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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$(A)\subset(F)$

L F
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  • I'm confused by "any" closed subset is the boundary of "any" subset. Obviously [0,1] is not the boundary of Q. I'd think you meant every closed set is a boundary of some some set, but that's obviously not true. – fleablood Oct 01 '15 at 20:25
  • How do you say to the set of $a\in X$ such $forall \varepsilon>0, B(a,\varpsilon)\cap X\neq\emptyset \wedge B(a,\varpsilon)\cap X^c\neq\emptyset$ ? And yes, the questions says "exists A in R-n " – L F Oct 01 '15 at 20:30
  • D'oh. Of course what I said was obviously not true is obviously true. Well, I'd say "any closed set is the boundary of some closed set" but I'll just slither off because I'm embarrassed by my "obviously not true comment" when it clearly was. – fleablood Oct 01 '15 at 20:36

4 Answers4

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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 $x \in F \Rightarrow x \in \partial A$.

Souza
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N. S.
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  • the task is to find $A$ for $F$ given such that all the set-inclusions are in fact equalities. – Max Oct 01 '15 at 21:00
  • But I already prove that for this $A,~~\partial A\subset F$. This solution seems ok for me. It's ok the set that I'm taking as $A$? – L F Oct 01 '15 at 21:10
  • As I Wrote in my answer, this choice for $A$ might be empty: For instance take the closed set $F=\left{\sqrt{2}\right}$. Then $\mathbb{Q}\cap F=\emptyset$ and thus the boundary of $A$ must be empty. But $F$ is not empty. EDIT: now I see you chose the complement of $\mathbb{Q}^n$ - sorry for that! Your approach works, mine too. – Max Oct 01 '15 at 21:14
  • The boundary of any subset of a closed set is a subset of the closed set. That's basically by definition. This by no means proves the assertion, but this was the problem he asked. That ∂A⊂A¯¯¯¯⊂F¯¯¯¯=F does not have any restrictions on A and A can even be empty. The convers F ⊂∂A will most certainly have restrictions on A. – fleablood Oct 01 '15 at 21:36
  • @LuisFelipeVillavicencioLopez You said in your answer that for this construction (which works) you cannot prove this implication. If it is the other implication you cannot prove, I an add that to the answer, but I only tried to answer to the part you cannot prove ;) – N. S. Oct 02 '15 at 15:40
  • Yes, and this is all I needed. Awesome way to show this, thanks! – L F Oct 02 '15 at 15:55
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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.

Community
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Henno Brandsma
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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.

fleablood
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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.

Max
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