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Definition. A family $\mathcal N$ of subsets of a topological space $X$ is called

$\bullet$ a network if for any open set $U\subset X$ and point $x\in U$ there exists a set $N\in\mathcal N$ such that $x\in N\subset U$;

$\bullet$ a closed network if $\mathcal N$ is a network whose every set $N\in\mathcal N$ is closed in $X$;

$\bullet$ a Borel network if $\mathcal N$ is a network whose every set $N\in\mathcal N$ is a Borel subset of $X$.

Since each base of the topology is a network, each second countable space has a countable Borel network. On the other hand, a regular topological space has a countable network if and only if it has a closed (and hence Borel) network.

Problem. Is there a (desirably Hausdorff) topological space $X$ that has a countable network but does not have a countable Borel network?

Remark. Such space $X$, if exists, cannot be second countable or regular.

Added in Edit. Alex Ravsky informed me that the real line endowed with the cofinite topology $\tau=\{\emptyset\}\cup\{\mathbb R\setminus F:F$ is finite$\}$ is an example of a $T_1$-space with a countable network that does not have a countable Borel network (since each Borel subset of $(\mathbb R,\tau)$ is either countable or has countable complement in $\mathbb R$).

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  • $\begingroup$ Do you have an example of a space with a countable network but not a countable closed network? $\endgroup$
    – James E Hanson
    Commented Jan 9, 2019 at 16:12
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    $\begingroup$ @JamesHanson The two-point space $X=\{a,b\}$ with topology $\{\emptyset,\{a\},X\}$ has a countable open network but does not have countable closed network. But it is not $T_1$-space. Now I will think on a $T_1$-example. $\endgroup$
    – Taras Banakh
    Commented Jan 9, 2019 at 16:33
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    $\begingroup$ @JamesHanson The real line endowed with the cofinite topology has a countable network but does not have a countable closed network (since the closure of any infinite set is the whole line). This is a $T_1$-example, but it is not Hausdorff. $\endgroup$
    – Taras Banakh
    Commented Jan 9, 2019 at 16:38
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    $\begingroup$ @JamesHanson A Hausdorff example can be constructed as follows: Let $\tau_E$ be the standard Euclidean topology on the real line and $\tau$ be the topology generated by the subbase $\tau_E\cup\{\mathbb R\setminus \mathbb Q\}$. The real line endowed with the topology $\tau$ is a functionally Hausdorff second-countable space, which does not have a countable closed network (to prove this fact, apply the Baire Theorem). $\endgroup$
    – Taras Banakh
    Commented Jan 9, 2019 at 16:52

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