Closed set $F$ is the boundary of any subset of $\mathbb{R}^n$ I need show that any closed subset $F\subset\mathbb{R}^n$ is the boundary of some set $A$ in $\mathbb{R}^n$.
Intuition tells me to take $A=F\setminus(\mathbb{Q}^n\cap int(F))$ and $int(F)$ is the set of interior points of $F$ but I can't prove that boundary$(A)\subset(F)$

Your problem is, that $\mathbb{Q}^n\cap F$ might be not dense or even empty. Instead of the section with $\mathbb{Q}^n$ for $k\in\mathbb{N}$ take a set $N_{k}$ of nodes inside your set $F$ wich are at least $\frac{1}{2k}$ away from each other and for which each point of your set $F$ is maximally $\frac{1}{k}$ away from one of the nodes in $N_k$. Then take for $A$ the union $A:=\cup_{k}N_k$ and see that the boundary of that set is the closure of your original set $F$. (since the interior is empty and the closure is the closure of your set; since $F$ is closed it equals its closure; by construction of $N_k$ you can easily choose sequences converging to Points in the closure of $F$)
So the idea is to construct a dense, at most countable subset.
