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Recall that a nonempty set ${\cal I}\subseteq {\cal P}(\omega)$ is a (set) ideal if

  • $B\in{\cal I}$ and $A\subseteq B$ imply $A\in{\cal I}$, and
  • $A,B \in {\cal I}$ implies $A\cup B\in {\cal I}$.

By $\text{Id}(\omega)$ we denote the set of all ideals on $\omega$. The Katetov ordering on $\text{Id}(\omega)$ is defined by $${\cal I} \leq_K {\cal J} :\Leftrightarrow (\exists f:\omega\to\omega)(\forall I\in{\cal I}) f^{-1}(I)\in J .$$

It is easy to see that $\leq_K$ is reflexive and transitive, but not anti-symmetric. Set ${\cal I}\simeq_K {\cal J}$ if ${\cal I}\leq_K{\cal J}$ and ${\cal J}\leq_K{\cal I}$.

Questions.

1) What are the cardinalities of the equivalence classes $[{\cal I}]_{\simeq_K}$ for ${\cal I}\in\text{Id}(\omega)$?

2) What is the cardinality of the poset $\text{Id}(\omega)/\simeq_K$?

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    $\begingroup$ Have you looked at the papers "Katetov Order on Borel Ideals" by Hrusak and "Some Structural Aspects of the Katetov Order on Borel Ideals" by Guzman-Gonzalez, Meza-Alcantara? They give some results which give lower bounds to your questions I think. $\endgroup$
    – Apollo
    Oct 6, 2017 at 18:32
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    $\begingroup$ When restricted to maximal ideals, the Katetov ordering coincides with the Rudin-Keisler (RK) ordering. It is a matter of counting to see that there are as many RK-classes of maximal ideals as there can be (i.e. $2^{2^{\aleph_0}}$). This implies that the answer to 2) is $2^{2^{\aleph_0}}$. $\endgroup$ Oct 6, 2017 at 23:30
  • $\begingroup$ Thanks Ramiro, also for mentioning the Rudin-Keisler order! I would assume that the equivalence classes have size $2^{\aleph_0}$ as this is the cardinality of all functions $f:\omega\to\omega$, but maybe this reasoning is false. If it is correct, maybe you could put your comment in an answer? $\endgroup$ Oct 7, 2017 at 6:33
  • $\begingroup$ It is not hard to see that classes restricted to maximal ideals have size at most $2^{\aleph_0}$ (could be less; for instance the class of the principal ideals is countable) for the reason you mentioned. I'm not so sure about the classes on the full set of ideals. $\endgroup$ Oct 7, 2017 at 21:22

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