Let $A \subset S$ be a subset of a regular surface $S$. Prove that if $A$ is itself regular then $A$ is open in $S$.

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I am trying sho that for each $p \in A$ exist a open $V \subset R3$ tal que $p \in V \cap S$ and $V \cap S \subset A$. Please, help me.

Here is the definition of a regular surface from Differential Geometry of Curves and Surfaces by Manfredo do Carmo: A subset S⊂R3 is a regular surface if, for each p∈S, there exists a neighborhood V⊂R3 and a map f:U→V∩S of an open set U⊂R2 onto V∩S⊂R3such that

f is $C^{∞}$.

f is a homeomorphism.

Since f is continuous, this means that f has an inverse $f^{-1}:V∩S→U$ which is continuous; that is, $f^{-1}$ is the restriction of a continuous map F:W⊂R3→R2 defined on an open set W containing V∩S.

For each q∈U holds f′(q):R2→R3 is one-to-one.