Trajectories of a vector field defined on complete manifold. Let $M$ be a complete Riemannian manifold, and let $X$ be a differentiable vector field in $M$. Suppose that there exists $c>0$ s.t $|X(p)|<c$, for all $p \in M$. Prove that the trajectories of $X$ are defined for all $t$.
Well, there exists a unique trajectory $\phi:I\to M$, such that $X(\phi(t))=\phi'(t)$ and $\phi(0)=p$ by existence and uniqueness theorem. On the other hand,
$\phi(t)=\phi(0)+\int^{t}_{0} X(\phi(t)) dt$, then
$|\phi(t)-p|\leq \int^{t}_{0} |X(\phi(t))| dt\leq tc$.
Now, I suppose that $I=(-\infty,a]$, then $\phi(t) \in \bar{B}_{ca}(p)$, as $\bar{B}_{ca}(p)\subset M$ is closed and bounded, Hopf-Rinow asserts that $\bar{B}_{ca}(p)$ is compact, since $M$ is complete. My question is as I conclude this proof.
 A: Your estimates are true but a little careless. You don't have $\phi(t)=\phi(0)+\int_0^t X(\phi(t)) dt,$ that doesn't make sense.
However, you have that $d(\phi(t),\phi(0)) \leq \int_0^t |\dot{\phi}(t)|dt =\int_0^t |X(\phi(t))|dt \leq tc,$ since $\phi(t)$ is a path connecting $\phi(t)$ and $\phi(0)$.
You can make another estimate: that of $d(\phi(t'),\phi(t)) \leq (t'-t)c \quad (1)$.
Now, let $a=\sup\{t \in \mathbb{R} \mid \phi(t) \text{ is defined}\}$ (if this doesn't seem rigorous enough, take $\sup\{t \in \mathbb{R} \mid \text{ there exists solution $\psi$ of the ODE on (0,t)}\}$, and notice that the $\psi$ are always extensions). Suppose $a< \infty.$ Take any sequence $t_n$ converging to $a$. Estimate $(1)$ shows that $\phi(t_n)$ is Cauchy. Since the manifold is complete, it follows (by definition or by Hopf-Rinow) that $\phi(t_n)$ converges to a point $q$. This $q$ doesn't depend on the sequence $t_n$ (this follows from the estimate $(1)$ again). Therefore $\phi(t) \to q$. Since $X$ is continuous, it follows that $X(q)=\lim X(\phi(t_n))$. Now, note that we have a local solution of the ODE around $q$, let that be $\gamma:(a-\epsilon, a+\epsilon) \to M$, $\gamma(a)=q$. By making $\Psi:(0,a+\epsilon) \to M$ given by $\Psi|_{(0,t]}=\phi$ and $\Psi|_{[t,t+\epsilon)}=\gamma,$ the glueing lemma gives that $\Psi$ is a solution for the ODE on $(0,a+\epsilon)$. Therefore, $a$ was not $\sup$, a contradiction.
The same reasoning applies to show that it is defined for arbitrary negative values.
