Closed subset of $\mathbb{R}^n$ and frontier Let $F$ be a closed subset of $\mathbb{R}^n$. Show that there exists $X \subset \mathbb{R}^n$ such that $\partial X = F$ (Frontier of $X$ is equal to $F$). Is this fact is true in general, i.e., for an arbitrary metric space $M$?
 A: Let $F^\circ$ denote the interior of $F$. We claim that for $X=(F^\circ \cap \mathbb{Q}^n) \cup \partial F$, one has $\partial X=F$. One has
$$\partial X \subseteq \partial (F^\circ \cap \mathbb{Q}^n) \cup \partial \partial F=F \cup \partial F=F$$
For the converse, suppose $x\in F$ and we show that $x \in \partial X$. There are two cases. Either $x\in F^\circ$ or $x\in \partial F$. If $x\in F^\circ$, then there exists an open set $U\subseteq F$ containing $x$. Let $x_i \in U \cap \mathbb{Q}^n$ such that $x_i \rightarrow x$, and so there is a sequence in $X$ converging to $x$. Similarly one can find a sequence $y_i \in U \backslash \mathbb{Q}^n$ converging to $x$ so there is a sequence outside $X$ converging to $x$. It follows that $x\in \partial X$ in this case.
If $x\in \partial F$, there exists a sequence $x_i=x\in X$ converging to $x$. Also there exists a sequence $y_i \notin F$ converging to $x$. But then $y_i$ is a sequence outside $X$ converging to $x$. It follows that $x\in \partial X$ in this case as well.  
A: Considering the first nice answer of this question, for the general case of a metric space $M$, I think it suffices the $M$ to have at least two dense disjoint sets $A$ and $B$. The $A$ could play the role of $\mathbb{Q}^n$  and the other that of $\mathbb{R}^n\setminus\mathbb{Q}^n$ .The rest comes as I said like the first answer.
