I have several questions on Lindelöf property.

If every point countable open cover of $X$ has a countable subcover (Condition A), does $X$ have Lindelöf property? How far is having Condition A from Lindelöf property?

A space $X$ is called $\omega_1$-Lindelöf if every $\omega_1$-sized open cover of $X$ contains a countable subcover.

Can every $\omega_1$-Lindelöf space with Condition A be Lindelöf?

A space $X$ is called discretely Lindelöf if the closure of every discrete subspace of $X$ is Lindelöf.

Can every discretely Lindelöf space with Condition A be Lindelöf?

  • $\begingroup$ The name is Lindelöf, not Lindeloff. $\endgroup$ Jan 2 '12 at 17:12

A space is Lindelöf iff it satisfies condition A and is metaLindelöf (every open cover has a point-countable refinement), so one could say the difference is metaLindelöfness.


$X = \langle \omega_1,2^{\omega_1} \rangle$

The set of singleton subsets of $\omega_1$ is a point countable $\omega_1$-sized
open cover of $X$ that does not have a countable subcover.

I don't know the answers to your other two questions.

  • $\begingroup$ What's the topology on $X$? $\endgroup$
    – Paul
    Jan 2 '12 at 10:52
  • $\begingroup$ The discrete topology (which is why the second entry in the ordered pair is the set of all subsets of $\omega_1$). $\endgroup$
    – user5810
    Jan 2 '12 at 10:56
  • 2
    $\begingroup$ OK, you explained it. To me, $2^{\omega_1}$ is a certain calculation in ordinal arithmetic... For your example, I would say: a set of power $\aleph_1$ with discrete topology. That way, I even avoid confusing ordinals with cardinals. $\endgroup$ Jan 2 '12 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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