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 $A_{i_{j_1}(U)}\cap A_{i_{j_2}(U)}=\emptyset$

  • $\begingroup$ Sounds a bit like second-countable meets paracompactness in some strong form. $\endgroup$
    – Asaf Karagila
    Feb 6 '13 at 19:59
  • $\begingroup$ All separable metrizable zero-dimensional will do this. It seems pretty close to (second countable plus) zero-dimensionality ($\dim(X) = 0$ allows disjoint refinements of open covers...) $\endgroup$ Feb 6 '13 at 22:04

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.

  • $\begingroup$ Do you know if this is also true for open balls i.e. in a second countable space, every open set is the union of disjoint open balls? $\endgroup$
    – Robert
    Feb 16 '13 at 1:01
  • $\begingroup$ @Robert: You mean a second countable metric space (as otherwise balls doesn’t make much sense)? It’s false in $\Bbb R^2$ with the usual metric. $\endgroup$ Feb 16 '13 at 1:04

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