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) \}}. $

  • $\begingroup$ 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. $\endgroup$
    – tomasz
    Jul 29 '12 at 21:14
  • 1
    $\begingroup$ @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. $\endgroup$
    – user31373
    Jul 29 '12 at 23:48
  • $\begingroup$ @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. $\endgroup$
    – user31373
    Jul 30 '12 at 0:21
  • $\begingroup$ @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. :) $\endgroup$
    – tomasz
    Jul 30 '12 at 0:21
  • $\begingroup$ @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. $\endgroup$
    – 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 $F: A_{fin}\to B_{fin}$.


  • 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 $f$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.