Hereafter, "ultrafilter" is intended to mean "nonprinicpal ultrafilter on the set, $\omega$, of natural numbers."

An ultrafilter $U$ is Ramsey if for every partition $P$ of $\omega$ into sets not in $U$, there is a set $a\in U$ such that $\forall p\in P, \vert{a\cap p}\vert = 1$.

Assuming the CH, one can construct a Ramsey ultrafilter by enumerating all paritions $\langle P_\alpha : \alpha<\omega_1\rangle$, and iteratively choosing infinite generating sets, $X_{\alpha+1}\subseteq X_\alpha$ such that either $X_{\alpha+1}\subset p$ for some $p\in P_\alpha$ or $\vert X_{\alpha+1}\cap p\vert \leq 1$ for all $p\in P_\alpha$. Because each limit stage $\beta$ is only countable we can find a set $X_\beta$ that is almost contained in $X_\alpha$ for each $\alpha<\beta$.

In fact, the construction above can be used to produce $\aleph_1$ distinct Ramsey ultrafilters: for a suitable $P_0$, there are $\aleph_1$ choices for $X_1$ that lead to distinct ultrafilters.

On the other hand, I believe that there are at least $2^\frak{c}$ ultrafilters: pick any almost disjoint family $\cal{A}$ of size $\frak{c}$, then for each $S\in \mathscr{P}(A)$ let $U_S$ be any ultrafilter containing $\{A\in S\}\cup\{\omega\setminus A: A\in \cal{A}\setminus S\}$.

My question is: when Ramsey ultrafilters exist, how many exist? Specifically I'm interested in the situation under the (G)CH.

Can the construction above be observed to produce the maximum number of ultrafilters? Or at least more than $\aleph_1$?


2 Answers 2


The Ramsey ultrafilters on $\omega$ are precisely those that are Rudin-Keisler minimal in $\beta\omega \setminus \omega$, and under CH there are $2^{2^\omega}$ of them. See Theorems 4.5 and 5.5 in W.W. Comfort, Ultrafilters: some old and new results, Bull. Amer. Math. Soc. 83 (1977), 417-455, available as a PDF from here.

  • $\begingroup$ Yes that is the result I needed, and thanks for the useful link! $\endgroup$
    – ok now
    Jun 30, 2011 at 19:37

It is consistent with ZFC that there are "few" Ramsey ultrafilters. For example, in the model of ZFC obtained by iterating $\omega_2$-many Sacks reals with countable support forcing over a model of ZFC+CH the continuum = $\omega_2 = 2^{\omega_1}$ and every Ramsey ultrafilter is $\omega_1$-generated. Thus there are only continuum-many of them.

See article by James Baumgartner and Richard Laver "Iterated perfect set forcing" Annals of Mathematical Logic 17 391-397 (1979), DOI: 10.1016/0003-4843(79)90010-X.


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