# 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.

• Lots of definitions here. What's a proper extension? What's a regular surface? Apr 27, 2019 at 0:53
• I added the definition in the body of the question
– Tan
Apr 27, 2019 at 21:55
• By the definition of a regular surface, for any $p\in S$, there is an open set $U\subset S$ containing $p$. Note that any coordinate patch for $S$ is by definition a coordinate patch for $\bar S$, since $\bar S$ is likewise a regular surface. Apr 28, 2019 at 21:23

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.