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>)




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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 $\mathbb{Q}$?




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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?



On homeomorphic subsets of $\mathbb{R}^3$ with non-homeomorphic complements
Let $A=\mathbb Q^3$ and $B=\{0\}\times\mathbb Q^2$. It is a [classical result](https://mathoverflow.net/q/26001/30186) 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.




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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.


