Show that the solution of an IVP eventually leaves a compact set I am reading a proof which states that a solution $(t,x(t))$ to an IVP ($x' = f(t,x), x(t_0) = x_0$), where $f$ is a function from an open set $U$ to $\mathbb{R^n}$ that is locally Lipschitz continuous in its second argument, will leave any compact set $C$ (contained in $U$) as $t$ approaches $\beta$. Here $(\alpha, \beta)$ is the maximal interval where a solution exists.
There are a few things in the proof that confuses me. I understand (or at least I think I do) that it relates to proving that there exists a solution $ x(\beta)=\zeta$, so that by Picard-Lindelof there will exist a neighborhood of $\beta$ where a solution exists, contradicting the assumption of $(\alpha, \beta)$ being the maximal interval of existence.
So the proof is done by contradiction. We assume that there is a sequence $(t_k, x(t_k)) \in C, \forall k,$ where $t_k \rightarrow \beta$. And then there is also a convergent subsequence, which we will also denote by $(t_k, x(t_k))$, where $$(t_k, x(t_k)) \rightarrow (\beta, \zeta) \in C $$ as $k \rightarrow \infty$. Since $U$ is open, there exists an $\epsilon > 0$ such that $R = [\beta, \beta+\epsilon] \times [\zeta-\epsilon, \zeta+\epsilon]$ in $U$. We let $M$ denote the maximum value of $|f|$ in $R$.
So far I believe that I have been able to follow the proof. But then it says that by Picard-Lindelof, that the solution passing through $(t_k, x(t_k))$ exists as long as the cone $$|x-x(t_k)| \leq M|t-t_k|, t\geq t_k $$ remains in $R$. And that this is true for some $t > \beta$ as long as we take $(t_k, x(t_k))$ close enough to $(\beta,\zeta)$. And that this gives the contradiction.
I don't quite understand what is meant by the cone remaining in $R$. And since we already know that $x(t_k) \rightarrow \zeta$ as $k \rightarrow \infty$, isn't that enough to conclude that $x(t) \rightarrow \zeta$? 
 A: No, $x(t_k)\to\xi$ does not imply the convergence of the solution, as that could oscillate like for example $\sin(1/(β-t))$.

The proof of Picard-Lindelöf has 3 stages: In the first, some compact cylinder is fixed and on this the quantities $M=\max|f|$ and the Lipschitz constant $L$ "computed". In the second stage the "butterfly" set centered at the initial condition is constructed, in the third this is further constricted using the Lipschitz constant to gain contractivity of the Picard operator.
Here we consider simultaneously initial value problems for data $(t_k,x(t_k))$ for $k$ large enough. As first stage set take the rectangle $R=[β-δ,β+δ]×[ζ−ϵ,ζ+ϵ]$ inside the domain, note that the time interval is two-sided and in its size decoupled from the space extension. For the simultaneous second stage shrink $δ$ so that $Mδ\leϵ$. Then choose $N$ so that $t_k>β-δ/3$ and $|x(t_k)-ζ|<ϵ/3$ for $k\ge N$. Now every box $[t_k-δ/2,t_k+δ/2]×[x(t_k)−ϵ/2,x(t_k)+ϵ/2]$ for the $k$-th IVP satisfies the "butterfly" condition for the second stage. This means that the third stage gives an interval radius $h$ that is independent of $k\ge N$ so that a local solution of the $k$-th IVP on $[t_k-h,t_k+h]$ exists. Now let $k\to \infty$, $t_k\toβ$ to get a contradiction to the non-extensibility of the maximum domain.
