A connection between compactness and clustering

Question

Given a metric space $X$, a compact subset $S$ of $X$ is one in which every open cover has a finite sub-cover. This turns out to be equivalent to saying that every countable subset of $S$ has a limit point in $S$ (one can prove this by going through sequential compactness, which in metric spaces is equivalent to compactness)

The real numbers as a metric space have the Lindelöf property which is analogous to compactness, but instead states that every open cover of $\mathbb{R}$ has a countable subcover. It turns out that the real numbers also have the property that each of its uncountable subsets has a limit point. My question is, is this a coincidence, or does the pattern continue? I.e. is the following claim true:

Given a cardinal $\alpha$ and a metric space $X$ with the property that every open cover has a sub-cover with cardinality strictly less than $\alpha$, every subset of $X$ with cardinality $\alpha$ has a limit point

So for compact sets the relevant cardinality is $\aleph_0$, and for the real numbers, the relevant cardinality is $\aleph_1$ (assuming GCH).

My first attempt was to start with a metric space $X$ having the above property for some cardinal $\alpha$, and then given any subset $S$ of $X$ of cardinality $\alpha$, if we assume it is discrete, we can try to create a cover for $X$ which has cardinality larger than or equal to $\alpha$, but is also irreducible. I am struggling however to create this cover, we can find pairwise disjoint open balls around each point in $S$, but it isn't clear how to extend this to a cover for the entirety of $X$.

Answer 1

Let $A$ be a subset of $X$ of cardinality $\alpha$. Let $$\mathscr{F}=\{\operatorname{closure}(A\setminus F)\colon F\subset A, F\text{ finite}\}$$ Then for any $\mathscr{G}\subset\mathscr{F}$ with $|\mathscr{G}|<\alpha$, $\bigcap\mathscr{G}\ne\emptyset$. By the assumption, $\bigcap\mathscr{F}\ne\emptyset$. Let $a\in\bigcap\mathscr{F}$. Then every neighborhood of $a$ intersects every cofinite subset of $A$ so $a$ is a limit point of $A$.

Edit:

• If $I$ is an index set, $|I|<\alpha$ and $\{F_i\}_{i\in I}$ is a collection of finite subsets of $A$ then $\bigcup_{i\in I}F_i$ does not exhaust $A$, so $\emptyset\ne\bigcap_{i\in I}A\setminus F_i\subseteq\bigcap_{i\in I}\operatorname{closure}(A\setminus F_i)$.
• If $\bigcap\mathscr{F}=\emptyset$ then $\{X\setminus C\colon C\in\mathscr{F}\}$ is an open cover of $X$. By the assumption, there is an index set $I$ with $|I|<\alpha$ and a collection $\{C_i\in\mathscr{F}\}_{i\in I}$ such that $\{X\setminus C_i\}_{i\in I}$ is an open cover of $X$. Then $\bigcap_{i\in I}C_i=\emptyset$.