How to show that this set isn't a regular surface?

I'm trying to solve this exercise from Do Carmo's Differential Geometry of Curves and Surfaces, and I want a hint on how to do it. The exercise is:

Is the set $S =\left\{(x,y,z)\in \mathbb{R}^3 \mid z=0, \ \ x^2+y^2\leq1 \right\}$ a regular surface ?

Well, for me this isn't a regular surface because the set $S$ is not relatively open in the $xy$ plane, however I'm not sure how to prove this formally. I thought on first proving that a subset of a regular surface is a regular surface itself if and only if it's open relative to the set we know it's a regular surface. Then I show that the $xy$ plane is a regular surface and then $S$ cannot be a regular surface because it's not open relative to the $xy$ plane.

Is this line of thought correct ? Thanks in advance for the help!

It looks to me like you're thinking too hard. For $S$ to be a regular surface in $\mathbb{R}^3$, every point of $S$ must have a neighborhood in $\mathbb{R}^3$ whose intersection with the surface is the regular image of an open set of $\mathbb{R}^2$. Points on the boundary of $S$ (i.e., satisfying the equality in the defining equation) don't have such neighborhoods.

• The geometric idea is clear, but I cannot prove it formally.
– Luks
May 25 at 16:30
• @Luks Start with proving that the set $\{ (x,y)\ |\ y \geq 0 \}$ is not homeomorphic to $\mathbb{R}^2$
– Neal
May 27 at 1:49

The geometric idea is clear, but I cannot prove it formally. I suppose $$S$$ is a regular surface, so there exists for the point, for example, $$p=(1, 0, 0) \in S$$, an open set $$V \subset \mathbb{R}^3$$ with $$(1, 0, 0) \in V$$ and a regular parametrization $$X : U \rightarrow S \cap V$$, with $$U \subset \mathbb{R}^2$$ an open set, such that $$X$$ is smooth, $$dX_q$$ is inyective (being $$q=X^{-1}(1, 0, 0)$$) and $$X$$ is a homeomorphism. I cannot find the contradiction formally. Thanks in advance for the help.

• This is not an answer. You should probably post a new question. Perhaps this hint will help: The map $X$ you have defined actually maps $U$ to $S\cap V\subset\Bbb R^2$. Since $dX_q$ is an isomorphism $\Bbb R^2\to\Bbb R^2$, the Inverse Function Theorem tells you that $X$ has a smooth local inverse and hence is an open mapping from $U$ to $\Bbb R^2$. May 25 at 18:01
• I am really sorry for posting my question where I should have posted an answer. I am new here. I´ll take your advice and make a new post for this question. Thank you very much and excuse me for my mistake posting here.
– Luks
May 25 at 20:12
• Sure. In the meantime, work out my hint. May 25 at 20:14

I don't quite agree with the answers provided above. Indeed, a homeomorphism maps open sets to open sets, but 'open' also relies on the topology you use. If we take $$U$$ be an open subset of $$\mathbb{R}^2$$ and $$V$$ be an open subset of $$\mathbb{R}^3$$, $$x: U \to V \cap S$$ be a homeomorphism, shouldn't we focus on the subspace topology induced by $$S$$? In that way, open sets intersect with $$S$$ are still open in S.

Below is a link that I think is reliable, which uses the concept of simply connected. https://math.stackexchange.com/a/3192522/1067823

If I had any mistake, welcome to point that out!