For any countable $ A$ , $B \subseteq A \implies B \cap B\space' \ne B $

Question

In which kind of metric spaces is the following true

For any non-empty countable set $A$ of the metric space , $B \subseteq A \implies B \cap B\space' \ne B $

Answer 1

I assume that $B'$ is the set of limit points of $B$. Note that $\varnothing'=\varnothing=\varnothing\cap\varnothing'$, so $A$ always has at least one subset $B$ such that $B\cap B'=\varnothing$. To avoid this triviality, I will assume that the desired property is that $B\cap B'\ne\varnothing$ for non-empty $B$. Note that $A$ serves no purpose: the desired property is simply that $B\cap B'\ne\varnothing$ whenever $B$ is a (non-empty) countable subset of the space.

the condition that $B\cap B'\ne B$ amounts to saying that $B$ has an isolated point in its relative topology. Recall that a scattered space is one in which every non-empty subspace has an isolated point. Clearly, then, every scattered metric space has the desired property. I’ll show that these are the only metric spaces with the property.

Suppose that $M$ is an uncountable metric space in which every countable subset has an isolated point, but $M$ is not scattered. Then $M$ has an uncountable subset $X$ with no isolated points such that every countable subset of $X$ has an isolated point. For $S\subseteq X$ let $I(S)$ be the set of isolated points of $S$.

Let $A_0$ be any non-empty countable subset of $X$. (E.g., fix any $x\in X$ and let $A_0=\{x\}$.) Suppose that $n\in\Bbb N$, and $A_n$ is a countable subset of $X$. $X$ has no isolated points, so for each $p\in I(A_n)$ there is a sequence $\langle x_k^p:k\in\Bbb N\rangle$ of distinct points of $X\setminus A_n$ converging to $p$; let

$$A_{n+1}=A_n\cup\bigcup_{p\in I(A_0)}\{x_k^p:k\in\Bbb N\}\;.$$

In this fashion we construct an increasing sequence $\langle A_n:n\in\Bbb N\rangle$ of countable subsets of $X$. Now let $A=\bigcup_{n\in\Bbb N}A_n$; clearly $A$ is countable. Let $p\in A$ be arbitrary; $p\in A_n$ for some $n\in\Bbb N$. If $p$ is not isolated in $A_n$, then clearly $p$ is not isolated in $A$. But if $p\in I(A_n)$, then by construction $p$ is not isolated in $A_{n+1}$ and hence is not isolated in $A$. Thus, $A$ has no isolated points. This contradiction shows that $X$ must in fact have an isolated point and hence that $M$ must be scattered.

Note that the argument does not require that $M$ be metric: it is sufficient that $M$ be first countable. Indeed, with a very minor change in the argument we need only assume that $M$ has countable tightness (equivalently, is countably generated).