TITLE: Extending homeomorphisms from closed countable sets to S^2
QUESTION [10 upvotes]: Let $A, B \subset S^2$ be closed, countable sets and $\phi \colon A \rightarrow B$ be a homeomorphism. Can we extend $\phi$ to a homeomorphism from $S^2$ to itself?
It is well-known that the answer is yes if $A$ and $B$ are finite. When $A$ and $B$ have only one accumulation point, the question is equivalent to the extension of a bijection between closed discrete sets in $\mathbb R^2$ to an homeomorphism of $\mathbb R^2$ to itself. This is also not too difficult to construct. Similary, I think I can show it in the case that $A$ and $B$ have only finitely many accumulation points.

REPLY [3 votes]: $A\cup B$ is countable and closed, so there is a point $n\in S^2\setminus (A\cup B)$ with an open disk centered at $n$ that does not meet $A\cup B$.  Using stereographic projection from the pole $n$, we consider $A$ and $B$ as closed, bounded subsets of the plane.  Moreover, each set is totally disconnected.  We then apply the following theorem:
Theorem 13.7 [page 93 of Moise's Geometric topology in dimensions 2 and 3]: 
Let $M$ and $M'$ be totally disconnected compact sets in $\mathbb{R}^2$, and let $f:M\to M'$ be a homeomorphism. Then $f$ has an extension $F:\mathbb{R}^2\to \mathbb{R}^2$.
The extension $\tilde{F}: S^2\to S^2$ of $F$ to the sphere is still a homeomorphism.