[proofplan] We prove $(1) \iff (2) \implies (3) \implies (4) \implies (5) \implies (1)$, then deduce the minimal-geodesic statement from $(3)$. The equivalence $(1) \iff (2)$ is the definition of geodesic completeness. Implication $(2) \implies (3)$ is logical specialisation. The crux is $(3) \implies (4)$: from $\exp_p$ defined globally at one point we know any $q$ is reachable by a minimal geodesic of length $d(p, q)$ (the [Minimal Geodesics From a Complete Point](/theorems/2724)), so closed bounded sets sit inside continuous images of compact balls and inherit compactness. Then $(4) \implies (5)$ is a Cauchy-sequence argument: a Cauchy sequence is bounded, lies in a closed bounded set, hence has a convergent subsequence, hence converges. Finally $(5) \implies (1)$ reverses an inextensible geodesic to produce a Cauchy sequence whose limit lets the geodesic continue, contradicting maximality. The closing claim about minimal geodesics is then exactly Theorem 2724. [/proofplan]

[step:Equivalence $(1) \iff (2)$ from the definition of geodesic completeness] By definition, $(M, g)$ is geodesically complete if every (maximal) geodesic $\gamma : I \to M$ has $I = \mathbb{R}$. By [Geodesic Rescaling](/theorems/2710), the rescaled geodesics $\gamma_p(\cdot, v)$ for $v \in T_p M$ exhaust the maximal geodesics through $p$ up to reparametrisation, and $\exp_p(v) = \gamma_p(1, v)$ is defined precisely when $\gamma_p(\cdot, v)$ extends to time $1$ — equivalently, by rescaling, when $\gamma_p(\cdot, v/n)$ extends to time $n$ for all $n$. Hence $\exp_p$ is defined on all of $T_p M$ for every $p \in M$ if and only if every maximal geodesic of $(M, g)$ has domain $\mathbb{R}$, i.e., $(M, g)$ is geodesically complete. This proves $(1) \iff (2)$. [/step]

[step:Implication $(2) \implies (3)$ by specialising to a single point] If $\exp_p$ is defined on all of $T_p M$ for every $p \in M$, then in particular it is defined on all of $T_p M$ for some $p \in M$, since $M \neq \varnothing$. This is the content of $(3)$. [/step]

[step:Implication $(3) \implies (4)$ via global reach of the exponential map]Assume $(3)$: fix $p \in M$ such that $\exp_p$ is defined on all of $T_p M$. Let $A \subseteq M$ be closed and bounded in $(M, d)$. We must show $A$ is compact. Since $A$ is bounded, there exists $R > 0$ with $A \subseteq \overline{B}_d(p, R) := \{x \in M : d(p, x) \leq R\}$. By the [Minimal Geodesics From a Complete Point](/theorems/2724) applied at $p$ (whose hypothesis "$\exp_p$ defined on all of $T_p M$" is exactly $(3)$, and connectedness of $M$ is given), every $q \in M$ is connected to $p$ by a minimal geodesic of length $d(p, q)$. In particular, for $q \in \overline{B}_d(p, R)$ there exists $v \in T_p M$ with $|v|_g = d(p, q) \leq R$ and $\exp_p(v) = q$. Therefore \begin{align*} \overline{B}_d(p, R) \subseteq \exp_p(\overline{B}_{T_p M}(0, R)), \end{align*} where $\overline{B}_{T_p M}(0, R) := \{v \in T_p M : |v|_g \leq R\}$. The closed Euclidean ball $\overline{B}_{T_p M}(0, R) \subseteq T_p M$ is compact (a closed bounded subset of the finite-dimensional inner product space $(T_p M, g_p)$). The map $\exp_p : T_p M \to M$ is continuous (smooth, in fact), so $\exp_p(\overline{B}_{T_p M}(0, R))$ is the continuous image of a compact set and is therefore compact, hence closed in $M$. Thus $\overline{B}_d(p, R)$ is contained in the compact set $\exp_p(\overline{B}_{T_p M}(0, R))$. Since $A \subseteq \overline{B}_d(p, R)$ is closed in $M$, it is closed inside this compact set, hence compact.[/step]

