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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) \}}. $

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    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.2012-07-29
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    @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.2012-07-29
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    @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.2012-07-30
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    @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. :)2012-07-30
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    @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.2012-07-30
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    @tomasz Indeed these are all isometries. If $x$ and $x'$ agree in the first $n$ digits, then their images under any isometry also agree in the first $n$ digits. Hence, the isometry must be of the kind described in the previous comment.2012-07-30
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    Oops, forgot that $ f_n$ must be linear in the last variable, that is $f_n=x_n+g_n (x_1,\dots x_{n-1})$.2012-07-30

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