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I was studying idempotent ultrafilters when I saw that no principal ultrafilter could ever be idempotent, because $\left\langle n \right\rangle \oplus \left\langle n \right\rangle = \left\langle 2n \right\rangle$.

A Ramsey ultrafilter is, by definition, never principal, so the question that arose to me was: do Ramsey idempotent ultrafilters exist?

First some explanation:

- Principal ultrafilter and free filter
- Injective function and ultrafilters
- Generalization of $f(\overline{S}) \subset \overline{f(S)} \iff f$ continuous
- Ultrafilters and measurability
- Which “limit of ultrafilter” functions induce a compact Hausdorff topological structure?
- Where has this common generalization of nets and filters been written down?

One of the characterisations of an ultrafilter to be Ramsey is:

For every partition $\{X_i \mid i \in \mathbb{N} \}$ where $X_i \notin \mathcal{F}$ for every $i \in \mathbb{N} $, there exists $X \in \mathcal{F}$ such that $\lvert X \cap X_i \rvert \leq 1$ for every $ i \in \mathbb{N} $.

An ultrafilter is called idempotent if $U \oplus U = U$, where

$U \oplus V = \{ A \subset \mathbb{N} \mid \{ m \in \mathbb{N} \mid A – m \in V \} \in U \}$ and $A – m = \{ n \in \mathbb{N} \mid m + n \in A \}$

(I was trying to construct a partition of $\mathbb{N}$ of infinitely many finite pieces, such that if I get an $ A \in \mathcal{U}$ that satisfies $\lvert X \cap X_i \rvert \leq 1$ for every $ i \in \mathbb{N} $, $A$ can not be an element of $ U \oplus U$ but I did not succeed).

I have a strong feeling that a Ramsey ultrafilter cannot be idempotent, but ‘a strong feeling’ is not a proof. Maybe someone can help me thinking about the subject? Do there exists proofs of the form I was trying to make it? Or maybe what I want to prove is not true? Can someone recommend literature about this subject or is it just irrelevant what I’m trying to do?

Thanks in advance!

- Existence of ultrafilters
- Algebra defined by $a^2=a,b^2=b,c^2=c,(a+b+c)^2=a+b+c$
- Ultrapower and hyperreals
- When an ideal is generated by idempotents
- Basic facts about ultrafilters and convergence of a sequence along an ultrafilter
- Idempotents in $\mathbb Z_n$
- every non-principal ultrafilter contains a cofinite filter.
- Ultrafilters and measurability
- Characterization of weak convergence in $\ell_\infty$
- Questions related to intersections of open sets and Baire spaces

The question seems (at least to some extent) to be answered in comments, so I am posting what was said in comments as a CW-answer.

From the above comment by Shawn Henry:

Ramsey ultrafilters are not idempotent. A nice topological proof can be put together from the material in section 5 of this paper of Blass: http://www.math.lsa.umich.edu/~ablass/ufdyn.pdf

First since $\mathcal U+\mathcal V=\operatorname{{\mathcal U}-lim}_n (n+\mathcal V)$, while P-points are, in view of their topological description, never limit points of a countable set of other ultrafilters, it follows that no P-point can be of the form $\mathcal U+\mathcal V$ with non-principal $\mathcal U$ and $\mathcal V$. In particular, no P-point can be recurrent. So the family of P-points and, a fortiori, the subfamily of selective ultrafilters are disjoint from the families of ultrafilters studied in the preceding sections (recurrent, idempotent, etc.).

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