AODE: Compact orbit iff periodic I've been trying to do the following question for a while and have tried following the hint. I have done everything apart from showing that the forward orbit of $z$ is equal to the orbit of $z$. Would it be possible to get another hint for this last step?
In the question $O$ stands for the orbit of a element and $Ω$ stands for the $\omega$-limit set of a point.
Let $f : E → R^N$ be a locally Lipschitz function on an open set $E ⊂ R^N$ . We
consider the autonomous ODE $x' = f(x)$.
Show that if $a ∈ E$ has a compact orbit, then it is periodic.
Hint: The standard properties of limit sets implies $O(a) = Ω(a) = \overline{O^+(a)} = O(z)$ for any $z ∈ Ω(a)$, which yields $O^+(z) = O(z)$. Conclude.
 A: From your last comment I guess that what you need is to show that
\begin{equation*}
O(a) = \Omega(a) = \overline{O^{+}(a)} = O(z) \quad \text{for any } z \in \Omega(a).
\end{equation*}
Denote by $a \cdot t$ the action of the flow of the ODE.
Since
$$
\Omega(a) = \bigcap_{t \ge 0} \overline{O^{+}(a \cdot t)}
$$
and $O(a)$ is compact, hence a closed set, we have
$$
\Omega(a) \subset O(a).
$$
But $O(a)$ is a minimal set (that is, a compact invariant set having no compact invariant proper subsets), so, by the invariance of $\Omega(a)$,
$$
\Omega(a) = O(a).
$$
Further,
$$
\overline{O^{+}(a)} = O^{+}(a) \cup \Omega(a) = O(a).
$$
Finally, take a $z \in \Omega(a)$.  As $\Omega(a) = O(a)$, there is $\tau \in \mathbb{R}$ such that $z = a \cdot \tau$.  Therefore $O(z) = O(a)$.
Notice that the above is indeed a proof in the theory of dynamical systems.  You should add the tag ds.dynamical-systems.
EDIT: I gave a proof of the auxiliary result, but now I am not sure how it contributes toward solving the original problem: notice that we have only that $O(a)$ is equal to the closure of $O^+(a)$, not to $O^+(a)$ itself.  I googled M. C. Irwin's Smooth Dynamical Systems: the proof on p. 45 seems to be much more complicated, using facts from general topology that are quite simple, but rather not generally known.
