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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

1. The problem statement, all variables and given/known data

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

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

Such that

1) x is differentiabel

2) x is an homomorphism.

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

2. Relevant equations

3. The attempt at a solution

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

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

condition 3)

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

[tex]dxq: \mathbb{R}^2 \rightarrow \mathbb{R}^3[/tex] and is thusly one-to-one.

How does this look???

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# Prove that subset of regular surface is also a regular surface

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