We have the following definitions:
Let $X$ be a metric space, $T:X\rightarrow X$ continuous. We say that $x\in X$ is periodic if there exists $n_0\in \Bbb{N}$ s.t. $T^n(x)=x$ .
We say $x\in X$ is uniformly recurrent if for all $\epsilon >0~$ $\exists ~n_k\rightarrow \infty$ as $k\rightarrow \infty$ s.t. $\sup_{k} |n_{k}-n_{k+1}|<\infty$ and $d(T^{n_k}x,x)<\epsilon$.
Now we have seen the so called Thue-Morse sequence and proved that this is a sequence which is uniformly recurrent but not periodic.
Then I asked myself if it is true that periodicity implies uniformly recurrence?
My guess would be no, since in the above definition of uniformly recurrence one need $n_k\rightarrow \infty$, I think this is the "problem" that we do not have "periodicity $\Rightarrow$ uniformly recurrence". Because if we wouldn't have this assumption one could take $n_k=n_0$ and would be done, but with this assumption it is not possible to prove that "periodicity $\Rightarrow$ uniformly recurrence" or am I wrong?
Thanks a lot for your help.
The answer is "yes, periodicity implies uniform recurrence".
It suffices to let $n_0$ be as in the definition of periodicity and then define $n_k = kn_0$. Then $|n_k - n_{k+1}| = n_0$ and $d(T^{n_k}x,x) = 0$.