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!


2 Answers 2


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.

  • $\begingroup$ 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... $\endgroup$ Commented Jun 23, 2023 at 16:01
  • $\begingroup$ Yes you are right. Thanks for pointing it out! @Stack_Underflow $\endgroup$
    – Lewis L
    Commented Jun 23, 2023 at 19:54
  • $\begingroup$ Thank you. Your updated answer seems good. Here is another possible method using the fundamental group math.stackexchange.com/a/3192522/1067823. $\endgroup$ 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.

  • 1
    $\begingroup$ I think this should be the correct answer. $\endgroup$
    – Tengu
    Commented Sep 10, 2019 at 18:02
  • $\begingroup$ 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'. $\endgroup$ Commented Jun 23, 2023 at 15:40

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