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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!

Old John
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3 Answers3

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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 $A$.

Chris Eagle
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    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. – Old John May 29 '12 at 19:26
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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$.

Some additional information:

The question probably came from Willard's General topology, problem 3 B.

As Henno Brandsma pointed out in a comment, a space that can be partitioned into two dense subsets is called resolvable. For some examples of irresolvable spaces, see the answer to "Is a perfect set a boundary?".

A somewhat stronger result can be found in "Any two disjoint open sets are the interior and exterior of some set".

José Carlos Santos
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Niels J. Diepeveen
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  • 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". – Chris Eagle May 31 '12 at 04:17
  • @Chris Eagle: That is a very nice way to put it. In fact it can be proved along the same lines that a closed subset of any boundary is also a boundary. – Niels J. Diepeveen Jun 01 '12 at 01:30
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    Such a space $Y$ is called resolvable. Just FYI. – Henno Brandsma May 04 '14 at 11:43
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    Dear @NielsJ.Diepeveen I'm reading your proof, could you give some help to understand it? I don't know how to prove that $\operatorname{Cl} V=X$ and $\operatorname{Int} V=\emptyset$. – Math Mar 29 '23 at 11:44
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@Math23: I also had problems convincing myself about those facts in the proof. I proceeded as follows:

A) Try to see that we can write $V=\left(\mathrm{Int\,}X\cap E\right)\cup\mathrm{Fr}\,X$. With this, convince yourself that $\mathrm{Cl}\,V\supset X$. Then, it's easy to see that $\mathrm{Cl}\,V\subset X$ (because $V\subset X$) and so $\mathrm{Cl}\,V=X$.

B) Notice that we can also write $V=X\cap\left[E\cup\mathrm{Cl}\left(Y\setminus X\right)\right]$. With this, convince yourself that $\mathrm{Int}\,V\subset \mathrm{Int}\,X\cap E$ and this implies $\mathrm{Int}\,V=\emptyset$. However, I now notice that Niels' argument on this is not that hard to see. What it says is that for every nhood of any point $y\in Y$ (regardless of $y$ being inside or outside $\mathrm{Int}\, X$) we can find points in that nhood that don't belong to $V$.

olaphus
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