[proofplan] The proof builds a line as a limit of long minimizing geodesics passing through a fixed compact set. Disconnectedness at infinity gives a compact $K \subseteq M$ such that $M \setminus K$ has two unbounded components $A_1, A_2$. We pick sequences $p_n \in A_1$, $q_n \in A_2$ tending to infinity. By completeness (Hopf–Rinow), each pair $p_n, q_n$ is joined by a unit-speed minimizing geodesic $\gamma_n$, which must traverse $K$ since paths between distinct unbounded components cross the boundary of $K$. After reparametrizing so that $\gamma_n(0) \in K$, the family $\{\gamma_n\}$ is uniformly bounded on compact sets and equicontinuous (unit speed). Arzelà–Ascoli on the compact-open topology gives a subsequence $\gamma_{n_k}$ converging on every compact subinterval of $\mathbb R$ to a unit-speed geodesic $\gamma_\infty : \mathbb R \to M$. The minimality of each $\gamma_n$ on its full interval, together with the unbounded growth of the interval lengths, passes to the limit to make $\gamma_\infty$ minimizing on every finite subinterval — hence a line. [/proofplan]

[step:Set up the data: compact $K$ separating two unbounded components $A_1, A_2$]Let $(M, g)$ be a complete connected Riemannian manifold of dimension $n$ that is **disconnected at infinity**: by definition, there exists a compact set $K \subseteq M$ such that $M \setminus K$ has at least two unbounded connected components. Fix two such components and call them $A_1$ and $A_2$. Each $A_j$ is unbounded as a subset of $(M, d_g)$, where $d_g$ is the Riemannian distance. Hence we can pick sequences \begin{align*} p_n \in A_1, \qquad q_n \in A_2 \qquad (n \in \mathbb{N}) \end{align*} with $d_g(p_n, K) \to \infty$ and $d_g(q_n, K) \to \infty$ as $n \to \infty$. (Concretely: fix a basepoint $x_0 \in K$; since $A_j$ is unbounded, there exist points in $A_j$ with $d_g(\cdot, x_0)$ arbitrarily large, and $d_g(\cdot, K) \geq d_g(\cdot, x_0) - \mathrm{diam}(K)$ on $A_j$.)[/step]

[guided]We begin by unpacking what the hypothesis "disconnected at infinity" gives us. By definition, there is a compact set $K \subseteq M$ such that $M \setminus K$ has at least two unbounded connected components. We fix two such components and call them $A_1$ and $A_2$. Geometrically, think of $K$ as a compact "neck" that separates $M$ into pieces, two of which (namely $A_1$ and $A_2$) extend off to infinity in distinct directions. Now we want to pick representatives from each unbounded piece, but we have to be careful about *which* representatives. Why? Looking ahead: minimizing geodesics from $p_n$ to $q_n$ must traverse $K$ (topologically, as we show in Step 2). We will then extract a limiting geodesic, and we want that limit to be defined on all of $\mathbb R$. For this, we need both "legs" of the geodesic — the part from $p_n$ to $K$ and the part from $K$ to $q_n$ — to grow without bound, so that the limiting geodesic extends in both directions to infinity. Each $A_j$ is unbounded in the metric space $(M, d_g)$, where $d_g$ is the Riemannian distance. By definition of unboundedness, we can pick sequences \begin{align*} p_n \in A_1, \qquad q_n \in A_2 \qquad (n \in \mathbb{N}) \end{align*} with $d_g(p_n, K) \to \infty$ and $d_g(q_n, K) \to \infty$ as $n \to \infty$. Concretely, fix a basepoint $x_0 \in K$. Since $A_j$ is unbounded, there exist points in $A_j$ with $d_g(\cdot, x_0)$ arbitrarily large. The triangle inequality gives $d_g(\cdot, K) \geq d_g(\cdot, x_0) - \mathrm{diam}(K)$ on $A_j$, so distance to $K$ is also unbounded along these sequences. Two design questions deserve highlighting: - *Why insist that $d_g(p_n, K) \to \infty$ rather than merely $p_n \to \infty$ in some other sense?