[guided]Assume $(3)$: fix $p \in M$ such that $\exp_p$ is defined on all of $T_p M$. Let $A \subseteq M$ be closed and bounded in $(M, d)$; we must show $A$ is compact. The natural strategy is to exhibit $A$ as a closed subset of a compact set — closed subsets of compact sets are compact in any Hausdorff space. Where do we get a compact set? From the exponential map. Since $A$ is bounded, there exists $R > 0$ with $A \subseteq \overline{B}_d(p, R) := \{x \in M : d(p, x) \leq R\}$. The closed Euclidean ball $\overline{B}_{T_p M}(0, R) := \{v \in T_p M : |v|_g \leq R\}$ is a compact subset of the finite-dimensional inner product space $(T_p M, g_p)$ (finite-dimensional Heine–Borel). The map $\exp_p : T_p M \to M$ is smooth, hence continuous, so $\exp_p(\overline{B}_{T_p M}(0, R))$ is the continuous image of a compact set, hence compact in $M$ (and therefore closed, since $M$ is Hausdorff). But we need to know this image *contains* the closed metric ball $\overline{B}_d(p, R)$. This is exactly where the [Minimal Geodesics From a Complete Point](/theorems/2724) comes in: that theorem says every $q \in M$ is reachable from $p$ by a minimal geodesic of length $d(p, q)$. Verifying its hypotheses: connectedness of $M$ is given (it is part of the standing hypothesis on $(M, g)$), and $\exp_p$ being defined on all of $T_p M$ is exactly condition $(3)$. The theorem produces, for each $q \in \overline{B}_d(p, R)$, a vector $v \in T_p M$ with $\exp_p(v) = q$ and $|v|_g = d(p, q) \leq R$, so $q \in \exp_p(\overline{B}_{T_p M}(0, R))$. We have therefore shown the inclusion \begin{align*} \overline{B}_d(p, R) \subseteq \exp_p(\overline{B}_{T_p M}(0, R)). \end{align*} Combining: $A \subseteq \overline{B}_d(p, R) \subseteq \exp_p(\overline{B}_{T_p M}(0, R))$, the rightmost set is compact, and $A$ is closed in $M$ (hence closed inside this compact set), so $A$ is compact. This is the only step in the cycle that uses Theorem 2724 — the rest are standard topology.[/guided]

[step:Implication $(4) \implies (5)$ via Cauchy sequences and Heine–Borel] Assume $(4)$. Let $(x_n)_{n \geq 1} \subseteq M$ be a Cauchy sequence in $(M, d)$. Choose $N$ with $d(x_n, x_m) < 1$ for $n, m \geq N$. Then for all $n \geq N$, \begin{align*} d(x_N, x_n) < 1, \end{align*} and the finitely many initial terms $x_1, \dots, x_{N-1}$ are at distance at most $\max_{1 \leq k \leq N-1} d(x_N, x_k) =: D$ from $x_N$. Therefore the entire sequence lies in $\overline{B}_d(x_N, R)$ where $R := \max(D, 1)$, a closed bounded set. By $(4)$, $\overline{B}_d(x_N, R)$ is compact. A Cauchy sequence in a compact metric subspace has a convergent subsequence; combined with the Cauchy property, the full sequence converges (a Cauchy sequence with a convergent subsequence converges to the same limit). Since the limit lies in $\overline{B}_d(x_N, R) \subseteq M$, we have shown $(M, d)$ is complete as a metric space. [/step]

