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$2.8$ 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.

Thanks in advance!

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Your proof is correct.

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).

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