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?

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    $\begingroup$ In the proof, Jech explicitly says that the existence of $X_{\alpha}$ depends on the fact that $\alpha$ is countable. $\endgroup$ Jan 29, 2019 at 0:05
  • $\begingroup$ There had been several questions about this already. Did you look around the site? $\endgroup$
    – Asaf Karagila
    Jan 29, 2019 at 8:17
  • $\begingroup$ Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer. $\endgroup$
    – user480881
    Feb 4, 2019 at 11:42

1 Answer 1


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 $n\in\mathbb{N}$.)

  • $\begingroup$ Dear Eric, Thank you for your answer and please excuse the late reply. To make my question more precise: I don't claim that $2^{\aleph_0} = \aleph_n$ is any way necessary. I am merely pondering the following thought: Suppose for example that $2^{\aleph_0} = \aleph_2$. What if I construct $X_{\omega_1}$ by applying the diagonal construction (of $X_\omega$) to $\{X_\omega, X_{\omega + 1}, \ldots\}$? As far as I can see, the finite-intersection property still holds, and so does the property that $|\{X_{\omega_1} - X_\beta\}| < \infty\, \forall \beta < \omega_1$. Am I missing something? $\endgroup$
    – user480881
    Feb 4, 2019 at 11:38
  • $\begingroup$ What do you mean by "applying the diagonal construction"? How would you apply it to $\aleph_1$ sets instead of countably many? The whole point of the "diagonal" is that it uses the fact that $\{X_\beta:\beta<\alpha\}$, like $\omega$, is countable. $\endgroup$ Feb 4, 2019 at 16:21
  • $\begingroup$ Dear Eric, Having returned to Jech's Set Theory after being occupied with different matters for a while, I now see that my misunderstanding stems from the terribly embarrassing fact of me for some reason believing that $\omega_1 = 2\omega$ (and hence assuming that the construction of $X_\omega$ can be continued...). Sorry for wasting your time. $\endgroup$
    – user480881
    Feb 6, 2019 at 11:52

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