* Because the distance $d_g(p_n, K)$ will become the length of the left half of our reparametrized geodesic. If this length stayed bounded, the limiting geodesic would be defined only on a bounded interval to the left of $0$, not on all of $(-\infty, 0]$, and we could not call it a line. - *Why both directions?* A line is, by definition, a unit-speed geodesic $\gamma : \mathbb{R} \to M$ that is *globally* minimizing — minimizing on every finite subinterval. If only one direction grew without bound, we would extract only a ray. Disconnectedness at infinity is exactly what supplies two distinct unbounded directions to pull on.[/guided]

[step:Use Hopf–Rinow completeness to find minimizing geodesics from $p_n$ to $q_n$]By [Hopf–Rinow](/theorems/2719) (the equivalence of geodesic completeness, metric completeness, and the property that closed bounded sets are compact, and the existence of minimizing geodesics between any two points on a complete Riemannian manifold), for each $n$ there exists a minimizing geodesic from $p_n$ to $q_n$: \begin{align*} \gamma_n : [0, L_n] &\to M, \qquad L_n := d_g(p_n, q_n) > 0, \\ \gamma_n(0) &= p_n, \quad \gamma_n(L_n) = q_n, \quad |\dot\gamma_n| \equiv 1, \quad L(\gamma_n|_{[0, L_n]}) = L_n. \end{align*} Here $L(\cdot)$ denotes the Riemannian arclength, and $|\dot\gamma_n| \equiv 1$ is unit speed. [claim:The geodesic $\gamma_n$ traverses $K$: there exists $t_n \in [0, L_n]$ with $\gamma_n(t_n) \in K$] [proof] The image $\gamma_n([0, L_n])$ is a path-connected subset of $M$ joining $p_n \in A_1$ and $q_n \in A_2$. The components $A_1, A_2$ of $M \setminus K$ are disjoint open sets, and the path goes from one to the other. If $\gamma_n([0, L_n]) \subseteq M \setminus K$, then the path would lie in $M \setminus K$ and would connect $p_n$ and $q_n$ within a single connected component of $M \setminus K$, since path-connected subsets of $M \setminus K$ lie in a single component. But $p_n, q_n$ are in different components $A_1 \neq A_2$, a contradiction. Hence $\gamma_n([0, L_n]) \cap K \neq \varnothing$, i.e., there exists $t_n \in [0, L_n]$ with $\gamma_n(t_n) \in K$. [/proof] [/claim] We now reparametrize $\gamma_n$ so that the chosen point in $K$ is at parameter $0$. Define \begin{align*} \tilde\gamma_n : [-t_n, L_n - t_n] &\to M, \\ \tilde\gamma_n(s) &:= \gamma_n(s + t_n). \end{align*} Then $\tilde\gamma_n$ is a unit-speed minimizing geodesic with $\tilde\gamma_n(0) = \gamma_n(t_n) \in K$, $\tilde\gamma_n(-t_n) = p_n$, and $\tilde\gamma_n(L_n - t_n) = q_n$. The defining lengths are \begin{align*} a_n &:= t_n = d_g(\gamma_n(t_n), p_n) \geq d_g(p_n, K), \\ b_n &:= L_n - t_n = d_g(\gamma_n(t_n), q_n) \geq d_g(q_n, K). \end{align*} The first inequality uses that the segment $\gamma_n|_{[0, t_n]}$ from $p_n$ to $\gamma_n(t_n) \in K$ has length $t_n$, and minimizing distance to $K$ is bounded above by this. The second is symmetric. By the choice of $p_n, q_n$ in Step 1, \begin{align*} a_n \to \infty \quad \text{and} \quad b_n \to \infty \quad \text{as } n \to \infty. \end{align*}[/step]

