Corollary 2.11 of Do Carmo's Riemannian Geometry states the following:
Suppose that $M$ is a complete riemannian manifold and that there exists $p \in M$ which has a cut point for every geodesic starting from $p$. Then M is compact.
The beginning of the proof states that $$M=\cup \{ \gamma(t): t \leq f(p, \gamma'(0)) \} $$ where $f$ is a function from the couples (point in $M$, unit vector in $T_p M$) to the extended reals $\mathbb{R} \cup \{ \infty \}$ which gives the cut time of the geodesic starting from $p$ with a given velocity, and gives $\infty$ if such time does not exists.
Now, I find no way of proving that $M$ is in fact equal to the set on the RHS, and have come to the point of thinking that the correct statement of the corollary should be that the cut point exist $\textit{for every}$ point in $M$, not just for a single point. I have however not been able to provide a counterexample either, so I am asking you, is the statement of the corollary correct? If so, how to prove that $M$ is that set?
Let $q\in M$. Then since $M$ is complete, there is a shortest geodesic $\gamma(t)$ so that $\gamma(0) = p$ and $\gamma(d) = q$, where $d =d(p,q)$. So it suffices to show that $d(p,q) \le f(p, \gamma'(0))$. This is trivial if $f(p, \gamma'(0)) = \infty$.
When $f(p,\gamma'(0))<\infty$, note that $d(p,q) > f(p, \gamma'(0))$ this contradicts that $\gamma$ is a shortest geodesic joining $p$ and $q$ (by definition of $f$).