According to the countable generator theorem, every ergodic invertible measure-preserving transformation has a generating partition. 


What are the generating partitions of the dyadic odometer ? I don't find the answer in textbooks. Is there a "canonical" one among them ?



What are the generating partitions of the odometer?
Quite a nice one is the two-set partition $A,A^c$, where $A$ is the set of points with an even number of terminal 0's:
$$
A = \bigcup\_{k \geq 0} A\_k
$$
where $A\_k = \bigl\{(x\_1, x\_2, \ldots) \mid x\_1=\ldots=x\_{2k}=0 \text{ and } x\_{2k+1}=1 \bigr\}$. 
Or, if you work with the odometer acting on the space $[0,1[$:
$$
A = \bigcup\_{k \geq 0} \left[\frac{1}{2^{2k+1}}, \frac{1}{2^{2k}} \right[ 
$$


This partition is generating for the following reason. Code a trajectory $x, Tx, T^2x, \ldots$ by a sequence $(v\_0, v\_1, \ldots)$ of $a$'s and $b$'s according to whether it's in $A$ or not. 


Then you can get the first digit $x\_1$ of $x$ by looking at the blocks of four consecutive terms of the $v$ sequence:


* if the first digit of $x$ is $x\_1=1$, then the block $(v\_0,v\_1,v\_2,v\_3)$ of four consecutive terms is one of $aaab$, $abaa$ or $abab$ (that is $(v\_0,v\_1,v\_2,v\_3)$ is $a\*a\*$, with at least one of the stars a $b$);
* if $x\_1=0$, then $(v\_0,v\_1,v\_2,v\_3)$ is $\*a\*a$ with at least one of the stars a $b$.


In particular, the sets of possible codes are disjoint, so that this information suffices to determine the first digit of $x$.


To get the second digit of $x$, look at blocks of eight consecutive terms:


* if the first two digits of $x$ are 00, then the code is $\*aba\*aba$ with at least one of the $\*$'s an $a$;
* if the first two digits of $x$ are $01$, then the code is $ba\*aba\*a$ with at least one of the $\*$'s an $a$;
* if the first two digits of $x$ are 10, then the code is $aba\*aba\*$ with at least one of the $\*$'s an $a$;
* if the first two digits of $x$ are 11, then the code is $a\*aba\*ab$ with at least one of the $\*$'s an $a$;


Again, these sets of possibilities are disjoint, so that one can recover the first and second digits of $x$ from 8 terms of the $v$ sequence.


And so on, looking at the block of $2^{n+1}$ consecutive terms of the $v$ sequence determines the first $n$ digits of $x$. 


