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.