Every closed set in $\mathbb R^2$ is boundary of some other subset

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!

• Does "frontier" mean boundary? May 29 '12 at 19:14
• Yes, I am sure the book used "frontier" to mean "boundary". May 29 '12 at 19:19

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 $A$.

• That would seem to do it - many thanks. I think I was somewhat blinded by the fact that the problem appeared in one of the first chapters of the book, well before concepts such as second countable had been introduced. May 29 '12 at 19:26

There is a 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 $$\operatorname{Fr} V \subset X$$ 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 $$\operatorname{Fr} V = X$$.

• Another way to phrase your result is that in a topological space $Y$, "every closed set is a boundary" is equivalent to "$Y$ is a boundary". May 31 '12 at 4:17
• Such a space $Y$ is called resolvable. Just FYI. May 4 '14 at 11:43