# Extending isometries between compact subspaces of Cantor space

Let $\omega$ be the set of natural numbers. $2^\omega$ is the Cantor space.

Suppose $K$, $L \subset 2^\omega$ are compact, and there is an isometry $f: K \to L$. Then how could one extend $f$ to an isometry from $2^\omega$ to $2^\omega$? Here we are considering $2^\omega$ with the minimum difference metric, which gives the standard product topology; i.e.

$d(x,y) = 2^{-\min \{ n : x(n) \neq y(n) \}}.$

-
Can someone give an example of an exotic isometry of the Cantor space (that is, one which isn't just a translation)? If there were none, it should be very easy to extend the isometry, if at all possible. – tomasz Jul 29 '12 at 21:14
@tomasz All isometries that I know arise as compositions of partial reflections. By which I mean the following construction: pick a finite binary sequence $a_1,\dots, a_n$. If $x\in \{0,1\}^{\omega}$ begins with this sequence, then flip all of its digits after $n$th. Otherwise leave it as it was. This can be visualized by reflecting a part of infinite binary tree around the axis of symmetry of that part. – user31373 Jul 29 '12 at 23:48
@tomasz Oh, I did not realize that by translation you meant group translations. I'm not used to thinking of this space as a group. – user31373 Jul 30 '12 at 0:21
@LeonidKovalev: yeah, you're right, I didn't think of that. But you need not flip all the bits after $n$th, you may flip only some arbitrary ones. (In the previous comment I did not notice that you flip bits only for SOME sequences, that's why I deleted it). Still, that doesn't strike me as very exotic. I wonder if that's all of them. :) – tomasz Jul 30 '12 at 0:21
@tomasz OK, let's summarize: for any collection of functions $f_n : \{0,1\}^n\to \{0,1\}$ we get an isometry $F$ such that the $n$th digit of $F(x)$ is $f_n$ applied to the beginning of $x$. Any map not of this kind must change some digit based on a later digit... I doubt this could be isometric. – user31373 Jul 30 '12 at 4:12

When $u$ is a finite binary word, $u\cdot 2^\omega$ means the interval of $2^\omega$ formed by infinite words with prefix $u$ ($\cdot$ denotes word concatenation).

Let $A_{fin}$ be the set of finite prefixes of $A\subseteq 2^\omega$.

We say that $F: A_{fin}\to B_{fin}$ is an isometry if:

• $|F(x)|=|x|$;
• $x\preceq y$ implies $F(x)\preceq F(y)$ where $\preceq$ is the prefix relation;
• $F$ is injective, or equivalently $F(x0)\ne F(x1)$ when both are in the domain of $F$.

Side-note: When $A_{fin}=B_{fin}$ (or more generally, when the multiset of word lengths of $A_{fin}$ and $B_{fin}$ coincide), the last condition can also be replaced by "$F$ is bijective". The first condition then becomes redundant.

Lemma: The relation $$\forall u\in A_{fin}, \quad f(u\cdot 2^\omega)\subseteq F(u)\cdot 2^\omega$$ is a bijection between isometries $f: A\to B$ and isometries $F: A_{fin}\to B_{fin}$.

Proof:

• When $uv\in A$, $F(u)$ must be the first $|u|=n$ symbols of $f(uv)$. This constructs $F$ in a well-defined way because $d(f(uv),f(uw))=d(uv,uw)\le 2^{-n}$. When $x\ne y$, if $|x|\ne|y|$ then of course $F(x)\ne F(y)$; otherwise $d(f(x),f(y))=d(x,y)>2^{-|x|}$ therefore $F(x)\ne F(y)$.
• Conversely, $f(x)=\lim_{u\preceq x} F(u)$ is well-defined by monotony of $F$, and if $d(x,y)=2^{-n}$ we can let $u0,u1$ be the prefixes of length $n+1$ of $x,y$: $2^{-n}=d(u0,u1)=d(F(u0),F(u1))=d(f(x),f(y))$.

Side-note: this also gives the following statement, which was not a priori obvious:

Corollary: An isometry $f: A\to A$ on $2^\omega$ is a bijection.

Corollary: Isometries $A_{fin}\to B_{fin}$ as defined above are precisely the isometries $A_{fin}\to B_{fin}$ under the minimum difference metric and under word length, where the metric is extended to finite words by $d(x,u)=\max d(x,u0^\omega),d(x,u1^\omega)$. This justifies the use of the word "isometry".

Theorem: Let $A,B$ be arbitrary subsets of $2^\omega$. An isometry $f: A\to B$ can be extended to an isometry $g: 2^\omega\to 2^\omega$. As could also be shown by topological arguments, the extension is unique if and only if $A_{fin}=(2^\omega)_{fin}$, that is iff $A$ is dense in $2^\omega$.

Proof: We just have to extend an isometry $F:A_{fin}\to B_{fin}$. When $u$ is non-empty, let $\sigma_F(u)$ be the last symbol of $F(u)$. We have that $\sigma:\{0,1\}^+\to\{0,1\}$ is the last symbol of an isometry $(2^\omega)_{fin}\to (2^\omega)_{fin}$ if and only if $\sigma(x1)=\neg \sigma(x0)$. So we can define $$\sigma(x0)=\begin{cases} \sigma_F(x0) & x0\in A_{fin}\\ \neg\sigma_F(x1) & x1\in A_{fin}\\ 0 & \text{else} \end{cases}$$ so that the relations $\sigma(x1)=\neg \sigma(x0)$ and $G(ua)=G(u)\sigma(ua)$ uniquely define an isometry $G: (2^\omega)_{fin}\to (2^\omega)_{fin}$. $G$ extends $F$, so the isometry $g: 2^\omega\to 2^\omega$ extends $f$.

-
Thanks for the very thorough explanation! (You did not define $2^*$, but I understand it is $(2^\omega)_{fin}$). – user31373 Jul 30 '12 at 20:36