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)


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


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?

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

1 Answer 1


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.

  • $\begingroup$ @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. $\endgroup$
    – Wojowu
    Jun 1, 2022 at 22:50
  • $\begingroup$ @bof : good question, I'm editing the main question to include it. $\endgroup$
    – Agelos
    Jun 2, 2022 at 7:44
  • $\begingroup$ @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... $\endgroup$
    – Agelos
    Jun 2, 2022 at 9:48
  • 2
    $\begingroup$ @bof The answer is still negative, see updated answer. $\endgroup$
    – Wojowu
    Jun 2, 2022 at 10:09
  • 1
    $\begingroup$ 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 $\endgroup$
    – Saúl RM
    Jun 2, 2022 at 12:50

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