[guided]The workhorse here is the [Hopf–Rinow theorem](/theorems/2719), which packages three equivalent properties of a Riemannian manifold and one important consequence: (i) geodesic completeness (every geodesic extends to all of $\mathbb{R}$, i.e., the exponential map is defined on all of $T_pM$) is equivalent to metric completeness as a metric space, (ii) closed bounded subsets are compact (the Heine–Borel property), and (iii) any two points are joined by a *minimizing* geodesic. Our hypothesis is metric completeness, so all three properties are available. We use (iii) right now to produce the joining geodesics; (ii) will reappear in Step 3 for Arzelà–Ascoli. Applied with the points $p_n, q_n \in M$, Hopf–Rinow yields, for each $n$, a unit-speed minimizing geodesic from $p_n$ to $q_n$: \begin{align*} \gamma_n : [0, L_n] &\to M, \qquad L_n := d_g(p_n, q_n) > 0, \\ \gamma_n(0) &= p_n, \quad \gamma_n(L_n) = q_n, \quad |\dot\gamma_n| \equiv 1, \quad L(\gamma_n|_{[0, L_n]}) = L_n. \end{align*} Here $L(\cdot)$ denotes Riemannian arclength, the length of the parameter interval coincides with the distance $L_n$ (this is what "minimizing" means), and unit speed is achieved by the standard arclength reparametrization. Next, we observe that $\gamma_n$ must cross $K$. Why? This is purely topological: $A_1, A_2$ are disjoint open subsets of $M \setminus K$. The image $\gamma_n([0, L_n])$ is a path-connected subset of $M$ joining $p_n \in A_1$ and $q_n \in A_2$. If $\gamma_n([0, L_n])$ were disjoint from $K$, it would lie in $M \setminus K$ and, being path-connected, would lie inside a single connected component. But its endpoints are in different components, a contradiction. Hence there exists $t_n \in [0, L_n]$ with $\gamma_n(t_n) \in K$ — this is the content of the claim above. Now we *center* the family at $K$ by reparametrizing so that the crossing point sits at parameter $0$: \begin{align*} \tilde\gamma_n : [-t_n, L_n - t_n] &\to M, \\ \tilde\gamma_n(s) &:= \gamma_n(s + t_n). \end{align*} This is just a shift of the parameter; minimality, unit speed, and arclength are unchanged. We have $\tilde\gamma_n(0) = \gamma_n(t_n) \in K$, $\tilde\gamma_n(-t_n) = p_n$, and $\tilde\gamma_n(L_n - t_n) = q_n$. Set \begin{align*} a_n &:= t_n = d_g(\gamma_n(t_n), p_n) \geq d_g(p_n, K), \\ b_n &:= L_n - t_n = d_g(\gamma_n(t_n), q_n) \geq d_g(q_n, K). \end{align*} The first equality holds because $\tilde\gamma_n|_{[-a_n, 0]}$ is a unit-speed minimizing path of length $a_n$ from $p_n$ to $\gamma_n(t_n)$; minimizing means its length equals the distance between its endpoints. The first inequality holds because $\gamma_n(t_n) \in K$, so $d_g(p_n, K) \leq d_g(p_n, \gamma_n(t_n)) = a_n$ (the distance to a set is the infimum of distances to points in the set, and $\gamma_n(t_n)$ is one such point). Symmetrically for $b_n$. Combining with the choice of $p_n, q_n$ from Step 1, \begin{align*} a_n \to \infty \quad \text{and} \quad b_n \to \infty \quad \text{as } n \to \infty. \end{align*} This is the key geometric fact we have established: as $n \to \infty$, we have longer and longer unit-speed minimizing geodesics passing through a fixed compact set $K$ at parameter $s = 0$, with the parameter interval $[-a_n, b_n]$ extending to infinity in *both* directions. This two-sided unboundedness is exactly what disconnectedness at infinity bought us.[/guided]

