Ramsey ultrapowers: any interesting properties? A Ramsey ultrafilter is a free ultrafilter $D$ on $\omega$ such that for any partition of $\omega$ into subsets, none of which is in $D$, there exists a set in $D$ that intersects each member of the partition in exactly one place.  
In Jech's Set Theory, Exercise 7.9, we have the following equivalent formulation:
An ultrafilter $D$ on $\omega$ is Ramsey iff every function $f:\omega\to\omega$ is either one-to-one on a set in $D$, or constant on a set in $D$.
This seems to indicate that in the ultrapower $\omega^\omega/D$, equivalence classes of constant functions are "separated" from the other equivalence classes (more than usual). Are there any order properties that might reflect this? 
I do not really know if this is an interesting question, but I would appreciate any ideas or references.
 A: I’m not sure to what extent this addresses your question, but since no one has suggested anything better, I’ll offer it. I’d say that it’s more a matter of all of the elements of the ultrapower being separated from one another more than usual. The property that I have in mind isn’t an order property, and it doesn’t require the full strength of the Ramsey hypothesis, but it is a nice property of Ramsey ultrafilters.
If $p\in\beta\omega\setminus\omega$, and we take $\{S^\omega/p:S\subseteq\omega\}$ as the base for a topology on $\omega^\omega/p$, it turns out that this topology is Hausdorff if $p$ is Ramsey.
Let $p$ be a free ultrafilter on $\omega$, and let $f,g\in{}^\omega\omega$; there are two ways in which $p$ might fail to distinguish $f$ and $g$. One is combinatorial: it may be that $f\equiv g\pmod{p}$, meaning that $\{n\in\omega:f(n)=g(n)\}\in p$ and hence that $f$ and $g$ represent the same element of the ultrapower $\omega^\omega/p$. The other is topological: it may be that $\bar f(p)=\bar g(p)$, where $\bar f:\beta\omega\to\beta\omega$ is the continuous extension of $f$ to the Čech-Stone compactification of $\omega$. It’s not hard to show that $\bar f(p)=\{U\subseteq\omega:f^{-1}[U]\in p\}$, and it’s then clear that if $f\equiv g\pmod p$, then $\bar f(p)=\bar g(p)$. However, in general it’s possible to have $\bar f(p)=\bar g(p)$ even when $f$ and $g$ are combinatorially distinct.
Sometime back in the late $1970$s I was briefly interested in what I called separative ultrafilters: free ultrafilters $p$ on $\omega$ with the converse property that if $f,g\in{}^\omega\omega$ and $\bar f(p)=\bar g(p)$, then $f\equiv g\pmod p$. Suppose that $p$ is such an ultrafilter, and $f,g\in{}^\omega\omega$ with $f\not\equiv g\pmod p$. Then $\bar f(p)\ne\bar g(p)$, so there is an $A\subseteq\omega$ such that $A\in\bar f(p)$ and $\omega\setminus A\in\bar g(p)$, so $f^{-1}[A]\in p$ and $g^{-1}[\omega\setminus A]\in p$. If $\hat f$ is the equivalence class of $f$ in the ultrapower $\omega^\omega/p$, $\hat f\in A^\omega/p$ and $\hat g\in(\omega\setminus A)^\omega/p$. Thus, $A^\omega/p$ and $(\omega\setminus A)^\omega/p$ are disjoint open nbhds of $\hat f$ and $\hat g$, whence the name separative. This topology is Hausdorff precisely when $p$ is separative. For some reason that I’ve long since forgotten, I never published the work; others did so starting a little later (and going much further), and these ultrafilters seem now to be generally known as Hausdorff ultrafilters. You can find a bit on them if you search the web, though I can’t give you specific references for the basic results that follow, since I’m working from my old notes.

Proposition. Every Ramsey ultrafilter is Hausdorff.
Proof. If $p$ is Ramsey, then $\text{mod }p$ every $f\in{}^\omega\omega$ is either injective or constant. Suppose that $f,g\in{}^\omega\omega$, and $\bar f(p)=\bar g(p)$. If $f$ is constant $\text{mod }p$, then $\bar f(p)=n$ for some $n\in\omega$; but then $\bar g(p)=n$, $$\{k\in\omega:f(k)=g(k)=n\}\in p\;,$$ and $f\equiv g\pmod p$. If $f$ is injective $\text{mod }p$, then $\bar f(p)$ is equivalent to $p$ in the Rudin-Keisler order on $\beta\omega$, and therefore so is $\bar g(p)=\bar f(p)$, and it follows that $g$ is also injective $\text{mod }p$. Thus, we may without loss of generality assume that $f$ and $g$ are bijections. Clearly $\overline{(g^{-1}\circ f)}(p)=p$, so $g^{-1}\circ f\equiv\mbox{id}_\omega\pmod p$, and $f\equiv g\pmod p$. It follows that $p$ is Hausdorff, as desired. $\dashv$

However, a Hausdorff ultrafilter need not be Ramsey: it turns out, for instance, that if $p$ and $q$ are Ramsey ultrafilters on $\omega$, the product ultrafilter
$$p\times q=\Big\{A\subseteq\omega\times\omega:\{n\in\omega:\big\{m\in\omega:\langle n,m\rangle\in A\}\in q\big\}\in p\Big\}$$
is Hausdorff iff $p\ne q$, and in that case it is not even a $p$-point, let alone Ramsey.
By the way, your definition of Ramsey ultrafilter is actually the traditional definition of selective ultrafilter; the name Ramsey ultrafilter on $\omega$ comes from the fact that these are precisely the ultrafilters with the property that for every $2$-coloring $\chi:[\omega]^2\to 2$ there is a $\chi$-monochromatic $U\in\mathscr{U}$. If you’ve not seen a proof that the two definitions are equivalent, there’s one here.
