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Let $X=\{0,1\}^{\mathbb{N}}$ and $\xi_n$ be the partition of $X$ defined by the equivalence relation $x \sim_n x' \Leftrightarrow (x_{n}, x_{n+1}, \ldots) = (x_{n}', x_{n+1}', \ldots)$. The sequence of partitions $(\xi_n)$ is decreasing and we introduce the intersection partition $\theta= \cap \xi_n$. I'm loooking for a group $G$ of transformations of $X$ such that $\theta$ is the partition into the orbits of $G$.

Maybe my question is somewhat unprecise, I am rather new in ergodic theory. Any comments are welcomed.

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Isn't it true that $\xi_n\subset \xi_{n+1}$? Why do you call this sequence decreasing? –  Mark Sapir May 4 '12 at 21:40
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Provided you mean to union up the equivalence relations, this equivalence relation is often called $E_0$ and can be easily realized as the orbit equivalence relation of the coordinate-wise action of $(\mathbb{Z}/2\mathbb{Z})^{< \omega}$ on $(\mathbb{Z}/2\mathbb{Z})^\omega$, or as the equivalence relation generated by addition by one in the $2$-adic integers (except you need to deal with the eventually constant sequences separately). –  Clinton Conley May 4 '12 at 22:10
    
The ordering between $\xi_n$ and $\xi_{n+1}$ is not the inclusion. Any element of $\xi_{n}$ is a subset of an element of $\xi_{n+1}$, that is, $\xi_n$ is finer than $\xi_{n+1}$, or $\xi_{n+1}$ is coarser than $\xi_n$. It is sensible to denote $\xi_n \geq \xi_{n+1}$ because it is more naturel to say that the smallest (coarsest) partition is the trivial partition $\{\varnothing, X\}$ and the biggest (finest) partition is ${\cal P}(X)$. –  Stéphane Laurent May 4 '12 at 22:17
    
@Stephane: Yes, I figured that out. That means I understood your question correctly. –  Mark Sapir May 4 '12 at 22:19
    
@Clinton: That was my first choice too, but dealing with eventually constant sequences is not easy. –  Mark Sapir May 4 '12 at 22:20

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up vote 1 down vote accepted

Consider the group $G_n$ of all permutations of strings $\{0,1\}^{\{1,2,...,n\}}$ of length $n$. That group acts on the set of all strings $\{0,1\}^{\mathbb N}$ in the natural way (it changes the $n$-prefix only). Now clearly $G_n\subset G_{n+1}$ for every $n$. Let $G=\cup G_n$. It is a (locally finite) group which acts on $\{0,1\}^{\mathbb{N}}$ and the orbit equivalence relation is your $\theta$ (if I understood the question correctly): two strings are in the same orbit iff they coincide almost everywhere.

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This does not seem quite right; for example $01^\infty$ is not in the same orbit as $1^\infty$ under the action you propose. The orbit equivalence relation generated by your action is strictly finer than eventual agreement. –  Clinton Conley May 4 '12 at 22:12
    
Yes, it is. $G_n$ is not the permutation group of indices, it is the permutation group of the set of strings of length $n$. –  Mark Sapir May 4 '12 at 22:15
    
So $|G_n|=(2^n)!$. –  Mark Sapir May 4 '12 at 22:16
    
Oops, sorry for misreading that. You're absolutely right. –  Clinton Conley May 4 '12 at 22:17
    
@Mark: what do you mean by two strings coinciding almost everywhere ? –  Stéphane Laurent May 4 '12 at 22:55

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