Not coplanar surface Is there exist smooth surface (moreover, graph of some function $f(x,y)$) every 4 points of which are not coplanar?
I think it's true (parabola is the example for less dimension), but can't find the example. Also I think there is algebraic example exist.
Upd: So the answer is No, and then I have additional question here.

I actually think there might not be. Consider the four points described by $f(\cos(\theta+i\cdot\frac{\pi}{2}),\sin(\theta+i\cdot\frac{\pi}{2})),$ with $i$ from $0$ to $3$, and consider the signed volume of the tetrahedron they define. Increasing $\theta$ by $\frac{\pi}{2}$ leads to that volume being negated. By the intermediate value theorem, then, there should be a value of $\theta$ which gives a volume of $0$, indicating that the four points are coplanar.
