# Omega limit set of omega limit set $\omega(\omega(a))$

Consider a dynamical system with a flow $\phi(t;a)$, and let $A\subset \mathbb{R}^n$. The omega limit set of $A$ is defined as the union of all $\omega(a)$ over all $a\in A$. Since for a given $a$, $\omega(a) \subset \mathbb{R}^n$, you can define $\omega(\omega(a))$.

I'm asked if it's true that $\omega(\omega(a)) = \omega(a)$, and to give either a proof or a counter example.

Intuitively, I don't think this is true, and I think a counter example lies with an orbit that is a loop, however, I can't seem to come up with an example.

Edit: I just found this similar question on Mathoverflow, and it has an answer that seems to be correct. However, the picture that is linked to doesn't work anymore, and I can't seem to figure out what it looked like. Does anyone else?

If discontinuities are allowed, consider the one-dimensional dynamical system $\dot x=F(x)$ where, for every integer $n$, $F(x)=n-x$ for every $n\lt x\leqslant n+1$. Then $\omega(n)=\{n-1\}$ for every integer $n$, hence $\omega(\omega(2))=\{0\}\ne\{1\}=\omega(2)$.

If one imposes that $F$ is continuous, one must go to two-dimensional systems. Consider the part $C$ of the curve $(x^2+y^2)^{3/2}=xy$ in the positive quadrant, see figure below.

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Assume that:

• The origin $O$ is a fixed point
• For every starting point on $C\setminus\{O\}$, the dynamical system follows $C$ clockwise at positive speed except at $O$ where the speed is zero
• There is a fixed point $Q$ inside $C$
• The trajectories starting from any point inside $C$ except $Q$ accumulate on $C$, spiralling clockwise

Then, for any $P\ne Q$ inside the loop $C$, $\omega(P)=C$ and $\omega(C)=\{O\}$ hence $\omega(\omega(P))\ne\omega(P)$.

• Oops, I'm sorry I forgot to mention that discontinuities aren't allowed. – RBS Dec 1 '14 at 14:27