Let be $T:X\to X$ a topological dynamical system, $X$ a compact space and $T$ is also a isometry. Let be $\mathcal{R}(T)$ the chain recurrent set of $T$.


**Theorem**: $\mathcal{R}(T)=X$


There is a simple demonstration of this fact?
I know a proof of this result, which I read in the Terence Tao's blog:


<http://terrytao.wordpress.com/2008/01/24/254a-lecture-6-isometric-systems-and-isometric-extensions/>


but hoped that this demonstration was simpler.


I also would like a simple example, if possible, a compact dynamical system $ T $ satisfying:


$$
\varnothing \neq \overline{Per(T)}\subsetneq \mathcal{R}(T)=X
$$


**Obs**: $Per(T)$= Periodic points of $T$



Chain Recurrent Set of a Isometry
Given $x \in X$, consider the bi-infinite sequence $T^n(x)$. Given $\epsilon>0$, the space $X$ is covered by a finite collection of sets of diameter $<\epsilon$, and by the pigeonhole principle one of those sets contains infinitely many entries of the sequence $T^n(x)$. It follows that for every $M > 0$ there exists $i < j$ such that $j-i \ge M$ and $d(T^i(x),T^j(x)) < \epsilon$, so $d(x,T^{j-i}(x)) < \epsilon$ and so
$$x, T(x), T^2(x), \ldots, T^{j-i}(x)
$$
is an $\epsilon$-chain from $x$ to itself.


