## Existance of bijective function which maps tensor product of subsets of a selective ultrafilter into the ultrafilter

In the answer on this question Andreas Blass had shown that for any selective ultrafilter $$\scr{U}$$ on $$\omega$$ and for any free subfilter $$\scr{F}\subset{U}$$ doesn’t exist bijection $$\varphi:\omega^2\to\omega$$ such that $$\varphi(\scr{F}\otimes\scr{F})\subset\scr{U}$$. Thus I am trying to weaken the conditions.

Question: Does there exist a pair of subsets $$\scr{A},\scr{B}$$ of selective ultrafilter $$\scr{U}$$ on $$\omega$$ and a bijection $$\varphi:\omega^2\to\omega$$ with following properties:

1. $$\scr{A}$$ and $$\scr{B}$$ have finite intersections property and $$\cap\scr{A}=\cap\scr{B}=\varnothing$$
2. $$\varphi(\scr{A}\otimes\scr{B})\subset\scr{U}$$ ?

## Dense subfilter of selective ultrafilter

Given selective ultrafilter $$\mathcal{U}$$ on $$\omega$$ and dense filter $$\mathcal{F_1}=\{A\subset\omega~|~\rho(A)=1\}$$, where $$\rho(A)=\lim_{n\to\infty}\frac{|A\cap n|}{n}$$ if the limit exists. Let $$\mathcal{F}=\mathcal{F_1}\cap\mathcal{U}$$.

Question: Does there exist a family $$\{A_i\subset\omega\}_{i<\omega},~A_i=\{a_{ik}\}_{k<\omega}$$ of pairwise disjoint subsets such that for any $$B\in\mathcal{F}$$ we have: $$\{a_{ik}~|~i,k\in B\}\in\mathcal{U}$$

Remark: The question is equivalent formulation of this one which still has no answer.

## 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: Is there exists (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, explane what is wrong if something is wrong

## The property of a 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 $$\varphi(\mathcal{F}\otimes\mathcal{F})\subset\mathcal{U}$$

## A question on the ultrafilter number

Let $$\mathfrak{u}$$ denote the ultrafilter number, which is defined to be the minimum cardinality of a subset of $$\mathcal{P}(\mathbb N)$$ which is a base for a nonprincipal ultrafilter on $$\mathbb{N}$$. Clearly $$\aleph_1\leq \frak{u}\leq 2^{\aleph_0}$$, so it is only interesting to study $$\frak{u}$$ under the negation of CH. Kunen proved that it is consistent that CH fails and that $$\frak{u}=\aleph_1$$. Martin’s axiom implies that $$\frak{u}=2^{\aleph_0}$$.

Is it consistent that $$\aleph_1<\frak{u}<2^{\aleph_0}$$? If so, can I please have a reference?

## Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$.

I have been stuck with this problem for a while now. I have a proof that letting $$U$$ be an ultrafilter, exactly one of $$x,x^*$$ belongs to $$U$$ for all $$x$$ in $$B$$, I did this by showing that both belong to $$U$$ implies that $$U=B$$ which cannot be the case, and if neither long to $$U$$ then a contradiction can be derived by de-Morgan’s laws. However, I’m stuck with the reverse implication, I need to prove that:

“If $$U$$ is a subset of a Boolean algebra $$B$$ such that $$\forall x \in B$$, exactly one of $$x \in U$$ and $$x^* \in U$$ is true.

Is this even the case? Any help would be greatly appreciated, thanks in advance.