# Prove that if $(O,g)$ is complete, then $O=M$ such that $O$ is a subset to the Riemannian manifold$(M,g)$.

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 $$O = M$$.

• 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) Commented Oct 23, 2023 at 18:13
• @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? Commented Oct 23, 2023 at 18:28
• 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? Commented Oct 23, 2023 at 18:34
• Why can you assume that $B_{\varepsilon}(p)$ is disjoint from $O$? Nothing allows you to do so Commented Oct 23, 2023 at 20:28
• 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. Commented Oct 24, 2023 at 15:33

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.