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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!

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  • $\begingroup$ Does "frontier" mean boundary? $\endgroup$ May 29 '12 at 19:14
  • $\begingroup$ Yes, I am sure the book used "frontier" to mean "boundary". $\endgroup$
    – Old John
    May 29 '12 at 19:19
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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 $A$.

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    $\begingroup$ 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. $\endgroup$
    – 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".

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  • $\begingroup$ 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". $\endgroup$ May 31 '12 at 4:17
  • $\begingroup$ @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. $\endgroup$ Jun 1 '12 at 1:30
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    $\begingroup$ Such a space $Y$ is called resolvable. Just FYI. $\endgroup$ May 4 '14 at 11:43

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