Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Sounds a bit like second-countable meets paracompactness in some strong form. – Asaf Karagila Feb 6 '13 at 19:59
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...) – Henno Brandsma Feb 6 '13 at 22:04
Thank you very much for the answer! – Andreas Rasmusson Feb 7 '13 at 6:01

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.

share|cite|improve this answer
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? – Robert Feb 16 '13 at 1:01
@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. – Brian M. Scott Feb 16 '13 at 1:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.