Let $S$ be a closed set of $\mathbb R^2$ and $f : S \to \mathbb R$ a continuously differentiable function. Show that the graph of $f$, $G(f) =$ {$(x, y, z) \in \mathbb R^3 : z = f(x, y) $} is the image of a parametrized surface.
First I don't know if there is a typo: shouldn't $f: S \to \mathbb R^3$?
And here's what I got:
A parametrized surface can be written as $r(u, v) = (x(u, v), y(u, v), z(u, v)) $, so its image by $f$ is $f(r(u, v)) = f((x(u, v), y(u, v), z(u, v))) $
How does this tell me what I want to reach? Any help please?
This is false. Take $S=\{y=0\}\subset \Bbb R^2$. This is a closed set. Let $f(x,y) = y$, so for $(x,y)\in S$ we have $f(x,y)=0$. $G(f)$ is a parametrized curve, not a parametrized surface.