Symbolic dynamics

Remarks

The itinerary map $x \mapsto (a_n)$ turns $T$ into the shift: the itinerary of $T(x)$ is the shifted itinerary of $x$. A coding is rarely continuous, since a connected space cannot be split into finitely many disjoint nonempty open pieces. Instead one often has a continuous surjection $\pi$ from a shift space onto $M$ with $\pi \circ \sigma = T \circ \pi$, a semiconjugacy that sends a sequence to a point with that itinerary. Example: for the doubling map $T(x) = 2x \bmod 1$ on $[0, 1)$ with $P_0 = [0, 1/2)$ and $P_1 = [1/2, 1)$, the itinerary of $x$ is its binary expansion $x = \sum_{n \ge 0} a_n 2^{-n-1}$ that does not end in all $1$s. Indeed $T$ deletes the first binary digit, and the first digit is $0$ exactly when $x \in P_0$. Here $\pi(a) = \sum_{n \ge 0} a_n 2^{-n-1}$ is a semiconjugacy from the one-sided full shift on $\set{0, 1}$ onto the circle $\R / \Z$. Non-example: a coding with the single region $P_0 = M$ gives the constant itinerary $000\ldots$ for every point, so it records nothing about the dynamics.