[proofplan] We prove the three claims in order. For (1), given $q = \exp_p(a) \in U$ with $|a| = r < \varepsilon$, the radial geodesic $\gamma(t) = \exp_p(ta)$ is a smooth curve from $p$ to $q$ of length $r$; uniqueness comes from $\exp_p$ being a diffeomorphism on $B(0, \varepsilon)$. The length-minimization step lifts any other curve $\tilde\gamma \in \Omega(p, q)$ to $T_pM$ (on a maximal initial segment that stays in $B(0, \varepsilon)$) and applies [Local Length Minimization](/theorems/2719) to conclude $\operatorname{length}(\tilde\gamma) \geq r$. For (2), the contrapositive: if $q \notin U$ we show $d(p, q) \geq \varepsilon$ by lifting any curve from $p$ to $q$ to $T_pM$ on its maximal initial segment in $B(0, \varepsilon)$ and observing it must reach radius $\varepsilon$ before exiting. For (3), if $\operatorname{length}(\gamma) = d(p, q) < \varepsilon$, then by (2) $q \in U$, by (1) the unique length minimizer is the radial geodesic $\exp_p(ta)$, and so $\gamma$ must coincide with it (up to reparametrization, which (1) excludes by uniqueness in the parametrization class). [/proofplan] [step:Establish notation and the candidate geodesic for part (1)] Let $q \in U$. Since $\exp_p\big|_{B(0, \varepsilon)}$ is a diffeomorphism onto $U$, there is a unique $a \in B(0, \varepsilon) \subseteq T_pM$ with $\exp_p(a) = q$. Set $r := |a|_{g_p} \in [0, \varepsilon)$. Define \begin{align*} \gamma: [0, 1] &\to M \\ t &\mapsto \exp_p(ta). \end{align*} This is the radial geodesic with initial point $p$ and initial velocity $a \in T_pM$, evaluated on $[0, 1]$. By [Geodesics Have Constant Speed](/theorems/2709), $|\dot\gamma(t)|_{g_{\gamma(t)}} = |a|_{g_p} = r$ for all $t \in [0, 1]$. Therefore \begin{align*} \operatorname{length}(\gamma) = \int_0^1 |\dot\gamma(t)|_{g_{\gamma(t)}} d\mathcal{L}^1(t) = \int_0^1 r\, d\mathcal{L}^1(t) = r < \varepsilon. \end{align*} Moreover, $\gamma(0) = \exp_p(0) = p$ and $\gamma(1) = \exp_p(a) = q$, so $\gamma \in \Omega(p, q)$. [/step] [step:Show $\gamma$ is the unique geodesic in $\Omega(p, q)$ of length less than $\varepsilon$] Suppose $\eta \in \Omega(p, q)$ is another geodesic with $\operatorname{length}(\eta) < \varepsilon$. Reparametrise $\eta$ to be defined on $[0, 1]$ with constant speed; this preserves the geodesic property by linear reparametrization invariance of geodesics, and preserves the length. Then $\eta$ is a constant-speed geodesic with $|\dot\eta(t)| = \operatorname{length}(\eta) < \varepsilon$ for all $t$. Set $b := \dot\eta(0) \in T_pM$, so $|b|_{g_p} = \operatorname{length}(\eta) < \varepsilon$. Hence $b \in B(0, \varepsilon)$. We claim $\eta(t) = \exp_p(tb)$ for all $t \in [0, 1]$. Since $|tb|_{g_p} \le |b|_{g_p} < \varepsilon$ for $t \in [0, 1]$, the curve $t \mapsto \exp_p(tb)$ is defined on $[0, 1]$. ODE uniqueness for the geodesic equation gives equality only locally; we extend it to all of $[0, 1]$ by a connectedness argument. Define \begin{align*} S := \{t \in [0, 1] : \eta(t) = \exp_p(tb)\}. \end{align*} (i) **Non-empty.** $0 \in S$, since $\eta(0) = p = \exp_p(0)$. (ii) **Open in $[0, 1]$.** Let $t_0 \in S$. Both $\eta$ and $t \mapsto \exp_p(tb)$ are geodesics in $M$, and at $t_0$ they pass through the same point with the same velocity. By the uniqueness of geodesics with prescribed initial data applied at $t_0$, the two curves agree on an open interval $J \ni t_0$. Hence $J \cap [0, 1] \subseteq S$. (iii) **Closed in $[0, 1]$.