# A niggling problem about frontier in topology

A problem is bugging me many years after I first met it:

Prove that any closed subset of $\mathbb{R}^2$ is the boundary of some set in $\mathbb{R}^2$.

I have toyed with this problem several times over the last 20 years, but I have never managed to either prove it, or conversely prove that the question as posed is in some way wrong.

I can’t remember which book I found the question in originally (but I am pretty sure that is exactly the question as it appeared in the book).

Any help, with either the topology or the source would be gratefully received!

#### Solutions Collecting From Web of "A niggling problem about frontier in topology"

There is an very elementary way to solve this, that is also much more widely applicable.

Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X \subset Y$ is closed, then there is a $V \subset X$ such that $\operatorname{Fr} V = X$.

Take $V = X \setminus (D \cap \operatorname{Int} X)$. Then since $V \subset X$ we have $\operatorname{Cl} V \subset \operatorname{Cl} X$, and $\operatorname{Fr} X \subset V$ and $E \cap \operatorname{Int} X$ dense in $\operatorname{Int} X$, therefore $\operatorname{Cl} V = X$. On the other hand $Y \setminus X$ is dense in $Y \setminus \operatorname{Int} X$ and $D \cap \operatorname{Int} X$ is dense in $\operatorname{Int} X$, therefore $\operatorname{Int} V = \emptyset$. It follows that $\operatorname{Fr} V = X$.

Any subspace of $\mathbb{R}^2$ is second countable, and hence separable. So if $X$ is your closed subspace, then let $A$ be a countable subset of $X$ that is dense in $X$. Then the closure of $A$ (in $\mathbb{R}^2$) is $X$, while the interior of $A$ is empty (since any nontrivial open set in $\mathbb{R}^2$ is uncountable). Thus $X$ is the boundary of $A$.