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(This has been edited)

I need a proof or a counter-example of this proposition:

Let $\phi$ be a flow in $\mathbb{R}^2$, such that $\phi$ has at least one non-wandering point, then $\phi$ has at least one bounded orbit.

Thanks in advance.

Ok... So that before was my question. I think I have a proof of this, please let me know what you think. Let $\phi$ be a flow in $\mathbb{R}^2$ such that $\phi$ has a non-wandering point $z$. I am going to prove that $\phi$ has a bounded orbit.

Let $S$ be a transversal section through $z$, since $z$ is non-wandering, there exists $y$ such that $O(y,\phi)$ (y's orbit ) intersect $S$ at $t_0$ and $t_1$ such that $t_0 < t_1$. Let $p_0 = \phi_{t_0}(y)$ and $p_1 = \phi_{t_1}(y)$.

We can now construct a curve $\alpha$ by concatenating $(\phi_t(y))_{t\in[t_0,t_1]}$ and the segment $[p_0,p_1]$ at $S$. By Jordan's Theorem we have that $\alpha$ divides the plane in two connected components $U_0,U_1$ and lets say $U_0$ is the one that contains $(\phi_t(x))_{t \in (0,\epsilon)}$ for all $x \in (p_0,p_1)$ and a very small $\epsilon$ (I can assure this statement makes sence because of $\phi$'s continuity).

If $y$ is periodic then $y$ has a bounded orbit and we are done. if not, having the previous definitions it is clear that $O^+(p_1,\phi) \subset U_0$.

Then $\overline{O^+(p_1,\phi)} \subset \overline{U_0}$ is a compact set, then we have that $\omega(p_1,\phi)$ is connected. Then it is either a point or a closed curve, if it is a point then it has to be a singular point, the its orbit is bounded.

If it is a closed curve $\beta$ then it divides the plane in two connected components $V_0,V_1$, lets say $V_0$ is the one included in $U_0$, then any orbit of a point in $V_0$ has to be bounded. This completes the proof.

Hope it makes sense, looking forward to any feedback.

Bajo Fondo
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1 Answers1

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Your proof is mostly correct, but it has a small hole and a small inefficiency.

First the hole: the Jordan curve theorem tells you that one of $\overline{U_0}$, $\overline{U_1}$ is a bounded set and the other is unbounded, and you made an unwarranted assumption that $\overline{U_0}$ is the bounded one. But that is not hard to repair: if $\overline{U_0}$ is the unbounded one, simply reverse the direction of the flow and then the roles of $U_0,U_1$ are switched; the reversed flow has a bounded orbit if and only if the original flow does.

Second the inefficiency: once you know that $\overline{O^+(p_1,\phi)} \subset \overline{U_0}$, it follows that $\omega(p_1,\phi)$ is bounded (you must also show it is nonempty). Also, $\omega(p_1,\phi)$ is a union of orbits (this is always true, for any flow and any point). Hence, the orbit of any point in the set $\omega(p_1,\phi)$ is bounded.

Lee Mosher
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