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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?

Agelos
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  • If your first question is answered satisfactorily, it would be nicer to accept the answer and post new questions, if such should arise, separately. – Jukka Kohonen Jun 03 '22 at 14:05
  • I did, the new question is here: https://mathoverflow.net/questions/423898/do-any-two-closed-countable-homeomorphic-subsets-of-mathbbr3-have-homeom. – Agelos Jun 03 '22 at 14:13

1 Answers1

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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.

Wojowu
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  • @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. – Wojowu Jun 01 '22 at 22:50
  • @bof : good question, I'm editing the main question to include it. – Agelos Jun 02 '22 at 07:44
  • @bof : there are advantages and disadvantages to the two options, and you may be right that it is better to start a new question. I'll wait a couple of days to see if there is a quick answer, and if not I could accept the answer and start a new post. An advantage of my approach is that we have all the discussion in one place... – Agelos Jun 02 '22 at 09:48
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    @bof The answer is still negative, see updated answer. – Wojowu Jun 02 '22 at 10:09
  • @Wojowu: that's another good example, thanks! What if we require that both $A,B$ be closed subspaces? – Agelos Jun 02 '22 at 10:52
  • ... under which assumption we could again ask if $A\cong B$ must contain a Cantor set. – Agelos Jun 02 '22 at 10:59
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    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 – Saúl RM Jun 02 '22 at 12:50
  • Perhaps an induction on Cantor Bendixon rank could work to prove that if $A,B\subseteq R^3$ are closed and countable (so they are dispersed and have countable Cantor-Bendixon rank) and $A$ is homeomorphic to $B$, then $\mathbb{R}^3\setminus A$ is homeomorphic to $\mathbb{R}^3\setminus B$. – Saúl RM Jun 02 '22 at 14:52