# is this set a regular surface?

I'm reading "Differential Geometry of Curves and Surfaces of Manfredo Docarmo" I'm doing the exercises of the chapter 2. Here is the definition of regular surface that we are following:

I have problems with this exercise:

Is with the first part, I think that it's not true that it's a regular surface (the second it is). I don't know how to prove that something is not a regular surface.

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What happens if you take the point $p$ on the boundary of the unitary disc? How would you construct the homeomorphism? – Sigur Jan 24 '13 at 23:50

You're correct that the first one is not a surface. Here's how to check that condition $2$ fails. Take a point $(x,y,0)$ in the boundary, that is where $x^2 + y^2 = 1$. Take any open neighborhood $V$ of this, and intersect $V \cap S$. This set $V \cap S$ needs to be homeomorphic to an open set $U$. This would imply that $V \cap S$ was itself open. Can you see why this cannot be the case?
I think that your idea is good, but not your conclusion, because you are saying that only because $V\cap S$ is homeomorphic to $U$ then $V\cap S$ is open in $\Bbb R^3$, but that it's not true, because the homeomorphism it's between the sets $U, V\cap S$ and the range is not $\Bbb R^3$ so the map is effectively an open map, but only between those two sets. (So $\phi(U)= V\cap S$ is open in $V\cap S$ but that is obvious in this case , where $\phi$ is the homeomorphism) – Daniel Jan 25 '13 at 21:08
For example, take the sphere $S^1$ , we now that this set it's a regular surface, and also we know that has empty interior. Let's take $p\in S^1$ and take a neighborhood $V$ of $p$ and then consider $V\cap S^1$ clearly this set it's no open in $\Bbb R^3$ (in fact it has empty interior), so under that conclusion the sphere would not be a regular surface. At least that it's how I understood your answer. – Daniel Jan 25 '13 at 21:36
$V \cap S$ is open in its own topology -- the subset topology. We can forget about $\mathbb{R}^3$ here. – Isaac Solomon Jan 26 '13 at 0:29
If you check, you'll see that $V \cap S$ cannot actually be open in its own topology. – Isaac Solomon Jan 27 '13 at 2:16