Question about entropy I see this question on the math stack exchange. I found it interesting and still there is no solution there
Let $(X,A,\nu)$ be a probability  space and $T:X\to X$ a  measure  preserving transformation $\nu$. Take a measurable partiton $P=\{P_0,...,P_{k-1}\}$. Let I be a set of all possible itineraries, that is,  I=$\{(i_1,...,i_n,...)\in k^{N}$; there is a $x\in X$, such that $T^n(x)\in P_{i_n}$ for all n$\in$N}
Suppose that I is  countably infinite.
Is true that the entropy of T with respect to P is $0$  ($h(T,P)=0)$?
 A: This condition is much more restrictive than just having zero entropy. In fact you are talking about a purely atomic invariant measure of the shift on $A^{\mathbb Z_+}$, where $A$ is a finite alphabet. Any such measure is a (possibly infinite) convex combination of measures equidistributed on orbits of periodic points. 
EDIT As the exposition of Daniel contains a number of mistakes, let me expand my answer.


*

*The partition $P$ gives rise to a symbolic coding of the original transformation by assigning to any point $x\in X$ the sequence of $P$-names of $(T^n x)_{n\ge 0}$, i.e.,
$$
x \mapsto (i_0,i_1,\dots) \;,
$$
where $i_n$ is determined by the condition $T^n x \in P_{i_n}$. The image of the original measure on $X$ under this transformation is a shift invariant measure. 

*Now one can forget about the original measure and deal just with the quotient shift invariant measure on $A^{\mathbb Z_+}$ (where $A$ is the set of elements of the partition $P$). By the assumption it is purely atomic. I claim that any purely atomic probability measure $m$ invariant under a certain transformation $T$ is a convex combination of uniform measures on periodic orbits. Indeed, let $x$ be a maximal weight atom of $m$. Then $m(Tx)\le m(x)$ because the weight of $x$ is maximal; on the other hand, $m(Tx)\ge m(x)$ because $x\in T^{-1}(Tx)$ and $T$ is measure preserving. Therefore $m(Tx)=m(x)>0$, so that $x$ must be a periodic point. Now one can remove the orbit of $x$ and apply this argument again, etc.

*The entropy of a periodic measure preserving transformation is zero. 
A: I think that I can expand on RW's answer
If $I$ is countably infinite, then we can cover $X$ with the countable partition
$$ X = \vee_{n=0}^\infty T^{-n}P $$
Take any element of this partition:
$$Q = P_{n_0} \vee T^{-1}P_{n_1} \vee \cdots \vee T^{-i}P_{n_i} \vee \cdots $$
where $n_i \in [1,k]$. Then 
$$
TQ = TP_{n_0} \vee P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$
is contained within partition element
$$ P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$ 
so $TQ$ (or more generally $T^iQ$) is either disjoint from $Q$ (when it is contained within another partition), or equal to $Q$
There are two possibilities for the measure of $Q$:
(1) the measure of $Q$ is zero, or
(2) the measure of $Q$ is positive, and $T^nQ=T^{n+k}Q$ for some $n,k < \infty$. Otherwise $\{T^iQ\}$ is an infinite sequence of disjoint sets, each of constant measure $\mu(Q)$, and
$$ \infty = \sum_{i=1}^\infty \mu(Q) = \sum_{i=1}^\infty \mu(T^iQ) \leq \mu(X) = 1$$
Hence there are (possibly infinitely many) $Q$'s of finite measure, each with finite orbit.
Each $Q$ represents the set of points that are indistinguishable in your topology, so we can say that the space you are talking about is really just a (countable) union of finite orbits with an atomic probability measure.
