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I'm studying Hopf-Rinow theorem and I don't see a step in the proof. Could someone help me, please?

(Definition) Let's $(M, \langle,\rangle)$ an ANII(axiom numerability 2) and Hausdorff Riemannian manifold. If $M$ is connected and $p,q \in M$. We define: $$d_L:M\times M \longrightarrow [0,\infty)$$ $$(p,q)\longmapsto inf\{l(C)\}$$ where C is a piecewise differentiable curve joining $p$ and $q$ and $l$ is the length of the curve.

I've proved that $d_L$ is a distance and that the topology induced by $d_L$ is the original topology on $M$.

Considering $(M, \langle,\rangle)$ an ANII and Hausdorff Riemannian manifold. If $M$ is connected and $p \in M$. Then prove that the following statements are equivalent:

(a) If $A$ is a closed and bounded subset of $M$ then $A$ is compact.

(b) $\exists \{K_n\}_{n\in \mathbb{N}}, K_n \subset M$ compact and $K_n \subset K_{n+1}, \forall n \in \mathbb{N}$ and $\bigcup_n K_n=M$ with the following property: If $(q_n)_{ n \in \mathbb{N}}\subset M$ sequence / $(q_n)\notin K_n, \forall n \in \mathbb{N}\Rightarrow lim_{n\rightarrow \infty} d_L(q_n,p)=\infty$.

Thanks.

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    $\begingroup$ Does ANII mean "Axiom of Numerability number II"? Also, I don't see where the question is. Are you asking why those two statements are equivalent? $\endgroup$ – Bruno Stonek Apr 23 '14 at 17:20
  • $\begingroup$ @Bruno Stonek Thank you, I'll edit it now. $\endgroup$ – miku Apr 23 '14 at 17:26
  • $\begingroup$ Have you checked do Carmo's "Riemannian geometry" book? The Hopf-Rinow theorem is there on pages 146-148, and he proves the equivalence of these two statements. $\endgroup$ – Bruno Stonek Apr 23 '14 at 20:20
  • $\begingroup$ @Bruno Stonek I study on my own with this book, but in that equivalence says: "It's obvious by general topology", buy I've been trying to solve it and I couldn't. $\endgroup$ – miku Apr 23 '14 at 21:03
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$(\Rightarrow)$ Assume (a) is true. Fix a point $o\in M$ and let $K_n$ be the closed unit ball of radius $n$. That is

$$K_n = \{ x\in M: d(x, o)\leq n\}\ .$$

Then each $K_n$ is closed and bounded, thus they are all compact by (a). Let $y\in M$. Then $d(o, y) \leq n$ for some $n$, so $y\in K_n$. Thus $\bigcup_n K_n = M$.

Let $p\in M$. Then as $q_n \notin K_n$, $d(o, q_n)> n$. By triangle inequality,

$$n< d(o , q_n) \leq d(p, o) + d(p, q_n) \Rightarrow d(p, q_n) > n-d(o, p) \to \infty\ ,$$

hence $\lim_n d(p, q_n) = \infty$.

$(\Leftarrow)$ Assume that $(b)$ is true and $A$ is closed and bounded. As $A$ is bounded, there is $p\in M$ and $C>0$ such that $d(x, p \leq C$ for all $x\in A$. First we show that there is $N\in \mathbb N$ such that

$$A \subset K_N\ .$$

Assume the contrary that such a $N$ does not exist. Then for all $n\in\mathbb N$, there is $q_n\in A$ such that $q_n \notin K_n$. By (b), we have $d(p, q_n) \to \infty$. But that contradicts the fact that $A$ is bounded.

Thus $A\subset K_N$ for some $N$. As $K_N$ is compact and $A$ is closed, $A$ is also compact.

Thus we have shown that (a) and (b) are equivalent. Note that it has nothing to do with manifold. That is true for any metric space.

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    $\begingroup$ Thank you very much. It's a fantastic answer. $\endgroup$ – miku Apr 24 '14 at 6:40
  • $\begingroup$ @John First of all thanks for your great answer. Now, how do we know $d(o,y)<\infty$? $\endgroup$ – Srinivasa Granujan May 8 '15 at 7:16
  • $\begingroup$ @Gauloises: That's because $M$ is connected. (Thus is also path connected as $M$ is a manifold). Actually you need to show that when you want to show that $d$ is a metric. $\endgroup$ – user99914 May 8 '15 at 20:17
  • $\begingroup$ Why do we need to prove that there is a $N$ s.t. $A \subset K_n$? Doesn't this follow directly form the fact that $A$ is bounded and $\{K_n\}_n$ is exhausting? In other words: why do we need the condition with the sequence $\{q_n\}_n$? $\endgroup$ – user247741 Jun 4 '16 at 15:18
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    $\begingroup$ Closed and boundedness is not a very restrictive condition, for example, if $M$ is the unit disk with standard euclidean metric, and $A=M$ is closed and bounded. If $K_n = \{ x \in D : |x|\le 1-\frac 1n\}$, then $\cup K_n = D$ with none of them contains $A =D$. So the condition on $\{q_n\}$ is necessary @JohnS. $\endgroup$ – user99914 Jun 4 '16 at 21:56

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