# $\omega(x)=\mathbb{R}^2$ is false.

Let $\phi$ be a flow on $\mathbb{R}^2$ coming from $C^1$ class function. I want to prove that there is no point $x \in \mathbb{R}^2$ such that $\omega(x)=\mathbb{R}^2$ (where $\omega (x) := \{y \in \mathbb{R}^2 \ | \ \exists t_{n} \to \infty : \text{ }\phi(t_{n},x) \to y \}$).

It is an implication from Poincare-Bendixson theorem?
(I've studied only this version:
$\emptyset \neq \omega (x)$ - bounded $\Rightarrow \omega (x)$ has an equilibrium point or is a closed orbit.)

Thank you for help.

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The definition of $\omega(x)$ is not clear to me. Can you write it in words ? Is it the set of all limit points of x under the action of $\phi$ –  nonlinearism Feb 20 '13 at 16:10
Yes_______________ –  user63123 Feb 20 '13 at 17:00

Chapter VII of Ordinary Differential Equationa by Hartman is a very thorough account of the Poincaré-Bendixson theory. You can certainly find the (expected) answer there. But it may be more instructive to prove it directly, not as a part of a more complicated construction.

Pick a point where the field does not vanish. Choose a neighborhood of this point in the form of a quadrilateral where the top and bottom sides are trajectories of the field, and all trajectories in between traverse the box from left to right. The trajectory we are interested in is marked in blue below. We can assume that it enters the box at some moment (this is the curve marked 1), otherwise there is nothing to prove. The curve marked 2 shows the first time the trajectory re-enters the box after traveling through 1.

(Of course the picture could be upside down.)

The trajectory must connect the exit point of 1 to the entrance point of 2, without entering the box in the meantime. There are two ways to do so: here is one

and here is the other.

Now look at the space left in the box between the curves 1 and 2. In either scenario, the trajectory can neither enter this space after traveling through 2, nor could it be there before traveling through 1.

What this shows is that every trajectory is nowhere dense. Consequently, $\omega(x)$ has empty interior.

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Great! Thank you very much. –  user63123 Feb 20 '13 at 16:58