Nowhere dense sets and metric space

Question

Excuse me can you see this question Prove that in a metric space the frontier of an open set is the set of accumulation points of a discrete set ... It wrote that " this requires the axioms of choice and is difficult "

Answer 1

I’ve left some details to be checked, but here’s an approach that appears to work.

Let $\langle X,d\rangle$ be a metric space, let $U$ be a non-empty open set in $X$, and let $F=\operatorname{bdry}U$. For $n\in\Bbb N$ let $V_n=U\cap\bigcup_{x\in F}B(x,2^{-n})$ and $R_n=V_n\setminus V_{n+1}$. For $\epsilon>0$ say that a set $D\subseteq X$ is $\epsilon$-discrete if $d(x,y)\ge\epsilon$ whenever $x,y\in D$ and $x\ne y$.

Proposition. Let $A\subseteq X$ and $\epsilon>0$. Then there is an maximal $\epsilon$-discrete $D\subseteq A$, i.e., one such that $A\subseteq\bigcup_{x\in D}B(x,\epsilon)$.

Proof. Construct $D$ by (possibly transfinite) recursion. Choose $x_0\in A$ arbitrarily. Given an ordinal $\eta$ and points $x_\xi\in A$ for all $\xi<\eta$, let $D=\{x_\xi:\xi<\eta\}$ if $\bigcup_{\xi<\eta}B(x_\xi,\epsilon)\supseteqq A$, and otherwise choose $x_\eta\in A\setminus\bigcup_{\xi<\eta}B(x_\xi,\epsilon)$ and continue. This must stop at some point. (Alternatively, this could be accomplished with Zorn’s lemma.) $\dashv$

For $n\in\Bbb N$ let $D_n$ be a maximal $2^{-(2n+2)}$-discrete subset of $R_{2n}$, and let $D=\bigcup_{n\in\Bbb N}D_n$. Then $D$ is discrete, and $F$ is the set of accumulation points of $D$.