Let $S \subset \mathbb{R}^n$ be a bounded set and let $B$ be the set of isolated points of $S$ ($x \in S$ is isolated if there exists an open neighbourhood $U$ of $x$ such that $U \cap S = \{x\}$. I claim $B$ is countable.
First I will assume there exists a countable basis $\beta$ for the topology on $\mathbb{R}^n$. Then for each $x \in B$, there exists an open set, call it $U_{x}$, in $\beta$ such that $U_{x} \cap S = \{x\}$. Then we can see that $B$ is in bijective correspondence with the collection of sets $\{U_{x} \}_{x \in B } $. Since $\{U_{x} \}_{x \in B }$ is a subset of the countable set $\beta$, it is countable and hence $B$ is countable.
Questions about my proof: Is my proof correct?
Also, did I use the axiom of choice in choosing the $U_{x}$? If so, is there a proof without it?
Best Answer
Your argument is fine, and you don’t need the axiom of choice.
If $\beta$ is countably infinite, by definition there is a bijection from $\Bbb N$ to $\beta$, so we can enumerate $\beta=\{B_k:k\in\Bbb N\}$, and for each isolated point $x$ of $S$ we can let
$$k(x)=\min\big\{k\in\Bbb N:B_k\cap S=\{x\}\big\}\,.$$
Your $U_x$ is then $B_{k(x)}$, and $k(x)$ is a well-defined natural number, not an arbitrary choice.