Consider the flow $\varphi: \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}^n$ and $L_{\omega}(x)$ the $\omega$-limit set of a point $x \in \mathbb{R}^n$. How can I show that if $L_{\omega}(x)$ is compact, then it is connected? I think one should assume it is connected and then get an absurd. Some help here would be nice.
Also, how can a limit set not be compact or connected? All the (common) examples I can think are compact and connected. Can someone give an exemple?
So... After some more thinking I got a proof.
Assume $L_{\omega}(x)$ is not connected. So there are disjoint open sets $A,B\in \mathbb{R}^n$ such that $L_{\omega} \subset A \cup B$ and $A \cap L_{\omega},B\cap L_{\omega}$ are non empty. Therefore, there are sequences $\{t_n\},\{s_n\}$ such that $\displaystyle\lim_{n \to \infty} t_n=\lim_{n \to \infty} s_n=\infty$, $t_n<s_n<t_{n+1}$ and $\varphi_{t_n}(x)\in A, \varphi_{s_n}(x)\in B$ for all $n$. Now $\{\varphi_t(x); t\in (t_n,s_n)\}$ is a curve joining a point in $A$ to a point in $B$. Then there is a $r_n\in (t_n,s_n)$ such that $\varphi_{r_n}(x) \notin A\cup B$. Note that $\displaystyle \lim_{n \to \infty}r_n=\infty$.
We know that $\{\varphi_{r_n}(x)\}\subset L_{\omega}(x)$. Since it is compact, there is a subsequence of $\{\varphi_{r_n}(x)\}$ converging to a point $y \notin A \cup B$. But that means that $y \in L_{\omega}(x)$. This contradiction completes the proof.