Local Isometry of Sphere How does one show that there exists no neighborhood of a point on a sphere that may be isometrically mapped into a plane?  I understand that I can find the first fundamental form of the sphere $(u, v, \sqrt{r^2 - u^2 - v^2})$, for fixed $a>0$, which is given by: $E = 1 + \frac{u^{2}}{r^2 - u^2 - v^2}$, $F = \frac{u v}{r^2 - u^2 - v^2}$, and $G = 1 + \frac{v^2}{r^2 - u^2 - v^2}$.  Meanwhile, for the plane $(u,v,0)$, the first fundamental form is given by: $E = 1$, $F = 0$, $G = 1$.  Since these first fundamental forms are not equal, the isometry is impossible.
But is this enough?  How do I account for the "neighborhood" condition?
In some sense, this is a follow up question to ones like that which is given here: There is no isometry between a sphere and a plane., but it is not truly derived therefrom.  The actual problem statement reads, in its entirety as follows: "Show that no neighborhood of a point on a sphere may be isometrically mapped into a plane"; this claim is made in utter isolation.  In particular, no mention of a metric is made.

If you mean the great circle distances on the sphere, it's fairly simple to brute force it.
Consider a small region on the sphere bounded by a circle.  If this were mapped isometrically to the plane, since the center of that spherical region is equidistant from the bounding circle, the same would be true in the plane: a simply connected region bounded by a circle, ie. a disk.
But now check the distance around the circle.  If the mapping were isometric, the distance around the circle would be preserved, but for the given radius (great circle distance to boundary from center of region), the flat disk will have a greater circumference.
