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.



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> **Problem 1.** How close are filters with countable network to filter with a countable base? 
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A more concrete question:



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> **Problem 2.** Is any filter with countable network diagonalizable?
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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](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.30.755&rep=rep1&type=pdf) 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. 



On filters possessing a countable network
Maybe I posed this question too quickly: for the 6 hours that passed since the time of asking this question I have found a (relatively simple) counterexample to my Problem 2.


**Example.** There exists a non-diagonalizable free filter with countable network on a countable set.


*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 $s\in 2^{<\omega}$. 


