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:


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 $T$

  • $\begingroup$ You should include a link to Tao's proof. You should also define Per(T). $\endgroup$ – Bill Johnson Mar 27 '13 at 16:31
  • $\begingroup$ Sorry, $Per(T)=$ periodic points of $T$ $\endgroup$ – user11178 Mar 27 '13 at 17:15

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.

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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 $x$.

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