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  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?

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  • $\begingroup$ What is a discrete family of open sets? $\endgroup$
    – LSpice
    Mar 6 at 20:15
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    $\begingroup$ @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. $\endgroup$
    – Nur Alam
    Mar 6 at 20:32
  • $\begingroup$ @AlessandroCodenotti Your space is not a counterexample: it is compact even and all discrete families of open sets have cardinality at most one. $\endgroup$
    – KP Hart
    Mar 9 at 7:35
  • $\begingroup$ good point @KPHart, I misread the definition of discrete family $\endgroup$ Mar 9 at 8:18

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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.

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    $\begingroup$ $\mathcal{U}_1$ covers $X\setminus A$ and $\mathcal{D}$ covers $A$. $\endgroup$
    – KP Hart
    Mar 10 at 10:21

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