I'm reading the book Differential Geometry of Curves and Surfaces written by do Carmo. And there is one theorem I'm trying to prove. Here is the statement:

If $S$ be a compact, connected, regular surface with constant Gaussian curvature $K$, then $S$ is a sphere.

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In the proof, do Carmo claims that $S$ is open in the sphere $\Sigma$. But I'm quite vague about his argument. Why is $S$ open in $\Sigma$ if $S$ is to be a regular surface? Thanks.


1 Answer 1


Let $p\in S$. Since $S$ is a regular surface, there exists an open subset $U\subseteq\mathbb R^2$ and a homeomorphism $\psi: U\to \psi(U)\subseteq S$ such that $\psi(U)$ is an open subset of $\mathbb R^3$ containing $p$. So $\Sigma\cap \psi(U)$ is an open subset of $\Sigma$ containing $p$.

  • $\begingroup$ Thanks. But, is $U$ an open set in $\mathbb R^2$? Also, do you conclude open-ness of $S$ by letting $p$ range over $S$ (thus, $S$ can be written as a union of open sets)? $\endgroup$
    – Steve
    Jul 5, 2020 at 15:12
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    $\begingroup$ Yes, you're right, that was a typo. And your argument is correct. $\endgroup$
    – cqfd
    Jul 5, 2020 at 15:47
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    $\begingroup$ Be aware that, according to the definition of parameterization, $\psi(U)$ is open in $S$, i.e., there exists $V$ open in $\mathbb R^3$ such that $\psi(U)=V\cap S$. See this answer for another proof of this fact. $\endgroup$ Feb 21 at 3:15

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