# On homeomorphic subsets of $\mathbb{R}^3$ with non-homeomorphic complements

Let $$A,B$$ be two homeomorphic topological subspaces of $$\mathbb{R}^3$$ such that their complements $$\mathbb{R}^3 - A, \mathbb{R}^3 - B$$ are not homeomorphic to each other. Must $$A \cong B$$ contain a homeomorphic image of the Cantor set?

(It is known that there are homeomorphic images $$A,B$$ of the Cantor set such that $$\mathbb{R}^3 - A, \mathbb{R}^3 - B$$ are not homeomorphic, see e.g. https://www.sciencedirect.com/science/article/pii/016686418690060X)

UPDATE 1:

The answer is no, as Wojowu's answer shows. This leads to Question 2: Must $$A \cong B$$ contain a homeomorphic image of $$\mathbb{Q}$$?

UPDATE 2:

After Wojowu's answer, the interesting question remaining is

Question 3: Let $$A,B$$ be two closed, countable, topological subspaces of $$\mathbb{R}^3$$ homeomorphic to each other. Must their complements $$\mathbb{R}^3 - A, \mathbb{R}^3 - B$$ be homeomorphic to each other?

• If your first question is answered satisfactorily, it would be nicer to accept the answer and post new questions, if such should arise, separately. Jun 3, 2022 at 14:05
• I did, the new question is here: mathoverflow.net/questions/423898/…. Jun 3, 2022 at 14:13

Let $$A=\mathbb Q^3$$ and $$B=\{0\}\times\mathbb Q^2$$. It is a classical result that they are homeomorphic (both homeomorphic to $$\mathbb Q$$), and their complements are not homeomorphic as $$\mathbb R^3-B$$ contains a subset homeomorphic to an open ball, while $$\mathbb R^3-A$$ doesn't (e.g. by the invariance of domain theorem).
However, since $$A,B$$ are countable, they don't contain a copy of the Cantor set.
The answer to the new question is also negative. Indeed let $$A=\{(0,0,n)\mid n\in\mathbb N\}$$ and $$B=\{(0,0,1/n)\mid n\in\mathbb N\}$$. Both of these are discrete and countable, so are homeomorphic. On the other hand, the complement of $$A$$ is a manifold, while $$(0,0,0)$$ in $$\mathbb R^3-B$$ has no Euclidean neighbourhood.
• @bof It should be, yeah - in $\mathbb R^3-A$ and $\mathbb R^3-B$ you can consider the subsets of points which have a Euclidean neighbourhood, and see that those will not be homeomorphic. Jun 1, 2022 at 22:50
• If $A,B$ are closed we can suppose they are countable, because if not, they contain perfect sets (their sets of condensation points) which in turn contain Cantor sets Jun 2, 2022 at 12:50