# Nonisomorphic free ultrafilters on $\omega$

Any bijection from $$\Bbb N$$ to itself transforms an ultrafilter on $$\Bbb N$$ to another (isomorphic) ultrafilter. Any two principal ultrafilters are isomorphic in that sense.

For free ultrafilters on $$\Bbb N$$, there are $$2^{2^{\aleph_0}}$$ of them. Since there are $$2^{\aleph_0}$$ bijections of $$\Bbb N$$ to itself, there are also $$2^{2^{\aleph_0}}$$ isomorphism classes of free ultrafilters on $$\Bbb N$$. So lots of free ultrafilters must be nonisomorphic to each other.

Question: Can you give an explicit example or construction of two free ultrafilters on $$\Bbb N$$ that are not isomorphic? Assume ZFC.

(Added at the suggestion of @bof in the comments below, in case the question proves too difficult to answer directly):

1. Give an explicit example of two filters such that no free ultrafilter extending one of them can be isomorphic to a free ultrafilter extending the other.
2. Can you state a property, preserved by isomorphism, possessed by some but not all free ultrafilters?

(1) is as good as the original question as far as I am concerned.

• @bof . The existence of a free ultrafilter on $\Bbb N$ cannot be done without the Axiom of Choice, so in a sense there is no explicit example, as it is also consistent with ZF that they don't exist. – DanielWainfleet Jun 24 at 7:51
• I don't know what you mean by an explicit example since even exhibiting a single nonprincipal ultrafilter on $\omega$ is rather nonexplicit, but what you call being isomorphic is usually called being Rudin-Keisler equivalent. The RK-order has been (is being) extensively studied and there are plenty of results from the 70/80s constructing large families of RK-incomparable ultrafilters on $\omega$. In J. van Mill's chapter on The Handbook of Set Theoretic Topology you can find the construction of two RK-incomparable ultrafilters on $\omega$. Would you count it as explicit? – Alessandro Codenotti Jun 24 at 9:34
• To add to the comment by @AlessandroCodenotti, you might also want to look at combinatorial properties such as p-points, q-points, etc., as well as rapid filters for (1), While it is consistent that no such ultrafilter exists, it is still a way to find a reasonable condition. (I should add that I know very little about these questions, but these are keyword for you to start your journey with.) – Asaf Karagila Jun 24 at 17:22
• @AlessandroCodenotti and Asaf: I was hoping there would something simple, but that may turn out to be elusive. Thanks for the references and background info. – PatrickR Jun 24 at 17:56
• There is a product operation on ultrafilters. Ultrafilters that have square roots are not isomorphic to those that haven't. – JCAA Jun 24 at 18:17

The simplest property that I can think of (right now) that provably (in ZFC) distinguishes some non-principal ultrafilters on $$\mathbb N$$ from others is "weak P-point", which means "not in the closure in $$\beta\mathbb N$$ of a countable set of other non-principal ultrafilters." The existence of weak P-points is a theorem of Kunen; the existence of non-principal ultrafilters that are not weak P-points is trivial (take any countably infinite set of non-principal ultrafilters and take any other point in their closure).

• Thanks! Would you have a reference for weak P-points? – PatrickR Jun 25 at 4:17
• ktiml.mff.cuni.cz/~verner/download/… has a good overview – PatrickR Jun 25 at 8:22
• If I understand you correctly, since $\beta \Bbb N$ \ $\Bbb N$ has no isolated points, this result of Kunen is that $\beta \Bbb N$ \ $\Bbb N$ is not a space of countable tightness. – DanielWainfleet Jun 26 at 6:21
• @DanielWainfleet Kunen's result looks a bit stronger than "not of countable tightness" to me. A counterexample to countable tightness would be a point p that is in the closure of a set A but not in the closure of any countable subset of A. Such a point p could still be in the closure of a countable set disjoint from A, so it might fail to be a weak P-point. – Andreas Blass Jun 26 at 13:05
• Yes I see. Stronger. – DanielWainfleet Jun 26 at 19:11