Subsets of boundaries of regular open subsets of separable metric spaces

Question

Let $X$ be a separable metric space. Prove that for any nowhere dense set $A$ in $X$, exist a regular open set $U$ such that $A$ is in the boundary of $U$. Can we have the same result without separability of $X$?

Answer 1

$\newcommand{\cl}{\operatorname{cl}}$Let $A$ be a nowhere dense set in $X$, and let $U=X\setminus\cl A$; clearly $\cl U=X$. Suppose that we can find open sets $V$ and $W$ such that $V,W\subseteq U$, $V\cap W=\varnothing$, and $A\subseteq\cl V\cap\cl W$; then just as in my answer to this question $\operatorname{int}\,\cl V$ will be a regular open set whose boundary contains $A$.

By the Bing metrization theorem $X$ has a base $\mathscr{B}=\bigcup_{n\in\Bbb N}\mathscr{B}_n$ such that each $\mathscr{B}_n$ is discrete. If $B\in\mathscr{B}_0$ and $|B\cap U|>1$, choose open sets $V(B),W(B)$ such that $\cl V(B),\cl W(B)\subseteq B\cap U$ and $\cl V(B)\cap\cl W(B)=\varnothing$. Let $$V_0=\bigcup_{B\in\mathscr{B}_0}V(B)\quad\text{ and }\quad W_0=\bigcup_{B\in\mathscr{B}_0}W(B)\;.$$ The families $\{V(B):B\in\mathscr{B}_0\}$ and $\{W(B):B\in\mathscr{B}_0\}$ are discrete and therefore closure-preserving, so $\cl V_0=\bigcup_{B\in\mathscr{B}_0}\cl V(B)$ and $\cl W_0=\bigcup_{B\in\mathscr{B}_0}\cl W(B)$, and hence the sets $\cl V_0$, $\cl W_0$, and $\cl A$ are pairwise disjoint.

Suppose that $V_n$ and $W_n$ are open, and $\cl V_n$, $\cl W_n$, and $\cl A$ are pairwise disjoint. If $B\in\mathscr{B}_{n+1}$ and $|B\cap U|>1$, choose open sets $V(B),W(B)$ such that $\cl V(B),\cl W(B)\subseteq(B\cap U)\setminus(\cl V_n\cup\cl W_n)$ and $\cl V(B)\cap\cl W(B)=\varnothing$. Let $$V_{n+1}=V_n\cup\bigcup_{B\in\mathscr{B}_{n+1}}V(B)\quad\text{ and }\quad W_{n+1}=W_n\cup\bigcup_{B\in\mathscr{B}_{n+1}}W(B)\;.$$ As before, the families $\{V(B):B\in\mathscr{B}_{n+1}\}$ and $\{W(B):B\in\mathscr{B}_{n+1}\}$ are closure-preserving, so

$$\cl V_{n+1}=\cl V_n\cup\cl\bigcup_{B\in\mathscr{B}_{n+1}}V(B)=\cl V_n\cup\bigcup_{B\in\mathscr{B}_{n+1}}\cl V(B)\;,$$

$$\cl W_{n+1}=\cl W_n\cup\cl\bigcup_{B\in\mathscr{B}_{n+1}}W(B)=\cl W_n\cup\bigcup_{B\in\mathscr{B}_{n+1}}\cl W(B)\;,$$

and $V_{n+1}$ and $W_{n+1}$ are open sets such that $\cl V_{n+1}$, $\cl W_{n+1}$, and $\cl A$ are pairwise disjoint.

Let $V=\bigcup_{n\in\Bbb N}V_n$ and $W=\bigcup_{n\in\Bbb N}W_n$; clearly $V$ and $W$ are disjoint open subsets of $U$. Suppose that $x\in A$, and $G$ is an open nbhd of $x$. There are $n\in\Bbb N$ and $B\in\mathscr{B}_n$ such that $x\in B\subseteq G$. Suppose that $|B\cap U|\le 1$; then $B\cap U$ is closed, so $B\setminus U$ is an open nbhd of $x$ contained in $A$, and $A$ is not nowhere dense. Thus, $|B\cap U|>1$, $G\cap V\supseteq B\cap V_n\supseteq V(B)\ne\varnothing$, and $x\in\cl V$. Similarly, $x\in\cl W$. That is, $V$ and $W$ are disjoint open sets such that $A\subseteq\cl V\cap\cl W$, and we’re done.

Added: I wondered for a while why this was asked first for separable metric spaces, and I think that I’ve figured it out. If $X$ is separable, then $X$ is second countable, and we can start with a countable base $\mathscr{B}=\{B_n:n\in\Bbb N\}$. We can then set $\mathscr{B}_n=\{B_n\}$ for $n\in\Bbb N$ and carry out the foregoing construction one basic open set at a time instead of one discrete subcollection at a time. This also avoids having to know the Bing metrization theorem and the fact that discrete families are closure-preserving.