[step:Apply Arzelà–Ascoli to extract a limiting unit-speed curve $\gamma_\infty : \mathbb R \to M$]We extract a limit. The geodesics $\tilde\gamma_n$ are unit-speed, so on any finite subinterval $[-T, T] \subseteq [-a_n, b_n]$ (which holds for all sufficiently large $n$ since $a_n, b_n \to \infty$), the family is uniformly Lipschitz (Lipschitz constant $1$ in arclength) and uniformly bounded — bounded because every $\tilde\gamma_n(s) \in \overline{B}(\gamma_n(t_n), |s|) \subseteq \overline{B}(K, T)$ for $s \in [-T, T]$, and $\overline{B}(K, T)$ is closed and bounded, hence compact by Hopf–Rinow on the complete manifold $(M, g)$. By the Arzelà–Ascoli theorem applied to maps $[-T, T] \to \overline{B}(K, T)$ (a compact metric space), the family $\{\tilde\gamma_n|_{[-T,T]}\}$ has a uniformly convergent subsequence on $[-T, T]$. By a diagonal argument, passing to a single subsequence (still indexed by $n$ for brevity), we obtain a continuous limit \begin{align*} \gamma_\infty : \mathbb R \to M \end{align*} with $\tilde\gamma_n \to \gamma_\infty$ uniformly on every compact subinterval $[-T, T] \subset \mathbb R$. [claim:The limit $\gamma_\infty$ is a unit-speed geodesic] [proof] Geodesics are stable under uniform-on-compacts limits in the following sense: a uniform-on-compacts limit of unit-speed geodesics $\tilde\gamma_n : [-T, T] \to M$ is itself a unit-speed geodesic on $[-T, T]$. Concretely, fix $s_0 \in (-T, T)$ and consider a small neighbourhood $\overline{B}(\gamma_\infty(s_0), \rho)$ that is geodesically convex (such a neighbourhood exists by the standard convexity radius construction at any point of a Riemannian manifold). For all sufficiently large $n$, $\tilde\gamma_n(s) \in \overline{B}(\gamma_\infty(s_0), \rho)$ for $s$ in a neighbourhood of $s_0$. In a normal coordinate chart at $\gamma_\infty(s_0)$, the geodesic equation $\ddot\gamma + \Gamma_{ij}^k(\gamma) \dot\gamma^i \dot\gamma^j = 0$ is a second-order ODE with smooth coefficients depending only on the position $\gamma(s)$. Geodesics are determined locally by initial position and initial velocity; the unit-speed condition fixes $|\dot\gamma| = 1$. For each $\tilde\gamma_n$, write $\tilde\gamma_n$ as the unique solution of the geodesic ODE with initial data $(\tilde\gamma_n(s_0), \dot{\tilde\gamma}_n(s_0))$ at $s = s_0$. Since the velocities $\dot{\tilde\gamma}_n(s_0)$ are unit vectors in $T_{\tilde\gamma_n(s_0)} M$ and $\tilde\gamma_n(s_0) \to \gamma_\infty(s_0)$, by passing to a further subsequence we may assume $\dot{\tilde\gamma}_n(s_0) \to v_\infty \in T_{\gamma_\infty(s_0)} M$ with $|v_\infty| = 1$ (using that the unit tangent bundle is a smooth manifold and unit vectors at nearby points converge along the convergence of basepoints). By smooth dependence of geodesics on initial data, $\tilde\gamma_n(s) \to \gamma_v(s)$ uniformly on a neighbourhood of $s_0$, where $\gamma_v$ is the unit-speed geodesic with $\gamma_v(s_0) = \gamma_\infty(s_0)$ and $\dot\gamma_v(s_0) = v_\infty$. By uniqueness of uniform