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Let $\sigma$ be a homeomorphism of $S^1$. Then the following statements are equivalent;

(1) O(z) is dense in S for some z in S,

(2) O(z) is dense in $S^1$ for every z in $S^1$,

(3) $\sigma$ is conjugate to $R_{\theta}$ for some irrational number $\theta$, (0<$\theta$<1/2).

When $\sigma$ satisfies the condition (1) or (2), the rotation $R_{\theta}$ in (3) is uniquely determined.

(where for a real number $\theta$, we denote by $R_{\theta}$ the rotation: $R_{\theta}(e^{2\pi ix})=e^{2\pi i(x+\theta)}$ on $S^1$ and orbit O(z) :={$\sigma^n(z):n \in \mathbb{Z}$}).

In this theorem I have proved that (1) implies (3) and (2) implies (1) but unable to prove (3) implies (1) since to prove that these are equivalent, we need to prove that (1)=>(3), (3)=>(2) and (2)=>(1).

And the uniqueness of $R_{\theta}$ can be proved by using the set {$e^{2\pi in\theta}|n\in\mathbb{Z}$} of eigenvalues of $R_{\theta}$ but unable to get the result. Please help me in proving the result.

Priya
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  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Did Jul 03 '14 at 07:28
  • sorry sir, i have edited the statement now and also mention the part in which i am facing problem. – Priya Jul 03 '14 at 08:56
  • Sir, please reconsider this mail and reopen it. – Priya Jul 03 '14 at 09:04
  • Done. (Next time, motivate your question from the start.) – Did Jul 03 '14 at 09:18
  • @Did thanks sir. – Priya Jul 03 '14 at 10:28
  • Duplicate: http://math.stackexchange.com/q/282102/ – Did Jul 04 '14 at 09:25
  • @Did sir in this answer, $\sigma$ is not used anywhere. Is there not any effect of this homeomorphism on the denseness of orbits? – Priya Jul 04 '14 at 10:58
  • Are you asking me a question about user @frogpie's answer? What is going on? Is this pure spam? – Did Jul 04 '14 at 12:16
  • @Did sorry sir, but I am just asking it to you because the address math.stackexchange.com/q/282102 send by you is the same as that by frogpie. – Priya Jul 04 '14 at 17:23

1 Answers1

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If I understand you correctly, you have proved (1) implies (3) and (2) implies (1), so what remains is (3) implies (2). For this, have a look at this answer.

Community
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froggie
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  • sir in this answer, $\sigma$ is not used anywhere. Is there not any effect of this homeomorphism on the denseness of orbits? – Priya Jul 04 '14 at 10:57
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    @Priyanka: The given answer proves that the group $H_1=\langle R_\theta\rangle$ has only dense orbits. If you assume (3) then the group $H_2=\langle \sigma\rangle$ is conjugate to $H_1$ by some homeomorphism $\tau$ such that $R_\theta=\tau\sigma\tau^{-1}$. It is easy to see that the orbit $H_2(z)=\tau^{-1}(H_1(\tau(z)))$, which is dense as a homeomorphic image of a dense subset. – Jyrki Lahtonen Jul 05 '14 at 12:35
  • @Priyanka: density is a topological property, and all topological properties are preserved under homeomorphism. – froggie Jul 06 '14 at 09:12
  • how are dense subsets related to irrational rotations of pi on the circle? –  Jun 21 '17 at 16:45