# P-filter property?

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 ?

• 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? – Andreas Blass Aug 27 at 16:04
• Yes, i have range over a countably infinite set. Isomorphism of filters is bijection between the underlying sets, sending one filter to the other. – Alexander Osipov Aug 28 at 14:41
• 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? – Andreas Blass Aug 28 at 21:17
• 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. – Alexander Osipov Aug 31 at 16:59

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.