In discussion of following questions [question1](https://mathoverflow.net/questions/324254/selective-ultrafilter-and-bijective-mapping), [question2](https://mathoverflow.net/questions/324466/on-infinite-combinatorics-of-ultrafilters), [question3](https://mathoverflow.net/questions/324835/the-example-of-the-idempotent-filter-or-subsets-family-with-finite-intersections) became clear (see definitions in [question3](https://mathoverflow.net/questions/324835/the-example-of-the-idempotent-filter-or-subsets-family-with-finite-intersections) ) that for the Frechet filter $\mathcal{N}$ we have $\mathcal{N}\nsim\mathcal{N}\otimes\mathcal{N}$ and for any ultrafilter $\mathcal{U}$ the situation is the same $\mathcal{U}\nsim\mathcal{U}\otimes\mathcal{U}$. But for any bijective mapping $\varphi:\omega\times\omega\to\omega$ there exists the filter $\mathcal{F}$ with $\mathcal{N}\subset\mathcal{F}$ such that $\varphi(\mathcal{F}\otimes\mathcal{F})=\mathcal{F}$ and thus $\mathcal{F}\sim\mathcal{F}\otimes\mathcal{F}$.


So, my following question has chance to receive positive answer. 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 $\mathcal{N}\subset\mathcal{F\_1}$. 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 
$$
\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}
$$



Some kind of idempotence of dense filter
The answer is negative. See proof given by Andreas Blass [here](https://mathoverflow.net/questions/325925/the-property-of-the-dense-subfilter-of-a-selective-ultrafilter). Main idea: filter $\cal{F}\otimes\cal{F}$ must contain $\cal{N}\otimes\cal{N}$ because any free filter contains $\cal{N}$ and thus $\cal{N}\subset\cal{F}$. Then $\varphi^{-1}(\cal{U})$ is selective ultrafilter in $\omega^2$ as bijective image of selective ultrafilter. But selective ultrafilter on $\omega^2$ can not contain $\cal{N}\otimes\cal{N}$.


