I'm trying to solve this exercise from Do Carmo's Differential Geometry of Curves and Surfaces, and I want a hint on how to do it. The exercise is:
Is the set $S =\left\{(x,y,z)\in \mathbb{R}^3 \mid z=0, \ \ x^2+y^2\leq1 \right\}$ a regular surface ?
Well, for me this isn't a regular surface because the set $S$ is not relatively open in the $xy$ plane, however I'm not sure how to prove this formally. I thought on first proving that a subset of a regular surface is a regular surface itself if and only if it's open relative to the set we know it's a regular surface. Then I show that the $xy$ plane is a regular surface and then $S$ cannot be a regular surface because it's not open relative to the $xy$ plane.
Is this line of thought correct ? Thanks in advance for the help!
Best Answer
It looks to me like you're thinking too hard. For $S$ to be a regular surface in $\mathbb{R}^3$, every point of $S$ must have a neighborhood in $\mathbb{R}^3$ whose intersection with the surface is the regular image of an open set of $\mathbb{R}^2$. Points on the boundary of $S$ (i.e., satisfying the equality in the defining equation) don't have such neighborhoods.