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



The property of the dense subfilter of a selective ultrafilter
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.


