I am confused about the limit set of ODEs on bounded sets. In particular, I dont understand the positive invariance of a limit set if the domain of the ODE is not closed. Consider the following:
Let $U \subset \mathbb{R}^n$ be a bounded set and $f:U\to\mathbb{R}^n$ be differentiable on $U$. Let $x(t,x(0))$ be the unique solution to the initial value problem $\dot{x} = f(x), x(0) \in U$. $x(t,x(0))$ is assumed to be positively invariant on $U$. That is, $x(0)\in U \implies x(t)\in U\ \forall t > 0$.
It can be shown that any solution to the IVP, $x(t,x(0))$, exhibits a nonempty, compact limit set which we will denote as $L^+$. Furthermore, $x(t,x(0)) \to L^+$ as $t\to\infty$ (see Nonlinear systems, Khalil).
Positive invariance of the limit set can be argued as follows (see also Omega limit set is invariant):
Consider a solution to the IVP, $x(t;x(0))$ and a sequence $t_n$ such that $t_n\to\infty$, $x(t_n,x(0))\to y\in L^+$ as $n\to\infty$. Due to uniqueness of the solution, $x(t+t_n;x(0)) = x(t,x(t_n))$ for all $n$. As $x(t) \to L^+$, $x(t+t_n;x(0))\to L^+$ as $n\to\infty$. Taking the limit as $n\to\infty$, \begin{equation} \lim_{n\to\infty}x(t+t_n;x(0)) = \lim_{n\to\infty}x(t,x(t_n)) = x(t,y) \in L^+. \end{equation}
My question is this; If $U$ is not closed and $L^+$ is contained within $\bar{U}\setminus U$, where does the above argument break down? It doesn't make sense to me that $L^+$ is positively invariant if the ODE is not defined on $L^+$.
Thanks in advance for any help!
The argument presented does not hold as the flow of the ODE is not necessarily continuous with respect to $x$. This can be the case if $f$ is discontinuous on the closure of $U$. Thus, it is not correct to suppose that \begin{equation} \lim_{n\to\infty}x(t,x(t_n)) = x(t,\lim_{n\to\infty}x(t_n)). \end{equation}