If $M$ is complete is the closed ball compact?

Question

Let $M$ be a Riemannian manifold and $p,q \in M$. Let $\Omega=\Omega(M;p,q)$ be the set of piecewise $C^\infty$ paths from $p$ to $q$. Let $\rho$ denote the topological metric on $M$ coming from its Riemann metric. Let $S$ denote the ball $\{x \in M : \rho(x,p) \le \sqrt{c}\}$ with $c >0$. Why if $M$ is complete is $S$ compact?

Answer 1

Since $M$ is complete, the exponential map $\exp_p$ from $T_pM$ to $M$ is well defined. Denote the norm on $T_pM$ induced by the Riemannian metric of $M$ by $\|\cdot\|_p$. Then $$K:=\{v\in T_pM:\|v\|_p\le \sqrt{c}\}$$ is a compact subset of $T_pM$, and hence $\exp_p(K)$ is a compact subset of $M$. By the definition of $\rho$, $S$ is a closed subset of $\exp_p(K)$, so $S$ is compact.

Answer 2

This is a simple application of the Hopf-Rinow theorem: if $M$ is complete, then every closed and bounded subset is compact; since $\{x \in M : \rho(x,p) \le \sqrt c \}$ is clearly closed and bounded, it is compact.