Open sets as disjoint unions from a countable collection

Question

What are sufficient conditions on a topological space $X$ for there to exist a countable collection $C=\{A_i\}_{i=1}^{\infty}$ of subsets of $X$ such that for any open set $U\subseteq X$, we may find a sequence $\{i_j(U)\}_{j=1}^{\infty}$ of natural numbers having the property that:

1. $U=\bigcup_{j=1}^{\infty}A_{i_j(U)}$

2. For $j_1\neq j_2$, we have $A_{i_{j_1}(U)}\cap A_{i_{j_2}(U)}=\emptyset$

Answer 1

The property is equivalent to $nw(X)=\omega$, where $nw(X)$ is the net weight of $X$. A family $\mathscr{N}$ of subsets of $X$ is a net for $X$ if every non-empty open set in $X$ is a union of members of $\mathscr{N}$; $$nw(X)=\omega+\min\{|\mathscr{N}|:\mathscr{N}\text{ is a net for }X\}\;.$$

It’s clear that a space with the property has net weight $\omega$, since the collection $C$ is a countable net. Conversely, suppose that $X$ has a countable net $\mathscr{N}$. Let

$$\mathscr{N}^*=\left\{N\setminus\bigcup\mathscr{F}:N\in\mathscr{N},\text{ and }\mathscr{F}\in\left[\mathscr{N}\right]^{<\omega}\right\}\;;$$ clearly $\mathscr{N}^*$ is also a countable net for $X$. Let $U$ be a non-empty open subset of $X$. Then for some $\alpha\le\omega$ there is a family $\mathscr{N}_U=\{N_k:k<\alpha\}\subseteq\mathscr{N}$ such that $\bigcup\mathscr{N}_U=U$. For $n<\alpha$ let $$M_n=N_n\setminus\bigcup_{k<n}N_k\in\mathscr{N}^*\;;$$ clearly $$U=\bigsqcup_{k<\alpha}M_k\;.$$

Thus, $\mathscr{N}^*$ is a countable family of sets such that every non-empty open set in $X$ is the union of a pairwise disjoint subfamily.

In particular, every second countable space has the property. It’s also clear that every space with the property is hereditarily separable and hereditarily Lindelöf.