# Existance of bijective function which maps tensor product of subsets of a selective ultrafilter into the ultrafilter

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. $$\varphi(\scr{A}\otimes\scr{B})\subset\scr{U}$$ ?
• 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. – Andreas Blass Mar 28 at 2:06
• @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. – ar.grig Apr 5 at 5:20
• 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] – Andreas Blass Apr 5 at 12:56
• 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 $\phi(\mathcal U)\supset\mathcal N\otimes\mathcal N$. – Andreas Blass Apr 5 at 13:01