Show that if $X$ has a countable basis, a collection $\mathcal{A}$ of subsets of $X$ is countably locally finite if and only if it is countable.
I am going to give my proof of the $\impliedby$ implication first. Suppose that $\mathcal{A}$ is a countable collection of subsets of $X$. Then we can enumerate them by $\Bbb{N}$ likeso: $\mathcal{A} = \{A_n \mid n \in \Bbb{N}\}$. But then $\mathcal{A} = \bigcup_{n=1}^\infty \{A_n\}$, where $\{A_n\}$ is locally finite since any given open set can intersect at most one set in $\{A_n\}$, and hence, intersect finitely many sets in $\{A_n\}$. This means that $\mathcal{A}$ is countably locally finite.
For the other direction, I am having a little trouble. A countably locally finite set can be written as a countable union of locally finite sets, so it seems that it suffices to show any locally finite collection is countable. Let $\mathcal{A}$ be any such collection. As with most arguments about cardinality, it seems that my best option is to construct a function $f : \mathcal{B} \to \mathcal{A}$ that is surjective. Here's an attempt. Assume that $A \in \mathcal{A}$ is not empty. Then there exists $a \in A$, and since $\mathcal{B}$ is a basis $B \in \mathcal{B}$ such that $a \in B$, so that $A \cap B \neq \emptyset$. So we might define $f(B) = A$ such that $A \cap B \neq \emptyset$ (note: there could be may sets in $\mathcal{A}$ satisfying this, so perhaps this won't work…?) Is it well-defined? Well, if $B=B'$, then $\emptyset \neq A \cap B = A \cap B'$, which I guess means that $f(B) = f(B')$ (?). But by definition, this is surjective, so $|\Bbb{N}| = |\mathcal{B}| \ge |\mathcal{A}|$, implying that $\mathcal{A}$ is countable.
As the periodic question marks in the above paragraph indicate, I don't feel terribly comfortable with some parts of the proof. Another thing is, I don't seem to use $\mathcal{A}$'s local finiteness anywheree. Is there a way of clarifying the proof? If it's incorrect, is there a way of fixing it? If not, I could use a hint on how to prove the theorem.
EDIT: Hold on! I may have a proof. Just give me a few minutes to type it up.
Okay. My proof is slightly inspired by Henno's answer given here, although I didn't read the entire proof until I came up with my own, because I wanted to try to solve the problem own my own as much as possible (indeed, the only thing I saw in the link was his definition of $I(n)$). Here is my second attempt at proving the "$\implies$" implication, which is I believe is slightly different from Henno's since I don't do a proof by contradiction.
First, let $\mathcal{A} = \{A_i \mid i \in I\}$ be some locally finite collection. We want to show that $I$ is countable. For each $n \in \Bbb{N}$, define $I(n) = \{i \in I \mid A_i \cap B_n \neq \emptyset \}$. Clearly the union of these sets is contained in $I$. Suppose that $i \in I$. Then given $A_i \neq \emptyset$, there exists $a_i \in A_i$; and since $\mathcal{B} = \{B_n \mid n \in \Bbb{N}\}$ is a basis, there exists an $n \in \Bbb{N}$ such that $a_i \in B_n$, which implies that $i \in I(n)$. Hence $I = \bigcup_{n=1}^\infty I(n)$, so that we need only prove that $I(n)$ is countable. To this end, let $n \in \Bbb{N}$ and let $x \in B_n$. Since $\mathcal{A}$ is locally finite, there exists a basis element $B_{x_k}$ containing $x$ that intersects only finitely many elements in $\mathcal{A}$. Hence, $\{B_{x_k}\}$ is a cover of $B_n$, each of which intersects only finitely many sets in $\mathcal{A}$; moreover, since $\mathcal{B}$ is countable, $\{B_{k_x}\}$ must also be countable and hence can be enumerated likeso: $\{B_{\alpha_k} \mid k \in \Bbb{N} \}$. Now, if $A_i \cap B_n \neq \emptyset$, then it must also intersect a set in the cover of $B_n$, which means that $I(n) \subseteq \bigcup_{k=1}^\infty I(\alpha_k)$. Seeing as the latter set is a countable union of finite sets, which is makes it countable, $I(n)$ is als countable.
How does this sound?
Best Answer
A slight refinement of the proof of mine you linked to:
If $\mathcal{A}$ is locally finite (and consisting of non-empty sets WLOG) and $\{B_n: n \in \mathbb{N}\}$ is a countable base for $X$, define
$$I = \{n \in \mathbb{N}: \mathcal{A}_n:= \{A \in \mathcal{A}: B_n \cap A \neq \emptyset \} \text{ is finite }\}$$
This is a subset of $\mathbb{N}$ so certainly a countable set. Then I claim that $$\mathcal{A} = \bigcup_{n \in I} \mathcal{A}_n$$
which is a countable union of finite sets making $\mathcal{A}$ at most countable.
All $\mathcal{A}_n \subseteq \mathcal{A}$ by definition so one inclusion is obvious. If $B \in \mathcal{A}$, then choose $x \in B$ and we have a neighbourhood $O_x$ of $x$ such that $\{A \in \mathcal{A}: O_x \cap A \neq \emptyset \}$ is finite by local finiteness of $\mathcal{A}$. We can find some $n_x$ such that $x \in B_{n_x} \subseteq O_x$ as the $B_n$ form a base, and so then $\mathcal{A}_{n_x} = \{A \in \mathcal{A}: A \cap B_{n_x} \neq \emptyset \}$ is finite. This means that $n_x \in I$ by definition of $I$, and $B \in \mathcal{A}_{n_x}$ as well, showing the other inclusion.
Variations of this idea will show that also a point-countable family is at most countable e.g.