The $\omega$-limit set is defined in this wikipedia article.
My question is: If we have an orbit $x_t$ and the $\omega$-limit set of this orbit contains one element $w^*$, does this imply that the orbit tends to $w^*$ as $t\to\infty$?
In general, I think, this is not correct. I did not find a counter example however.
Anyway, for a bounded orbit $x_t$ this should be fulfilled:
Suppose $x_t$ is a bounded orbit and its $\omega$-limit set contains one element, i.e. $\Omega(x_t)=\left\{\omega^*\right\}$. I think we can assume that the orbits/ solutions live in $\mathbb{R}^n$, i.e. $x_t\subset\mathbb{R}^n$. Since $x_t$ is bounded and closed, it should be sequentially compact, hence each sequence has a convergent subsequence. In particular, each subsequence has a convergent subsequence which has to converge to $\omega^*$ since if it converges to $\omega\neq\omega^*$, then we would have $\omega\in\Omega(x_t)$.
In general, a sequence $(a_n)$ converges to $a$ iff each subsequence of $(a_n)$ has a subsequence that converges to $a$.
Using this statement, we have $x_t\to\omega^*$ as $t\to\infty$ by above.
Do you think this makes sense?