# 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!

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 continuous function, where $$U$$ is any open set in $$R^2$$

To case 2),you can choose the map as $$id_{R^2}$$

Updated June 23: The previous answer is indeed not sufficient and contains some mistake pointed out by Stack_Underflow. So here is an updated version for the proof of Case 1).

The main goal is to show that $$V\bigcap B$$ cannot be homeomorphic to any open subset of $$\mathbb{R}^2$$, which is done by noting that a homeomorphism always maps interior points to interior points, but the boundary points of $$V\bigcap B$$ are mapped to interior points if there exists a homeomorphism between $$V\bigcap B$$ and some open set in $$\mathbb{R}^2$$.

One can also use the connectedness to proceed as in the answer given by JHL.

• Don't we use the subspace topology induced by $B$ in terms of a homeomorphism? In this case, $V \cap B$ is open in $B$ and nothing seems wrong... Commented Jun 23, 2023 at 16:01
• Yes you are right. Thanks for pointing it out! @Stack_Underflow Commented Jun 23, 2023 at 19:54
• Thank you. Your updated answer seems good. Here is another possible method using the fundamental group math.stackexchange.com/a/3192522/1067823. Commented Jun 24, 2023 at 2:29

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.

• I think this should be the correct answer. Commented Sep 10, 2019 at 18:02
• Yes I think this is the correct one. Using fundamental groups to illustrate that. But I would like to see more details about 'open sets with one point being removed is not simply connected'. Commented Jun 23, 2023 at 15:40