Dense filter and selective ultrafilter

We say that $$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$ is the density of subset $$A\subset\omega$$ if the limit exists. Let us define the filter $$\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$$.

Question: Does there exist (in ZFC & CH) selective ultrafilter $$\mathcal{U}$$ and a bijection $$\varphi:\omega\to\omega$$ such that $$\varphi(\mathcal{F_1})\subset\mathcal{U}$$ ?

Remark: Someone had downvoted two similar questions. Please, explain what is wrong if something is wrong

• I would assume that people downvoted because they missed the word "selective". This turns the question from a triviality into something pretty difficult. Please, read carefully before you downvote! – Jan-Christoph Schlage-Puchta Mar 26 '19 at 20:37

1 Answer

No, there are no such $$\mathcal U$$ and $$\varphi$$.

Let me start the proof with two simplifying observations. First, $$\varphi$$ is irrelevant, because bijections of $$\omega$$ to itself preserve selectivity of ultrafilters.

Second, every selective ultrafilter $$\mathcal U$$ is a P-point, which means that, given any countably many sets $$A_n\in\mathcal U$$, there is a set $$B\in\mathcal U$$ almost included in all of the $$A_n$$"s, i.e., $$B-A_n$$ is finite for all $$n$$. To prove that selective ultrafilters $$\mathcal U$$ have this property, let sets $$A_n\in\mathcal U$$ be given. If $$\bigcap_nA_n\in\mathcal U$$, then this intersection serves as the required $$B$$, so assume $$\bigcap_nA_n\notin\mathcal U$$. Then we can partition $$\omega$$ into the pieces $$\omega-A_0, A_0-A_1, A_1-A_2, A_2-A_3, \dots$$ and $$\bigcap_nA_n$$, none of which are in $$\mathcal U$$. Selectivity provides a set $$B\in\mathcal U$$ that intersects each of these pieces in at most one point. Then $$B-A_n$$ is finite (in fact it has at most $$n+1$$ elements) for each $$n$$, as required.

So now it suffices to show that no P-point ultrafilter $$\mathcal U$$ can include the density-$$1$$ filter $$\mathcal F_I$$. For each natural number $$n\geq 2$$, partition $$\omega$$ into $$n$$ sets each of which has density $$\frac1n$$; for example, take the congruence classes modulo $$n$$. Being an ultrafilter, $$\mathcal U$$ must contain one of these $$n$$ sets, say $$A_n$$. If $$\mathcal U$$ is a P-point, it contains a set $$B$$ almost included in all of these $$A_n$$'s. But from almost inclusion it easily follows that $$B$$ has upper density at most the density $$\frac1n$$ of $$A_n$$. (Upper density is defined like density but with $$\limsup$$ instead of $$\lim$$.) So $$B$$ has upper density $$0$$, which means it has density $$0$$. Thus, $$\omega-B$$ has density $$1$$ and is therefore in $$\mathcal F_I$$, but it is not in $$\mathcal U$$ because $$B\in\mathcal U$$.

• Thank you. Really, $\varphi(\mathcal{F_1})\subset\mathcal{U}\implies\mathcal{F_1}\subset\varphi^{-1}(\mathcal{U})$ which is impossible because $\varphi^{-1}(\mathcal{U})$ is also selective an thus contains sets with density 0. Weak question. But what about this one? – ar.grig Mar 24 '19 at 3:11