Suppose I have a discrete dynamical system with a finite set X of states, and suppose I want to prove that every state of X ends up, sooner or later, in a subset Z under the dynamics of the system. Then a natural proof strategy is to break this "convergence" statement into two parts, by first showing that every state of X eventually ends up in a subset Y, and then showing that every state in Y ends up in Z (with Z $\subset$ Y $\subset$ X). This might simplify the problem quite a bit. 


My question is,



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> Under what conditions will the above proof strategy work for a continuous dynamical system $\dot{x}=f(x)$ ($x \in R^n$)? 
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The main issue I have is that it might take infinite time to reach Y. To see how things may go wrong if $f$ is not continuous, consider the following example: 


* $\dot{x} = -1-x$, $\quad$ for $x<-1$,
* $\dot{x}=-x$, $\quad\quad$ for $-1 \le x \le 1$, $\quad$ and
* $\dot{x}=1-x$, $\quad$ for $x>1$.


In this case every $x$ in $[-1,1]$ converges to zero, and every point outside $[-1,1]$ converges to either $-1$ or $1$, but it takes infinite time to do so and it is not true that all points in $R$ converge to zero. 



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> Smoothness of $f$ might be enough to rule out this behavior; I am unable to construct a counterexample with $f$ continuous. Is it actually enough, and if so, why? 
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When is convergence transitive? 
If the equation is defined in $\mathbb{R}$, then $f$ continuous is enough. It is clear that if a point of $X$ does not reach the set $Y$ in finite time, then $f$ vanishes in this limit point. Since every point of $Y$ converges to $Z$, you get that this limit point must thus belong to $Z$. 


In higher dimensions, I would say that the same holds only that there is the subtelty that you don't have a priori a dynamical system defined (due to lack of uniqueness of solutions, the group condition fails) however, if $\dot x = f(x)$ integrates into a flow (which must then be a continuous flow, this will hold for example if $f$ is locally Lipchitz) an analogous argument works (now $f$ may not vanish at the omega-limit set of a trayectory, but if $x$ does not enter $Y$, then it will converge to a point in $Z$ anyway). 


