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If the Riemannian exponential map at one point $p$ of a connected finite-dimensional Riemannian manifold is defined on all of $T_{p}M$, every point can be joined to $p$ by a minimizing geodesic. The distance splitting through a small geodesic sphere selects a radial direction attaining $d(p,q)=r+d(\gamma (r),q)$. If this equality stopped before the target, repeat the splitting at its last point. A broken minimizing path cannot have a corner, so the new segment continues the same geodesic, contradicting maximality. This is the minimizing-geodesic lemma used in the radial continuation proof of Hopf-Rinow.

// Target: riemannian-geometry.bigb

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.