Geodesic completeness means compactness

The identity chart and the constant coefficient $g_{11}=1$ make $(M,g)$ a nonempty smooth boundaryless Riemannian $1$-manifold by [F1]. Since $\mathbb{R}$ is convex, [F1] also makes it connected.

Fix $x,y\in \mathbb{R}$. If $x=y$, the constant curve has length zero. If $x\mathrm{\ne }y$, let $u=(y-x)\mathrm{/}\mathrm{\mid }y-x\mathrm{\mid }\in \{-1,1\}$. For any piecewise-$C^1$ curve $\sigma$ from $x$ to $y$, the constant unit field $u$ is the gradient of $s\mapsto us$, so [F2] evaluates its line integral as $u(y-x)=\mathrm{\mid }y-x\mathrm{\mid }$ and bounds this by the Euclidean length of $\sigma$, which equals its $g$-length because $g=d{x}^2$. Conversely the segment $c(t)=(1-t)x+ty$ on $[0,1]$ has constant speed $\mathrm{\mid }y-x\mathrm{\mid }$ and length $\mathrm{\mid }y-x\mathrm{\mid }$. Taking the infimum in [F2] therefore gives $d_g(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ in both cases.

By step 1.2, the Riemannian metric space $(M,d_g)$ is exactly the usual metric real line, so [F3] makes it complete. The hypotheses checked in step 1.1 let [F4] apply under [A1], and metric completeness then makes $(M,g)$ geodesically complete. In particular every maximal geodesic, including the constant one, has domain all of $\mathbb{R}$.

The whole space $M$ is closed in itself but is not bounded: for any centre $a\in \mathbb{R}$ and radius $r>0$, the point $a+r+1$ has $d_g(a,a+r+1)=r+1>r$ by step 1.2. Hence [F5] says that $M$ is not compact. Together with step 2.1, this is the required geodesically complete noncompact counterexample.

The exact failed inference is now visible in [F4]: completeness makes every closed bounded subset compact, but the whole Euclidean line is not bounded, so that clause cannot be applied to $M$ itself. Andrews proves the completeness equivalences and the minimizing-geodesic conclusion in the cited Hopf--Rinow theorem, printed pp. 106--108; it does not assert compactness of the whole manifold and does not supply this Euclidean-line counterexample.

The empty manifold is compact and supplies no noncompact witness; a connected nonempty zero-manifold is a compact singleton. The Euclidean-line witness is genuinely one-dimensional, nonempty and boundaryless. Its distance calculation includes $x=y$, its unboundedness uses positive radii, and the completeness conclusion covers zero-velocity as well as nonconstant geodesics with full parameter domain $\mathbb{R}$, so there is no finite endpoint or degenerate-interval omission. Every displayed witness is explicit. The direct distance calculation [F2], Euclidean completeness [F3], Heine--Borel [F5] and the unboundedness calculation are choice-free. Assumption [A1] is used only when invoking the current Hopf--Rinow interface [F4] to pass from metric to geodesic completeness; no full choice axiom is used. The proof refutes the forward implication; no converse is asserted here.