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I am currently working on Peter Petersen´s book: Riemannian Geometry (2016). I am asking if I proved the statement correctly or not.

Exercise: Let $O \subset (M,g)$ be an open susbet of a Riemannian manifold. Show that if $(O,g)$ is complete, then $O = M$.

My proof: By way of contradiction, assume $p \in M \setminus O$. s.th. $\exists \epsilon > 0$ for an open ball $B_{\epsilon}(p)$ with $$ B_{\epsilon}(p) \cap O = \varnothing. $$ Since $O$ is open, $$ \forall n \in \mathbb{N}, \exists p_n \in O \quad \text{s.th.} \quad d(p,p_n)<\frac{1}{n}, $$ by convergence it implies $p_n \to p$ as $n \to \infty$ and $\frac{1}{n} \to 0$. For every $n$, there exists a geodesic $\gamma_n : [0,1] \to M$ s.th. $$ \gamma_n(0)= p_n, \gamma_n(1)= p_{n+1} $$ due to completness of $(O,g)$.

Again by completness, $\forall \gamma_n$ is extended to a geodesic $\Gamma_n : \mathbb{R} \to O$. By the geodesic extension and convergence, $$ \forall \delta>0, \exists N\in\mathbb{N} \quad \text{s.th.} \quad n \geqslant N \implies d(p,p_n)< \delta = \frac{\epsilon}{2}. $$

Now for $\Gamma_n(t)$ and applying the triangle inequality, we have $$ d(p,\Gamma_n(t)) \leqslant d(p,p_n) + d(p_n, \Gamma_n(t)) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \\ \implies \Gamma_n(t) \in B_\epsilon(p). $$ A contradiction arises because $B_{\epsilon}(p) \cap O = \varnothing$, thus $O = M$.

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    $\begingroup$ Why would the very first sentence of your proof be true? + you need to ask for $M$ to be connected (if not, there are obvious counter-examples) $\endgroup$
    – Didier
    Commented Oct 23, 2023 at 18:13
  • $\begingroup$ @Didier So by holding connectedness in mind. My assumption for contradiction will be more valid since there exists a seqeunce $\{p_n\}$ by the fact of $M$ being connected and continuing on with geodesic extension and pointwise convergence? $\endgroup$ Commented Oct 23, 2023 at 18:28
  • $\begingroup$ One counter-example would be $\mathbb{R}^n \setminus \{0\}$ becuase not assuming connectedness will not allow geodesics to be extended to all the reals? $\endgroup$ Commented Oct 23, 2023 at 18:34
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    $\begingroup$ Why can you assume that $B_{\varepsilon}(p)$ is disjoint from $O$? Nothing allows you to do so $\endgroup$
    – Didier
    Commented Oct 23, 2023 at 20:28
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    $\begingroup$ I realized my mistake now! You`re right. I belive that in order to prove the statement, I will have to use the Hopf-Rinow thm. $\endgroup$ Commented Oct 24, 2023 at 15:33

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OP noticed by themselves that they need to use Hopf-Rinow's Theorem at some point. Here is a complete proof.

Let $p\in O\subset M$. Up to considering the connected component of $p$ in $O$ (which is closed in $O$, and hence complete whenever $O$ is complete), we assume that $O$ is connected. Notice that $T_pO = T_pM$. Let $U_p^O\subset T_pM$ be the domain of $\exp_p^O$, and $U_p^M\subset T_pM$ be that of $\exp_p^M$. Since $O\subset M$, we have $U_p^O \subset U_p^M$. Now, $(O,g|_O)$ is complete, and Hopf-Rinow's Theorem ensures that $U_p^O = T_pO = T_pM$. From $U_p^O \subset U_p^M$, it follows that $U_p^M = T_pM$. Therefore, $\exp_p^M$ is defined on all of $T_pM$. The point $p$ is therefore a pole of $(M,g)$. The result now reduces to the following classical exercise.

Let $(M,g)$ be a connected Riemannian manifold with a pole. Then $(M,g)$ is complete.

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  • $\begingroup$ I got the picture! Thanks for helping! $\endgroup$ Commented Oct 24, 2023 at 16:36

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