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Let $\mathcal{F}$ be a $P$-filter on $\omega$. Denote by $\Omega=\bigsqcup \omega_i$ where $\omega_i=\omega$. Consider the $P$-filter $\mathcal{S}$ on $\Omega$ whose base is as follows $(\bigsqcup_i F_i, F_i\in \mathcal{F})$.

$\mathcal{F}$ filter is isomorphic to $\mathcal{S}$ filter ?

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  • $\begingroup$ Is $i$ intended to range over a countably infinite set? Are you asking about the usual notion of isomorphism of filters (bijection between the underlying sets, sending one filter to the other), or about order-isomorphism of the filters considered just as partially ordered sets, or about another notion of isomorphism? $\endgroup$
    – Andreas Blass
    Commented Aug 27, 2020 at 16:04
  • $\begingroup$ Yes, i have range over a countably infinite set. Isomorphism of filters is bijection between the underlying sets, sending one filter to the other. $\endgroup$
    – Alexander Osipov
    Commented Aug 28, 2020 at 14:41
  • $\begingroup$ I've seen "P-filter" defined to mean a filter $\mathcal F$ such that, for any countably many sets $A_n\in\mathcal F$, there exists $B\in\mathcal F$ with all $B-A_n$ finite. I've also seen it defined with some additional requirements, for example that $\mathcal F$ contains all cofinite sets. What additional requirements do you intend your P-filter $\mathcal F$ to satisfy? $\endgroup$
    – Andreas Blass
    Commented Aug 28, 2020 at 21:17
  • $\begingroup$ P-filter to mean a filter F such that, for any countably many sets An∈F, there exists B∈F with all B−An finite. $\endgroup$
    – Alexander Osipov
    Commented Aug 31, 2020 at 16:59

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After the clarifications in comments of OP (August 28 and today), here's a counterexample. Let $\mathcal F$ be the filter consisting of only $\omega-\{0\}$ and $\omega$. Then $\mathcal S$ (as defined in the question) contains a set whose complement is infinite, whereas $\mathcal F$ does not. So they cannot be isomorphic.

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