Is there a locally countable and weakly Lindelöf space which is not ccc?

A space $X$ is locally countable if for each point $x\in X$ there is an open neighbourhood $O_x$ of $x$ such that $|O_x| < \omega_1$;

Here ccc denotes the countable chain condition;

A space $X$ is called weakly Lindeöf if for any open cover $\mathcal U$ of $X$ there is a counable subset $\mathcal V \subset \mathcal U$ such that $\bigcup \mathcal V$ is dense in $X$.

Thank you.


1 Answer 1


No, because "locally countable + weakly Lindelöf" $\Rightarrow$ separable $\Rightarrow$ ccc.

For the first implication, use local countability to choose for each $x \in X$ some countable open neighborhood $O_x \ni x$. Then $\mathcal U = \{O_x \,:\, x \in X\}$ is an open cover of $X$, and we may use the weakly Lindelöf property to find a countable $\mathcal V \subseteq \mathcal U$ such that $\bigcup \mathcal V$ is dense in $X$. But $\bigcup \mathcal V$ is countable, so this shows $X$ is separable.

For the second implication, let $D$ be any countable dense subset of $X$. If $\mathcal U$ is a collection of pairwise disjoint open subsets of $X$, then each $U \in \mathcal U$ contains some $d_U \in D$ (because $U$ is open) and the function $U \mapsto d_U$ is an injection (because of "pairwise disjoint"). Thus $\mathcal U$ is countable.


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