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Let us define the density of subset $A\subset\omega$ : $$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$ if the limit exists. Let $\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$. $\mathcal{F_1}$ is the filter and for the Frechet filter we have $\mathcal{N}\subset\mathcal{F_1}$. For arbitrary selective ultrafilter $\mathcal{U}$ let $\mathcal{F}=\mathcal{F_1}\cap\mathcal{U}$.

Question: is there exists a bijection $\varphi:\omega\times\omega\to\omega$ such that $$ \varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U} $$

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  • $\begingroup$ I don't understand the reason of downvote. Can anyone explain what is the problem with question? $\endgroup$
    – ar.grig
    Commented Mar 21, 2019 at 14:15
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    $\begingroup$ I conjecture there's no reason for the downvote. Someone has also downvoted my answer here, my answers to your earlier questions, and my answer to a question about the possible simultaneous existence of P-points and Q-points. In no case was a reason for the downvote given. The simplest explanation seems to be that somebody (I hope just one person) doesn't like ultrafilters. $\endgroup$
    – Andreas Blass
    Commented Mar 24, 2019 at 19:35
  • $\begingroup$ May be my bad English is the problem. The other reason is the questions quality. I am newbie to ultrafilters and sets theory but the questions are coming from research in other sphere. $\endgroup$
    – ar.grig
    Commented Mar 25, 2019 at 7:03
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    $\begingroup$ I don't think English is the problem. I just got another unexplained downvote on another ultrafilter question, mathoverflow.net/questions/326274 . $\endgroup$
    – Andreas Blass
    Commented Mar 25, 2019 at 17:30
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    $\begingroup$ I have been getting many unexplained downvotes and delete votes recently too from people who don't like set theory and similar areas. mathoverflow.net/a/320749/22277 mathoverflow.net/q/321898/22277 mathoverflow.net/q/321894/22277 mathoverflow.net/q/326224/22277 mathoverflow.net/q/321504/22277 mathoverflow.net/q/326024/22277 $\endgroup$
    – Joseph Van Name
    Commented Mar 26, 2019 at 3:32

1 Answer 1

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No, there do not exist such selective $\mathcal U$ and bijection $\varphi$. Since selectivity is preserved by bijections and since the Fréchet filter $\mathcal N$ is included in $\mathcal F_I\cap\mathcal U$, it suffices to show that no selective ultrafilter $\mathcal U$ on $\omega\times\omega$ includes $\mathcal N\otimes\mathcal N$.

Suppose, toward a contradiction, that we had a selective ultrafilter $\mathcal U\supseteq\mathcal N\otimes\mathcal N$. Partition $\omega\times\omega$ into the columns $\{n\}\times\omega$. Such a column cannot be in $\mathcal U$ because its complement $(\omega-\{n\})\times\omega$ is in $\mathcal N\otimes\mathcal N$ and therefore in $\mathcal U$. So, by selectivity, $\mathcal U$ contains a set $A$ that meets each column at most once. But the complement $(\omega\times\omega)-A$ of this $A$ is in $\mathcal N\otimes\mathcal N$, because it contains a cofinite (in fact at least co-singleton) part of every column. So $(\omega\times\omega)-A\in\mathcal N\otimes\mathcal N\subseteq\mathcal U$, which, together with $A\in\mathcal U$, contradicts the fact that $\mathcal U$ is a filter.

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    $\begingroup$ The argument is easily modified to show that no P-point includes $\mathcal N\otimes\mathcal N$. And in this form, there's a converse: An nonprincipal ultrafilter on a countable set is a P-point if and only if no isomorphic copy of it on $\omega^2$ extends $\mathcal N\otimes\mathcal N$ $\endgroup$
    – Andreas Blass
    Commented Mar 24, 2019 at 19:31
  • $\begingroup$ Thus none of Katetov's idempotent filters can be subfilter of a selective ultrafilter. What about existing of non-principal subfilter $\mathcal{F}$ of selective ultrafilter $\mathcal{U}$ with the property $\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}$? $\endgroup$
    – ar.grig
    Commented Mar 25, 2019 at 7:24
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    $\begingroup$ @ar.grig I assume you still want $\varphi$ to be a bijection (if it's merely a function, take it to be either projection $\omega^2\to\omega$ and take $\mathcal F=\mathcal U$) and that "non-principal" means containing all cofinite sets (there are at least two meanings of "non-principal" for filters; they agree for ultrafilters but not in general). Then the answer is no. As in my earlier answer, $\varphi$ is irrelevant, $\mathcal F$ extends the Fréchet filter $\mathcal N$, and no selective ultrafilter extends $\mathcal N\otimes\mathcal N$. $\endgroup$
    – Andreas Blass
    Commented Mar 25, 2019 at 12:32
  • $\begingroup$ I want $\varphi$ to be bijection and non-principal filter in meaning of empty intersection of filter. $\endgroup$
    – ar.grig
    Commented Mar 25, 2019 at 13:21
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    $\begingroup$ @ar.grig Then my previous comment provides a negative answer to your next-to-previous comment. A filter whose intersection is empty must contain all cofinite sets. $\endgroup$
    – Andreas Blass
    Commented Mar 25, 2019 at 14:10

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