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 $X$.



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> Is there a Hausdorff weakly Lindelof space which is not DCCC?
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Is there a Hausdorff weakly Lindelof space which is not DCCC?
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 $O\_x$.


