TITLE: Unit Disk Regular Surface?
QUESTION [6 upvotes]: I am having trouble proving these two problems:
1) is $\{(x,y,z)\in \mathbb{R}|z=0, x^2+y^2\leq1\}$ a regular surface?  I say no because the closed unit disk is a closed surface, so we cannot differentiate on the edges.  But how can I define a function to prove this or is this enough explanation?
2) is $\{(x,y,z)\in \mathbb{R}|z=0, x^2+y^2<1\}$ a regular surface?  I say yes because it is an open interval, so we can differentiate everywhere. But how can I define a function to prove this?  Perhaps $f(x,y)=x^2+y^2$ as a level curve at $f=1$ and show that its gradient is nonzero?  Also, saying $f=1$ wouldn't give me the correct closed interval, because it would contain points like (1,0) and (0,1), for example.  Help!

REPLY [3 votes]: picking any connected neighborhood of point $(0,1,0) \in \mathbb R^3$, we can show that
the intersection with the closed disk $\overline D = \{(x, y, z) \in \mathbb R^3 | z = 0, x^2 + y^2 \le 1\}$ is not homeomorphic to an open set in $\mathbb R^2$.
the set of intersection is simply connected even if the point $(0, 1, 0)$ is removed, while any open set in $\mathbb R^2$ cannot be simply connected after removing a point from it.
notice that the image of embedding from $\mathbb R^2$ into $\mathbb R^3$ is always not open. in fact the property of being open can be changed depending on what space the set lies in.