Unit Disk Regular Surface? 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!
 A: To case 1),according to the definition, if $p\in\partial B$,where B is the closed disk,then for any neighborhood $V$ of $p$,$V\bigcap B$ is not an open set in $R^3$,so you cannot find any continues function $x:U\rightarrow V\bigcap B$ according to the topological definition of a continues function,where $U$ is any open set in $R^2$
To case 2),you can choose the map as $id_{R^2}$
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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.
