# On filters possessing a countable network

Let $\mathcal F$ be a free filter on $\omega$ and $$\mathcal F^+:=\{E\subset \omega:\forall F\in\mathcal F\;E\cap F\ne\emptyset\}.$$ A family $\mathcal N$ of subsets of $\omega$ is called a network for $\mathcal F$ if for any $F\in\mathcal F$ and $E\in\mathcal F^+$ there exists a set $N\in\mathcal N$ such that $N\subset F$ and $N\cap E\in\mathcal F^+$.

It is clear that each base of a filter is a network, so each filter with a countable base has countable network.

Problem 1. How close are filters with countable network to filter with a countable base?

A more concrete question:

Problem 2. Is any filter with countable network diagonalizable?

We recall that a filter $\mathcal F$ on $\omega$ is diagonalizable if $\mathcal F$ has an infinite pseudointersection $I$, which is an infinite set $I\subset\omega$ such that for any $F\in\mathcal F$ the set $I\setminus F$ is finite.

Remark 1. The filter $$\mathcal F_1=\{E\subset \omega\times\omega: \forall n\in\omega\; |F\setminus(\{n\}\times\omega)|<\infty\}$$does not have countable base but has a countable network $\mathcal N=\{\{n\}\times[m,\omega):n,m\in\omega\}$. For every $n\in\omega$ the infinite set $\{n\}\times\omega$ is a pseudointersection of $\mathcal F$.

Remark 2. The filter $$\mathcal F_2=\{E\subset \omega\times\omega: \exists n\in\omega\;\forall m\ge n\; |F\setminus(\{m\}\times\omega)|<\infty\}$$does not have countable network and is not diagonalizable.

Remark 3. Each filter $\mathcal F$ with countable network is $\omega$-+diagonalizable, which means that there exists a countable family $\mathcal D\subset\mathcal F^+$ such that each set $F\in\mathcal F$ contains some set $D\in\mathcal D$. In this paper Laflamme constructs Example 1.3 of an $\omega$-+diagonalizable filter which is not diagonalizable. Unfortunately, this filter has no countable network, so does not provide a counterexample to Problem 2.

• Isn't $\mathcal{N}$ also a network for the extension of $\mathcal{F}_1$ by the sets $[n,\omega)\times\omega$? So that would be a non-diagonizable filter with character $\mathfrak{d}$ and a countable network. – KP Hart Sep 22 '18 at 19:56
• @KPHart No, $\mathcal N$ is not a network for $\mathcal F_1$ as $\mathcal N\not\subset\mathcal F_1^+$. – Taras Banakh Sep 22 '18 at 20:15
• OK, thjat clarifies a few things – KP Hart Sep 22 '18 at 20:23

Proof. Consider the space $X=2^{<\omega}\cup 2^\omega$ endowed with the standard compact metrizable topology in which $2^{<\omega}$ is a dense discrete subspace of $X$. Let $\mathcal U$ be the family of open sets $U\subset X$ such that $2^\omega\setminus U$ is finite. The family $\mathcal U$ induces a filter $\mathcal F=\{U\cap 2^{<\omega}:U\in\mathcal U\}$ on the countable set $2^{<\omega}$.
It is easy to see that the filter $\mathcal F$ is not diagonalizable. On the other hand it has a countable network $\mathcal N$ consisting of the sets ${\uparrow}s:=\{t\in 2^{<\omega}:s\le t\}$ with $s\in 2^{<\omega}$.