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My question is that

A double cone ( also named as "circular cone") is not a surface.

I know its reason. But I cannot show this mathematically.


Suppose $\sigma : U \to S\cap W$ Is a surface patch.

Because the vertex $(0,0,0,)$ is a problem, S is not a surface.

I can see it. But I cannot express it in the mathematical way.

When I remove the vertex point $(0,0,0)$, the double cone is a surface.

Please can someone show/write these mathematically?

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2 Answers 2

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If you remove a point from an open subset in $\mathbb{R}^2$, it remains connected. A surface patch around the vertex has to give a homeomorphism to an open subset of $\mathbb{R}^2$. However, any open neighborhood of the vertex has the property that if you remove the vertex it becomes disconnected. Thus it cannot be homeomorphic to a subset of $\mathbb{R}^2$.

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  • $\begingroup$ Then, I define an open set $U=\Bbb R^2 /\{(0,0)\}$. And $\sigma : U \to \Bbb R^3$ Such that $\sigma(u,v)=(u,v,- or + \sqrt{u^2+v^2})$ $\endgroup$
    – 1190
    Commented Nov 23, 2013 at 14:37
  • $\begingroup$ I do not really know how to express my ideas for this proof. $\endgroup$
    – 1190
    Commented Nov 23, 2013 at 14:40
  • $\begingroup$ Can you help me more explicitly? $\endgroup$
    – 1190
    Commented Nov 23, 2013 at 14:41
  • $\begingroup$ Hmm, is this answer enough to get successful point for a homework? I know all these. But I cannot express properly. So i have asked Dear I.Solomon. $\endgroup$
    – 1190
    Commented Nov 23, 2013 at 14:51
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Note that if S where to be a regular surface, then it would be, in an open nhood of $(0,0,0)$ in $S$, the graph of differentiable function of the form: $x=f(y,z)$, $y=g(x,z)$ or $z=h(x,y)$. But that obviously cannot be the case, once the projections of S onto the coordinate planes aren't injective.

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    $\begingroup$ for future readers, this is proposition 2.23 in Ros-Montiel Curves and Surfaces (the explanation comes from lemma 2.21) $\endgroup$ Commented Apr 22, 2018 at 20:53

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