Is a $\sigma$-locally finite collection of open sets locally countable?

Question

Problem

I encountered this statement on nLab, which says that weakly Lindelöf spaces with a $\sigma$-locally finite basis are second-countable. The original proof given below the statement is

Proof. Let $\mathcal{V}$ be a $\sigma$-locally finite basis. For each $x\in X$, there is a neighborhood $N_{x}$ meeting countably many members of $\mathcal{V}$. If $X$ is weakly Lindelöf, there is a countable $\{N_n\}_{n=1}^\infty$ which covers a dense subset of $X$. Then $\mathcal{U}=\{V\in\mathcal{V}\mid N_n\cap V\neq\varnothing ~\text{for some}~n\}$ is a countable basis for $X$.

The proof is extremely brief, and I couldn't understand the italicized part, which I refer to as the collection $\mathcal{V}$ being locally countable for the time being. I believe it is not a very common property since I've searched $\pi$-Base and Wikipedia but couldn't find anything about this locally countable property. The proof simply stated it as if it is an easy corollary. I know that $\mathcal{V}$ being a $\sigma$-locally finite basis implies that $X$ is first-countable, but this doesn't take me any further.

My main question is, how could I prove that $\mathcal{V}$ is locally countable? Would $\mathcal{V}$ being a $\sigma$-locally finite basis alone suffice? In fact, as I claim later, $\mathcal V$ itself is countable, but it requires the additional assumption that the space $X$ is weakly Lindelöf. And this sounds like circular reasoning to me. I would like to know if the locally countable property holds without the weakly Lindelöf assumption.

My thoughts

In my opinion, the statement itself is true, as $\mathcal{V}$ itself is countable, which follows easily from the following claim which I believe is true:

Claim. Every locally finite family of nonempty open subsets of a weakly Lindelöf space is countable.
Proof. (This is an imitation of the proof of theorem 5.1.24 of Engelking's General Topology) Let $\mathcal{A}$ be a locally finite family of nonempty open subsets of a weakly Lindelöf space $X$. For every $x\in X$ choose a neighborhood $U_x$ of $x$ that intersects only finitely many members of $\mathcal{A}$ and take a countable subcover $\mathcal{U}$ that covers a dense subset of $X$. Since every member of $\mathcal{A}$ meets some $U\in \mathcal{U}$, it follows that $\mathcal{A}$ is countable.

I think the situation is a bit strange, since second countability is a very desirable property, but I couldn't find this statement or the locally countable property anywhere else. I checked $\pi$-base and there are no counterexamples to the statement.

Answer 1

You're correct; we should add σ-Locally Finite Base + Weakly Lindelöf => Second Countable to the pi-Base. https://github.com/pi-base/data/issues/584

First, I'll confirm that your claim that every locally finite (or even locally countable) collection in a weakly Lindelöf space is countable. So a σ-locally finite base would be a countable union of countable sets, showing second-countability.

But to your original question: is every σ-locally finite base $\mathcal U=\bigcup_{n<\omega}\mathcal U_n$ a locally countable base? Naively one might say: take a neighborhood of a point, it hits finitely-many of each $\mathcal U_n$, so it hits countably-many overall. But that's no good: to get the finite collection for each $\mathcal U_n$, a different neighborhood may be chosen, and the intersection of $\omega$-many neighborhoods need not be a neighborhood.

My assumption is that there must be some σ-locally finite base that's not locally countable, but it's late and I'm not coming up with one on the fly at the moment.

EDIT: Okay, here's an example. Take the post office metric on $\mathbb R$ where $0$ has its usual neighborhoods and each other point is isolated. As a metrizable space it has a σ-locally finite base, for example, $\mathcal U_n=\{\{x\}:x\not\in(-2^{-n},2^{-n})\}\cup\{(-2^{-n},2^{-n})\}$.

But note that any base for this space contains $\{\{x\}:x\not=0\}$, showing that it cannot be locally countable: every neighborhood of $0$ contains uncountably-many isolated points and thus meets uncountably-many basic open sets.