Chain Recurrent Set of a Isometry 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 $T$
 A: 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.
A: An example of the type you're asking for is given by the following transformation of the circle (considered as the set $[0,1]/\sim$, where $\sim$ identifies 0 and 1: $T(x)=x^2$. 
The only periodic point is 0. On the other hand, every point is chain recurrent, because given $x\in(0,1)$, and $\epsilon > 0$, there exist $m$ and $n$ such that $T^m(x)=x^{2^m}<\epsilon/2$ and $T^{-n}x=x^{2^{-n}} > 1-\epsilon/2$.
Now the orbit $\overline{(x,Tx,\ldots,T^mx,T^{-n}x,\ldots,T^{-1}x)}$ is $\epsilon$-chain recurrent and includes the point $x$.
