I was thinking about the fact that if a surface in $\mathbb{R}^3$ is locally isometric to the plane, then its Gaussian curvature is $K=0$ (which is a direct consequence of Theorema Egregium)
I have read on Wikipedia (https://en.m.wikipedia.org/wiki/Gaussian_curvature) that if a surface has a Gaussian curvature $K=0$, then it is developable.
I have the understanding that a surface is developable $\iff$ it is locally isometric to a plane. Is that correct ?
If so, a little help on the proof that "$K=0$ implies developable" would be appreciated. If not a little clarification or a counterexample would be much helpful.
Thanks a lot