Let $X$ be a compact metrizable space which is perfect (i.e. no point is isolated). Let $f$ be a topologically transitive homeomorphism, meaning there is some point $x$ such that the set $\{ f^n (x) ; n \in \mathbb{Z} \}$ is dense. Then there is some $y$ such that the set $\{ f^n (y) ; n \geq 0 \}$ is dense.

This exercise is from Katok and Hasselblatt's book "Introduction to the modern theory of dynamical system". I've been trying hard but I couldn't work it around. Could you give me some tip?


HINT: Let $X$ be such a space, and let $f$ be a transitive homeomorphism. Let $\mathscr{B}$ be a countable base for $X$, and for each $B\in\mathscr{B}$ let

$$U(B)=\bigcup_{n\ge 0}f^{-n}[B]\;;$$

clearly $U(B)$ is open. Let $G=\bigcap_{B\in\mathscr{B}}U(B)$; $\mathscr{B}$ is countable, so $G$ is a $G_\delta$-set in $X$.

  • Show that if $x\in G$, then the positive semiorbit of $x$ under $f$ is dense in $X$.

  • Show that each $U(B)$ is dense in $X$, and apply the Baire category theorem to conclude that $G\ne\varnothing$. A further hint for this part is spoiler-protected below.

For any $B_0,B_1\in\mathscr{B}$, there must be an $n\ge 0$ such that one of $f^n[B_0]\cap B_1$ and $B_0\cap f^n[B_1]$ is non-empty. There must be an $m>n$ such that $x$ and $f^m(x)$ both belong to that set, since every non-empty open set must contain infinitely many points of any dense set.

  • $\begingroup$ @timofei: You’re very welcome. $\endgroup$ – Brian M. Scott Jan 25 '13 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.