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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?

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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.

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

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