Is the Rudin-Keisler order of ultrafilters linear? - MathOverflow

Is the Rudin-Keisler order of ultrafilters linear?

2011-02-27T13:49:20Z

Is the Rudin-Keisler order of ultrafilters linear?

Is it a well ordering?

Answer by Martin Sleziak for Is the Rudin-Keisler order of ultrafilters linear?

2011-02-27T14:43:04Z

This would be more suitable as a comment, but I do not have enough reputation points. Among the first results that google spits out for Rudin Keisler is the paper Anatoly Gryzlov: On the Rudin-Keisler order on ultrafilters; http://dx.doi.org/10.1016/S0166-8641(96)00109-5 It claims that there are incomparable ultrafilters (with respect to R-K order) and provides several further references.

Answer by Joel David Hamkins for Is the Rudin-Keisler order of ultrafilters linear?

2011-02-27T14:56:38Z

One often considers the Rudin-Keisler order in the large cardinal context of a measurable cardinal $\kappa$, where one considers only $\kappa$-complete nonprincipal ultrafilters on $\kappa$. So let me provide the answer for this context, which is that it is independent of ZFC whether the Rudin-Keisler order on such ultrafilters is linearly ordered.

Two ultrafilters are Rudin-Keisler equivalent if and only if they are isomorphic, if and only if they give rise to the same ultrapower embedding of the universe (I believe this was a topic of some of your previous MO questions). One ultrafilter $\mu$ is RK-below another $\nu$ if and only if the ultrapower $j_\nu$ by $\nu$ can be factored as $j_\nu=h\circ j_\mu$ for some elementary embedding $h:M_\mu\to M_\nu$. It follows that the Rudin-Keisler minimal ultrafilters are precisely the normal measures.

On the one hand, in the canonical inner model $L[\mu]$ having one measurable cardinal $\kappa$, it turns out that every ultrapower embedding by a $\kappa$-complete ultrafilter on $\kappa$ is equivalent to a finite iteration of $\mu$. That is, every $\kappa$-complete nonprincipal ultrafilter is Rudin-Keisler equivalent to a finite product $\mu^n$ of $\mu$ with itself. Such measures form an increasing $\omega$-sequence, and so in this model, the Rudin-Keisler order on measures on $\kappa$ is a linear order isomorphic to $\omega$.

On the other hand, in contrast, one can perform forcing so as to create many normal measures. In such a model, the Rudin-Keisler order cannot be linear, since normal measures, being minimal, are incomparable with respect to the Rudin-Keisler order.

Update. Meanwhile, the Rudin-Keisler order on this collection of ultrafilters is well-founded, which fulfills part of what you had requested. The reason is that we can associate to each $\kappa$-complete ultrafilter $\mu$ on $\kappa$ the least ordinal $\delta$ generating the whole embedding, that is, for which $M_\mu=\{j_\mu(f)(\delta)\mid f:\kappa\to V\}$. It turns out that $\mu\lt_{RK}\nu$ implies $\delta_\mu\lt \delta_\nu$, and so the Rudin-Keisler order is well-founded.

Answer by Stefan Geschke for Is the Rudin-Keisler order of ultrafilters linear?

2011-02-27T18:10:15Z

As Martin Sleziak already pointed out, there are Rudin-Keisler incomparable ultrafilters on $\omega$ (while Joel is talking about ultrafilters on a measurable cardinal). This is provable in ZFC.

Andreas Blass showed that under Martin's Axiom, there are actually $2^{\mathfrak c}$ pairwise R-K incomparable ultrafilters, i.e., as many as you could possibly have. Here $\mathfrak c$ is $2^{\aleph_0}$. He also showed that the real line can be embedded into the R-K order of ultrafilters on $\omega$ (this again assumes Martin's Axiom).
Blass' results actually talk about $P$-points, which need not exist at all by a result of Shelah.