I have the problems 2.2 and 2.9 of Differential Geometry of Curves and Surfaces by Manfredo do Carmo and this problems give me a doubt. They are:
Is the set $\{(x,y,z)\in\mathbb{R}^3;z=0\qquad x^2+y^2\leq 1\}$ a regular surface? Is the set $\{(x,y,z)\in\mathbb{R}^3;z=0\qquad x^2+y^2< 1\}$ a regular surface?
and
Let $V$ be an open set of plane $xy$. Show that:
$$\{(x,y,z)\in\mathbb{R}^3;z=0\qquad (x,y)\in V\}$$
is a regular surface.
Well, I thinked as follow:
Given any $V\in \mathbb{R^3}$ (open or not), there is $V^*\supset V$ open in $\mathbb{R^3}$. Consider $f(u,v)=(u,v,0)$ from $V^*$ to $\mathbb{R}^3$.
This is a differenciable function, an homeomorphism, and its derivative is injective, so I thinked that this proves that any set of form
$$\{(x,y,z)\in\mathbb{R}^3;z=0\qquad (x,y)\in V\}$$
is a regular surface.
By the questions above, I think I am wrong, but I could not understand yet.
Thanks in advance for a help.
Best Answer
Fix an open subset $V \subseteq \mathbb{R}^2$, then indeed for every $k \in \mathbb{R}$ $$\{(x,y,z) \in \mathbb{R}^3: z = k, (x,y) \in V\}$$ is a regular surface parametrized by $\phi: V \to \mathbb{R}^3$, $\phi(x,y) = (x,y,0)$. The open hypothesis can't be removed: for example, your set $$\{(x,y,z) \in \mathbb{R}^3: z = 0, x^2+y^2 \leq 1\}$$ is not a regular surface. For example, you won't find a neighbourhood of any point on the border that is homeomorphic to an open subset of $\mathbb{R}^2$.