Is every paracompact CCC space Lindelöf?

Question

In Every paracompact space with the Suslin property is Lindelöf and How to prove that every Paracompact space with the Suslin property is Lindelof it's observed that every preregular paracompact space with the Suslin property, a.k.a. the countable chain condition or CCC, is Lindelöf. (Note that the first link uses preregular implicitly to obtain a characterization of "paracompactness", that is, using a locally finite refinement, as "fully normal", that is, using a $\sigma$-discrete open refinement.)

Can this be achieved without the preregular assumption?

Answer 1

We may see this is true by an argument from nlab suggested in this $\pi$-Base discussion. Note that we weaken the assumption of paracompact to "para-Lindelöf".

Given an open cover $\mathcal U$ of para-Lindelöf CCC $X$, take a locally countable refinement $\mathcal V$. Then each point has an open neighborhood that intersects countably-many sets from $\mathcal V$.

Let $\alpha<\omega_1$ be an ordinal. Suppose pairwise-disjoint open sets $W_\beta$ for $\beta<\alpha$ are defined such that each intersects countably-many sets from $\mathcal V$. If $\bigcup_{\beta<\alpha} W_\beta$ is not dense, choose a point with an open neighborhood missing $\bigcup_{\beta<\alpha} W_\beta$; this neighborhood can be chosen to also intersect countably-many sets from $\mathcal V$. We let $W_\alpha$ be this neighborhood. (If $\bigcup_{\beta<\alpha} W_\beta$ is dense, let $W_\alpha=\emptyset$.)

Note that there must be some $\alpha<\omega_1$ with $\bigcup_{\beta<\alpha}W_\beta$ dense; otherwise $\{W_\alpha:\alpha<\omega_1\}$ would violate the CCC.

Then every member of $\mathcal V$ intersects $\bigcup_{\beta<\alpha}W_\beta$, but each $W_\beta$ intersects only countably-many members of $\mathcal V$. This proves that $\mathcal V$ is countable.

Since $\mathcal U$ has a countable refinement $\mathcal V$, $\mathcal U$ has a countable subcover. Therefore $X$ is Lindelöf.

Answer 2

Here is another way to show the result, making use of the comments in this answer.

Every CCC space is weakly Lindelöf.

This was shown in Is every ccc space weakly Lindelöf?. So the desired result will follow from this in combination with the following more general result.

Theorem: Every paracompact weakly Lindelöf space is Lindelöf.

A space $X$ is called weakly Lindelöf if every open cover of $X$ has a countable subcollection whose union is dense in $X$.

Proof of theorem:

Let $\mathcal U$ be an open cover of $X$. Let $\mathcal V$ be a locally finite open refinement of $\mathcal U$ covering $X$. (Wlog, $\emptyset\notin\mathcal V$.)

Each $x\in X$ has an open nbhd $W_x$ meeting only finitely many elements of $\mathcal V$. By the weakly Lindelöf property, the collection $\{W_x:x\in X\}$ has a countable subcollection $\mathcal W$ with $\bigcup\mathcal W$ dense in $X$.

Now each $V\in\mathcal V$ meets $\bigcup\mathcal W$ by density, so intersects some element of $\mathcal W$. Since $\mathcal W$ is countable and each element of $\mathcal W$ intersects only finitely many elements of $\mathcal V$, the collection $\mathcal V$ is itself countable.

That is enough to conclude that $\mathcal U$ has a countable subcollection covering $X$, i.e., that $X$ is Lindelöf.

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Note: Basically the same argument (replace "finite/finitely" with "countable/countably") shows:

Every para-Lindelöf weakly Lindelöf space is Lindelöf.

Answer 3

This originally adapted an argument from https://dantopology.wordpress.com/2009/10/18/ccc-paracompact-lindelof/ that seemed to provide an answer, but as noted by David Gao and PatrickR, it seems to have an error.

Let $X$ be paracompact and CCC. Given an open cover $\mathcal U$, apply paracompact to obtain a locally-finite refinement $\mathcal W$.

Ma notes that we may refine $\mathcal W$ further to some $\mathcal V$ so that each member of $\mathcal V$ intersects countably-many other members of $\mathcal W$. For each $x\in X$, choose $U_x$ open that intersects countably-many members of $\mathcal W$, then pick $W_x\in\mathcal W$. We set $V_x=U_x\cap W_x$ and $\mathcal V=\{V_x:x\in X\}$. For each $x\in X$, let $C_x=\{W\in\mathcal W:V_x\cap W\not=\emptyset\}$. Then $C_x\subseteq\{W\in \mathcal W:U_x\cap W\not=\emptyset\}$, which we see is countable by the choice of $U_x$.

Consider the equivalence class on $\mathcal V$ where $V_0\sim V_1$ provided there is a chain of sets $W_0,\dots,W_n$ in $\mathcal V$ with $V_0=W_0$, $V_1=W_n$, and $W_m\cap W_{m+1}\not=\emptyset$. This partitions $\mathcal V$ such that the union of each part is disjoint from the other unions: if they intersected, that would witness a finite chain connecting a member of each part.

The union of each part is an open set, and by the CCC we see there are only countably-many parts.

Ma then suggests that each part is itself countable: choose some member of the part, note it intersects finitely-many other members of the part (*), and is chained to every other member of the part by a finite $W_0,\dots,W_n$.

If so, we could conclude that $\mathcal V$ is a countable refinement of $\mathcal U$, so $\mathcal U$ has a countable subcover (take each member of $\mathcal V$ and find a member of $\mathcal U$ that contains it).

However, (*) is not necessarily true: members of $\mathcal V$ only have intersection with finitely-many members of $\mathcal W$, not $\mathcal V$. Originally, this answer tried to rectify this by setting $C_x=\{V\in\mathcal V:V_x\cap V\not=\emptyset\}$ and claiming $|C_x|\leq|\{W\in\mathcal W:U_x\cap W\not=\emptyset\}|$ but it was pointed out in the comments that this is not necessarily true: the mapping $V_y\mapsto W_y$ may not be injective.