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Let $X$ be a Lindelöf space, and $E \subset X$.

I am interested in the following property.

(1) $E$ is countable $\Leftrightarrow$ $\forall x \in X, \exists U_x$ open set such that $x \in U_x$ and $U_x \cap E$ is countable

Which in case $E$ is closed leads to :

(2) $E$ is countable $\Leftrightarrow$ $\forall x \in E, \exists U_x$ open set such that $x \in U_x$ and $U_x \cap E$ is countable

(by taking $U_x = X \backslash E$ for $x \in X \backslash E$)

For example, (2) can be used to prove that a discrete and closed subspace of a Lindelöf space (like $\mathbb{R}$ or any $\sigma$-compact space) is countable.

Here is the proof for ($\forall x \in X, \exists U_x$ open set such that $x \in U_x$ and $U_x \cap E$ is countable $\Rightarrow E$ is countable) :

We have $X = \cup _{x \in X} U_x$. Because $X$ is Lindelof, we have a countable family $\{x_i\}_{i \in \mathbb{N}}$ such that $X = \cup _{i \in \mathbb{N}} U_{x_i}$. Then $E = X \cap E = \cup _{i \in \mathbb{N}} (U_{x_i} \cap E )$ is a countable union of countable subsets, so it is countable.

My questions are :

  1. Does this property have a name ?

  2. If not, is it linked to some property of Lindelöf spaces (like a countability axiom)

  3. If not, is it an ingredient to prove some known thing related to Lindelöf spaces ?
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The property $(1)$ is equivalent to the requirement that if $A\subseteq X$ is uncountable, then there is an $x\in X$ such that $U\cap A$ is uncountable for each open nbhd $U$ of $x$. In other words, each uncountable subset of $X$ has a condensation point in $X$. As you observe, every Lindelöf space has this property; the converse, however, is not true.

Example. Let $X$ be $\omega_2$ with the order topology. For $\alpha\in\omega_2$ let

$$U_\alpha=[0,\alpha)=\{\xi\in\omega_2:\xi<\alpha\}\;,$$

and let $\mathscr{U}=\{U_\alpha:\alpha\in\omega_2\}$; $\mathscr{U}$ is an open cover of $X$ with no countable subcover. (Indeed, $\mathscr{U}$ has no subcover of cardinality $\omega_1$, either.) Now suppose that $A\subseteq X$ is uncountable; there is a least $\alpha\in\omega_2$ such that $U_\alpha\cap A$ is uncountable. If $\alpha$ were not a condensation point of $A$, there would be a $\beta<\alpha$ such that $(\beta,\alpha]\cap A$ was countable, but then $U_\beta\cap A$ would be uncountable, contradicting the choice of $\alpha$. Thus, $\alpha$ is a condensation point of $A$.

The fact that $\mathscr{U}$ in the example has no subcover of cardinality $\omega_1$ is not an accident.

Theorem. If $X$ has property $(1)$, and every open cover of $X$ has a subcover of cardinality at most $\omega_1$, then $X$ is Lindelöf.

Proof. If $X$ is not Lindelöf, it has an open cover $\mathscr{U}$ with no countable subcover, and we may assume that $|\mathscr{U}|=\omega_1$ and enumerate $\mathscr{U}=\{U_\xi:\xi<\omega_1\}$. For each $\eta<\omega_1$ let $$F_\eta=X\setminus\bigcup_{\xi<\eta}U_\xi\;;$$ each $F_\eta$ is a non-empty closed set, and $\bigcap_{\eta<\omega_1}F_\eta=\varnothing$. For $\eta<\omega_1$ let $x_\eta\in F_\eta$, and let $A=\{x_\eta:\eta<\omega_1\}$. The points $x_\eta$ may not all be distinct, but I claim that $A$ is uncountable.

Suppose not; then there is an $x\in A$ such that $E=\{\eta<\omega_1:x_\eta=x\}$ is uncountable. But for each $\eta\in E$ we have $x=x_\eta\notin\bigcup_{\xi<\eta}U_\xi$, so $$x\notin\bigcup_{\eta\in E}\bigcup_{\xi<\eta}U_\xi=\bigcup\mathscr{U}=X\;,$$ which is impossible.

Now let $x\in X$; clearly $x\in U_\eta$ for some $\eta<\omega_1$. Let $V=X\setminus F_{\eta+1}=\bigcup_{\xi\le\eta}U_\xi$; $V$ is an open nbhd of $x$, and $V\cap A\subseteq\{x_\xi:\xi\le\eta\}$, so $V\cap A$ is countable, and $x$ is not a condensation point of $A$. Thus, $A$ has no condensation point, contradicting the hypothesis that $X$ has property $(1)$. It follows that $X$ must be Lindelöf after all. $\dashv$

I’ve not seen property $(1)$ used much, but a strengthening of it is equivalent to $X$ being hereditarily Lindelöf.

Theorem. A space $X$ is hereditarily Lindelöf if and only if each uncountable subset of $X$ contains a condensation point of itself.

This is not hard to prove, and you can also find a proof in Dan Ma’s Topology Blog.

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