Extending homeomorphisms from closed countable sets to S^2 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.
 A: Yes. It follows from the Cantor case. First, it is no restriction to assume that $A,B$ are compact totally disconnected (e.g., countable) subsets of the plane and we want to prove that any homeomorphism $A\to B$ extends to a self-homeomorphism of the plane.
Lemma: every compact totally disconnected subset $K $ of $\mathbf{R}^k$ is contained in a Cantor subset of $\mathbf{R}^k$.
Proof: we can suppose $K$ nonempty. If $x$ is an isolated point in $K$, let $N_x$ be the distance of $x$ to its complement in $K$. Then choose a Cantor subset $L_x$ in $\mathbf{R}^k$ containing $x$, of diameter $<N_x/2$. Then $L=\bigcup L_x$ is nonempty compact, totally disconnected, contains $K$, and has no isolated point. (Totally disconnected requires a little argument: the point is to check that for every partition of $K$ into 2 clopen subsets $U,V$, there exists only finitely many $x$ such $L_x$ meets both $U$ and $V$, using that $N_x$ tends to zero and $U,V$ are at positive distance. Using this, it's easy to extend continuously every continuous map $K\to\{0,1\}$ to $L$). So it's a Cantor set. $\Box$
Since it is known that every Cantor subset of the plane can be mapped by a self-homeomorphism of the plane onto the standard dyadic Cantor set in the line $\mathbf{R}\times\{0\}$ (This follows from the Denjoy-Riesz theorem, see this MathSE post), we can suppose $A$, $B$ are contained in this given dyadic Cantor set.
So now consider $A,B$ in $\mathbf{R}\times\{0\}$. Then if $Q$ is a Cantor set, clearly every homeomorphism $A\to B$ extends to a homeomorphism $A\times Q\to B\times Q$. Assuming $A$ nonempty, $A\times Q$ is a Cantor set. Hence if we have proved the extension result in the Cantor case, we are done. 
So after this second reduction, we can assume $A=B$ is the standard dyadic Cantor set in $\mathbf{R}\times\{0\}$. This case is not obvious, but is already answered in this other MathSE post, with a reference to Edwin E. Moise's book Geometric Topology.
A: $A\cup B$ 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.
A: The answer is yes in higher dimensions too: see
Georgakopoulos - Every countable compact subset of S^n is tame.
In two dimensions, the generalisation to totally disconnected compact sets (Theorem 13.7 in David Snyder's answer) can also be deduced from Richards' classification of non-compact surfaces.
