# Do any two closed, countable, homeomorphic subsets of $\mathbb{R}^3$ have homeomorphic complements?

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.)

• It should be true and the proof (I hope) can be done by induction on the scattered height of those countable closed subspaces. Jun 3, 2022 at 18:06
• 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? Jul 18, 2022 at 16:43
• @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 $(1/n,0,0)$. Jul 19, 2022 at 15:39
• 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 $A$. Jul 19, 2022 at 15:42
• ... I forgot to say that $B$ also contains the point $(0,0,0)$. Jul 19, 2022 at 15:45

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.