0

Maybe this is a foolish question but if we have a dynamical system, for example an ODE, and are talking about orbits (or trajectories), aren't bounded orbits $x_t$ automatically compact sets in $\mathbb{R}^n$?

In other words: Are (bounded) orbits closed sets?

Rhjg
  • 2,029
  • 1
    No, orbits are "usually" not closed. Think of irrational rotation on a torus, the Lorenz system, etc. If instead you're talking about discrete dynamics, an infinite bounded orbit is never closed. – Andrew D. Hwang Jan 06 '16 at 15:11
  • Ok, thanks. And if we add the property that the $\omega$-limit set contains exactly one element? Does this make the orbit closed? – Rhjg Jan 06 '16 at 15:12
  • The limit set is not part of a non-recurrent orbit, but is in the closure. – Andrew D. Hwang Jan 06 '16 at 15:14
  • I am just wondering since I was told to use compactness of the bounded orbit $x_t$ to show that $x_t\to\omega^$ in case the $\omega$-limit set contains one element, i.e. $\Omega(x_t)=\left{\omega^\right}$. Now I do not know how to show that the bounded orbit is closed (in order to have compactness = bounded + closed). – Rhjg Jan 06 '16 at 15:18
  • 2
    Use the compactness of the closure of the orbit. – Daniel Fischer Jan 06 '16 at 15:19
  • I try: By $\bar{x_t}$ denote the closure of the orbit $(x_t){t\in\mathbb{R}}$. $\bar{x_t}$ is bounded (since $x_t$ is bounded) and closed (by definition). Hence it is a compact subset of $\mathbb{R}^n$ and, in particular, sequentially compact. That is, each sequence in $\bar{x_t}$ has a convergent subsequence; in particular, each subsequence of $(x_t){t\in\mathbb{R}}$ has a convergent subsequence which has to converge to $\omega^$ since otherwise there would be another element in $\Omega(x_t)$. From this it follows that $(x_t)_{t\in\mathbb{R}}$ converges to $\omega^$ (by analysis 1). Ok? – Rhjg Jan 06 '16 at 15:28
  • @AndrewD.Hwang I always thought of orbits as closed sets: If I consider an open ball around a point on an orbit, the full open ball does not belong to the orbits... where is my mistake? – Rhjg Jan 06 '16 at 15:34
  • 1
    @Rhjg: For your first question, you need to consider sequences for which $t \to \infty$. Otherwise your argument looks good. To address your second question, think of an interval, say $(0, 1)$, embedded in the plane. The interval is, in a sense, "locally closed". The problem is the missing endpoints. – Andrew D. Hwang Jan 06 '16 at 15:47
  • But if I think f.e. of the orbits of harmonic oscillator they are closed sets? – Rhjg Jan 06 '16 at 16:18

0 Answers0