# Open set of positive curvature exists in $S \subset \mathbb{R}^3$ regular, compact, connected, orientable.

In answering the following question from Do Carmos book on curves and surfaces

Let $S \subset \mathbb{R}^3$ be regular, compact, connected, orientable surface which is not homeomorphic to a sphere. Prove that there are points on $S$ where the Gaussian curvature is positive, negative and zero.

I know that we have Gauss-Bonnet: $\int_S K=2\pi\chi(S)$ and that for such a surface not a sphere $\chi(S) < 0$ so clearly there must be a open set with negative curvature. But why must there be a open set with positive curvature? (I know the zero part will follow from the intermediate value theorem).

Intuitively I know that $K$ is the product of the principle curvatures and therefore any open set with negative curvature must locally look like a 'saddle', and I cannot fathom a surface that looks like that everywhere, but this doesn't seem rigorous.

Can anybody help?

Every compact connected surface in $\mathbb{R}^3$ has points of positive curvature. Consider a family of spheres at the origin you expand, until you have a tangent point with one such sphere. This is a point of positive curvature.
Observe you don't need a region of positive curvature, only a non-zero number of points. It turns out that you get a region for free by continuity, if the surface is at least $C^2$.