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.(adsbygoogle = window.adsbygoogle || []).push({});

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

Problem: Let A[itex]\subset[/itex]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[itex]\cap[/itex]S, where U is an open set in R[itex]^{3}[/itex]

A regular surface is a subset S[itex]\subset[/itex]R[itex]^{3}[/itex] such that for each p[itex]\in[/itex]S, there exists a neighborhood V in R[itex]^{3}[/itex] and a map x: U -> V[itex]\cap[/itex]S of an open set U[itex]\subset[/itex]R[itex]^{2}[/itex] onto V[itex]\cap[/itex]S such that:

1. x is differentiable

2. x is a homeomorphism

3. dx[itex]_{p}[/itex]:R[itex]^{2}[/itex]->R[itex]^{3}[/itex] is one-to-one

3. The attempt at a solution

I have solved the implication: A open in S [itex]\Rightarrow[/itex] 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 [itex]\subset[/itex]A, and another way is to show that A=U[itex]\cap[/itex]S, where U is and open set in R[itex]^{3}[/itex].

If A and S are regular surfaces, with A[itex]\subset[/itex]S and p[itex]\in[/itex]A, then I can find a map x[itex]_{s}[/itex]: U->V[itex]_{s}[/itex] and a map x':U'->V[itex]_{a}[/itex] where U and U' are open sets in R[itex]^{2}[/itex] and V[itex]_{s}[/itex] and V[itex]_{a}[/itex] 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[itex]\in[/itex]V and V [itex]\subset[/itex]A, or and open set in R[itex]^{3}[/itex] such that A=U[itex]\cap[/itex]S.

I would like some tips (not the solution) on how to "attack" the problem. Thanks.

(Sorry if my english is bad).

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# Homework Help: Question about regular surfaces

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