In ODE where $x'=f(x(t))$ and $x(0)=x_0$,
Let $S_\delta$= connected component of the set {x in S|$V(x)\leq \delta$} that contains $x_0$. It's also closed.
Lemma: for every $\epsilon>0$ there exists a $\delta>0$ such that:
(1) $B(x_0,\epsilon)$ $\subseteq$ $S_\delta$.
(2)$S_\epsilon$ $\subseteq$ $B(x_0,\delta)$
I want to prove (1) by contradiction using Lyapunov Function $V(x(t))$. My professor asked me to assume that $V'(x)<0$ so that $x_0$ is asymptotically stable and that it's a strictly Lyapunov function ($V(x(t_1))$ $<$ $V(x(t_0))$). He also said $V(x_0)=0$ and that i should fix $\epsilon=1/n$. Because he proved (2) exactly that way!
But I'm still confused as both parts are incomparable. (I don't know about this material outside ODE!)
Thank you for the help!