Prove that subset of regular surface is also a regular surface

[AlphaMale28]

Prove that subset of regular surface is also a regular surface(updated)

Dear Sirs and Madam's

I have following problem which I hope you go assist me in. I have been recommended this forum because I heard its the best place with the best science experts in the world.

Anyway the problem is as follows

Homework Statement

Let $$A \subset S$$ be a subset of regular surface S. Prove that A itself is a regular surface iff A is open in S. Where $$A = U \cap S$$ and where U is open in $$\mathbb{R}^3$$

I am using a pretty old book by a guy named Do Carmo so just as now. on page 52 there is a definition of a regular surface:

A subset of $$S \subset \mathbb{R}^3$$ is a regular surface if for eac $$p \in S$$ there exist a neighbourhood V in $$\mathbb{R}^3$$ and a map $$x: U \rightarrow V \cap S$$ of a open set $$U \subset \mathbb{R}^2$$ onto $$V \cap S \subset \mathbb{R}^3$$

Such that

1) x is differentiabel

2) x is an homomorphism.

3) For each q in U the differential $$dx_q : \mathbb{R}^2 \rightarrow \mathbb{R}^3$$ is onto-one.

Homework Equations

The Attempt at a Solution

condition 2) By the definition above let $$p \in A$$. Next assume that $$x: U \rightarrow S$$. Where U is open subset of $$\mathbb{R}^3$$. Then

$$x^{-1}(A \cap x(U)) \subset U$$ is a regular surface and by the definition its open in S.

condition 3)

Again we assume that $$p \in A$$ Next $$x: u \rightarrow A$$ where U is a subset of $$\mathbbb{R}^3$$. next we assume that x(q) = p and that

$$dxq: \mathbb{R}^2 \rightarrow \mathbb{R}^3$$ and is thusly one-to-one.

How does this look?

 

[lippyka]

Is it correct? Thank you for your help! Your Student Answer:Yes, your solution looks correct. To prove that a subset of a regular surface is also a regular surface, we need to show that the subset satisfies the definition of a regular surface; namely, that there exists a neighbourhood V in $\mathbb{R}^3$ and a map $x: U \rightarrow V \cap S$ of an open set $U \subset \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that:1) x is differentiable; 2) x is an homomorphism;3) for each q in U, the differential $dx_q : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is one-to-one.In your proof, you assumed that $x: U \rightarrow S$. Since $A \subset S$, it follows that $x: U \rightarrow A$. Thus, your proof implies that there exists a neighbourhood V in $\mathbb{R}^3$ and a map $x: U \rightarrow V \cap A$ of an open set $U \subset \mathbb{R}^2$ onto $V \cap A \subset \mathbb{R}^3$ such that:1) x is differentiable; 2) x is an homomorphism;3) for each q in U, the differential $dx_q : \mathbb{R}^2 \rightarrow \mathbb{R}^3$ is one-to-one. Therefore, since the map $x$ satisfies all the conditions of a regular surface, we can conclude that A is a regular surface.