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.