## The thing about Hopf Rinow Theorem

Lemma 1: ${M}$ is a Riemannian manifold. ${\forall p\in M}$, ${\exists\,}$ neighborhood ${N}$ of ${p}$ and ${\epsilon>0}$ such that ${\exp_p:B_\epsilon(0)\rightarrow N}$ is a diffeomorphism and any two points in ${N}$ can ve connected by a unique geodesic with length smaller than ${\epsilon}$.

Lemma 2: ${p\in M}$, ${r}$ is small enough such that ${\exp_p:B_r(0)\rightarrow N_r(p)=\exp(B_r(0))}$ is a differeomorphism, then for any ${q\not \in N_r(p)}$, ${\exists \,q'\in \partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Proof: ${d(\cdot,q)}$ is a continuous function on ${\partial N_r(p)}$. Since ${\partial N_r(p)}$ is compact, then ${\exists\, q'\in \partial N_r(p)}$ such that

$\displaystyle d(q',q)=\inf\{d(\tilde{q},q)|\tilde{q}\in\partial N_r(p)\}$

Suppose ${\gamma_n}$ is a minimizing sequence of ${d(p,q)}$. For any ${\gamma_n}$, there exists ${q_n\in \partial N_r(p)\cap \gamma_n}$, then

$\displaystyle L(\gamma_n)\geq d(p,q_n)+d(q_n,q)\geq r+d(q',q)$

Letting ${n\rightarrow \infty}$, ${d(p,q)\geq r+d(q',q)}$. By the triangle inequality,

$\displaystyle d(p,q)= r+d(q',q)$

$\Box$

Thm(Hopf-Rinow) The following statements are equivalent for Riemannian manifold ${M}$:
(1) ${M}$ is a complete metric space, where the metric is induced by

$\displaystyle d(p,q)=\inf\{L(\gamma)|\gamma\text{ is a curve connects } p,q\}$

(2)The closed and bounded sets of ${M}$ are compact.
(3)${\exists \,p\in M}$ such that ${\exp_p}$ is defined on all of ${T_pM}$.
(4) ${\forall\,p\in M}$, ${\exp_p}$ is defined on ${T_pM}$
Furthermore, each of the statements ${(1-4)}$ implies
(5) Any two points ${p,q\in M}$ can be connected by a geodesic of shortest length. Proof: Let us prove ${(3)\Rightarrow (5)}$ firstly.
By lemma 2, there exists ${r>0}$ and ${q'}$ on ${\partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Suppose ${q'=\exp_ptv}$ for some ${v\in T_pM}$, ${||v||=1}$, then ${\gamma(t)=\exp_ptv}$ is defined on ${[0,\infty)}$ by assumption. Consider ${\{t|t\in [r,d(p,q)]\}}$ satisfies

$\displaystyle d(p,q)=t+d(\gamma(t),q),$

denote such points as ${I}$. ${I}$ is nonempty as ${r\in I}$ and it is closed by the sake of continuity.
Suppose ${t_0=\max_{t\in I}t}$. If ${t_0=d(p,q)}$, we are done. Otherwise consider ${q_0=\gamma(t_0)}$, by lemma 2, ${\exists\,\epsilon>0}$ small enough, ${q''\in\partial N_\epsilon(q_0)}$ such that

$\displaystyle q\not\in N_{\epsilon}(q_0),\quad d(q_0,q)=\epsilon+d(q'',q)$

then

$\displaystyle d(p,q)=t_0+\epsilon+d(q'',q)$

We are going to prove ${q''=\gamma(t_0+\epsilon)}$. From the triangle inequality

$\displaystyle d(p,q'')\geq d(p,q)-d(q'',q)=t_0+\epsilon$

$\displaystyle d(p,q'')\leq d(p,q_0)+d(q_0,q'')=t_0+\epsilon$

then we must have ${d(p,q'')=t_0+\epsilon}$. Then the union of ${\gamma([0,t_0])}$ and the geodesic from ${q_0}$ to ${q''}$ constitutes a shortest curve from ${p}$ to ${q''}$, therefore it must be a geodesic. By lemma 1, such geodesic is unique with given initial values. So it must concider with ${\gamma(t)}$ in the ${N_\epsilon(q_0)}$, which means ${q''=\gamma(t_0+\epsilon)}$. Then

$\displaystyle d(p,q)=t_0+\epsilon+d(\gamma(t_0+\epsilon),q)$

This contradicts the fact ${t_0}$ is maximal. So ${t_0}$ must be ${d(p,q)}$, and the curve ${\gamma([0,t_0])}$ is just the minimal geodesic connects ${p}$ and ${q}$.

${(4)\Rightarrow (3)}$ Evidently

${(3)\Rightarrow (2)}$ Suppose ${K}$ is a bounded set in ${M}$. From ${(5)}$ we know, ${K}$ can be contained in ${\exp_p B_r(0)}$ where ${B_r(0)\subset T_pM}$ for some ${r}$ large enough. Since ${B_r(0)}$ is compact and ${\exp_p}$ is a continuous mapping, then ${\exp_p B_r(0)}$ is compact, therefore its closed subset ${K}$ is compact.

${(2)\Rightarrow(1)}$ Any cauchy sequence in ${M}$ is bounded, its closure is compact by ${(2)}$. Then it must have a converge subsequence and being cauchy, it has to converge itself.

${(1)\Rightarrow (4)}$ For any ${p\in M}$, we need to prove ${\exp_ptv}$ is defined on ${[0,\infty)}$ for any ${v\in T_pM}$. Consider ${t_n\nearrow T<\infty}$ and ${\exp_p tv}$ is defined on each ${t_n}$, denote ${\exp_pt_nv=q_n}$.

Since ${d(q_n,q_m)\leq |t_n-t_m|}$, ${\{q_n\}}$ is a cauchy sequence, ${\exists\, q\in M}$ such that ${q_n\rightarrow q}$. Since there exists a neighborhood of ${q}$, say ${N_\epsilon(q)}$, such that ${\forall z\in N_\epsilon(q)}$, any geodesic starting from ${z}$ can be extended at least up to length ${\rho_0>0}$.

For sufficiently large ${n}$, ${q_n\in N_\epsilon(p)}$ and ${d(q_n,q)<\rho_0}$. ${\exp_ptv|_{[t_n,t_{n+1}]}}$ is a curve starting from ${\gamma(t_n)}$, then it can be extended at least up to length ${\rho_0}$. By the uniqueness of geodesic, ${\exp_ptv}$ can be extened beyond ${q}$, namely ${q=\exp_pTv}$ is well defined. So ${\exp_p}$ is defined on whole ${[0,\infty)}$.

$\Box$

Remark:  Should thank Bin Guo’s picture