extenstion theorem of solution for ode in hartman 
This is first theorem of extension theorem of solution of ode in hartman.
I wonder the exact meaning of last part "thus $(b_1,y(b_1)),(b_2,y(b_2)),...$ is unbound or has a cluster point in $\partial E$".
I read this part as "if the sequence $(b_1,y(b_1)),(b_2,y(b_2)),...$ is bounded, they have convergent subsequence, so this part imply each convergent subsequence can be converge other points, but their limit point should be in $\partial E$ ".
but I cannot have convince about my understanding.
"Maybe that sequence converges?" if we can say  $(b_1,y(b_1)),\cdots$ convergent sequecne,
the story is completely different.
the above sequence is convergent sequence? or Is my understanding correct?
corollary 2.1.
Let $f(t,y)$ be continuous on an open $(t,y)$-set $E$ and satisfy $|f(t,y)|\leq M$. Let $E_0$ be a compact subset of $E$. Then there exists an $\alpha>0$, depending on $E,E_0,M,$ with the property that if $(t_0,y_0)\in E_0$, then $(1.1)$ has a solution on $|t-t_0|\leq \alpha$.

There is a nested sequence of subsets $E_n\subset E_{n+1}\subset\dots\subset E$ that exhausts $E$ and where every $\bar E_n$ is compact. Also, $(b_k,y(b_k))$ are outside $E_n$ for $k\ge n$.
Any cluster point of the sequence can not be an inner point of $E$, as it would be then also an inner point of some subset $E_n$. Which is impossible as the subsequence converging to the cluster point is mainly outside of $E_n$.
So the only location that remains for the cluster point is the boundary of $E$.
