The $\omega$-limit set of a trajectory of an autonomous ODE is an invariant set for the ODE Consider the following ODE 
$$x'(t) = h(x(t)), x(0) = \bar{x} $$ where $h$ is lipschitz
For some solution trajectory $x(t)$ its omega limit set is defined as $=\bigcap_t \overline{\{x(s): s> t\}}$. I want to prove that it is an invariant set for the ODE.
i.e. for any other solution of the ODE $y(.)$ if there exist an increasing sequence $\{t_n\}$ tending to infinity as $n$ tends to infinity and $x(t_n) \to  y(0)$ then I need to find an increasing sequence $\{t'_n\}$ (depending on $t$) tending to infinity as $n$ tends to infinity such that  $ x(t'_n) \to y(t) $ . $x(.)$ and $y(.)$ are unrelated other than they are solutions to the same ODE. How to proceed ?
 A: You need to show that, given the system $\dot x=h(x)$ its omega limit set is invariant. Literally it means that if $y\in \omega$-limit set than the solution $x(t;y)$ is also in omega limit set. We have that there exists a sequence $(t_k)$ such that $x(t_k;x_0)\to y$. Now for a fixed $t$ consider $x(t+t_k;x_0)$. By the properties of the solution $x(t+t_k;x_0)=x(t;x(t_k;x_0))\to x(t;y)$, which actually implies that $x(t;y)$ is in omega limit set.
A: Edit: I doubt that this is the cleanest solution, but it seems to work.
Here I assume, by the way, that $h$ is Lipschitz, which by Picard-Lindelof is enough to guarantee existence and uniqueness for any given initial condition; in fact I don't think this hypothesis can be lifted wholly.
Let $\bar{y}$ be an element of the omega limit set $S_x := \bigcap_t \overline{\{x(s) : s > t\}}$, say, it is the limit $\lim_{n \to \infty} y(t_n)$, and let $y(t)$ be the solution such that $y_0 = \bar{y}$.
First, if $h(\bar{y}) = 0$, then $y(t) = \bar{y}$ is a (constant) solution, and trivially the solution stays in $S_t$ for all $t > 0$.
So, henceforth suppose not. If $h(\bar{y}) > 0$ is positive, so that $h$ is positive on some interval $I := (\bar{y} - \epsilon, \bar{y} + \epsilon)$. Then, for sufficiently large $n$, $x(t_n) \in I$; for each $n$, if $x(t_n) \leq \bar{y}$, then there is some small $\Delta_n \geq 0$ such that $x(t_n + \Delta_n) = \bar{y}$. If the reverse inequality holds, the same argument applies flowing backward in time. A similar argument holds if $h(\bar{y}) < 0$. Either way, we get a sequence of times $\tau_n := t_n + \Delta_n$ that tends to infinity and such that $x(\tau_n) = \bar{y}$. 
Now, any value $y$ achieves after $0$ can be written as $y(\alpha)$ for some $\alpha$. Since the differential equation is homogeneous, $x(t + \tau_n)$ is also a solution to the IVP with $y(0) = \bar{y}$ for all $n$; in particular $x(\alpha + \tau_n) = y(\alpha)$ for all $n$, but the times $\alpha + t_n$ also tend to infinity, and so $y(\alpha) \in S_x$. Since $\alpha$ is arbitrary, $y(t) \in S_x$ for all $t \geq 0$.