** If $t_n \in S$ and $t_n \to t_*$, then by continuity $\eta(t_*) = \lim_n \eta(t_n) = \lim_n \exp_p(t_n b) = \exp_p(t_* b)$, so $t_* \in S$. Since $[0, 1]$ is connected and $S$ is non-empty, open, and closed, $S = [0, 1]$. Hence $\eta(t) = \exp_p(tb)$ for all $t \in [0, 1]$. Then $\eta(1) = q$ implies $\exp_p(b) = q = \exp_p(a)$. Since $\exp_p$ is injective on $B(0, \varepsilon)$ and $a, b \in B(0, \varepsilon)$, we conclude $b = a$. Hence $\eta(t) = \exp_p(ta) = \gamma(t)$ for all $t$, proving uniqueness. [/step] [step:Show $\operatorname{length}(\tilde\gamma) \geq r$ for every $\tilde\gamma \in \Omega(p, q)$] Let $\tilde\gamma: [0, 1] \to M$ be any piecewise $C^1$ curve in $\Omega(p, q)$. We lift $\tilde\gamma$ to $T_pM$ on a maximal initial segment. **Case A: $\tilde\gamma([0, 1]) \subseteq U$.** Since $\exp_p\big|_{B(0, \varepsilon)} \to U$ is a diffeomorphism, define \begin{align*} \psi: [0, 1] &\to B(0, \varepsilon) \subseteq T_pM \\ t &\mapsto (\exp_p\big|_{B(0, \varepsilon)})^{-1}(\tilde\gamma(t)). \end{align*} Then $\psi$ is piecewise $C^1$ (as the composition of a piecewise $C^1$ curve with a diffeomorphism), $\psi(0) = (\exp_p\big|_{B(0, \varepsilon)})^{-1}(p) = 0$ (since $\exp_p(0) = p$ and $0 \in B(0, \varepsilon)$), and $\psi(1) = (\exp_p\big|_{B(0, \varepsilon)})^{-1}(q) = a$. By [Local Length Minimization](/theorems/2719) applied to $\psi$, \begin{align*} \operatorname{length}(\tilde\gamma) = \operatorname{length}(\exp_p \circ \psi) \ge |a|_{g_p} = r. \end{align*} **Case B: $\tilde\gamma([0, 1]) \not\subseteq U$.** Let \begin{align*} T := \sup\{s \in [0, 1] : \tilde\gamma([0, s]) \subseteq U\}. \end{align*} Since $\tilde\gamma(0) = p \in U$ and $U$ is open in $M$, the set $\{s : \tilde\gamma([0, s]) \subseteq U\}$ is non-empty and contains a neighbourhood of $0$, so $T > 0$. By definition of $T$ and continuity, $\tilde\gamma([0, T)) \subseteq U$ but $\tilde\gamma$ does not stay in $U$ past $T$. We argue that $\tilde\gamma(T) \in \partial U$ (the topological boundary of $U$ in $M$), or $T = 1$ and $\tilde\gamma(1) = q \in U$ (which puts us back in Case A). If $T < 1$: by maximality of $T$ and continuity of $\tilde\gamma$, $\tilde\gamma(T) \in \overline{U} \setminus U = \partial U$. The boundary $\partial U$ is the image under $\exp_p$ of $\partial B(0, \varepsilon) = \{v \in T_pM : |v|_{g_p} = \varepsilon\}$ — but $\exp_p$ may not be injective on $\overline{B(0, \varepsilon)}$, so we argue more carefully. Define \begin{align*} \psi: [0, T) &\to B(0, \varepsilon) \subseteq T_pM \\ t &\mapsto (\exp_p\big|_{B(0, \varepsilon)})^{-1}(\tilde\gamma(t)), \end{align*} which is piecewise $C^1$. We claim $\lim_{t \to T^-} |\psi(t)|_{g_p} = \varepsilon$. Since $\psi([0, T)) \subseteq B(0, \varepsilon)$, the curve $\psi$ is bounded in $T_pM$, so every sequence $t_n \to T^-$ admits a subsequence along which $\psi(t_{n_k})$ converges to some $w \in \overline{B(0, \varepsilon)}$. If $w \in B(0, \varepsilon)$, then by continuity \begin{align*} \tilde\gamma(T) = \lim_k \tilde\gamma(t_{n_k}) = \lim_k \exp_p(\psi(t_{n_k})) = \exp_p(w) \in U, \end{align*} contradicting $\tilde\gamma(T) \notin U$. Hence every subsequential limit $w$ lies in $\overline{B(0, \varepsilon)} \setminus B(0, \varepsilon) = \partial B(0, \varepsilon)$, so $|w|_{g_p} = \varepsilon$. Since every subsequential limit of $|\psi(t)|_{g_p}$ as $t \to T^-$ equals $\varepsilon$, we obtain a full limit: \begin{align*} \lim_{t \to T^-} |\psi(t)|_{g_p} = \varepsilon. \end{align*} In particular, for any prescribed $\delta > 0$ pick $T' \in (0, T)$ with $|\psi(T')|_{g_p} \ge \varepsilon - \delta$. Apply [Local Length Minimization](/theorems/2719) to the curve $\psi\big|_{[0, T']}$ from $0$ to $\psi(T') \in T_pM$, after rescaling its parameter to $[0, 1]$: \begin{align*} \operatorname{length}(\tilde\gamma\big|_{[0, T']}) \ge |\psi(T')|_{g_p} \ge \varepsilon - \delta. \end{align*} Letting $\delta \to 0^+$, \begin{align*} \operatorname{length}(\tilde\gamma) \ge \operatorname{length}(\tilde\gamma\big|_{[0, T']}) \ge \varepsilon - \delta, \end{align*} hence $\operatorname{length}(\tilde\gamma) \ge \varepsilon > r$. If $T = 1$: then $\tilde\gamma([0, 1)) \subseteq U$. If additionally $\tilde\gamma(1) = q \in U$, then $\tilde\gamma([0, 1]) \subseteq U$ and we are in Case A. (The endpoint $\tilde\gamma(1) = q \in U$ holds by definition of $\Omega(p, q)$.) So in this sub-case Case A applies and we obtain $\operatorname{length}(\tilde\gamma) \ge r$. Combining the cases, $\operatorname{length}(\tilde\gamma) \ge r$ for all $\tilde\gamma \in \Omega(p, q)$. [/step] [step:Establish equality and the reparametrization characterisation: $\gamma$ is the unique length-minimising curve] We have $\operatorname{length}(\gamma) = r$ and $\operatorname{length}(\tilde\gamma) \ge r$ for all $\tilde\gamma \in \Omega(p, q)$. Hence $d(p, q) = \inf_{\tilde\gamma \in \Omega(p, q)} \operatorname{length}(\tilde\gamma) = r$, and $\gamma$ is a length minimizer. For uniqueness: suppose $\tilde\gamma \in \Omega(p, q)$ has $\operatorname{length}(\tilde\gamma) = r$. By Case B above, this forces Case A — i.e., $\tilde\gamma([0, 1]) \subseteq U$ — because Case B yielded $\operatorname{length}(\tilde\gamma) \ge \varepsilon > r$. So the lift $\psi: [0, 1] \to B(0, \varepsilon)$ exists. The statement of [Local Length Minimization](/theorems/2719) includes an equality clause: the bound $\operatorname{length}(\exp_p \circ \psi) \ge |a|$ is attained with equality if and only if $\psi(t) = \rho(t)\, v_0$, where $v_0 = a/|a|$ and $\rho: [0, 1] \to [0, |a|]$ is a continuous non-decreasing surjection with $\rho(0) = 0$ and $\rho(1) = |a| = r$. Applying this clause directly to our $\psi$ (which satisfies $\operatorname{length}(\exp_p \circ \psi) = \operatorname{length}(\tilde\gamma) = r = |a|$), we obtain $\psi(t) = \rho(t)\, v_0$ for such a $\rho$. Then \begin{align*} \tilde\gamma(t) = \exp_p(\psi(t)) = \exp_p(\rho(t)\, v_0) = \gamma(\rho(t)/r), \end{align*} which is a monotone reparametrization of $\gamma$. The set of monotone reparametrizations of $\gamma$ defines a single curve up to reparametrization, namely the geodesic segment from $p$ to $q$. So $\tilde\gamma$ is unique up to reparametrization, and the unique constant-speed reparametrization on $[0, 1]$ is $\gamma$ itself. This completes the proof of (1). [guided] The crucial input here is the equality case of [Local Length Minimization](/theorems/2719). That theorem's proof established \begin{align*} |\dot\sigma(t)|^2 = \dot\rho(t)^2 + \rho(t)^2 |B(t)|^2, \end{align*} where $B(t) = (d\exp_p)_{\psi(t)}(\dot{v}(t))$ is the angular term. Equality $\operatorname{length}(\sigma) = |a|$ requires three things: (i) $|B(t)| = 0$, i.e., $\dot{v}(t) = 0$ pointwise a.e. on the open set where $\psi$ avoids $0$; (ii) $\dot\rho(t) \ge 0$ a.e. (so the absolute value can be dropped); (iii) the radial speed $|\dot\rho|$ integrates from $0$ to exactly $|a|$. Condition (i) means $v$ is locally constant, which on a connected component of $\psi^{-1}(T_pM \setminus \{0\})$ forces $v \equiv v_0$. Condition (ii) ensures $\rho$ is non-decreasing. Condition (iii) plus $\psi(0) = 0, \psi(1) = a$ forces $\rho(0) = 0$ and $\rho(1) = r$. Thus the lift $\psi$ is exactly the radial path $t \mapsto \rho(t) v_0$ with $\rho$ a non-decreasing piecewise $C^1$ surjection from $[0, 1]$ onto $[0, r]$. Composing with $\exp_p$, the image curve $\tilde\gamma$ is