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.
 A: I guess it is not possible. We will use implicit function theorem. 
Suppose that we have such a $f(x,y)$.
If there is such $c,(x_0,y_0)$ that $f(x_0,y_0)=c$ and $\frac{\partial f}{\partial y}(x_0,y_0) \neq 0 $ than there is function $g(x)$ defined in neighbourhood of $x_0$. That $f(x,g(x))=c$. So you get infinity points which are coplanar.
Therefore $\frac{\partial f}{\partial y} = 0 $ everywhere. Exchange $x$ and $y$ and you get the same for $x$ coordinate.
So $f(x,y)$ has to be constant function. But on this function all points are coplanar.
So there is no such a surface.
A: 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.
