# Re. Jech Set Theory, Theorem 7.8 ($2^{\aleph_0} = \aleph_1 \implies \exists\: \text{Ramsey ultrafilter}$)

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?

• In the proof, Jech explicitly says that the existence of $X_{\alpha}$ depends on the fact that $\alpha$ is countable. – Karl Kroningfeld Jan 29 '19 at 0:05
• Dear Karl and Asaf, Thank you for your comments. I have clarified my question in a comment to Eric's answer. – user480881 Feb 4 '19 at 11:42

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

• 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? – user480881 Feb 4 '19 at 11:38
• 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. – Eric Wofsey Feb 4 '19 at 16:21
• 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. – user480881 Feb 6 '19 at 11:52