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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.

THIG
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    Take an element $x(-t)$ ($t>0$) in the backward orbit. As the orbit is equal to the forward orbit, this implies that there exists $T>0$ such that $x(T)=x(-t)$. Therefore, the orbit is periodic. – Severin Schraven Mar 31 '19 at 16:44
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    Thank you so much. But I think my question was a bit misleading I was able to conclude that but it was the step before i was unable to conclude. – THIG Mar 31 '19 at 17:13

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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.

user539887
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    Is it easy to see that the orbit is a minimal set? Anyway +1 for such a nice answer. – Severin Schraven Mar 31 '19 at 23:04
  • @SeverinSchraven Thank you. The orbit is a compact set, by assumption. For any proper subset $A$ of $O(a)$ there are $t_1,t_2\in\mathbb{R}$ such that $a \cdot t_1\in A$ and $a \cdot t_2\in O(a)\setminus A$. But $a\cdot t_2=(a \cdot t_1) \cdot (t_2-t_1)$, so $A$ cannot be invariant. – user539887 Apr 01 '19 at 06:53
  • So compactness is only needed to show $\Omega(a)\subseteq O(a)$. In the minimality we only need invariance of $\Omega(a)$, right? – Severin Schraven Apr 01 '19 at 08:27
  • A minimal set is, by definition, a compact invariant set such that there does not exist any proper compact invariant subset of it. – user539887 Apr 01 '19 at 11:05
  • @SeverinSchraven See my edit. – user539887 Apr 02 '19 at 07:11
  • Can't we do it this way: If the forward orbit is not closed, then there exists a monotone increasing sequence of times $(t_n)$ with $t_n \rightarrow \infty$ such that $t_n \cdot a \rightarrow b\notin O^+(a)$. However, then $b\in O(a)$ (as $O(a)$ is closed) is a fixed point by the semigroup property of the flow. Which implies that the orbit actually consist only of this point. – Severin Schraven Apr 02 '19 at 12:24
  • Why should $b$ be a fixed point? And we have that any $c \in O^-(a)$ is the limit of a sequence $a \cdot t_n$. If $t_n \to \tau\in[0,\infty)$ we are done. But if $t_n\to\infty$ I don't see contradiction here. – user539887 Apr 02 '19 at 18:29