limits, $\gamma_\infty(s) = \gamma_v(s)$ near $s_0$. So $\gamma_\infty$ is a unit-speed geodesic on a neighbourhood of every interior point of $[-T, T]$, hence on $(-T, T)$, and by continuity at the endpoints, on $[-T, T]$. Since $T$ was arbitrary, $\gamma_\infty$ is a unit-speed geodesic on all of $\mathbb R$. [/proof] [/claim] So $\gamma_\infty : \mathbb R \to M$ is a unit-speed geodesic with $\gamma_\infty(0) = \lim_n \tilde\gamma_n(0) \in K$ (since $K$ is compact and $\tilde\gamma_n(0) \in K$, the limit lies in $K$).[/step]

[guided]The Arzelà–Ascoli step is the analytic heart of the construction. We need to extract a limit from the family $\{\tilde\gamma_n\}$, but each $\tilde\gamma_n$ is defined on a different interval $[-a_n, b_n]$. The idea is to fix a compact subinterval $[-T, T]$, observe that for $n$ large the restriction $\tilde\gamma_n|_{[-T, T]}$ is well-defined (since $a_n, b_n \to \infty$), apply Arzelà–Ascoli on this fixed interval, and then use a diagonal argument to obtain a single limit on all of $\mathbb{R}$. To apply Arzelà–Ascoli we need to verify two hypotheses: - *Uniform boundedness on compacts*. For $s \in [-T, T]$, since $\tilde\gamma_n(0) \in K$ and $\tilde\gamma_n$ has unit speed, the triangle inequality gives $d_g(\tilde\gamma_n(s), K) \leq d_g(\tilde\gamma_n(s), \tilde\gamma_n(0)) \leq |s| \leq T$. So $\tilde\gamma_n(s) \in \overline{B}(K, T) := \{x \in M : d_g(x, K) \leq T\}$. The set $\overline{B}(K, T)$ is closed and bounded (bounded because $K$ is bounded), so by [Hopf–Rinow](/theorems/2719) — using property (ii), closed bounded sets are compact in a complete manifold — it is compact. - *Equicontinuity*. Each $\tilde\gamma_n$ has unit speed, hence is $1$-Lipschitz: $d_g(\tilde\gamma_n(s), \tilde\gamma_n(s')) \leq |s - s'|$. The Lipschitz constant is the same for every $n$, which is precisely uniform equicontinuity. Both hypotheses of Arzelà–Ascoli are satisfied for the family $\{\tilde\gamma_n|_{[-T, T]}\} \subset C^0([-T, T]; \overline{B}(K, T))$, so a subsequence converges uniformly on $[-T, T]$. We now apply the diagonal trick: extract a subsequence converging uniformly on $[-1, 1]$, refine to a sub-subsequence converging on $[-2, 2]$, and so on; take the diagonal subsequence (still indexed by $n$ for brevity). This produces a single sequence converging uniformly on every $[-T, T]$. The pointwise limit defines a continuous map \begin{align*} \gamma_\infty : \mathbb R \to M \end{align*} with $\tilde\gamma_n \to \gamma_\infty$ uniformly on every compact subinterval $[-T, T] \subset \mathbb{R}$. So far $\gamma_\infty$ is just continuous. The substantive question is whether it is a *geodesic*. This is the content of the claim: a uniform-on-compacts limit of unit-speed geodesics is itself a unit-speed geodesic. We sketch the argument. Geodesics are characterized as solutions of a second-order ODE — the geodesic equation $\ddot\gamma + \Gamma_{ij}^k(\gamma) \dot\gamma^i \dot\gamma^j = 0$ in normal coordinates — with smooth coefficients depending only on position. A geodesic is uniquely determined by its initial data $(\gamma(s_0), \dot\gamma(s_0))$. Pick any $s_0 \in \mathbb{R}$ and a geodesically convex neighbourhood of $\gamma_\infty(s_0)$. For $n$ large, $\tilde\gamma_n(s_0)$ lies inside this