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View Full Version : Prove that subset of regular surface is also a regular surface


AlphaMale28
Apr14-09, 12:14 PM
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 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.
2. Relevant equations



3. 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???