A regular surface can't cover any open set of $\mathbb{R}^3$.

Question

This problem has brought me several problems because I do not know where to start to solve it, if you have any ideas, it's welcome:

Let $S \subset\mathbb {R}^3$ be a regular surface. Prove that $\mathbb{R}^ 3\setminus S$ is dense in $\mathbb{R}^3.$

The only thing I can think of is that assuming $\mathbb{R}^3\setminus S$ is not dense in $\mathbb{R}^3$ and taking any parametrization $f$ of $S$, maybe it implies that ${\rm d}f(p)$ is not injective for some $p\in S.$

Answer 1

What you have to show is, that each $p\in S$ can be approximated by points in $\mathbb R^3\setminus S$.

You could use the following definition/characterization of a regular surface $S\subset\mathbb R^3$:

For all $p\in S$ there are open subsets $U,V\subseteq \mathbb R^3$ with $p\in U$ and a diffeomorphism $\phi:U\to V$ such that $\phi(U\cap S)=V\cap(\mathbb R^2\times\{0\})$.

Then show that $\phi(p)$ can be approximated by points in $V\setminus (\mathbb R^2\times\{0\})$ and conclude by applying $\phi^{-1}$ that $p$ can be approximated by points in $U\setminus S$.