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

by AlphaMale28
Tags: prove, regular, subset, surface
 P: 1 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???