neighbourhood. The unit tangent vectors $\dot{\tilde\gamma}_n(s_0)$ live in the unit tangent bundle, which is a smooth manifold; passing to a further subsequence, they converge to a unit vector $v_\infty \in T_{\gamma_\infty(s_0)} M$. By smooth dependence of geodesic flow on initial data, $\tilde\gamma_n \to \gamma_v$ uniformly near $s_0$, where $\gamma_v$ is the unit-speed geodesic with initial data $(\gamma_\infty(s_0), v_\infty)$. By uniqueness of uniform limits, $\gamma_\infty = \gamma_v$ near $s_0$, i.e., $\gamma_\infty$ is a unit-speed geodesic on a neighbourhood of $s_0$. Since $s_0$ was arbitrary, $\gamma_\infty$ is a unit-speed geodesic on all of $\mathbb{R}$. Finally, $\gamma_\infty(0) \in K$. Why? Each $\tilde\gamma_n(0) \in K$, the sequence $\tilde\gamma_n(0)$ converges to $\gamma_\infty(0)$ by uniform convergence at $s = 0$, and $K$ is compact (in particular closed), so the limit lies in $K$.[/guided]

[step:Show that $\gamma_\infty$ is minimizing on every compact subinterval]It remains to verify that $\gamma_\infty$ is a *line*: minimizing on every finite subinterval $[s_1, s_2] \subset \mathbb R$. Fix $s_1 < s_2$ in $\mathbb R$. For all sufficiently large $n$, we have $-a_n < s_1$ and $s_2 < b_n$, so $[s_1, s_2] \subseteq [-a_n, b_n]$, the domain of $\tilde\gamma_n$. Each $\tilde\gamma_n$ is minimizing on its full domain $[-a_n, b_n]$ (it is a reparametrization of the minimizing $\gamma_n$). In particular, the restriction $\tilde\gamma_n|_{[s_1, s_2]}$ is minimizing between its endpoints: \begin{align*} L(\tilde\gamma_n|_{[s_1, s_2]}) = s_2 - s_1 = d_g(\tilde\gamma_n(s_1), \tilde\gamma_n(s_2)). \end{align*} Indeed, restrictions of minimizing geodesics are minimizing (any shorter path between $\tilde\gamma_n(s_1)$ and $\tilde\gamma_n(s_2)$ would yield a shorter path between $p_n$ and $q_n$ when concatenated with the unaffected ends, contradicting minimality of $\tilde\gamma_n$ on $[-a_n, b_n]$). Now pass to the limit. The lengths satisfy $L(\tilde\gamma_n|_{[s_1, s_2]}) = s_2 - s_1$ identically (no limit needed for length, since unit speed). The endpoints converge uniformly: $\tilde\gamma_n(s_1) \to \gamma_\infty(s_1)$, $\tilde\gamma_n(s_2) \to \gamma_\infty(s_2)$. The Riemannian distance $d_g$ is continuous as a map $M \times M \to \mathbb R$, so \begin{align*} d_g(\tilde\gamma_n(s_1), \tilde\gamma_n(s_2)) \to d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)). \end{align*} Combining with $d_g(\tilde\gamma_n(s_1), \tilde\gamma_n(s_2)) = s_2 - s_1$ for all $n$: \begin{align*} d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)) = s_2 - s_1 = L(\gamma_\infty|_{[s_1, s_2]}). \end{align*} This is exactly the condition for $\gamma_\infty|_{[s_1, s_2]}$ to be minimizing between its endpoints. Since $s_1 < s_2$ were arbitrary, $\gamma_\infty$ is minimizing on every compact subinterval. By definition, $\gamma_\infty : \mathbb R \to M$ is a **line** in $M$, i.e., a unit-speed geodesic $\mathbb R \to M$ such that $d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)) = |s_2 - s_1|$ for all $s_1, s_2 \in \mathbb R$. This completes the proof.[/step]

