# Does there exist a strongly star-Lindelöf space which is not DCCC?

1. A space $$X$$ is said to be strongly star-Lindelöf if for every open cover $$\mathcal U$$ of $$X$$ there exists a countable subset $$A$$ of $$X$$ such that $$St(A,\mathcal U)=X$$.

2. A space $$X$$ has discrete countable chain condition (DCCC) if every discrete family of nonempty open sets is countable.

Every strongly star-Lindelöf $$T_1$$-space is DCCC. Let $$\mathcal{U}$$ be a discrete family of non-empty open sets in the space $$X$$, and pick a point $$x_U\in U$$ for each $$U\in\mathcal{U}$$. The set $$F=\{x_U:U\in\mathcal{U}\}$$ is closed (by discreteness of $$\mathcal{U}$$). Then $$\mathcal{V}=\mathcal{U}\cup\{X\setminus F\}$$ is an open cover and for every point $$x$$ in $$X$$ the star $$\operatorname{St}(x,\mathcal{V})$$ meets $$F$$ in at most one point. By strong star-Lindelöfness $$F$$ is countable and hence so is $$\mathcal{U}$$.
• @Hart: Why $F$ is closed? Can you expalin it? Sep 2, 2021 at 7:54
• I tend to assume that the spaces I work with are Hausdorff (or at least $T_1$). In that case if $x\in X\setminus F$ then $x$ has a neighbourhood $O$ that meets at most one $U\in\mathcal{U}$. Deleting that one point from $O$ leaves a neighbourhood disjoint from $F$. I added $T_1$ to my answer. Sep 2, 2021 at 8:14