Question about regular surfaces

[IvanT]

Hello, I have been trying to solve the following problem about regular surfaces from Do Carmo's book of differential geometry of curves and surfaces, section 2-3, exercise 14.

Homework Statement

Problem: Let A$\subset$S be a subset of a regular surface S. Prove that A is a regular surface if and only if A is open in S; that is, A=U$\cap$S, where U is an open set in R$^{3}$

A regular surface is a subset S$\subset$R$^{3}$ such that for each p$\in$S, there exists a neighborhood V in R$^{3}$ and a map x: U -> V$\cap$S of an open set U$\subset$R$^{2}$ onto V$\cap$S such that:
1. x is differentiable
2. x is a homeomorphism
3. dx$_{p}$:R$^{2}$->R$^{3}$ is one-to-one

The Attempt at a Solution

I have solved the implication: A open in S $\Rightarrow$ A regular surface, my problem is with the other part of the implication. One way to show that A is open in S, is to show that for each point in A there is an open set V of S, containing the point and such that V $\subset$A, and another way is to show that A=U$\cap$S, where U is and open set in R$^{3}$.
If A and S are regular surfaces, with A$\subset$S and p$\in$A, then I can find a map x$_{s}$: U->V$_{s}$ and a map x':U'->V$_{a}$ where U and U' are open sets in R$^{2}$ and V$_{s}$ and V$_{a}$ are open sets in S and A respectively, such that conditions 1,2 and 3 are satisfied. However, I don't know how to use this to guarantee the existence of an open set V in S such that p$\in$V and V $\subset$A, or and open set in R$^{3}$ such that A=U$\cap$S.

I would like some tips (not the solution) on how to "attack" the problem. Thanks.
(Sorry if my english is bad).

 

[lippyka]

My solution to this problem would be to first show that A open in S implies A is regular surface, which follows from the fact that open subsets of regular surfaces are regular surfaces. To prove that A regular surface implies A open in S, I would use the fact that for each point p in A, there exists a neighbourhood V in S and a map x:U->V such that x is differentiable, x is a homeomorphism, and dx_{p}:R^{2}->R^{3} is one-to-one. This implies that each point in A has an open neighbourhood in S and thus A is open in S.