Is a regular surface an open set if it's contained in another regular surface? I am reading do Carmo's differential geometry. In a proof he claims the following without proving: if regular surface $\bar{S}$ is a proper extension of a regular surface $S$, then $S$ is an open set in $\bar{S}$. I'm wondering if anyone knows how to prove that. 
I only consider surface in $R^3$. A surface $S$ is regular if for every $p\in S$, one can define a smooth coordinate patch on a neighborhood of $p$ and one can define a normal at $p$. 
A proper extension here means that $S$ is a proper subset of $\bar{S}$. It doesn't really matter though.
 A: The challenge here is that to establish that a parameterization (a.k.a. patch) for the sub-surface is also a parameterization for the containing surface, one should show that the image of said parameterization is open in the surface, knowing that it is open in the sub-surface. This creates a circularity in the argument because what we are trying to prove is that the sub-surface is open in the surface.
There is a more general result [cf. Shastri - Elements of Differential Topology, Lemma 8.1.2] which, applied to the case of surfaces, states that for every $p\in\bar S$ there are two natural projections, say $x_1,x_2\colon\mathbb R^3\to\mathbb R$ such that the restriction $\chi$ of $(x_1,x_2)$ to $\bar V\cap\bar S$ is a chart around $p$ in $\bar S$ for some $\bar V$ open in $\mathbb R^3$.
On the other hand, since every curve landing in $S$ also lands in $\bar S$, we deduce that $T_p(S)\subseteq T_p(\bar S)$. Then $T_p(S) = T_p(\bar S)$ because both spaces have the same dimension. This means that $d\chi|_S(p)=d\chi(p)$ is a linear isomorphism. By the Inverse Function Theorem, it follows that $\chi|_{V\cap S}$ is a chart around $p$ in $S$ for some $V$ open in $\mathbb R^3$ (Note that $\chi|_S$ is smooth because it is the restriction of $(x_1,x_2)$ to $S$. This observation is key because it breaks the circularity seen in arguments that rely on the unproven smoothness of the inclusion $S\hookrightarrow\bar S$).
Now take $W=V\cap\bar V$. The restrictions of $\chi$ to $W\cap S$ and $W\cap\bar S$ are charts (respectively in $S$ and $\bar S$) too. In particular $\chi(W\cap S)$ is open in $\mathbb R^2$. Therefore, $W\cap S=\chi^{-1}(\chi(W\cap S))$ is open in $W\cap\bar S$, which means that $W\cap S$ is an open nbh of $p$ in $\bar S$ included in $S$.
(The same argument can be used to prove that if $Y\subseteq X$ are two embedded submanifolds of $\mathbb R^n$ with $\dim Y= \dim X$, then $Y$ is open in $X$. In particular, if $X$ is connected and $Y$ is closed then $Y=X$.)

Remark: Since we cannot justify the smoothness of $S\hookrightarrow\bar S$, we cannot say that an arbitrary chart in $\bar S$ remains smooth when restricted to $S$. This is why we have to go up to $\mathbb R^3$ to obtain a chart in $\bar S$ from a smooth map defined on $\mathbb R^3$: in this case the restriction of the chart to $S$ is smooth because the inclusion $S\hookrightarrow\mathbb R^3$ is known to be smooth.
