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?

(The analogous statement in 2 dimensions is true, by Richards' classification of non-compact surfaces: https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf.)

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    $\begingroup$ It should be true and the proof (I hope) can be done by induction on the scattered height of those countable closed subspaces. $\endgroup$ Jun 3, 2022 at 18:06
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    $\begingroup$ Also nontrivial: If $A$ and $B$ are two closed, countable, nonhomeomorphic subspaces of $\mathbb R^3$, does this imply that $\mathbb R^3 \setminus A$ and $\mathbb R^3 \setminus B$ are not homeomorphic? $\endgroup$
    – Will Brian
    Jul 18, 2022 at 16:43
  • $\begingroup$ @WillBrian: the answer is no: let $A$ be the set of points $(n,0,0)$ where $n\in \mathbb{N}$. Let $B$ be the set of points $(1/n,0,0)$ where $n\in \mathbb{N}_{>0}$. To see that $\mathbb{R}^3 \setminus A$ is homeomorphic with $\mathbb{R}^3 \setminus B$, think of $\mathbb{S}^3$ as the 1-point compactification of $\mathbb{R}^3$, and find a homeomorphism of $\mathbb{S}^3$ that maps the point at infinity to 0 and maps $(n,0,0)$ to $(1/n,0,0)$. $\endgroup$
    – Agelos
    Jul 19, 2022 at 15:39
  • $\begingroup$ But I think that the answer becomes positive if you replace $\mathbb{R}^3$ with $\mathbb{S}^3$. To see this, notice that the Freudenthal boundary of $\mathbb{S}^3 \setminus A$ coincides with $A$ for any closed countable $A$. $\endgroup$
    – Agelos
    Jul 19, 2022 at 15:42
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    $\begingroup$ ... I forgot to say that $B$ also contains the point $(0,0,0)$. $\endgroup$
    – Agelos
    Jul 19, 2022 at 15:45

1 Answer 1


The answer is yes, and it is now given in the preprint Georgakopoulos. Every countable compact subset of S^n is tame.

The proof exploits the fact every closed, countable, subspace of $\mathbb{S}^3$ has a basis $\mathcal{O}$ induced by metric balls of $\mathbb{S}^3$, such that for any two such balls $X,Y$ we have either $X \cap Y = \emptyset$, or $X\subseteq Y$ or $B\subseteq Y$. These bases allow decomposing each of $\mathbb{S}^3 - A$ and $\mathbb{S}^3 - B$ into an infinite family of finitely punctured spheres. One then constructs a homeomorphism between $\mathbb{S}^3 - A$ and $\mathbb{S}^3 - B$ as a union of homeomorphisms between those punctured spheres in a ping-pong fashion similar to Richards' approach.


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