# Is every regular, separable, first countable, pseudocompact, linearly Lindelöf Moore space Lindelöf?

Suppose that $X$ is regular, separable, first countable, pseudocompact, linearly Lindelöf and it is a Moore space. Is it Lindelöf?

$X$ is linearly Lindelöf if every open cover of $X$ that is a chain with respect to $\subseteq$ has a countable subcover.

• What are your thoughts/attempts? – Math1000 Jul 18 '15 at 7:27
• @Math1000 It's mostly my thoughts and attempts, but you have not enough reputation to view my three deleted answers. I started to think about this question about two years ago and then I obtained a first result, stating if such a space $X$ is Lindelöf then $X$ is a metrizable compact and then I got stuck. But today the question was edited and I continued my attempts. I obtained a positive answer under CH, and a proof (with a gap) of a positive answer in general.Now I am trying to fill this gap, so I deleted my answers (I am mainly checking if each Moore space with countable extent is Lindelöf) – Alex Ravsky Jul 18 '15 at 16:00
• It seems that I fixed the gap, so I am going to correct my answer later. – Alex Ravsky Jul 18 '15 at 17:57
• @AlexRavsky: every Moore space with countable extent is Lindelof. – Paul Jul 19 '15 at 13:53
• @Paul Thanks. I already know it, but I need some time to assemble my answer. – Alex Ravsky Jul 19 '15 at 15:32

It is easy to check that each linearly Lindelöf space has countable extent. We recall that a space $X$ is a $\sigma$-space if $X$ has a $\sigma$-discrete network. [Gru, 4.3] Every Moore space $X$ is a $\sigma$-space. [Gru, 4.5] Moreover, a space $X$ is a Moore space iff $X$ is a $\sigma$-space and a $p$-space (or a $w\Delta$-space). [Gru, 4.7.(i)]. If $X$ is a $\sigma$-space and $e(X)=\omega$ then $nw(X)=\omega$, see the proof. So each linearly Lindelöf $\sigma$-space has countable network, and, therefore, it is Lindelöf, which already answers positively your question. But we shall continue. If $X$ is a $p$-space, then $nw(X)=w(X)$. [Gru, 4.2] In particular, each linearly Lindelöf Moore space is a second countable regular space. By Nagata-Smirnov Theorem [Eng, 4.4.7], a topological space $X$ is metrizable iff $X$ is regular and has a $\sigma$-locally finite base. In particular, each linearly Lindelöf Moore space is a second countable regular space and is metrizable. (Another proof of this fact is the following: Let $X$ be a linearly Lindelöf Moore space. Then, by the above, the space $X$ is Lindelöf. By [Eng, Theorem 5.1.2], each regular Lindelöf space is paracompact. By [Gru, Cor. 3.4] a paracompact Moore space is metrizable. Clearly, each metrizable Lindelöf space is second countable). Next, if $X$ is a linearly Lindelöf Moore pseudocompact space then $X$ is a metrizable compact. Indeed, by the above, the space $X$ metrizable, and, therefore, normal. By [Eng, Th. 3.10.21] every pseudocompact normal space is countably compact. And a countably compact Lindelöf space is compact. So the space from your question is a metrizable compact.

Remark 1. See Propositions 1.2.(7) and Remark 1.3 from [BR] about similar relations between cardinal invariants of Moore spaces. Which I, after writing this answer, think to rewrite, because it seems that $e(X)=w(X)$ for each Moore space $X$ and, moreover, if $e(X)=\omega$ then the space $X$ is metrizable, so $dc(X)=w(X)=\omega$.

Remark 2. Under CH there is an other approach for you question, because this assumption implies that each separable linearly Lindelöf regular space is second countable. Indeed, it is easy to prove that any open cover of cardinality $\omega_1$ of a linearly Lindelöf space has a countable subcover. By easily proved Theorem 1.5.7 from [Eng], $w(X)\le 2^{d(X)}$ for each regular space $X$. In particular, each separable linearly Lindelöf regular space is Lindelöf.

Remark 3. A space $X$ is subparacompact if every open cover of $X$ has a $\sigma$-discrete closed refinement. See Sections 3 and 4 of [Bur] about definitions of and relationships among covering properties.

