# The property of the dense subfilter of a selective ultrafilter

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 $$\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}$$

• I don't understand the reason of downvote. Can anyone explain what is the problem with question? – ar.grig Mar 21 '19 at 14:15
• 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. – Andreas Blass Mar 24 '19 at 19:35
• 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. – ar.grig Mar 25 '19 at 7:03
• I don't think English is the problem. I just got another unexplained downvote on another ultrafilter question, mathoverflow.net/questions/326274 . – Andreas Blass Mar 25 '19 at 17:30
• 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 – Joseph Van Name Mar 26 '19 at 3:32

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.
• 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 $\mathcal N\otimes\mathcal N$ – Andreas Blass Mar 24 '19 at 19:31
• 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 $\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}$? – ar.grig Mar 25 '19 at 7:24
• @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 $\mathcal N\otimes\mathcal N$. – Andreas Blass Mar 25 '19 at 12:32
• I want $\varphi$ to be bijection and non-principal filter in meaning of empty intersection of filter. – ar.grig Mar 25 '19 at 13:21