Does Transitivity imply Recurrence?

102 Views Asked by At

Consider the usual Poincaré recurrence theorem, and the notion of topological transitivity e.g. from here.

Does transitivity (on a compact measurable space) imply recurrence?

Due to different uses of the terms transitivity and recurrence in literature, I could not find a proper answer

1

There are 1 best solutions below

2
On BEST ANSWER

Within the usual definitions in topological dynamics the answer is yes:

Let $(X,d)$ be a complete separable metric space ($X$ compact is allowed here) and let $X$ be perfect (that is, $X$ has no isolated points). Moreover let $f:X \to X$ be continuous, and let $O(x)$ and $\omega(x)$ denote the orbit and the omega limit set of $x \in X$ (with respect to $f$). Within this setting the following theorem of Birkhoff is known:

Theorem: Equivalent are:

  1. $f$ is topological transitive;
  2. $D(f):=\{x \in X \mid \overline{O(x)}=X\} \not= \emptyset$;
  3. $D(f)$ is a dense $G_\delta$-set.

Next, $x \in X$ is called recurrent if $x \in \omega(x)$. As each $x \in D(f)$ satisfies $\omega(x)=X$ we have that each $x \in D(f)$ is recurrent. A dense $G_\delta$-set is "big" in the topological sense (its complement is of first category in the sense of Baire). Having a complement of first category can be considered as the topological analogue to "having a complement of measure $0$". Thus, in view of Birkhoff's Theorem one can say that if $f$ is topological transitive then most points are recurrent.