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I've seen an exercise in which I am to prove that $$\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2=z^2\}$$ is not a submanifold of $\mathbb{R}^3$. I've done a little bit of research and found these answers

Showing that a level set is not a submanifold

How to show a level set isn't a regular submanifold

However I still don't understand, how do I exactly prove that a neighboorhood of the origin is not homeomorphic to any open set in $\mathbb{R}^3$ ? I didn't get the 2 vs 4 components thing

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user116708
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  • Are you sure you do not know how to verify that it is not locally homeomorphic to $R^3$? Did you try the implicit function theorem to conclude that this set if 2-dimensional? – Moishe Kohan May 02 '14 at 16:42

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Can you see that the set you are describing is a double cone? Indeed, for any fixed $z$ the equation describes a circle, and as $z$ varies the radii vary. Thus any neighbourhood $U$ of the origin with the origin removed will be not connected. However if $U$ is homeomorphic to some open $V\subset \mathbb R^2$, then $U-\{0\}$ will be isomorphic to $V-\{v\}$ for some point $v$. However the latter is always connected.

Simon Markett
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