a monotone reparametrization of the radial geodesic $\gamma$. The reparametrization is the function $s = \rho(t)/r: [0, 1] \to [0, 1]$. This is the standard rigidity statement: minimization in $\Omega(p, q)$ characterises the radial geodesic up to monotone reparametrization. The constant-speed condition then pins down $\gamma$ uniquely. [/guided] [/step] [step:Prove part (2) — the metric ball of radius $\varepsilon$ is contained in $U$] Let $q \in M$ with $d(p, q) < \varepsilon$. We show $q \in U$. Suppose for contradiction $q \notin U$. By definition of $d$ as infimum of lengths, \begin{align*} d(p, q) = \inf_{\tilde\gamma \in \Omega(p, q)} \operatorname{length}(\tilde\gamma). \end{align*} Choose any $\tilde\gamma \in \Omega(p, q)$ with $\operatorname{length}(\tilde\gamma) < \varepsilon$. Apply the analysis of Case B from Step 3 to $\tilde\gamma$. Since $q = \tilde\gamma(1) \notin U$, we have $T < 1$ in the notation of Step 3, where $T$ is the supremum of $s$ with $\tilde\gamma([0, s]) \subseteq U$. Then by the argument of Step 3 (Case B), $\operatorname{length}(\tilde\gamma) \ge \varepsilon$ — contradicting $\operatorname{length}(\tilde\gamma) < \varepsilon$. Hence $q \in U$. [/step] [step:Prove part (3) — the unique length minimizer of length $< \varepsilon$ is the radial geodesic] Let $q \in M$ and $\gamma' \in \Omega(p, q)$ with $\operatorname{length}(\gamma') = d(p, q) < \varepsilon$. By part (2), $q \in U$. By part (1), the unique length-minimising curve in $\Omega(p, q)$ (up to reparametrization, with constant-speed parametrization on $[0, 1]$) is the radial geodesic $\gamma(t) = \exp_p(ta)$ where $a = (\exp_p\big|_{B(0, \varepsilon)})^{-1}(q)$, which is itself a geodesic. So $\gamma'$ is a (monotone) reparametrization of $\gamma$. Reparametrizations of geodesics by piecewise $C^1$ monotone functions remain geodesics if and only if the reparametrization is affine (i.e., $\rho(t) = ct$ for some $c > 0$). For a curve $\gamma'$ that is a priori only piecewise $C^1$, the conclusion is: $\gamma'$ has the same image as a geodesic and traces it monotonically. Whether one calls this "a geodesic" depends on the parametrisation convention. Under the convention that "geodesic" requires constant-speed parametrisation, $\gamma'$ is a geodesic if and only if it has constant speed; if not, it is a reparametrization of one. In the typical formulation of this theorem (3) — which assumes $\gamma'$ has constant speed implicitly — $\gamma'$ is a (constant-speed) geodesic. [guided] The statement (3) is somewhat subtle because of the reparametrization issue. Stripped to its essentials, the claim is: any curve realising the distance $d(p, q)$ within the open ball of radius $\varepsilon$ is the radial geodesic, possibly reparametrized. A curve $\gamma'$ that achieves the minimum length within $\Omega(p, q)$ is, as proved in part (1), a reparametrization of the radial geodesic $\gamma$. If we wish to call $\gamma'$ "a geodesic" in the strong sense (constant covariant acceleration $\nabla_{\dot{\gamma'}}\dot{\gamma'} = 0$), this requires $\gamma'$ to have constant speed, i.e., the reparametrization to be affine. The image and trajectory of $\gamma'$, however, are uniquely determined. In practice, theorem (3) is invoked with the understanding that "geodesic" means "geodesic up to reparametrization", and in that sense $\gamma'$ is always a geodesic. The strict interpretation requires the additional hypothesis that $\gamma'$ has constant speed, which combined with the rigidity from part (1) forces $\gamma' = \gamma$ (the affine reparametrization on $[0, 1]$). [/guided] [/step]