Theorem 7.8 in Jech's Set Theory states that if $2^{\aleph_0} = \aleph_1$, there exists a Ramsey ultrafilter. The proof is constructive: We enumerate all partitions of $\omega$ (denoted $\mathcal{A}_\alpha$, where $\alpha = 1, 2, \ldots < \omega_1$) and define $X_{\alpha + 1} \subseteq X_\alpha$ as either a subset of some $A \in \mathcal{A}_\alpha$ or such that $|X_{\alpha+1} \cap A| \leq 1 \forall A \in \mathcal{A}_\alpha$. If $\alpha$ is a limit ordinal, then $X_\alpha$ is such that $X_\alpha – X_\beta$ is finite for all $\beta < \alpha$. The desired Ramsey ultrafilter is then given by $\{X: X_\alpha \subseteq X\, \text{for some $\alpha$}\}$.
My question is regarding the assumption that $2^{\aleph_0} = \aleph_1$; I don't see for example why the suggested construction could not also be applied if $2^{\aleph_0} = \aleph_n$, where $n \in \mathbb{N}$. Am I right in assuming that the proposed assumption is just a (weak?) sufficient condition?
Best Answer
The construction very crucially uses the assumption that $2^{\aleph_0}=\aleph_1$ in limit steps. To choose $X_\alpha$ such that $X_\alpha-X_\beta$ is finite for all $\beta<\alpha$, you must use the fact that there are only countably many such $\beta$, so that you can build $X_\alpha$ by a diagonal construction so that it is infinite and yet eventually contained in each $X_\beta$.
(The argument can be generalized to use weaker assumptions than $2^{\aleph_0}=\aleph_1$; for instance, Martin's axiom suffices. But the result is not provable in ZFC alone, and I don't know why you think it would be relevant to assume something like $2^{\aleph_0}=\aleph_n$ for $n\in\mathbb{N}$.)