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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?

edited Jan 2 2012 at 17:11

Dmitri Pavlov
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asked Jan 2 2012 at 9:37

Paul
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The name is Lindelöf, not Lindeloff. – Dmitri Pavlov Jan 2 2012 at 17:12

Henno Brandsma · Accepted Answer · 2012-01-04 21:45:03Z UTC

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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.

answered Jan 4 2012 at 21:45

Henno Brandsma
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Ricky Demer · Answer 2 · 2012-01-02 10:34:13Z UTC

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

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.

answered Jan 2 2012 at 10:34

Ricky Demer
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What's the topology on $X$? – Paul Jan 2 2012 at 10:52

The discrete topology (which is why the second entry in the ordered pair is the set of all subsets of $\omega_1$). – Ricky Demer Jan 2 2012 at 10:56

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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. – Gerald Edgar Jan 2 2012 at 15:47

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