[guided]We have produced a unit-speed geodesic $\gamma_\infty : \mathbb{R} \to M$. To upgrade this to a *line*, we must verify that $\gamma_\infty$ is minimizing on every finite subinterval. The strategy is a continuity argument: each $\tilde\gamma_n$ is minimizing on its full domain $[-a_n, b_n]$, the unboundedness $a_n, b_n \to \infty$ ensures any fixed subinterval is eventually inside this domain, and minimality passes to the limit. Fix $s_1 < s_2$ in $\mathbb{R}$. Since $a_n \to \infty$ and $b_n \to \infty$, for all sufficiently large $n$ we have $-a_n < s_1$ and $s_2 < b_n$, so $[s_1, s_2] \subseteq [-a_n, b_n]$. (This is exactly where the *two-sided* unboundedness from Step 1 is consumed: if $a_n$ stayed bounded, the inclusion would fail for very negative $s_1$, and the argument below would only give a ray.) Each $\tilde\gamma_n$ is minimizing on its full domain $[-a_n, b_n]$ — it is a reparametrization of the minimizing $\gamma_n$ from Step 2. Restrictions of minimizing geodesics are themselves minimizing: any shorter path between $\tilde\gamma_n(s_1)$ and $\tilde\gamma_n(s_2)$ could be concatenated with the unaffected segments $\tilde\gamma_n|_{[-a_n, s_1]}$ and $\tilde\gamma_n|_{[s_2, b_n]}$ to give a strictly shorter path from $p_n$ to $q_n$, contradicting minimality of $\tilde\gamma_n$ on $[-a_n, b_n]$. Hence \begin{align*} L(\tilde\gamma_n|_{[s_1, s_2]}) = s_2 - s_1 = d_g(\tilde\gamma_n(s_1), \tilde\gamma_n(s_2)). \end{align*} The first equality is unit speed; the second is the minimizing property of the restriction. We now pass to the limit $n \to \infty$. The arclength side requires nothing: it is identically $s_2 - s_1$ for every $n$. For the distance side, recall from Step 3 that $\tilde\gamma_n \to \gamma_\infty$ uniformly on $[s_1, s_2]$, so in particular $\tilde\gamma_n(s_1) \to \gamma_\infty(s_1)$ and $\tilde\gamma_n(s_2) \to \gamma_\infty(s_2)$. The Riemannian distance $d_g : M \times M \to \mathbb{R}$ is continuous (a metric is continuous in its arguments by the reverse triangle inequality), so \begin{align*} d_g(\tilde\gamma_n(s_1), \tilde\gamma_n(s_2)) \to d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)). \end{align*} Combining the two displays — the first holds for every $n$, the second is the limit — we conclude \begin{align*} d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)) = s_2 - s_1 = L(\gamma_\infty|_{[s_1, s_2]}). \end{align*} This is the minimizing condition: arclength equals distance between endpoints. Since $s_1 < s_2$ were arbitrary, $\gamma_\infty$ is minimizing on every compact subinterval. By the definition of a line — a unit-speed geodesic $\mathbb{R} \to M$ such that $d_g(\gamma_\infty(s_1), \gamma_\infty(s_2)) = |s_2 - s_1|$ for all $s_1, s_2 \in \mathbb{R}$ — this completes the proof. A final retrospective: completeness of $(M, g)$ has been used essentially three times. (i) Hopf–Rinow in Step 2 to produce the minimizing geodesics $\gamma_n$ between $p_n$ and $q_n$. (ii) Hopf–Rinow's "closed bounded sets are compact" in Step 3, which gave compactness of $\overline{B}(K, T)$ for Arzelà–Ascoli. (iii) Implicitly, completeness ensures the limiting geodesic $\gamma_\infty$ extends to all of $\mathbb{R}$: in an incomplete manifold, the geodesic flow could escape to infinity in finite time, and the limiting curve would have a strictly bounded interval of definition. All three uses are essential — none can be removed without breaking the construction.[/guided]