A basic question on omega limit sets equilibrium points Consider the following O.D.E $$\dot{x}(t)=h(x(t))$$ with $h$ being lipschitz. consider a trajectory of it. Assume that its omega limit sets are finite. Then I have read in a paper that its  omega limit set necessarily consists of equilibrium points of the O.D.E. Why ?
 A: Suppose the $\omega$ limit set of a trajectory has an isolated point $p$.  I claim $p$ is an equilibrium point.  
Proof by contradiction: 
Suppose $p$ is not an equilibrium point.  Take some $q \ne p$ such that the solution with $x(0) = p$ has $x(T) = q$, where $T > 0$.  Take $0 < \epsilon < |p - q|/2$, and $0 < \delta < \epsilon/2$ such that if $|x(0) - p| < \delta$ then $|x(T) - q| < \epsilon$.
Thus every solution starting (at time $0$) inside the circle $C$ of radius $\delta$ centred at $p$ is outside $C$ at time $T$.  By the Intermediate Value 
Theorem, it must be on the circle at some time in the interval $(0,T)$.
Now since $p$ is an $\omega$ limit point of your trajectory $x(t)$, there are arbitrarily large times at which the trajectory is inside the circle.
By the previous paragraph, there are arbitrarily large times $t_n$ at which 
the trajectory is on the circle.  A limit point of $x(t_n)$, which exists
by compactness of $C$, is then an $\omega$ limit point of the trajectory.
So the trajectory contains other $\omega$ limit points arbitrarily close to $p$, contradicting the assumption that $p$ is isolated.
