I know that the Cantor set is countable dense homogeneous. My question is: if A,B,C,D are countable dense subsets of the Cantor set such that the pairs A and B and C and D are disjoint, there exists a homeomorphism f of the Cantor set such that f(A) = C and f(B) = D?



Is the Cantor set countable dense homogeneous in pairs?
The answer is yes, unless I'm missing something, via a back-and-forth argument.


For simplicity I'll view the Cantor set as the set of infinite binary sequences, $2^\omega$, and think of this as the set of paths through the infinite binary tree $2^{<\omega}$.


Any rooted-tree automorphism $\alpha$ of $2^{<\omega}$ induces an autohomeomorphism $h\_\alpha$ of $2^\omega$. Call the autohomeomorphisms so induced the *standard* ones.


Let $A\cup B=X$, $C\cup D=Y$. Call a finite partial bijection $p: X\rightarrow Y$ *good* if


* $p: A\rightarrow C$, $p: B\rightarrow D$, and
* There is some standard autohomemorphism $h$ of $2^\omega$ such that $p\prec h$.


Then:


* The empty map is good.
* **(Forth)** For any good map $p$ and any $x\in X$, there is some $y\in Y$ such that $p\cup\{(x, y)\}$ is again good.
* **(Back)** For any good map $p$ and any $y\in Y$, there is some $x\in X$ such that $p\cup\{(x, y)\}$ is again good.


*Why? Well, the first point is trivial, and the second and third are true for the same reason, so let's talk about the **forth** part. Think of a good map $p$ as labelling each node of $2^{<\omega}$ which is an initial segment of an element of the domain of $p$, **or the immediate successor of such an initial segment**, with either "$0$" or "$1$". For instance, if $p$ is the map which sends $00000...$ to $10101010...$ and does nothing else, then $p$ would label $0$ with $1$, $00$ with $0$, $000$ with $1$, $0000$ with $0$, and so on; and also labels $1$ with $0$ (because that's forced by where $p$ sends $0$), but would not label $11$ anything (because that's not forced by $p$ at all). Now, if $x\in 2^\omega\setminus dom(p)$, then at some point $x$ "leaves the labelled part" of $2^{<\omega}$. As soon as that happens, we're free to label the remaining nodes of $x$ however we like; so, just pick some appropriate element of $Y$ that hasn't been used yet, and go!*


This lets us build, via the usual back-and-forth method, a bijection $B: X\rightarrow Y$ such that every finite submap of $B$ is good.


Now consider the map $$b: 2^{<\omega}\rightarrow 2^{<\omega}: \sigma\mapsto \tau\iff \exists \sigma\prec x\in X, p\prec B(x\in dom(p), p(x)\upharpoonright\vert\sigma\vert=\tau).$$ This is a rooted-tree automorphism of $2^{<\omega}$, and so induces a standard autohomeomorphism $h\_b$ of $2^\omega$. And it's easy to check that $h\_b$ extends $B$.


