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I'm having an introductory chair of dynamical systems at my faculty. I've read the teacher's notes and searched the internet, but can't seem to find if the following affirmations are true or false:

(1) If a homeomorphism of a compact metric space has a dense orbit, then it has no periodic orbits.

(2) There exists a homeomorphism of $\mathbb{R}$ with dense orbit.

(3) If $f:\mathbb{S}^1 \longrightarrow \mathbb{S}^1$ is a homeomorphism that inverts orientations, then $f$ has two fixed points.

Any help would be greatly appreciated.

Jean Marie
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Diogo Santos
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1 Answers1

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  1. False. Just take $f : \lbrace 0 \rbrace \rightarrow \lbrace 0 \rbrace$ defined by $f(0)=0$.

  2. False. A homeomorphism $f$ of $\mathbb{R}$ is monotonic. If $f$ is increasing, then any sequence of the form $(f^n(x))$ is monotonic and hence cannot be dense ; if $f$ is decreasing, then the sequences $(f^{2n}(x))$ and $(f^{2n+1}(x))$ are monotonic and hence the sequence $(f^n(x))$ cannot be dense.

  3. True. Consider a lift $F$ of $f$. You have $F(0)-0 = (F(1)-1) + 2$, so you get two fixed points.

TheSilverDoe
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  • In point 2, how can we conclude that $(f^n(x))$ is not dense by the monotonicity of the subsuccessions? – Diogo Santos Feb 04 '22 at 14:01
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    @DiogoSantos Just use the fact that if $A \cup B$ is dense in $E$, then either $A$ or $B$ is dense in $E$ (with $A=\lbrace f^{2n}(x), n\in \mathbb{Z} \rbrace $ and $B=\lbrace f^{2n+1}(x), n\in \mathbb{Z} \rbrace $). – TheSilverDoe Feb 04 '22 at 15:12