Closed set $F$ is the boundary of any subset of $\mathbb{R}^n$

Question

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$(A)\subset(F)$

Answer 1

Since $A \subset F$ you have $$\partial A \subset \overline{A} \subset \overline{F} =F$$

where $\partial A$ denotes the boundary of $A$ and $\overline{X}$ denotes the closure of $X$.

P.S. If it is the other implication you cannot prove, here is how you can prove it:

If $x \in F$ then we have two possibilities:

case 1: $x \in \text{Int}(F)$

In this case $B_r(x) \subset F$ for some $r$ and then $$ B_r(x) \cap \mathbb{Q} \subset A \\ B_r(x) \cap (\mathbb{R}\setminus\mathbb{Q}) \cap A =\emptyset $$

It is easy to show from here that $x$ is a boundary point of $A$.

case 2: $x \notin \text{Int}(F)$.

Then we have $x \in A$. Also as $A$ has no interior we have $\text{Int}(A)=\emptyset$ and hence $$A \subset \overline{A} = \overline{A}\setminus\text{Int}(A)=\partial A$$

This shows that $$A \subset \partial A$$ and hence, as $x \in A$ we get $x \in \partial A$ .

In both cases we showed that $x \in F \Rightarrow x \in \partial A$.

Answer 2

If x is in the boundary of A the it is an adherent point of A and thus an adherent point of F. F is closed so all adherent points of F are in F. So x is in F.

Answer 3

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.

Answer 4

This answer shows that if a space $X$ can be written as the union of two disjoint dense subsets (such a space is called resolvable), then every closed set in it is the boundary of some other set.

This certainly holds for all $\mathbb{R}^n$ (take all rational points, i.e. $\mathbb{Q}^n$ and its complement, e.g.).

The idea is quite general.