[step:Implication $(5) \implies (1)$ via extending an inextensible geodesic]Assume $(M, d)$ is complete. Let $\gamma : [0, b) \to M$ be a maximal geodesic with $b < \infty$; we derive a contradiction. Without loss of generality $\gamma$ has constant speed $|\dot\gamma|_g \equiv c \geq 0$ by [Geodesics Have Constant Speed](/theorems/2709). If $c = 0$, then $\gamma$ is constant: the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$ with $\dot\gamma \equiv 0$ admits the constant curve as a global solution defined on all of $\mathbb{R}$, so the maximal domain is $\mathbb{R}$ already, contradicting $b < \infty$. Hence assume $c > 0$. Pick any sequence $t_n \uparrow b$ with $t_n \in [0, b)$. For $m, n$ with $t_m, t_n$ close to $b$, \begin{align*} d(\gamma(t_m), \gamma(t_n)) \leq \ell(\gamma|_{[\min(t_m, t_n), \max(t_m, t_n)]}) = c \cdot |t_m - t_n|. \end{align*} Since $(t_n)$ is Cauchy in $[0, b)$ (it converges to $b$), $(\gamma(t_n))$ is Cauchy in $M$. By $(5)$, there exists $q \in M$ with $\gamma(t_n) \to q$. The limit $q$ does not depend on the sequence $(t_n)$: if $(t_n')$ is another sequence with $t_n' \uparrow b$, form the merged sequence $(s_k)$ defined by $s_{2k-1} := t_k$ and $s_{2k} := t_k'$. Both $(t_n)$ and $(t_n')$ converge to $b$, so $(s_k) \to b$, and in particular $(s_k)$ is Cauchy in $\mathbb{R}$. By the Lipschitz estimate from constant speed, $d(\gamma(s_j), \gamma(s_k)) \leq c\, |s_j - s_k|$, so $(\gamma(s_k))$ is Cauchy in $(M, d)$ and converges by (5) to some single limit $q^* \in M$. Both subsequences $(\gamma(t_n))$ and $(\gamma(t_n'))$ inherit this limit, forcing the two limits to coincide. Hence $\gamma$ extends continuously to $\bar\gamma : [0, b] \to M$ with $\bar\gamma(b) = q$. We invoke the following uniform local existence statement for the geodesic flow. [claim:Uniform local existence on a neighbourhood of $q$] There exist $\rho, \varepsilon > 0$ such that for every $r \in M$ with $d(r, q) < \rho$ and every $w \in T_r M$ with $|w|_g \leq 1$, the geodesic $\gamma_w$ with $\gamma_w(0) = r$ and $\dot\gamma_w(0) = w$ is defined on the entire interval $[-\varepsilon, \varepsilon]$. [/claim] [proof] The geodesic spray $G$ is a smooth vector field on $TM$: the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$, second-order on $M$, is the first-order ODE $\dot{(r,w)} = G(r,w)$ on $TM$. By the standard theorem on smooth flows of smooth vector fields, the maximal flow domain $\mathcal{D} \subseteq \mathbb{R} \times TM$ of $G$ is open in $\mathbb{R} \times TM$, and the flow $\Phi : \mathcal{D} \to TM$ is smooth. Fix $\rho_0 > 0$ small enough that $\overline{B}_d(q, \rho_0)$ is compact (such $\rho_0$ exists because, by [Exponential Map as a Local Diffeomorphism](/theorems/2712) applied at $q$, there is a normal neighbourhood of $q$ inside which sufficiently small closed metric balls have compact closure). The closed unit-disk bundle restricted to this ball, \begin{align*} K := \{(r, w) \in TM : r \in \overline{B}_d(q, \rho_0),\ |w|_g \leq 1\}, \end{align*} is compact in $TM$ (a fibre bundle with compact fibre $\overline{B}(0, 1) \subseteq \mathbb{R}^n$ over the compact base). The set $\{0\} \times K \subseteq \mathcal{D}$ since the flow is defined at time $0$ from any initial condition. By openness of $\mathcal{D}$, for each $(r, w) \in K$ there exist $\varepsilon_{(r,w)} > 0$ and an open neighbourhood $U_{(r,w)} \subseteq TM$ of $(r,w)$ such that $(-\varepsilon_{(r,w)}, \varepsilon_{(r,w)}) \times U_{(r,w)} \subseteq \mathcal{D}$. The collection $\{U_{(r,w)} : (r,w) \in K\}$ is an open