I know that $$S=\{(x,y,z)\in \mathbb R^3: z^2=x^2+y^2\}$$ is not a regular surface, because it has a vertex in $(0,0,0)$. But how to show it precisely? Maybe here is useful the theorem that a regular surface is locally a graph of infinite differentiable function of the form $z=f(x,y)$ or $y=g(x,z)$ or $x=h(y,z)$?
[Math] How to prove that the set is not a regular surface
calculusdifferential-geometry
Best Answer
Consider the upper cone $S_1$ where $$x^2 + y^2 = z^2$$ with $z\geq0$ and the lower cone $S_2$ where $$x^2 + y^2 = z^2$$ with $z\lt0$. Notice that any open set in $\mathbb{R^3}$ containing the vertex $(0,0,0)$ must also contain points of $S_1$(besides the vertex) and of $S_2$. We have that $S = S_1 \bigcup S_2$. Suppose there is an open set $U$ in $\mathbb{R^2}$, an open set $W$ in $\mathbb{R^3}$ containing $(0,0,0)$, and an homeomorphism $H:U\to S\cap W$. If $a,b$ and $c$ are three distinct points in $U$ such that $H(a)=u$, $H(b)=(0,0,0)$ and $H(c)=v$, with $u\in S_1$ and $v \in S_2$. You can find a path connecting $a$ and $c$ in $U$ such that $b$ is not in this path, but as $H$ is an homeomorphism, it means that the image of this path by $H$ does not pass through the vertex, i.e., the image of this path by $H$ would connect a point in $S_1$ (different from the vertex) and a point of $S_2$ but not contain the vertex. Contradiction.