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.



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


