Let $S = \{(x,y,z)\in \mathbb{R}^3 : z = 0, x^2+y^2\leq 1\}$ and $W = \{(x,y,z)\in \mathbb{R}^3 : z = 0, x^2+y^2< 1\}$ are those regular surfaces?
My attempt:
Suppose $S$ is a regular surfacem then for each $p\in S$ exist a neighborhood $V\in \mathbb{R}^3$ and a map $\mathbb{X} : U\rightarrow V\cap S$ with $U$ a open set in $\mathbb{R}^2$ such that $\mathbb{X}$ is a dipheomorfism and for each $q\in U$ the differential $d\mathbb{X}_q:\mathbb{R}^2 \rightarrow \mathbb{R}^3$ is injective.
Let $p$ in the border of the disk $S$, then by definition exists $V_p$ such that $p\in V_p\cap S$.
We know that $\mathbb{X}$ is a dipheomorfism, then $f$ is continous and $V_p$ is a closed set, this implies $U$ cannot be open (by topology definition of continuous function)
Is this correct?
For the second exercise:
Let $\mathbb{X}:U\subset \mathbb{R}^2 \rightarrow V\cap W$ such that $\mathbb{X}(x,y)=(x,y,0)$ with $U = \{((x,y)\in\mathbb{R}^2 : x^2 + y^2 < 1\}$, clearly $U$ is an open set.
$\mathbb{X}$ is a dipheomorfism because is continous, differentiable, with continous inverse.
Then, $W$ is a regular surface.
Are my answers correct? Thanks for read!