# Is there a linearly Lindelöf non-Lindelöf $P$-space?

A completely regular topological space $$(X,\tau)$$ is a $$P$$-space, if every $$G_\delta$$-subset of $$X$$ is open (i.e $$\tau$$ is closed under countable intersections).

A topological space $$X$$ is linearly Lindelöf if every open cover of $$X$$ which is linearly ordered by the subset relation has a countable subcover.

Is there a linearly Lindelöf non-Lindelöf $$P$$-space?

• math.stackexchange.com/questions/474376/… Please, do not double post your question on MSE and on MO. If you see that your question is still without an answer on one of these sites after a while, you can still post it on the other site. – Stefan Hamcke Aug 23 '13 at 15:02
• I don't know why people are voting to close this question. Perhaps at least one of the voters could give a reason? – Ramiro de la Vega Aug 24 '13 at 0:36
• Robson: In Arhangelski-Tkachenko's book, the following is listed as an open problem: "is every linearly Lindelof P-group Lindelof?". I've also seen "Is the product of two linearly Lindelof P-spaces linearly Lindelof?" listed as open, so my guess is that your question is also open, but I'm not sure. – Ramiro de la Vega Aug 24 '13 at 0:41
• Is it at least true that every discretely Lindelof P-space is Lindelof? "Discretely Lindelof" means "the closure of every discrete set is Lindelof". It's not too hard to see that every discretely Lindelof space is linearly Lindelof. – Santi Spadaro Nov 8 '17 at 18:23

It was shown by A.K. Misra in "A topological view of $P$-spaces" that any regular $\aleph_1$-Lindelöf $P$-space is normal (see Corollary 4.6). Also, it is not hard to see that any linearly Lindelöf space must be $\aleph_1$-Lindelöf.
Edit: As pointed out by Robson (see comment below), the notion of $\aleph_1$-Lindelöf used by Misra in his paper doesn't correspond to modern terminology. So I don´t know if a regular linearly Lindelöf $P$-space must be normal. However the following is question 486 in "Open problems in topology II":
(Arhangel´ski) Is the product of two linearly Lindelöf $P$-spaces Lindelöf?
So I still think that the OP´s question is open, since product of two Lindelöf $P$-spaces is Lindelöf (so it is the same question after all).
• Corolllary 4.6: Every regular $\aleph_1$-Lindelöf $P$-space is paracompact, hence normal. By the context, I think "$\aleph_1$-Lindelöf" in this corollary means "$L(X)\leq\aleph_1$". Note that if the corollary 4.6 was true for current definition of "$\aleph_1$-Lindelöf" then the question would be answered because: linearly Lindelöf + regular + paracompact $\Rightarrow$ Lindelöf – Robson Figueiredo Aug 26 '13 at 17:59