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Suppose $A \subset S$ and both $A$ and $S$ are regular surfaces. Show that $A$ is open in $S$ (w/ respect to subspace topology on $\mathbb{R}^3$.

Note that the definition of a regular surface $S$ is a subset of $\mathbb{R}^3$ such that for every point $p \in S$, there exists an open neighborhood $V$ in $\mathbb{R}^3$ of $p$ and an open set $U$ in $\mathbb{R}^2$ such that there is $x:U \to V \cap S$ is a homeomorphism, is differentiable, and its differential is one to one.

So I am trying to show that for every point $p \in A$, it is in the interior of $A$, with respect to subspace topology on $S$. That is $p$ is in the set $V^{'}\cap S \subset A$ where $V^{'}$ open in $\mathbb{R}^3$.

Since $A$ is regular surface, there is a open neighborhood $V$ of $p$ in $\mathbb{R}^3$ of $p$ and an open set $U$ in $\mathbb{R}^2$ such that there is $x:U\to V \cap S$ is a homeomorphism, etc.

I am referring to this link here where I am not quite sure how he concludes the last part: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_analyst_2002;task=show_msg;msg=0003.0001.0001.0001

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The Inverse Function Theorem tells you that a smooth map with invertible derivative is an open map. I'm skeptical about your definition, by the way; I've never seen a definition that says "differentiable" rather than $C^1$. The latter is absolutely needed for applications of the Inverse Function Theorem.

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  • $\begingroup$ how would I use inverse function theorem? Please elaborate. $\endgroup$
    – sarah
    Mar 14, 2014 at 1:30
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    $\begingroup$ Apply it to $x_S^{-1}\circ x_A$, mapping an open subset of $\Bbb R^2$ to $\Bbb R^2$. $\endgroup$ Mar 14, 2014 at 1:33
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    $\begingroup$ Because $x_S$ is a homeomorphism and $x_A(V)=x_S\big(x_S^{-1}\circ x_A(V)\big)$. $\endgroup$ Mar 14, 2014 at 2:13
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    $\begingroup$ Ok, I think I get it. So is $x_s^{-1} \circ x_a(V)$ open in the domain of $x_S$? $\endgroup$
    – sarah
    Mar 14, 2014 at 2:19
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    $\begingroup$ Yes. By the open mapping consequence of the IFT. $\endgroup$ Mar 14, 2014 at 2:23

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