# How much do idempotent ultrafilters generate in terms of semigroups?

It is known that the set of ultrafilters on, say, the natural numbers $\mathbb{N}$, can naturally be endowed with the structure of a compact topological left semigroup (which fails to be anything nicer in quite a spectacular fashion, e.g. it is badly uncommutative and right-discontinuous).

An very nice class of ultrafilters are the idempotent ones, i.e. such that $p + p = p$. That they exist is already non-trivial (Ellis theorem) and useful (an extremely elegant proof of Hindman's theorem). Since the situation is non-commutative, it cannot be hoped that if $p$ and $q$ are idempotent, then $p+q$ is again idempotent. However, there are a number of properties that are (1) interesting (in my humble opinion), (2) true of idempotents (3) preserved under sums.

Hence, the following question feels natural: How large is the family of ultrafilters generated by the idempotents by taking sums? What if you allow limits as well? Is it in any significant way larger than the idempotents themselves? What if instead of idempotents, one looks at other special ultrafilters?

Edit: Given that the question got a few upvotes, but no answers, I thought I should mention what I know about the problem.

A first motivation, and a partial (negative) result is the following. Suppose $f : \mathbb{N} \to \mathbb{N}$ is a map. For $\alpha \in \mathbb{T} = \mathbb{R}/\mathbb{Z}$, we can consider the sequence $(f(n)\alpha)_{n\in \mathbb{N}}$. For any ultrafilter $p$ there is then a well defined notion of the generalised limit (with respect to this ultrafilter) given by the rule that: $p\!-\!\lim_n f(n) \alpha = \gamma$ if and only if for any open neighbourhood $U \subset \mathbb{T}$ of $\gamma$ we have $\{ n \ : \ f(n) \alpha \in U \} \in p$ (so, $f(n) \alpha$ is close to $\gamma$ for $p$-many $n$).

It turns out that if $p$ is an idempotent and $f$ is a polynomial with $f(0) = 0$, then $p\!-\!\lim_n f(n) \alpha = 0$. (It is closely related to the other question asked here). However, this is not quite the end of the story. It is true that $$(p+q)\!-\!\lim_k f(k) \alpha = p\!-\!\lim_n q\!-\!\lim_m f(n+m) \alpha$$ which can be used to conclude that if the claim holds for two ultrafilters $p,q$, then it also holds for their sum. What is even more, the map $p \mapsto p\!-\!\lim_n f(n) \alpha$ is continuous. The bottom line is that the mentioned fact can be generalised from the idempotents to everything one can generate by taking sums and closures. There are the following two consequences. On one hand, if we knew something about how large is the closed sub-semigroup generated by the idempotents, there would be a result that would follow immediately. On the other hand, this shows that we surely can't generate everything, because there are ultrafilters $p$ with $p\!-\!\lim_n \alpha n \neq 0$ for any fixed $\alpha$.

I would be grateful for any of the following:

• another reason why the idempotents don't generate all ultrafilters.

• any description of how much is it that the idempotents generate.

• answer to whether there exist ultrafilters for which it holds that $p\!-\!\lim_n f(n) \alpha = 0$ under the assumptions above, but $p$ is not generated by the idempotents.

Edit: It turns out that the ultrafilters generated by the idempotents (allowing sums and limits) are still rather small fraction of the full space. In particular, if $A$ is the set of all such ultrafilters, then it takes at least $2^{\mathfrak{c}} = \# \beta \mathbb{Z}$ translates of $A$ (i.e. sets like $A+p$, $p+A$, or even $p+A+q$...). The argument is not difficult, but I am not sure if it makes sense to present it here.

Edit: This has been cross-posted to MathOverflow.

• For my edification alone, I'd love to see a link describing how to make the ultrafilters a compact topological left semigroup. :) – Thomas Andrews Mar 29 '13 at 20:56
• @Thomas: Compactness is easy: this is just $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$. The operation is described very nicely about two-thirds of the way down this page. You might also be interested in this question and my answer. – Brian M. Scott Mar 29 '13 at 21:13
• @ThomasAndrews: a good place to look is Vitaly Bergelson's "Ultrafilters, IP sets, Dynamics, and Combinatorial Number Theory" available at his personal page (as well as plenty of other excellent papers): math.osu.edu/~bergelson.1 – Jakub Konieczny Mar 29 '13 at 22:36
• Thanks Brian and Feanor! Love this sort of thing. – Thomas Andrews Mar 29 '13 at 22:38
• @MartinSleziak - Thanks for the advice! I'll make sure to follow it if I cross-post again. (In my defence, I was still young and naive at the time I posted this one). – Jakub Konieczny Sep 10 '16 at 19:35