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Given a bounded closed set $A$ in $\mathbb R^n$, can $A$ be uniquely determined by $\partial A$, except for the boundary itself?

Or, use it differently, given two bounded closed sets $A_1, A_2$ in $\mathbb R^n$ with $\partial A_1 = \partial A_2$, $A_1 \ne\partial A_1$, and $A_2 \ne\partial A_2$, is it true that $A_1 = A_2$?

Without the boundedness assumption the assertion is clearly false: the sphere $\{ x \in \mathbb R^n : |x|=1\}$ is the common boundary to $\{ x \in \mathbb R^n : |x| \ge 1\}$ and $\{ x \in \mathbb R^n : |x| \le 1\}$.

note) It was originally intended for the Euclidean compact(bounded and closed) set, but it was incorrectly modified as an open set. I am sorry.

Mohsen Shahriari
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Y choe
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    $[0,1]$ and ${0,1}$ are compact with the same boundary ${0,1}$ – Robert Z Jul 06 '20 at 11:35
  • I missed it. Thank you. If {0,1} is excluded, ie the boundary itself is excluded, is the subset unique? – Y choe Jul 06 '20 at 11:43
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    Hello, I have edited the question, I hope it reflects what you wish to ask, at least the example by Robert is excluded. You can aslo edit the question, by clicking the edit buttom. – Arctic Char Jul 06 '20 at 11:49
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    See Lakes of Wada : https://en.wikipedia.org/wiki/Lakes_of_Wada – Robert Z Jul 06 '20 at 12:28
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    See https://math.stackexchange.com/questions/1076066/two-disjoint-connected-and-bounded-open-sets-in-the-plane-that-shares-the-same-b – Robert Z Jul 06 '20 at 12:33
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    This question has been answered, but is there some study of this issue in a more systematic way? What if we enforce some regularity on the boundary? What properties of a boundary allow or limit the number of sets it can bound? – Nick Alger Jul 09 '20 at 09:05
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    I don't think the accepted answer is correct. – feynhat Jul 09 '20 at 13:10

3 Answers3

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Being bounded is not really an obstacle to your sphere example. You just need to tweak it a bit:

$$A_1=\big\{v\in\mathbb{R}^n\ \big|\ 1<\lVert v\rVert< 2\big\}$$ $$A_2=A_1\cup \big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert< 1\big\}$$ $$\partial A_1=\partial A_2=\big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert=1 \big\}\cup\big\{v\in\mathbb{R}^n\ \big|\ \lVert v\rVert=2 \big\}$$

freakish
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I think the other examples given so far are not correct. I think this works:

Take two disks in $\Bbb{R}^2$, and let one set be the union of one disk and the circle around the other, and let the other set be the union of the other disk with the circle around the former. Then they are closed, bounded, different than their boundaries, but have the same boundary while being different than one another.

I'll write a formula if this isn't clear.

EDIT. In fact, here's a much simpler example, in $\Bbb{R}$. Let $A_1=\{0,1\}\cup [2,3]$ and $A_2=[0,1]\cup \{2,3\}$.

EDIT 2. Like feynhat said, if you want a connected example, take my first example (in $\Bbb{R}^2$) and set $A_1=\{(x,y):x^2+y^2\leq 1\}\cup\{(x,y):((x-2)^2+y^2= 1\}$, $A_2=\{(x,y):x^2+y^2= 1\}\cup\{(x,y):((x-2)^2+y^2\leq1\}$, or something like this. The idea is to take two tangent disks. Perhaps someone can draw a figure and add it.

Cronus
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    Great example. We can even take the two disks to be touching each other (if OP demands the sets to be connected). – feynhat Jul 09 '20 at 07:52
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    @feynhat Right, this would be a little nicer. Although I still think that perhaps there is a way of strengthening a bit the requirement that $A_i$ is different than its boundary and get some meaningful (and true) statement... – Cronus Jul 09 '20 at 08:06
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Consider a torus $\mathbb{T}^2$ in $\mathbb{R}^3$ and a circle $S^1$ on it such that it is null homotopic in $\mathbb{T}^2$. Then you get a disc on one side of the circle and a surface with nontrivial fundamental group on the other which share the same border. So the answer to the question is negative.

astro
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  • Does that mean they share a circle-shaped border in $S^1$? Isn't the boundary of the disc in $\mathbb{R}^3$ different from the boundary of the surface in $\mathbb{R}^3$? – Y choe Jul 06 '20 at 16:56
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    I don't think this answers OP's question, since, they asked about subset of $\Bbb R^n$ (the boundary was being taken in $\Bbb R^n$). In your example, the boundary is computed in the subspace topology of $\Bbb T^2$ inherited from $\Bbb R^3$. If you compute the boundary of those sets in $\Bbb R^3$, then the subsets (the contractible disc on the torus and its complement) will be their own boundaries. – feynhat Jul 09 '20 at 06:07
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    I agree with @Feynhat, this isn't really a counterexample. (It works insite the torus, but not in the Euclidean plane) – Cronus Jul 09 '20 at 07:14