As we know, every regular weakly Lindelof space is DCCC. Here DCCC denotes discrete countable chain condition, a space $X$ has discrete countable chain condition if every discrete family of non-empty open sets of $X$ is countable.

A space $X$ is said to be weakly Lindelof if every open cover $\mathcal U$ of $X$ contains a countable subfamily $\mathcal V \subset \mathcal U$ such that $\bigcup \mathcal V$ is dense in $X$.

Is there a Hausdorff weakly Lindelof space which is not DCCC?


1 Answer 1


The answer to this problem is negative because of the following

Theorem. If a topological space $X$ is weakly Lindelof, then each discrete (more generally, locally countable) family of open sets in $X$ is at most countable.

Proof. Let $\mathcal U$ be a locally countable family of open subset of $X$. Then each point $x\in X$ has a neighborhood $O_x$ meeting at most countably many sets of the family $\mathcal U$. By the weak Lindelof property of $X$ the open cover $\{O_x:x\in X\}$ contains a countable subfamily $\{O_{x}:x\in C\}$ whose union $\bigcup_{x\in C}O_x$ is dense in $X$ and hence intersects each set $U\in\mathcal U$.

Assuming that $\mathcal U$ is uncountable and applying the Pigeonhole Principle, we can find a point $x$ of the countable set $C$ such that $O_x$ intersects uncountably many sets of $\mathcal U$. But this contradicts the choice of $O_x$.


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