• Fix .
• We will prove that is either constant on a set in or it is finite-to-one.
• Pick any with .
• Let be the minimum function analogous to .
• Now partition intoand
• Our strongly sumable ultrafilter will give us (with the usual growth condition) included in one of these two parts.
• If is included in the first part, then is bounded on . (In particular, is constant on a set in .)
• Consider and pick any other .
• Due to the growth condition, we have and reversely .
• In particular, .
• So is bounded on