# 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 classes ultrafilters?

Some motivation: 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 $$\alpha$$.

Partial result: 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 $$2^{\mathfrak{c}} = | \beta \mathbb{N}|$$ translates of $$A$$ (i.e. sets like $$A+p$$, $$p+A$$, or even $$p+A+q$$...). The argument is not difficult, and relies on the fact that one can set up a surjective (topological left semigroup)-morphism from $$\beta \mathbb{N}$$ to a topological group, whose cardinality is $$2^\mathfrak{c}$$.

(This is a slightly modified question from Stack Exchange that was left unanswered for a a long while, and also generated a number of upvotes. I hope might find an answer here. The original question can be found here)

• How many idempotents exist? – Monroe Eskew Mar 22 '17 at 23:41