In the answer on this question Andreas Blass had shown that for any selective ultrafilter $\scr{U}$ on $\omega$ and for any free subfilter $\scr{F}\subset{U}$ doesn't exist bijection $\varphi:\omega^2\to\omega$ such that $\varphi(\scr{F}\otimes\scr{F})\subset\scr{U}$, because $\scr{U}$ is P-point and P-point is exactly an ultrafilter which isomorphic image in $\omega^2$ can not contain $\scr{N}\otimes\scr{N}$ for Fréchet filter $\scr{N}$. Thus I am trying to weaken the conditions.

Question: Does there exist a pair of subsets $\scr{A},\scr{B}$ of selective ultrafilter $\scr{U}$ on $\omega$ and a bijection $\varphi:\omega^2\to\omega$ with following properties:

  1. $\scr{A}$ and $\scr{B}$ have finite intersections property and $\cap\scr{A}=\cap\scr{B}=\varnothing$
  2. $\varphi(\scr{A}\otimes\scr{B})\subset\scr{U}$ ?
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    $\begingroup$ I don't have time right now to check carefully, but it seems to me the argument I gave earlier (about $\mathcal N\otimes\mathcal N$) will show that, in order to satisfy condition 2, you need a singleton that is either not disjoint from any set in $\mathcal A$ or not disjoint from any set in $\mathcal B$. And that will contradict the "empty intersection" requirement in condition 1. $\endgroup$ – Andreas Blass Mar 28 at 2:06
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    $\begingroup$ @AndreasBlass: Where can I see proof of the fact that P-point is exactly an ultrafilter which isomorphic image in $\omega^2$ can not contain $\scr{N}\otimes\scr{N}$ ? I have understood one-way implication from your previous posts. $\endgroup$ – ar.grig Apr 5 at 5:20
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    $\begingroup$ My answer and comment at mathoverflow.net/questions/325925 give one direction of the equivalence. For the other direction, suppose $\mathcal U$ is a nonprincipal ultrafilter on $\omega$ that is not a P-point. So it contains sets $A_n$ such that no $B\in\mathcal U$ is almost included in all the $A_n$ (where "almost" means modulo finite sets). It is easy to arrange matters so $A_n\supset A_{n+1}$ and $A_n-A_{n+1}$ is infinite for all $n$; we can also arrange that $\bigcap_nA_n=\varnothing$. [continued in next comment] $\endgroup$ – Andreas Blass Apr 5 at 12:56
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    $\begingroup$ Also, without loss of generality, $A_0=\omega$. Choose for each $n$ some bijection $b_n:A_n-A_{n+1}\to\omega$ and define $\phi:\omega\to\omega^2$ by $\phi(x)=(n,b_n(x))$ where $n$ is the unique number such that $x\in A_n-A_{n+1}$. Then $\phi$ is a bijection and $\phi(\mathcal U)\supset\mathcal N\otimes\mathcal N$. $\endgroup$ – Andreas Blass Apr 5 at 13:01

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