cover of the compact set $K$; extract a finite subcover indexed by $(r_1, w_1), \dots, (r_N, w_N)$ and set $\varepsilon := \min_{1 \leq i \leq N} \varepsilon_{(r_i, w_i)} > 0$. Then $(-\varepsilon, \varepsilon) \times K \subseteq \mathcal{D}$. Choosing any $\rho \in (0, \rho_0)$ gives the claim. [/proof] Apply the claim with the unit vector $w := \dot\gamma(t_0) / c$ (recall $c = |\dot\gamma|_g > 0$): for any $r$ with $d(r, q) < \rho$, the geodesic at $r$ in any unit direction exists for time at least $\varepsilon$. By [Geodesic Rescaling](/theorems/2710), the geodesic at $r$ with initial velocity of norm $c$ exists for time at least $\varepsilon / c$. Choose $t_0 \in [0, b)$ close enough to $b$ that $d(\gamma(t_0), q) < \rho$ and $b - t_0 < \varepsilon / c$. Then the geodesic with initial conditions $(\gamma(t_0), \dot\gamma(t_0))$ — which is $\gamma$ itself by uniqueness — extends for time at least $\varepsilon / c > b - t_0$ past $t_0$. This means $\gamma$ extends past $b$, contradicting maximality of the domain $[0, b)$. By the symmetric argument with $\gamma : (a, 0] \to M$ (or equivalently reversing the parametrisation), the geodesic also extends past any finite left endpoint. Hence every maximal geodesic has domain $\mathbb{R}$, proving $(1)$.[/step]

[guided]Assume $(M, d)$ is complete. We prove $(1)$ by contradiction: suppose $\gamma : [0, b) \to M$ is a maximal geodesic with $b < \infty$. The strategy is "metric completeness fills in geodesic limits, then uniformity of the geodesic flow re-enters the manifold and continues the geodesic past the supposed endpoint." By [Geodesics Have Constant Speed](/theorems/2709), $|\dot\gamma|_g \equiv c \geq 0$ is constant. If $c = 0$, then $\dot\gamma \equiv 0$, so the geodesic equation $\nabla_{\dot\gamma}\dot\gamma = 0$ admits the constant curve as a global solution, and the maximal domain of $\gamma$ is already $\mathbb{R}$ — contradicting $b < \infty$. So we may assume $c > 0$. **(i) Cauchy from constant speed.** A constant-speed geodesic has Lipschitz parametrisation: by definition of $d$ as the infimum of path lengths, $d(\gamma(s), \gamma(t)) \leq \ell(\gamma|_{[\min(s,t), \max(s,t)]}) = c|s - t|$. Concretely, pick any sequence $t_n \uparrow b$ with $t_n \in [0, b)$. For all $m, n$, \begin{align*} d(\gamma(t_m), \gamma(t_n)) \leq \ell(\gamma|_{[\min(t_m, t_n), \max(t_m, t_n)]}) = c \cdot |t_m - t_n|. \end{align*} Since $(t_n)$ converges to $b$ in $\mathbb{R}$, it is Cauchy, so by the Lipschitz bound $(\gamma(t_n))$ is Cauchy in $(M, d)$. **(ii) Limit exists by $(5)$.** Metric completeness gives some $q \in M$ with $\gamma(t_n) \to q$. The limit is independent of the choice of sequence: if $t_n' \uparrow b$ is another sequence, interleave $s_{2k-1} := t_k$, $s_{2k} := t_k'$. Then $(s_k) \to b$, so $(s_k)$ is Cauchy in $\mathbb{R}$, the Lipschitz bound makes $(\gamma(s_k))$ Cauchy in $(M, d)$, and (5) gives a single limit $q^* \in M$. Both subsequences $(\gamma(t_n))$ and $(\gamma(t_n'))$ inherit this limit, so the two limits coincide. Hence $\gamma$ extends continuously to $\bar\gamma : [0, b] \to M$ with $\bar\gamma(b) = q$. **(iii) Re-enter the manifold.