1
$\begingroup$
  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?

$\endgroup$
4
  • $\begingroup$ What is a discrete family of open sets? $\endgroup$
    – LSpice
    Commented Mar 6, 2022 at 20:15
  • 1
    $\begingroup$ @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
    Commented Mar 6, 2022 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
    Commented Mar 9, 2022 at 7:35
  • $\begingroup$ good point @KPHart, I misread the definition of discrete family $\endgroup$ Commented Mar 9, 2022 at 8:18

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ $\mathcal{U}_1$ covers $X\setminus A$ and $\mathcal{D}$ covers $A$. $\endgroup$
    – KP Hart
    Commented Mar 10, 2022 at 10:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.