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Consider a dynamical system given by its flow $\phi(t,x)$, where $t \in R$, $x \in R^n$ and $\phi: R \times R^n \to R^n$ is (say) differentiable.

The $\omega$-limit set, $\omega(p)$, of a point $p \in R^n$ is the set of all $q \in R^n$ such that there exists a sequence $t_0,t_1,\dots$ with $t_n \to \infty$ and $$\lim_{n \to \infty} \phi(t_n, p) = q.$$

The $\omega$-limit set of a set $X$ is simply the union of the $\omega$-limit sets of the points in $X$.

It is not too hard to show that for any $X$, $\omega(\omega(X)) \subseteq \omega(X)$.

However in all examples that I can think of, it always holds that $\omega(\omega(X)) = \omega(X)$. Is it possible to prove that, or is it false?

I guess either a proof or a counterexample should be already known, but I can't locate either.

(Note: I already asked a related question here on MO, but the question above is more specific and not answered there.)

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up vote 5 down vote accepted

Stable homoclinic (or heteroclinic) loops in the plane provide a simple counterexample.

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See the flow in the pucture (at Let $X$ be the closed cornered ellipse. Then $\omega(X)$ is the union of two arcs and the center point $o$. And $\omega(\omega(X))$ consists of just three points. – Pengfei Jan 21 '11 at 12:28
I mean, that is one of the examples mentioned by Michael. – Pengfei Jan 21 '11 at 12:30
How smooth is the flow in this case? I.e. is it $C^1$? $C^2$? – Vincenzo Jan 24 '11 at 10:21

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