# Proof of Hopf-Rinow's theorem in Do Carmo's Riemannian Geometry

$$2.8$$ Theorem (Hopf and Rinow [HR]). Let $$M$$ be a Riemannian manifold and let $$p \in M$$. The following assertations are equivalent:

a) $$\exp_p$$ is defined on all $$T_p(M)$$.

b) The closed and bounded sets of $$M$$ are compact.

c) $$M$$ is complete as a metric space.

d) $$M$$ is geodesically complete.

e) There exists a sequence of compact subsets $$K_n \subset M$$, $$K_n \subset K_{n+1}$$ and $$\bigcup_\limits{n} K_n = M$$ such that if $$q_n \notin K_n$$, then $$d(p,q_n) \rightarrow \infty$$.

In addition, any of the statements above implies that

f) For any $$q \in M$$ there exists a geodesic $$\gamma$$ joining $$p$$ to $$q$$ with $$l(\gamma) = d(p,q)$$.

It was proved that $$a) \Longrightarrow b) \Longrightarrow c) \Longrightarrow d) \Longrightarrow a)$$. I'm trying understand the equivalence between $$b$$ and $$e$$. The proof is only this:

b) $$\iff$$ e). General topology.

I tried think about why this is direct, so below is my thoughts:

b) $$\Longrightarrow$$ e)

As $$a$$ is equivalent to $$b$$, there is a sequence of $$\overline{B}_n(0) \subset T_pM$$. Define $$K_n := \exp_p \left( \overline{B}_n(0) \right)$$, then $$K_n$$ is compact by the continuity of $$\exp_p$$ and compactness of $$\overline{B}_n(0)$$ in $$T_pM$$, $$K_n \subset K_{n+1}$$ by construction and $$\bigcup_\limits{n} K_n = M$$ because $$\bigcup_\limits{n} \exp_p^{-1} \left( K_n \right) = \bigcup_\limits{n} \overline{B}_n(0) = T_pM$$ and $$\exp_p$$ is surjective (recall that $$a$$ is equivalent to $$b$$). Thus, given $$n$$, if $$q_n \notin K_n$$, then $$d(p,q_n) > n$$, in other words, $$d(p,q_n) \rightarrow \infty$$.

e) $$\Longrightarrow$$ b)

Let be $$K$$ a closed and bounded set of $$M$$ and $$\{ K_n \}$$ the sequence of compact subsets of $$M$$ as in the hypothesis. We can suppose without loss of generality that $$p \in K$$ since it's arbitrary. Observe that $$K \subset K_n$$ for some $$n$$, otherwise, there would be $$q_n \in M \backslash K_n$$ for each $$n$$ and we would have $$d(p,q_n) \rightarrow \infty$$ by hypothesis, which contradicts the fact that $$K$$ is bounded, then $$K \subset K_n$$ for some $$n$$, therefore $$K$$ is compact in $$M$$ because is a closed set contained in a compact set.

I would like to know if I'm right and, if I'm not right, I would like to know why the equivalence between $$b$$ and $$e$$ is direct by General topology. I would also appreciate if there is another proof more direct just using General topology as suggested in the book.

For a more direct version of b)$$\implies$$e), just do the same in a general metric space:
Let $$K_n:=\bar B_n(p)$$, which is closed and bounded, hence compact by b).