Proposition. If $X$ is a subparacompact Hausdorff space then $e(X)=l(X)$.

Proof. Since $e(X)\le l(X)$, it suffices to prove that $l(X)\le e(X)$. Fix an open cover $\mathcal U$ of $X$. By Theorem 3.1 of [Bur], there is a sequence $\{\mathcal G_n\}$ of open refinements such that for any point $x\in X$ there exists $n$ such that $\operatorname{ord} (x,\mathcal G_n)=1$. For each $n$ put $X_n=\{x\in X: \operatorname{ord} (x,\mathcal G_n)=1\}$ and $\mathcal H_n=\{U\in\mathcal G_n: X_n\cap U\ne\varnothing\}$. For each element $U\in\mathcal H_n$ pick an arbitrary point $x(U)\in U\cap X_n$ and put $Y_n=\{x(U):U\in\mathcal H_n\}$. Since $Y_n\subset X_n$, the map $x:\mathcal H_n\to Y_n$ is injective, so $|\mathcal H_n|\le |Y_n|$. We claim that $Y_n$ is a closed and discrete subset of the space $X$, so $|Y_n|\le e(X)$. Indeed, let $x\in X$ be an arbitrary point. Pick an arbitrary member $U$ of the family $\mathcal G_n$ which covers the point $x$. Let $y\in U\cap Y_n$ be an arbitrary point. Since $y\in X_n$, $\operatorname{ord} (y,\mathcal G_n)=1$, hence $y=x(U)$, that is there is at most one such a point $y$. Next, $X_n\subset\bigcup\mathcal H_n$. Since $\mathcal H_n\subset \mathcal G_n$ and $\mathcal G_n$ is a refinement of the cover $\mathcal U$, there exists a subcover $\mathcal V_n$ of $\mathcal U$ of cardinality at most $|\mathcal H_n|\le |Y_n|$, which covers the set $X_n$. Put $\mathcal V=\bigcup\mathcal V_n$. Then $\mathcal V\subset\mathcal U$ is a cover of the set $\bigcup X_n=X$ and $|\mathcal V|\le \sum|\mathcal V_n|\le\omega\cdot e(X)=e(X)$.

Corollary 1. Let $X$ be a regular space. If $X$ is Moore, or paracompact, or a $\sigma$-space, or a strong $\Sigma$-space then $e(X)=l(X)$.

Proof. All these spaces are subparacompact, see [Gru].

Corollary 2. Each linearly Lindelöf subparacompact Hausdorff space is Lindelöf.

References

[BR] Taras Banakh, Alex Ravsky. Verbal covering properties of topological spaces // arXiv: 1503.04480 [math.GN], Topology Appl. (Proceedings of Lepanto Conference) (to be published).

[Bur] Dennis K. Burke, Covering properties, p. 347-422 in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

[Gru] Gary Gruenhage Generalized Metric Spaces, p. 423-501 in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.

• @Paul I was thinking about your question, but, unfortunately, I have a weak intuition about linearly Lindelöf spaces. And it seems that there exists a space $X$ satisfying all conditions of your question except linear Lindelöfness, see Example here. – Alex Ravsky Sep 13 '13 at 6:03
• I know this example. Thanks all the same! – Paul Sep 13 '13 at 8:48
• Alex. When the number of edits on a post exceeds 10, that triggers a system flag, and summons a moderator to the scene. Everything looks fine here. Just for your information: If you foresee the need for a lot of editing, then you can use the sandbox. The reason that is preferred is that each and every edit "bumps" the post and puts it on the front page. While that has it upsides it also has a downside in that if you do it a lot, that will begin to annoy other users competing for that front page time:) – Jyrki Lahtonen Jul 19 '15 at 18:14
• It seems clear to me that you were just polishing an old answer, and it would have been impractical for you to move the entire post to the sandbox for these edits. So no harm, no foul. I just give this same sermon to all the users who manage to trigger this flag :-) – Jyrki Lahtonen Jul 19 '15 at 18:16
• Last but not least, thank you for spelling Lindelöf correctly :-) – Jyrki Lahtonen Jul 19 '15 at 18:16