I have the following question,

If $S\subset\mathbb{R}^3$, is a regular surface, then interior of $S$, can be empty?

My approach: I have a list of ideas for this problem.

  • The definition, that I know, for regular surface is for each point in the surface, therefore the interior cannot be empty.
  • $A$ is a open subset of a regular surface if and only if, $A$ is a regular surface (this prop. is easy to demonstrate); So that, if $A$ is a open set, then it interior is not empty, and $\mbox{int}\{A\}\subset\mbox{int}\{S\}$. Then, interior of $S$ is not empty.
  • I try to think in what case I have empty interior of a surface. If the interior of surface is empty, then his points are in the boundary of $S$, but the boundary of this set is a closed set, therefore I think that, I would lose the condition of regularity for the homeomorphism.
  • On the other hand, If we consider $S$ as, the "shell" of the sphere, then $S$ is a regular surface, and the interior of $S$ (I have not demonstrated it) will be a empty set.
  • $\begingroup$ this is Montiel-Ros problem 1 in page 55 (topic of surfaces) $\endgroup$ Apr 23, 2018 at 19:13
  • $\begingroup$ What does "interior" mean for you here? $\endgroup$ Apr 24, 2018 at 12:35

1 Answer 1


Consider an interior point $p$. There exists an open ball $O$ in $\mathbb{R}^3$ such that $p \in O$ and $O \subseteq S$. Just use that the tangent plane operation is monotone so that $\mathbb{R}^3 = T_p(O) \subseteq T_p(S) \cong \mathbb{R^2}$ to get a contradiction: the space cannot be contained in a plane.


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