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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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    $\begingroup$ I'm not sure that your definition of $\omega$-limit set of a set is standard. As defined in, e.g., p. 29 of C. Conley's "Isolated invariant sets and the Morse index" (1978), the $\omega$-limit set of a set $X$ is not simply the union of the $\omega$-limit sets of the points in $X$; in general it strictly contains the union. $\endgroup$
    – Matthew Kvalheim
    Commented Apr 15, 2020 at 18:45

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

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    $\begingroup$ See the flow in the pucture (at freeimagehosting.net/image.php?9c8be08dae.jpg). 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. $\endgroup$
    – Pengfei
    Commented Jan 21, 2011 at 12:28
  • $\begingroup$ I mean, that is one of the examples mentioned by Michael. $\endgroup$
    – Pengfei
    Commented Jan 21, 2011 at 12:30
  • $\begingroup$ How smooth is the flow in this case? I.e. is it $C^1$? $C^2$? $\endgroup$
    – Vincenzo
    Commented Jan 24, 2011 at 10:21

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