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Let $S\subseteq\mathbb{R}^3$ be a smooth surface, no boundary (not necessarily compact). For any $r\in\mathbb{R}^3\backslash S$, define a smooth function $h_r:S\to\mathbb{R},q\mapsto |q-r|$

Let $p\in S$ be a critical point of $h_r$, then $\vec{pr}\perp T_p S$. So $p$ is non-degenerated if $\frac1{|p-r|}\neq k_1,k_2$. $k_1,k_2$ are two principal curvatures ($\vec{pr}$ be the normal direction at $p$, $k_{1,2}$ need not to be different).

Call $h_r$ a Morse function, if every critical point $p\in S$ of $h_r$ is non-degenerated.

Show that $\{r\in\mathbb{R}^3\backslash S|h_r$ is Morse funtion$\}$ is a open and dense set in $\mathbb{R}^3$.

I find little useful information of the set, possibly its complement is not a discrete set.

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HINT: Consider the sets $\{x\in\Bbb R^3: |x-p| = 1/k_1(p) \text{ for some } p\in S\}$ and $\{x\in\Bbb R^3: |x-p| = 1/k_2(p) \text{ for some } p\in S\}$.

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  • $\begingroup$ Since S is not compact, the set $\{p,p+\frac1{k_1},p+\frac1{k_2}|p\in S\}$ may have a boundary $\endgroup$ – yaoliding Jun 11 '16 at 3:00
  • $\begingroup$ No matter. It's still a surface (perhaps non-compact) and hence a set of measure zero. I guess you're right that the set of $r$ may not be open, but it definitely contains an open, dense set. $\endgroup$ – Ted Shifrin Jun 11 '16 at 6:54

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