Subparacompact spaces and paracompactness

Question

Let $X$ be a topological space.

$X$ is called paracompact if every open cover has a locally finite open refinement. (No separation axiom assumed here.)

$X$ is called subparacompact if every open cover has a $\sigma$-discrete closed refinement. Equivalently, every open cover has a $\sigma$-locally finite closed refinement. Equivalently, for every open cover $\mathscr U$ of $X$, there is a sequence $(\mathscr V_n)_n$ of open covers where each $\mathscr V_n$ is a refinement of $\mathscr U$ and, for each $x \in X$, there is some $n$ with $\operatorname{ord}(x,\mathscr V_n)=1$. (Here, $\operatorname{ord}(x,\mathscr V_n)=|\{V\in\mathscr V_n:x\in V\}|$.)

Various other characterizations are given in [B] (see Theorem 1 in Dan Ma's topology blog).

It is often claimed that paracompact spaces are subparacompact. That is true for paracompact Hausdorff spaces, but not in general (example below). For context,

$\quad$ paracompact $T_2$ $\Rightarrow$ paracompact regular $\Rightarrow$ paracompact normal $\Rightarrow$ paracompact

with paracompact normal $=$ fully normal (see for example https://math.stackexchange.com/a/4862626). The result with the weakest assumption in this chain is the following:

Proposition: Paracompact normal spaces are subparacompact.

I provide a proof further below.

Example of $X$ paracompact and not subparacompact: The particular point topology on a three-point set, $X=\{p,a,b\}$ with the open sets being the empty set and all sets containing $p$. This space is finite, hence compact and paracompact. Consider the open cover $\mathscr U=\{\{p,a\},\{p,b\}\}$. There is no closed refinement $\mathscr V$ of $\mathscr U$, since if $p\in V\subseteq U$ with $V\in\mathscr V$ and $U\in\mathscr U$, necessarily $X=\overline{\{p\}}\subseteq V\subseteq U$, which is impossible. So $X$ is not subparacompact.

Now my question. I'd like to add the subparacompact property to pi-base. Looking at the proposition and example above, what are other results that show when a paracompact space can be subparacompact, even if it is not normal? (See a list of sample spaces here if that helps. For example, I think one should be able to deduce the cofinite topology on $\mathbb N$ is subparacompact, but I am interested in general results showing this, not a proof for just that space.)

And in general, even without paracompactness, are there other simple results available to deduce that a space is or is not subparacompact? (other that the contrapositive of the obvious subparacompact $\Rightarrow$ submetacompact)

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Proof of Proposition: Let $X$ be paracompact normal and let $\mathscr U$ be an open cover of $X$. The cover $\mathscr U$ has a locally finite open refinement $\mathscr V=\{V_\alpha:\alpha\in A\}$. In a normal space, every point-finite open cover is shrinkable (see Theorem 1.5.18 in Engelking, or this Dan Ma's blog post). So there are open sets $W_\alpha$ with $\overline{W_\alpha}\subseteq V_\alpha$ and the $W_\alpha$ covering $X$. The collection $\mathscr W=\{\overline{W_\alpha}:\alpha\in A\}$ is a locally finite closed refinement of $\mathscr U$.

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[B] D. Burke, Covering properties, Chapter 9 of Handbook of set-theoretic topology.

Answer 1

As mentioned in the question, the particular point topology on a three-element set is not subparacompact. I found this is implied by the following (use the contrapositive):

Every Alexandrov connected subparacompact space is ultraconnected.

Let's denote the specialization preorder on $X$ by $x\le y$ if $x\in\overline{\{y\}}$. For $x,y\in X$, the subset $\{x,y\}$ is connected iff $x\le y$ or $y\le x$.

A space is Alexandrov (or Alexandrov-discrete) if an arbitrary intersection of open sets is open. See wikipedia for other characterizations. In particular, every point $x\in X$ has a smallest nbhd $O_x=\{y\in X:x\le y\}$.

An ultraconnected space is one in which any two nonempty closed sets intersect.

Lemma: For any two points $p,q$ in an Alexandrov connected space, there is a "connected chain" between them; i.e., a finite sequence $(p=a_0,a_1,\dots,a_n=q)$ with each $\{a_{i-1},a_i\}$ connected.

Proof: The relation "there is a connected chain between $x$ and $y$" is clearly an equivalence relation. The equivalence class of $x\in X$ contains its smallest nbhd $O_x$. So each equivalence class is open, and there is only one by connectedness of $X$.

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Proof of result: Let $X$ be Alexandrov and connected. Suppose $X$ is not ultraconnected and let us show it is not subparacompact. There are two disjoint nonempty closed sets; so we can find two point $x,y$ with $\overline{\{x\}}\cap\overline{\{y\}}=\emptyset$. By the Lemma, there is a connected chain between the two points. So assume $x$ and $y$ have been chosen such that their closures are disjoint and the length of a connected chain $(x,\dots,y)$ between them is as small as possible over the whole space.

By minimality of the chain and a case by case analysis, it is easy to check that the chain starts as $(x,w,\dots)$ with $x\le w$. And then the chain continues as $(x,w,z,\dots)$ with $w\ge z$. Necessarily $\overline{\{x\}}\cap\overline{\{z\}}=\emptyset$ (otherwise, if $t\in\overline{\{x\}}\cap\overline{\{z\}}$, we would have a strictly shorter chain starting with $(t,z,\dots)$ and ending in $y$, with also $\overline{\{t\}}\cap\overline{\{y\}}=\emptyset$ since $\overline{\{t\}}\subseteq\overline{\{x\}}$). One more application of minimality of the chain gives $z=y$.

Now consider the open cover $\mathscr U=\{O_t:t\in X\}$. If $\mathscr V$ is a closed refinement of $\mathscr U$, we have $w\in F$ for some $F\in\mathscr V$ and $F\subseteq O_t$ for some $t$. Hence $x,y\in\overline{\{w\}}\subseteq F\subseteq O_t$, and therefore $t\le x$ and $t\le y$. Consequently, $t\in\overline{\{x\}}\cap\overline{\{y\}}\ne\emptyset$. This contradiction shows that $X$ is not subparacompact.