# Is a compact, connected subset of $\Bbb{R}^n$ whose boundary has empty interior inside it determined by its boundary?

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 $$\mathrm{Int}_{A_i}(\partial A_i)= \varnothing$$.

• @mr_e_man I added an explanation. Jul 9, 2020 at 18:52

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.
• 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 $\partial A_2$. Jul 9, 2020 at 19:13