Euclidean space has zero curvature

In Cartesian coordinates, $(g_{ij})=({\delta }_{ij})$ is a constant smooth symmetric matrix, and $v^T({\delta }_{ij})v={\sum }_i(v^i{)}^2>0$ for every nonzero $v$; hence [F1] makes $g_{\mathrm{E}}$ a Riemannian metric.

Every derivative ${\mathrm{\partial }}_i{g}_{j\mathrm{\ell }}={\mathrm{\partial }}_i{\delta }_{j\mathrm{\ell }}$ is zero, so [F2] gives ${\mathrm{\Gamma }}^k{}_{ij}=0$ identically for all indices; consequently every derivative ${\mathrm{\partial }}_i{\mathrm{\Gamma }}^{\mathrm{\ell }}{}_{jk}$ is also zero.

Substitution of step 2.1 into [F3] makes both derivative terms and both quadratic terms zero, so $R^{\mathrm{\ell }}{}_{kij}=0$ for every $i,j,k,\mathrm{\ell }$. Because the Cartesian coordinate vectors form a basis at every point, this is exactly $R^{g_{\mathrm{E}}}=0$ on all of ${\mathbb{R}}^n$.

For $n=0$, every index range is empty and the unique curvature field on the one-point manifold ${\mathbb{R}}^0$ is zero; for $n=1$, antisymmetry is not needed because steps 1.1–3.1 still give the sole coordinate component zero. The standard metric is nondegenerate by step 1.1, the global chart has neither a boundary nor a parameter endpoint, and all coordinates and tensors are explicit, so no choice principle is used. The claim is an equality, not a biconditional.