A Riemannian manifold is flat iff it is locally isometric to Euclidean space

Suppose first that a neighborhood $U$ of $p$ has a local isometry $y$ to an open subset of Euclidean space. In the coordinate frame ${\mathrm{\partial }}_i$ induced by $y$, [F5] gives $g({\mathrm{\partial }}_i,{\mathrm{\partial }}_j)={\delta }_{ij}$. Write ${\mathrm{\Gamma }}_{ijk}=g({\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_j,{\mathrm{\partial }}_k)$. Torsion freeness in [F3] makes ${\mathrm{\Gamma }}_{ijk}={\mathrm{\Gamma }}_{jik}$, while metric compatibility and the constant metric coefficients make ${\mathrm{\Gamma }}_{ijk}=-{\mathrm{\Gamma }}_{ikj}$. Alternating these two relations around the three indices gives ${\mathrm{\Gamma }}_{ijk}=-{\mathrm{\Gamma }}_{ijk}$, so every ${\mathrm{\Gamma }}_{ijk}=0$.

Conversely suppose $\mathrm{Rm}=0$. Nondegeneracy of $g$ in [F2] gives $R=0$, so [F1] supplies near $p$ a parallel frame $(V_1,\dots ,V_n)$. Shrink its domain to a connected coordinate neighborhood. Metric compatibility [F3] gives $d(g(V_i,V_j))=0$; by [F7], every entry of this Gram matrix is constant there.

Thus every coordinate field in step 1.1 is parallel on $U$. Substitution in the curvature commutator gives $R({\mathrm{\partial }}_i,{\mathrm{\partial }}_j){\mathrm{\partial }}_k=0$; tensoriality and [F2] give $\mathrm{Rm}=0$ on $U$. Since such neighborhoods cover $M$, local Euclidean isometry implies $\mathrm{Rm}=0$ globally.

Apply [F6] to $(V_1(p),\dots ,V_n(p))$. The resulting orthonormal basis is obtained by an invertible constant matrix $C$; applying that same matrix to the fields defines a parallel frame $(E_1,\dots ,E_n)$. The Gram matrix is constant by step 1.2 and equals the identity at $p$, so this frame is orthonormal throughout the neighborhood.

Torsion freeness and parallelness give $[E_i,E_j]={\mathrm{\nabla }}_{E_i}{E}_j-{\mathrm{\nabla }}_{E_j}{E}_i=0$. By [F4], after shrinking again there are coordinates $(x^1,\dots ,x^n)$ with $E_i=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^i$. Consequently $g_{ij}=g(E_i,E_j)={\delta }_{ij}$, so the coordinate map is a local diffeomorphism satisfying $g=x^{\ast }{g}_{\mathrm{E}\mathrm{u}\mathrm{c}}$ and hence is a Riemannian local isometry by [F5]. This proves the reverse implication at the arbitrary point $p$.

In dimension zero, each point is itself an open neighborhood and is isometric to the unique open subset ${\mathbb{R}}^0$, while both curvature tensors vanish. In dimension one the same proof applies and the pair skews force curvature to vanish. The empty manifold satisfies both universal conditions. The boundaryless hypothesis is essential to the stated target: a boundary point cannot have a neighborhood locally diffeomorphic to an open subset of ${\mathbb{R}}^n$. No infinite or global selection is made.