# Omega-limit-set is non empty

The context is an autonomous ODE $\dot{x}=f(x)$.

Let $x_t$ be an orbit which does not tend to $+\infty$ as $t\to\infty$. Then its $\omega$-limit set $\Omega(x_0)$ is non empty.

Concerning the notation, I only can speculate that $x_t$ sometimes is meant as an orbit and sometimes as points on the orbit (such as $x_0$).

The proof we wrote down only was:

It exists a bounded sequence $\left\{x_{t_m}\right\}, t_m\to +\infty$.

That was all...

As far as I see we have to show that there is some sequence $\left\{t_m\right\}$ such that $$\lim_{m\to\infty}x_{t_m}=x_0,$$

where here $x_{t_m}$ and $x_0$ are meant as points on the orbit. But do not see how to show.

• Do you know the defintion of $\Omega(x_0)$? – Lee Mosher Oct 30 '15 at 14:33
• Concerning the notation, the orbit $(x_t)$ is often defined for $0 \le t < +\infty$, and so when $t=0$ one obtains $x_0$ which is the initial value of the orbit and is the same as the $x_0$ in the notation $\Omega(x_0)$ (another possibility is that $(x_t)$ is defined for $-\infty < t < +\infty$). – Lee Mosher Oct 30 '15 at 14:33
• Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. – Julián Aguirre Oct 30 '15 at 14:36
• We only defined the $\omega$-limit set for $x_t$, i.e. for the orbit I think. I do not know exactly how this is connected with $\Omega(x_0)$. Maybe when writing $\Omega(x_0)$ what one means is nothing else but the $\omega$-limit set of the orbit which goes through $x_0$? – M. Meyer Oct 30 '15 at 15:01