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

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

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

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

• What is a discrete family of open sets? Mar 6 at 20:15
• @LSpice: A family $\mathcal A$ of subsets of a space $X$ is said to be discrete if every point of $X$ has a neighborhood that intersects at most one member of $\mathcal A$. Such family of open sets of $X$ is called a discrete family of open sets. Mar 6 at 20:32
• @AlessandroCodenotti Your space is not a counterexample: it is compact even and all discrete families of open sets have cardinality at most one. Mar 9 at 7:35
• good point @KPHart, I misread the definition of discrete family Mar 9 at 8:18

1 Answer

There is no $$T_1$$-example: assume $$X$$ is $$T_1$$ and star-Lindelöf. Let $$\mathcal{D}$$ be a discrete family of open sets. Choose $$x_D\in D$$ for all $$D\in\mathcal{D}$$ and put $$A=\{x_D:D\in\mathcal{D}\}$$. Let $$\mathcal{U}_1$$ be the family of all open sets that meet at most one $$D$$ and that are disjoint from $$A$$. Then $$\mathcal{U}=\mathcal{U}_1\cup\mathcal{D}$$ is an open cover and every element of $$\mathcal{U}$$ intersects at most one member of $$\mathcal{D}$$.

Now let $$\mathcal{V}$$ be a countable subfamily of $$\mathcal{U}$$ whose start is equal to $$X$$. In order to cover $$A$$ every $$D\in\mathcal{D}$$ must intersect an element of $$\mathcal{V}$$. But every $$V\in\mathcal{V}$$ intersects at most one $$D$$, so $$\mathcal{D}$$ is countable.

• $\mathcal{U}_1$ covers $X\setminus A$ and $\mathcal{D}$ covers $A$. Mar 10 at 10:21