# Metrically Ramsey ultrafilters

## TL;DR

This paper investigates ultrafilters in metric spaces, introducing the concept of metrically Ramsey ultrafilters, and explores their properties, existence, and relation to classical Ramsey ultrafilters, especially on the natural numbers.

## Contribution

It establishes the existence of countable subsets in ultrametric spaces ensuring ultrafilters are metrically Ramsey and examines properties of metrically Ramsey ultrafilters on natural numbers.

## Key findings

- Every infinite ultrametric space has a countable subset related to metrically Ramsey ultrafilters.
- Every metrically Ramsey ultrafilter on natural numbers contains a member with no 2-term arithmetic progression.
- If such an ultrafilter has a thin member, it maps to a classical Ramsey ultrafilter via a specific function.

## Abstract

Given a metric space $(X,d)$, we say that a mapping $\chi: [X]^{2}\longrightarrow\{0.1\}$ is an isometric coloring if $d(x,y)=d(z,t)$ implies $\chi(\{x,y\})=\chi(\{z,t\})$. A free ultrafilter $\mathcal{U}$ on an infinite metric space $(X,d)$ is called metrically Ramsey if, for every isometric coloring $\chi$ of $[X]^{2}$, there is a member $U\in\mathcal{U}$ such that the set $[U]^{2}$ is $\chi$-monochrome. We prove that each infinite ultrametric space $(X,d)$ has a countable subset $Y$ such that each free ultrafilter $\mathcal{U}$ on $X$ satisfying $Y\in\mathcal{U}$ is metrically Ramsey. On the other hand, it is an open question whether every metrically Ramsey ultrafilter on the natural numbers $\mathbb{N}$ with the metric $|x-y|$ is a Ramsey ultrafilter. We prove that every metrically Ramsey ultrafilter $\mathcal{U}$ on $\mathbb{N}$ has a member with no arithmetic progression of length 2, and if $\mathcal{U}$ has a thin member then there is a mapping $f:\mathbb{N}\longrightarrow\omega $ such that $f(\mathcal{U})$ is a Ramsey ultrafilter.