** We need a *uniform* local existence statement: a single pair $(\rho, \varepsilon)$ with $\rho, \varepsilon > 0$ such that for every basepoint $r$ with $d(r, q) < \rho$ and every $w \in T_r M$ with $|w|_g \leq 1$, the geodesic $\gamma_w$ at $r$ in direction $w$ exists on $[-\varepsilon, \varepsilon]$. Note: the [Exponential Map as a Local Diffeomorphism](/theorems/2712) only gives existence of $\exp_q$ on a single ball at the one point $q$ — that is local existence at a fixed basepoint, not the *uniform-in-basepoint* statement we need. The correct tool is the openness of the flow domain of the geodesic spray combined with compactness — this is the content of the claim above. To produce $(\rho, \varepsilon)$: the geodesic spray $G$ is a smooth vector field on $TM$ (the geodesic equation, second-order on $M$, is the first-order ODE $\dot{(r,w)} = G(r,w)$ on $TM$). By the standard theorem on smooth flows, the maximal flow domain $\mathcal{D} \subseteq \mathbb{R} \times TM$ of $G$ is open, and the flow is smooth. Pick $\rho_0 > 0$ with $\overline{B}_d(q, \rho_0)$ compact (using Theorem 2712 at $q$), and form the compact unit-disk subbundle \begin{align*} K := \{(r, w) \in TM : r \in \overline{B}_d(q, \rho_0),\ |w|_g \leq 1\}. \end{align*} Since $\{0\} \times K \subseteq \mathcal{D}$ (the flow exists at time zero) and $\mathcal{D}$ is open, each $(r, w) \in K$ has a neighbourhood $(-\varepsilon_{(r,w)}, \varepsilon_{(r,w)}) \times U_{(r,w)} \subseteq \mathcal{D}$. The $U_{(r,w)}$ cover the compact $K$; extract a finite subcover and set $\varepsilon := \min_i \varepsilon_{(r_i, w_i)} > 0$. Then $(-\varepsilon, \varepsilon) \times K \subseteq \mathcal{D}$. Choose any $\rho \in (0, \rho_0)$. **Continuing the geodesic.** Pick $t_0 \in [0, b)$ close enough to $b$ that $d(\gamma(t_0), q) < \rho$ and $b - t_0 < \varepsilon / c$ (possible because $\gamma(t) \to q$ as $t \to b^-$ and $b$ is finite). Apply the uniform existence statement at $r := \gamma(t_0)$ with the unit vector $w := \dot\gamma(t_0) / c$ (which has $|w|_g = 1$ since $|\dot\gamma|_g \equiv c$): the geodesic at $\gamma(t_0)$ in direction $w$ exists on $[-\varepsilon, \varepsilon]$. By [Geodesic Rescaling](/theorems/2710), the geodesic at $\gamma(t_0)$ in direction $\dot\gamma(t_0) = c \cdot w$ exists for time at least $\varepsilon / c > b - t_0$ past $t_0$. By uniqueness of geodesics with prescribed initial conditions, this is $\gamma$ continued — so $\gamma$ extends past $b$, contradicting maximality of the domain $[0, b)$. The same argument runs with a finite left endpoint by reversing parametrisation. Hence every maximal geodesic has domain $\mathbb{R}$, proving $(1)$.[/guided]

[step:Deduce existence of minimal geodesics] By the cycle, all five conditions are equivalent. In particular, condition $(3)$ holds: there exists $p \in M$ with $\exp_p$ defined on all of $T_p M$. (Indeed under the equivalent condition $(2)$, *every* point has this property.) For any $q \in M$, the [Minimal Geodesics From a Complete Point](/theorems/2724) applied at $p$ — connectedness of $M$ is given, and $\exp_p$ is defined globally by $(3)$ — produces a minimal geodesic from $p$ to $q$. The "any two points" version is the same statement with the choice of base point $p$ replaced by an arbitrary point. Under condition $(2)$, every $p \in M$ has $\exp_p$ defined globally, so the [Minimal Geodesics From a Complete Point](/theorems/2724) applies at any $p$: any two points $p, q \in M$ are joined by a minimal geodesic. [/step]