Let $f:\mathbb R^n\to\mathbb R$ be complete and $\dot x=f(x)$ the associated dynamical system with flow $\Phi: \mathbb R\times\mathbb R^n\to\mathbb R$. Then, for $x_0\in\mathbb R^n$ we have:
If $\psi: t\mapsto\Phi(t,x_0)$ is bounded, then $\omega(x_0)$ is not empty.
How do I prove that?
I know that it is a direct proof using some kind of compact argument. How do I obtain such an argument, though, if $\|\Phi(t,x_0)\|<\infty\ \forall t\in\mathbb R,x_0\in\mathbb R^n?$
Hint: by contradiction.
Suppose that $\| \Phi(t, x_0) \| \leqslant R$ and for each $x$ with $\| x \| \leqslant R$ there is an open neighborhood $x \in U_x$ and $t_x \in \mathbb{R}$ such that $\psi \big[ [t_x, \infty) \big]$ is disjoint from $U_x$. Then $\{ U_x : \| x \| \leqslant R \}$ is a cover of $\{ x : \| x \| \leqslant R \}$. Take a finite subcover...