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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.

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    $\begingroup$ @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. $\endgroup$ – DanielWainfleet Jun 24 at 7:51
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    $\begingroup$ 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? $\endgroup$ – Alessandro Codenotti Jun 24 at 9:34
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    $\begingroup$ 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.) $\endgroup$ – Asaf Karagila Jun 24 at 17:22
  • $\begingroup$ @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. $\endgroup$ – PatrickR Jun 24 at 17:56
  • $\begingroup$ There is a product operation on ultrafilters. Ultrafilters that have square roots are not isomorphic to those that haven't. $\endgroup$ – JCAA Jun 24 at 18:17
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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).

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  • $\begingroup$ Thanks! Would you have a reference for weak P-points? $\endgroup$ – PatrickR Jun 25 at 4:17
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    $\begingroup$ ktiml.mff.cuni.cz/~verner/download/… has a good overview $\endgroup$ – PatrickR Jun 25 at 8:22
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Jun 26 at 6:21
  • $\begingroup$ @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. $\endgroup$ – Andreas Blass Jun 26 at 13:05
  • $\begingroup$ Yes I see. Stronger. $\endgroup$ – DanielWainfleet Jun 26 at 19:11

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