Poincare recurrence theorem and convergence on compact metric spaces

Question

I am looking for a proof (or a reference to a proof) of the following theorem:

Let $X$ be a compact metric space with metric $d$, endow $X$ with the Borel $\sigma$-algebra and a probability measure $\mu$. Let $T\colon X\to X$ be a continuous map which is $\mu$-preserving. Then for $\mu$-almost every $x\in X$ there is a sequence $n_k\to\infty$ in $\mathbb{N}$ such that $T^{n_k}x\to x$ as $k\to\infty$.

I tried already for some time and looked for references, unfortunately unsuccessful. It seems that one needs to find the right formulation to be able to use the Poincare recurrence theorem. Any ideas?

Answer 1

Let $\mathcal P_1,\mathcal P_2,\ldots$ be a sequence of refining partitions of decreasing diameter (converging to 0) of $X$. Let $\delta_n$ be the diameter of $\mathcal P_n$. Let $\mathcal P_n(x)$ be the element of $\mathcal P_n$ containing $x$. For $\mu$-a.e. $x$, $\mu(\mathcal P_n(x))\ne 0$.

Now the Poincaré recurrence theorem applies to each element (with positive measure) of $\mathcal P_n$, saying that for almost every $z\in B\in\mathcal P_n$, $z$ returns to $B$ infinitely often. Hence there exist infinitely many $k^{(n)}_i$ such that $d(T^{k^{(n)}_i}z,z)\le \delta_n$. Let the union of the exceptional sets for each $B$ in $\mathcal P_n$ be $E_n$. Let $E$ be the union of the $E_n$. Then $E$ has measure 0.

Now just put it all together. Let $E'$ be the set of $x'$ such that $\mu(\mathcal P_n(x))=0$ for some $n$. This is another set of measure 0. Now if $x\in X\setminus(E\cup E')$, we're in business:

$\mu(\mathcal P_n(x))>0$ for all $n$ and $x$ returns to $\mathcal P_n(x)$ infinitely often. A diagonal argument finishes things off.

There may be a reference for this in Furstenberg's book.

Answer 2

This is part of the statement of Proposition 4.1.18 in Introduction to the Modern Theory of Dynamical Systems by Katok and Hasselblatt. As far as I am aware there is essentially only one proof, which is that given by Anthony. It generalises to the following result which I have found useful once or twice but which is less frequently stated:

Proposition: Let $T$ be a measure-preserving transformation of the probability space $(X,\mathcal{F},\mu)$, $Y$ a second-countable topological space, and $f \colon X \to Y$ measurable. Then $$\mu\left(\left\{x \in X \colon f(x) \in \overline{\left\{f(T^nx) \colon n \geq 1\right\}}\right\}\right)=1.$$

Proof: Let $\{U_k\colon k \geq 1\}$ be a countable base for the topology of $Y$. For each $k \geq 1$ we have $$\mu\left(\left\{x \in f^{-1}(U_k) \colon f(T^nx ) \in U_k \text{ infinitely often}\right\}\right)=\mu\left(f^{-1}(U_k)\right)$$ by the Poincare recurrence theorem. Taking an appropriate intersection over all $k$ we find that there is a set of full measure on which every $x$ returns infinitely often to every $f^{-1}(U_k)$ to which $x$ originally belonged. Since $\{U_k\}$ is a base the result follows.

(Your question corresponds to the case $Y:=X$ and $f:=\mathrm{id}$.)

Answer 3

idea: By contradiction. Assume there exists a set $B$ of positive measure such that there is no sequence such that $T^{n_k}x→x$. Then there should be a set $B'$ of positive measure and an $\epsilon>0$ such that for all $x\in B'$, $T^nx$ does not come $\epsilon$ close to itself again. Pick a finite $\epsilon/4$ cover of $B'$. One of the intersections of $B'$ with one of the elements of this cover must have positive measure. Thus we can assume WLOG that $B'$ has diameter smaller than $\epsilon/2$.

Then by Poincare recurrence there exists an $n$ such that $$ B' \cap T^{-n}B' \neq \emptyset. $$ Assume $x\in B' \cap T^{-n}B' $. Then $T^{n}x \in B$, hence it is $\epsilon/2$ close to itself. Contradiction.