# Every hereditarily Lindelöf scattered space is countable

Fact: Every hereditarily Lindelöf scattered space is countable.

Here, a space is Lindelöf if every open cover of the space has a countable subcover, and a hereditarily Lindelöf space is one for which each of its subspaces is Lindelöf. A scattered space is one in which every nonempty subset $$A$$ has an isolated point in $$A$$.

The very interesting paper [1] (available here) studies some properties of Lindelöf scattered spaces. It mentions in its introduction the Fact above as a basic result in this area, apparently well-known. Can anyone provide a proof?

[1] Banakh, Taras; Brian, Will; Ríos Herrejón, Alejandro, First-countable Lindelöf scattered spaces, Topology Appl. 322, Article ID 108318, 29 p. (2022). ZBL1514.54003.

• The Lemma in the answer to math.stackexchange.com/questions/4954570 should probably be an ingredient. Commented Aug 6 at 18:18
• I see four downvotes on this question at the moment, two of which were cast after the question was edited to add a more complete reference and the key definitions (I would be a bit more comfortable if a definition of Lindelöf were included, but I recognize that this is my problem, since I haven't thought about Lindelöf spaces in at least a decade, and I'm too lazy to Google it---it isn't needed, but it would be a nice quality-of-life improvement). In my opinion, this question does meet the context requirements for the site. Commented Aug 10 at 15:42
• If a fifth close vote is cast, I intend to override that vote with a moderator reopen vote (given that two of the existing close votes are likely unrelated to the current version of the question). I won't take action after that, but I do want to make myself clear to the next potential close voter. Commented Aug 10 at 15:43
• @XanderHenderson Added definition of Lindelöf as suggested. Commented Aug 11 at 21:51

Yes, this is a well-known result. I guess it is contained in one of the books of Juhasz about cardinal functions. In fact, it holds for arbitrary cardinals, namely:

Let $$X$$ be an infinite, scattered topological space. Then $$hL(X) = |X|$$, where $$hL(X) = \text{sup}\{L(A): A \subset X\}$$ is the hereditarily Lindelöf degree of $$X$$.

The proof uses a well-known characterization of scattered spaces:

Definition
Let $$\le$$ be a well-ordering of the topological space $$X$$.
$$\le$$ is called right-separating, if for each $$x \in X$$, $$X^{\le x} = \{y\in X: y\le x\}$$ is open.
Then also $$\{y\in X: y < x\} = \bigcup_{y < x} X^{\le y}$$ is open.

Theorem
The topological space $$X$$ is scattered, if and only if there exists a right-separating well-ordering on $$X$$.

For the time being, I would prefer to omitt the proof, although I haven't found a (online) reference yet, only several places, where this result is mentioned. Please let me know, if further elaboration is needed.

Using the theorem, we can now prove the above statement:

"$$\le$$" is obvious.
"$$\ge$$": W.l.o.g. we may assume that $$X = \lambda \in$$ On and each $$\alpha$$, $$\alpha < \lambda$$, is open.
Assume that $$hL(X) < |X| = |\lambda|$$. Then $$\kappa := hL(X)^+ \le |\lambda|$$, and $$\kappa$$ is a regular cardinal. $$\{\alpha: \alpha < \kappa\}$$ is an open cover of $$\kappa$$. Hence there exists $$T \subset \kappa$$ with $$\bigcup_{\alpha \in T} \alpha = \kappa$$ and $$|T| \le L(\kappa) \le hL(X)$$. Hence $$T$$ is cofinal in $$\kappa$$, which implies $$\kappa = cf(\kappa) \le |T| \le hL(X) < \kappa$$.
Contradiction! Hence $$|X| \le hL(X)$$.

• The right-separated characterization of scattered seems very reasonable. Thanks for this general answer. Commented Aug 6 at 22:51

Here is my attempt after reasoning similarly to this answer: https://math.stackexchange.com/q/4954574, and making use of its Lemma.

Suppose $$X$$ is hereditarily Lindelöf and scattered.

Let $$B$$ be the set of points $$x \in X$$ at which $$X$$ is locally countable (i.e., such that $$x$$ has a countable nbhd). The set $$B$$ is open in $$X$$. Now, as $$X$$ is hereditarily Lindelöf, $$B$$ is Lindelöf. So, as $$B$$ is Lindelöf and locally countable, $$B$$ is countable.

Let $$A=X\setminus B$$. Suppose by contradiction that $$A\neq \emptyset$$. By the scattered property, $$\exists x \in A$$ such that there is an open set $$U \subseteq X$$ such that $$U\cap A=\{x\}$$. Since $$x$$ doesn't have any countable nbhds (it isn't an element of $$B$$), $$U$$ is necessarily uncountable. But $$U \setminus \{x\} \subseteq B$$ and $$U \setminus \{x\}$$ is uncountable, so this is a contradiction, as $$B$$ is countable. Therefore, $$A = \emptyset$$, and $$X = B$$, so $$X$$ is countable.