Boundaries of homeomorphic compact subspaces of $\mathbb R^n$ If $X$ and $Y$ are homeomorphic compact subspaces of $\mathbb R^n$, does it follow that their boundaries $\partial X$ and $\partial Y$ are also homeomorphic, preferably by the same homeomorphism?
This question has been asked here, but the comments discuss previous versions of that post.
It appears intuitive, though it may not be correct. If this is wrong in general, what conditions can be added to turn this statement true?
 A: It is true for any two closed $X,Y \subset \mathbb R^n$, but it is not trivial. We need "invariance of domain".
Since $X$ is closed, we have $\partial X = X \setminus \operatorname{int } X$, thus $X$ is the union of the disjoint sets $\operatorname{int } X$ and $\partial X$; similarly for $Y$. Let $h : X \to Y$ be a homeomorphism. Then $h$ maps $\operatorname{int } X$ injectively into $\mathbb R^n$ and by invariance of domain $h(\operatorname{int } X)$ is open in $\mathbb R^n$. Since $h(\operatorname{int } X) \subset h(X) = Y$, we see that $h(\operatorname{int } X)  \subset \operatorname{int } Y$. Considering the homeomorphism $h^{-1} : Y \to X$, we get $h^{-1}(\operatorname{int } Y)  \subset \operatorname{int } X$ which implies $\operatorname{int } Y = h(h^{-1}(\operatorname{int } Y)) \subset h(\operatorname{int } X)$. Therefore
$$h(\operatorname{int } X) = \operatorname{int } Y .$$
This shows that $h(\partial X) = \partial Y$, i.e. $h$ maps $\partial X$ homeomorphically onto $\partial Y$.
The counterexamples in the comments to Homeomorphic compact spaces have homeomorphic boundaries do not apply here because one of the spaces $X,Y$ is not closed.
