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Let $S\subset {\mathbb{R}}^3$ be a smooth surface. Suppose its Gaussian curvature tends to infinity in absolute value along every maximal geodesic with a finite affine-parameter endpoint. Then $S$ is an inextendible embedded surface: a smooth extension would let some geodesic reach the extension boundary in finite time while its curvature remained locally bounded.

Suppose that a connected smooth surface $\tilde{S}\subset {\mathbb{R}}^3$ properly contains $S$. Since $S$ and $\tilde{S}$ have the same dimension, $S$ is open in $\tilde{S}$. Choose a point of its relative boundary. Inside a sufficiently small compact normal neighbourhood in $\tilde{S}$, choose $q\in S$ and a closest point $p\in \tilde{S}\setminus S$. A minimizing $\tilde{S}$-geodesic from $q$ to $p$ stays in $S$ before its endpoint, since encountering the complement earlier would contradict the choice of $p$.

The restriction of this curve is consequently a geodesic of $S$ with a finite maximal endpoint at $p$. On the other hand, the Gaussian curvature of $S$ is the restriction of the smooth Gaussian curvature of $\tilde{S}$, so it remains bounded near $p$. This contradicts condition $(\ast )$. Henceso $S$ is an inextendible embedded surface. This is the curvature-blowup criterion for inextendibility of a surface.

The punctured circular coneis geodesically incomplete because a generator reaches the missing vertex in finite time. It is an inextendible embedded surface: any proper connected embedded extension would have to add the vertex, but its tangent planes have different limits as $\theta$ varies. Nevertheless its Gaussian curvature of a cone away from its vertex is identically zero.