We say that a sequence $(\mathcal X_n)$ of families of subsets of a topological space $X$ is a $\sigma$-disjoint cover of $X$ if every family $\mathcal X_n$ consists of mutually disjoint sets and $\bigcup\limits_n\bigcup\mathcal X_n=X$.

Let us say that a space $X$ is weakly Lindelof, if every open cover of $X$ admits a $\sigma$-disjoint subcover. Clearly, every Lindelof space is weakly Lindelof.

Question 1. Is there any well-known in the literature name for the class of "weakly Lindelof" spaces?

The following question concerns weaker property than "weak Lindeloffness".

Question 2. Does there exist a $\sigma$-disjoint cover of a Banach space $X$ by open balls of diameters $\le 1$?

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    $\begingroup$ I don’t know how the property you describe is called, but I’d say weakly Lindelof is already taken: any open cover of X contains a countable cover of a dense subset of X. $\endgroup$ Jul 14, 2018 at 20:11

1 Answer 1


I found HERE

A space is called screened if every open covering has a $\sigma$-disjoint open refinement

Do you think this is equivalent to weakly Lindelof: every open covering has a $\sigma$-disjoint subcover ?

That page refers to
D.K. Burke, "Covering properties" K. Kunen (ed.) J.E. Vaughan (ed.) , Handbook of Set-Theoretic Topology , North-Holland (1984) Chapt. 9; pp. 347–422

Screened is not equivalent to weakly Lindelof. An uncountable discrete space is screened: indeed every open cover has a disjoint open refinement consisting of singletons. But not every open cover has a sigma-disjoint subcover.
Let $X$ be our uncountable set. Choose one distinguished point $x_0$. Let us consider the open cover $\mathcal U := \{\;\{x_0,x\}\;: x \in X \setminus \{x_0\}\}$ made up of doubletons. A disjoint subfamliy of $\mathcal U$ can only have one set in it, so a sigma-disjoint subfamily of $\mathcal U$ can be at most countable.


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