Let $\text{NPU}(\omega)$ be the set of non-principal ultafilters on $\omega$. The Rudin-Keisler preorder on $\text{NPU}(\omega)$ is defined by $${\cal U} \leq_{RK} {\cal V} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall U\in{\cal U}) f^{-1}(U)\in {\cal V} .$$

It is easy to see that $\leq_{RK}$ is reflexive and transitive, but not anti-symmetric. Set ${\cal U}\simeq_{RK} {\cal V}$ if ${\cal U}\leq_{RK}{\cal V}$ and ${\cal V}\leq_{RK}{\cal U}$. So $\text{NPU}(\omega)/\simeq_{RK}$ is a poset with the Rudin-Keisler order applied to equivalence classes.

Question. If ${\cal C},{\cal D}$ are totally ordered subsets of $\text{NPU}(\omega)/\simeq_{RK}$ that are maximal with respect to $\subseteq$, do they necessarily have the same cardinality?


1 Answer 1


Yes. I claim that such a maximal chain must always have cardinality $\frak{c}^{+}$.

First of all, take notice that for each ultrafilter $U$, there are at most $\frak{c}$ many Rudin-Kiesler equivalence classes below $U$ simply because there are at most continuumly many functions $f:\omega\rightarrow\omega$ and every ultrafilter RK-below $U$ is of the form $\{R\subseteq\omega|f^{-1}[R]\in U\}$ for some function $f$.

Now observe that for every collection $\mathcal{C}$ of at most continuumly many ultrafilters, by my answer here there is an ultraftiler $U$ with $V\leq_{RK}U$ for each $V\in\mathcal{C}$.

Suppose now that $\mathcal{C}$ is a maximal chain in the collection of ultrafilters. Now, let $(U_{\alpha})_{\alpha<\lambda}$ be a strictly increasing cofinal sequence where $\lambda$ is some regular cardinal. Then since every collection of at most continuumly many ultrafilters on $\omega$ has a RK-upper bound, we conclude that $\lambda>\frak{c}$. On the other hand, since each $U_{\alpha}$ has at most continuumly many $RK$-smaller ultrafilters, we conclude that $\lambda=\frak{c}^{+}$. Since each $U_{\alpha}$ has at most continuumly many predecessors, we coonclude that $|\mathcal{C}|=\mathfrak{c}^{+}$ as well.

  • $\begingroup$ What happens if $\frak c^+$ is singular? $\endgroup$
    – Asaf Karagila
    Oct 20, 2017 at 7:01
  • $\begingroup$ Asaf Karagila. ? $\endgroup$ Oct 20, 2017 at 21:52

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