From the answer of Andreas Blass and comments of Ali Enayat on my question Selective ultrafilter and bijective mapping it became clear that for any free ultrafilter $\mathcal{U}$ we have $\mathcal{U}\nsim\mathcal{U}\otimes\mathcal{U}$. Where $\mathcal{F}\otimes\mathcal{G}=\{X\subset A\times B~|~\{a\in A~|~ \{b\in B~|~ (a,b)\in X\} \in\mathcal{G}\}\in\mathcal{F}\}$ for any two filters $\mathcal{F}, \mathcal{G}$ on sets $A, B$ respectively. And $\mathcal{F}\sim\mathcal{G}$ means existing of bijection $f:A\to B$ such that $B\in\mathcal{F}\iff f(B)\in\mathcal{G}$. From this link given by Ali Enayat I have also understood that $\mathcal{N}\nsim\mathcal{N}\otimes\mathcal{N}$ for the Freshet filter $\mathcal{N}$ but on any infinite set there exists a nontrivial idempotent filter $\mathcal{F}$, i.e. $\mathcal{F}\sim\mathcal{F}\otimes\mathcal{F}$. I couldn't get article referenced there as "On idempotent filters, Casopis Pest. Mat. 102 (1977), 412-418", in which the example of idempotent filter is given. So, I am asking the example of the idempotent filter or subsets family with finite intersections property. Can such a filter or family contain only infinite subsets?

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    $\begingroup$ The Katětov's paper you mention can be found at eudml or at dml.cz. (Both locations are easy to find using Google Scholar.) $\endgroup$ Mar 7 '19 at 8:31
  • $\begingroup$ Thank you. I have got the answer. $\endgroup$
    – ar.grig
    Mar 8 '19 at 14:20

Following the link given by Martin Sleziak in comment I have got the answer:

There exists filter $\mathcal{F}$ with the following properties:

  1. $\mathcal{N}\subset\mathcal{F}$
  2. $\mathcal{F}\sim\mathcal{F}\otimes\mathcal{F}$

Existence proof is easy. But construction of $\mathcal{F}$ is complicated.


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