Any two points admit a minimizing geodesic

The set $M$ is open: if $z\in M$, the Euclidean ball of radius $\parallel z{\parallel }_2\mathrm{/}2$ about $z$ misses the origin. Thus [F1] makes $M$ a smooth boundaryless $2$-manifold. The restricted tensor has the constant identity matrix, so it is smooth and positive definite and is a Riemannian metric by [F1]. The radius-zero clause of [F1] makes $M$ path-connected, hence connected; it is plainly nonempty.

Let $\sigma :[a,b]\to M$ be any piecewise-$C^1$ curve from $p$ to $q$, and write $\sigma (t)=(x(t),y(t))$. The function $x$ is continuous, with $x(a)=-1$ and $x(b)=1$, so [F4] supplies $t_0\in (a,b)$ with $x(t_0)=0$. Put $z=\sigma (t_0)=(0,y_0)$. Because $\sigma$ takes values in $M$, one has $y_0\mathrm{\ne }0$.

For every real $\epsilon >0$, define a two-segment curve $c_{\epsilon }:[0,2]\to M$ by $c_{\epsilon }(t)=(-1+t,\epsilon t)$ for $0\le t\le 1$ and $c_{\epsilon }(t)=(t-1,\epsilon (2-t))$ for $1\le t\le 2$. It joins $p$ to $q$ through $(0,\epsilon )$. On the first segment, a zero first coordinate forces $t=1$ and then the second coordinate is $\epsilon$; on the second it again forces $t=1$. Hence the curve never meets the origin. Its two constant velocities are $(1,\epsilon )$ and $(1,-\epsilon )$, so [F2] gives $L_g(c_{\epsilon })=2\sqrt{1+{\epsilon }^2}$. Moreover $\sqrt{1+{\epsilon }^2}\le 1+{\epsilon }^2\mathrm{/}2$, because squaring the nonnegative right side adds ${\epsilon }^4\mathrm{/}4$. Given $\delta >0$, taking $0<\epsilon <\sqrt{\delta }$ therefore gives $L_g(c_{\epsilon })<2+\delta$.

Refine a piecewise-$C^1$ subdivision to include $t_0$. On the first restriction, take the constant Euclidean unit vector $u=(z-p)\mathrm{/}\parallel z-p{\parallel }_2$. It is the gradient of the linear function $w\mapsto u\cdot w$, so [F3] evaluates its line integral as $u\cdot (z-p)=\parallel z-p{\parallel }_2$ and bounds this by the Euclidean length of the restriction. The restricted Riemannian metric is Euclidean, so [F2] identifies that length with its Riemannian length. Applying the same argument to the second restriction, with $(q-z)\mathrm{/}\parallel q-z{\parallel }_2$, and then using additivity gives $L_g(\sigma )\ge \parallel z-p{\parallel }_2+\parallel q-z{\parallel }_2=2\sqrt{1+y_0^2}>2$. Thus every competitor has length strictly greater than $2$.

Steps 2.1 and 1.3 show that the set of competitor lengths has infimum exactly $2$, so [F2] gives $d_g(p,q)=2$. Step 2.1 also shows that no competitor has length $2$. Consequently no length-minimizing curve, and in particular no minimizing geodesic, joins $p$ to $q$. This refutes the full universal claim.

The precise missing hypothesis is completeness, not connectedness. Indeed step 1.1 verifies all of [F5]'s manifold hypotheses, while step 3.1 contradicts the minimizing-geodesic conclusion; under [A1], [F5] therefore shows that this $M$ is neither metrically nor geodesically complete. Andrews proves the complete-manifold implication in the cited Hopf--Rinow theorem, printed pp. 106--108, but does not give this punctured-plane counterexample or its nonattainment calculation; those are proved in steps 1.1--3.1.

Empty manifolds have no pair of points to witness the failure, and a nonempty connected zero-manifold is a singleton, where the constant geodesic minimizes. A punctured line is disconnected, so dimension two is used to keep the counterexample connected. Here $p\mathrm{\ne }q$, the parameter intervals and both detour pieces are nondegenerate, $t_0$ is an interior parameter, and all curve endpoints are included. The witnesses $p,q,c_{\epsilon }$ are explicit and no family of them is selected. The intermediate-value construction [F4] is canonical and choice-free, and the explicit line-integral and detour calculations in steps 1.1--3.1 use no choice principle. Assumption [A1] is used only when invoking the current Hopf--Rinow interface [F5] for the completeness diagnosis in step 4.1; no full choice axiom is used. There is no iff claim to check.