# Some kind of idempotence of dense filter

In discussion of following questions question1, question2, question3 became clear (see definitions in question3 ) 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 $$\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}$$

The answer is negative. See proof given by Andreas Blass here. 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 $$\cal{N}\otimes\cal{N}$$.