Suppose $A_1,A_2$ are bounded, closed, connected subsets of $\Bbb{R}^n$, such that $\partial A_i$ has empty interior inside $A_i$ (for both $i$). Is it true that if $\partial A_1=\partial A_2$ then $A_1=A_2$?

This question is inspired by this one, in which instead of asking that $\partial A_i$ should have empty interior inside $A_i$, it was just required that $\partial A_i\neq A_i$.

If any of the assumptions above are removed, I can find a counterexample. But this seems more involved to me, if there is a counterexample at all.

In the title I asked this for general $n$, but really I am interested in determining the minimal $n$ for which there is such an example (if there is such $n$). Clearly this is impossible for $n=1$, but I'm not even sure about $n=2$.

EDIT. Let me explain what I mean by the sentence "$\partial A_i$ has empty interior inside $A_i$". Since $A_i$ is closed, we have that $\partial A_i\subseteq A_i$. Now, consider the subspace $A_i$ of $\Bbb{R}^n$ (with the subspace topology). $\partial A_i$ is a subset of this subspace, and one can look at its interior $\mathrm{Int}_{A_i}(\partial A_i)$, as a subset of the space $A_i$. I require that $\mathrm{Int}_{A_i}(\partial A_i)= \varnothing$.

  • $\begingroup$ @mr_e_man I added an explanation. $\endgroup$
    – Cronus
    Jul 9, 2020 at 18:52

1 Answer 1


Here's an example. It's a little tricky to explain, so I'll just upload a picture:

enter image description here

The subsets are inside $\Bbb{R}^2$. The idea is that there are infinitely many such tear-shaped 'layers', red and then blue and then red and then blue, with black in between. We take $A_1$ to be the red + black bits, and $A_2$ to be the blue + black bits. Then the boundary of both is the black, and they satisfy all the necessary conditions.

Obviously this isn't a rigorous proof, but I think writing this out in detail would be torture...

  • $\begingroup$ Should they be circles sharing a tangent line, instead of tear shapes? It looks like the corner point is in the interior (in $A_2$) of $\partial A_2$. $\endgroup$
    – mr_e_man
    Jul 9, 2020 at 19:13
  • 1
    $\begingroup$ @me_e_man Yeah this is just because the drawing is a bit crude, the idea is that each layer should go all the way to the corner point. I guess this could work with tangent circles too. $\endgroup$
    – Cronus
    Jul 9, 2020 at 19:23

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