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Does there exist a submersion $f:\mathbb{R}^{3}\setminus\{0\} \to \mathbb{R}$ for which there are $c_1$ and $c_2$ in $\mathbb{R}$ such that $f^{-1}(c_1)$ is compact and $f^{-1}(c_2)$ is non-compact.

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    Oh no xD, the formula I gave doesn't do the job... I'll think about it :D.2011-07-25

2 Answers 2

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Why not try something simpler, like $f=\frac{x^{2}+y^{2}+z^{2}}{1+z^{2}}$. Then it is easy to see that $\operatorname{grad}f$ is a non-zero vector at each point of $\mathbb{R}^{3}\backslash\{0\}$, so $f$ is a submersion; on the other hand

the level set $f^{-1}(1):~x^{2}+y^{2}=1$ is a cylinder, hence noncompact, while

the level set $f^{-1}(\frac{1}{2}):~x^{2}+y^{2}+\frac{1}{2}z^{2}=\frac{1}{2}$ is an ellipsoid, hence compact.

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This is a simpler example than what I had.

Let $U=\{(x,y)\in\mathbb R^2:0 and let $r:U\to\mathbb R$ be the function giving the distance to the origin. Pick any diffeo $\phi:\mathbb R^2\setminus\{0\}\to U$, and consider the function $r\circ\phi:\mathbb R^2\setminus\{0\}\to(0,1)$ which is a submersion with compact and non-compact fibers. Rotating this, we get an example of $\mathbb R^3\setminus\{0\}$.

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    I agree with you that the Reeb foliation of the solid torus cannot be obtained as the level sets of a submersion, but I don't understand why the foliation of $3$ space you describe cannot be. For the solid torus, a single leave revolves infinitely often and passes infinitely often by a *fixed* point in the boundary, and continuity would imply that the submersion would have to be constant on the solid torus, OK. But this argument fails with the foliation you describe. I can see that the further down you travel along the limit cylinder, the steeper the gradient must be...2011-07-25