My teacher handed out an excerpt from a book by Robinson on chaotic dynamical systems. The excerpt is from a chapter on Markov partitions, and the following part has me confused:
Let $$f(x)= \frac{4}{3}x \quad (\text{mod }1).$$ This map has a discontinuity at $x= 3/4$. It does not have a Markov partition because $(4/3)^n$ is never an integer, so $f^n(1)$ never returns to $0$.
I'm not entirely surely why this contradicts the existence of a Markov partition. Unfortunately the excerpt doesn't actually contain a definition of Markov partition, but I think I have a vague understanding of the concept from earlier reading.
Could someone please explain why $f^n(1)$ not returning to $0$ prevents the existence of a Markov partition?
EDIT: I'll elaborate a bit on what I understand Markov partitions to be. Assuming the function above is defined on $[0,1]$, we can partition the domain into intervals $I_1,I_2,\ldots I_n$. Then, to each $x \in [0,1]$ we assign an itinerary, a sequence $s_0 s_1 s_2 s_3 \ldots$ such that $s_k = i$ if $f^k(x) \in I_i$. The idea is now that we can define a shift map on the space of such sequences, and try to show that this shift map has the same dynamics as $f$. This practice is called symbolic dynamics.
I understand that not all partitions are Markov partitions, but what exactly are the properties that a partition must have in order to qualify as a Markov partition? Specifically, why does $f^n(1) \not= 0$ prevent a Markov partition from existing?
In the context of a partition into subintervals, one way to turn this into a Markov partition with symbolic dynamics as you describe is to require that the restriction of $f$ to the interior of each subinterval $I_m$ is a homeomorphism onto the interior of a finite union of one or more consecutive subintervals $I_j \cup I_{j+1} \cup \cdots \cup I_k$. This is the definition used, for example, in the paper of Milnor and Thurston "On iterated maps of the interval"; their context requires the map to be continuous, but having discontinuity at the endpoints is still okay.
The best I can do is to point out that in order for this definition to be satisfied, the sequence $f^n(1) = (4/3)^n$ would have to take on only finitely many values, namely the endpoints of the intervals $I_1,I_2,…,I_n$. That is clearly impossible for the sequence $(4/3)^n$.