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.

• I guess it's even true when $A,B$ are arbitrary totally disconnected closed subsets. It seems to follow from Gehman's 1936 paper, relying on earlier papers projecteuclid.org/download/pdf_1/euclid.bams/1183498682 but I haven't completely understood the theorem (in this paper, "sphere" means "2-sphere". Since $A,B$ have connected complements, we can add an isolated point to each of them, and thus reduce to the case when $A,B$ are compact subsets of the plane. Then it seems to follow from earlier papers by the same author, which I haven't accessed. – YCor Dec 13 '16 at 18:11
• You can view these sets as "almost" like ordinals: there are the isolated points, the cluster points of isolated points, the cluster points of those, etc. If this process terminates, there is an argument the homeomorphism extends. But there are closed countable subsets of $S^2$ where this process does not terminate, I'm thinking of the well-ordered subsets of $\mathbb R$ like the ordinal $\omega^\omega$. It's not immediately clear to me if such subsets of $S^2$ can be knotted. Probably not. . . – Ryan Budney Dec 13 '16 at 20:33

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.

• Brouwer, in "On the structure of perfect sets of points", first proved that any two Cantor sets are homeomorphic (the publisher has the article posted here ) and thus any Cantor set (whether a subset of the plane or not) can be mapped (but not necessarily equivalently) to the standard Cantor set in $\mathbb{R}\times \{0\}$. The only place the plane hypothesis is truly necessary is to prove the self-map on $A$ extends to the plane. – David Snyder Dec 15 '16 at 3:20
• @DavidSnyder I know, but I didn't use this. Reading again, I see I'm not clear (when I said "by a self-homeo" it could only mean "by a self-homeo of the plane"). Now edited. – YCor Dec 15 '16 at 3:24
• Moise's proof doesn't assume the two Cantor sets are standard, just compact, perfect,totally disconnected subsets of the plane, so the extra step to embed the sets in standard Cantor sets isn't strictly necessary. I presume that proof is due originally to Moise since Bing mentions the results in his "Tame Cantor sets in $\mathbb{R}^3$" but does not mention Moise (Bing and Moise were fierce competitors in those days; on the other hand, it may have been an exercise that RL Moore would give his grad students?). – David Snyder Dec 15 '16 at 3:57
• @DavidSnyder Yes, I don't use that the given Cantor set is standard, I just use that it's a single one. I could argue the same fixing a single Cantor set in the line and proceed. Is this what you mean? I think it's not important. – YCor Dec 15 '16 at 4:09
• @PietroMajer: thanks for letting me know, I hadn't noticed. We can erase these both comments now. – YCor Dec 17 '16 at 23:53

$A\cup B$ 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.