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Let $M$ be a smooth manifold with or without boundary and $A$ a compact subset of $M$, does there exist a compact smooth embedding submanifold $N\subset M$ with or without boundary such that $N\supset A$?

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    $\begingroup$ Certainly you cannot expect such an $N$ to not have boundary. $\endgroup$ – Eric Wofsey Jul 13 at 4:51
  • $\begingroup$ @EricWofsey $\Bbb S^2\subset \Bbb R^3$ $\endgroup$ – Born to be proud Jul 13 at 8:39
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Yes, there is always a regular domain (i.e., a smooth, codimension-$0$, closed, embedded submanifold with boundary) that contains $A$. Here's a proof. References are to my Introduction to Smooth Manifolds (2nd ed.).

First of all, Proposition 2.28 shows that there is a smooth positive exhaustion function $f\colon M\to (0,\infty)$. Because $A$ is compact, $f$ achieves its maximum on $A$ -- let $R$ be that maximum.

Next, by Sard's theorem (Thm. 6.10), there must be a number $b>R$ that is a regular value of $f$, and by Propposition 5.47, the set $N = f^{-1}\big( (-\infty,b]\big)$ is a regular domain in $M$ containing $A$.

EDIT: As Eric Wofsey pointed out, the argument above works when $M$ has empty boundary, but if $\partial M \ne \emptyset$, then the boundary of $N$ might intersect $\partial M$ in complicated ways, preventing $N$ from being a smooth submanifold with boundary. (Ironically, I recently made the same point in an answer to another MSE question.)

The basic idea still works if $\partial M$ is compact, because then we can just choose $b$ large enough that $f^{-1}(b)$ is disjoint from $\partial M$. It also works if $A$ is contained in the interior of $M$, because in that case we can let $f$ be an exhaustion function for $\operatorname{Int} M$ and end up with $N$ completely contained in the interior.

But if $A$ meets $\partial M$ and $\partial M$ is not compact, this simple argument won't work. I'm pretty sure it's possible to modify the argument by smoothing out the boundary of $N$ near points where it intersects $\partial M$, but I don't have time to work out the details.

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  • $\begingroup$ This doesn't quite work if $M$ has boundary, since $f^{-1}(\{b\})$ might intersect the boundary. $\endgroup$ – Eric Wofsey Jul 13 at 23:57
  • $\begingroup$ Oh, you're right. This only works in limited circumstances. I'll edit. $\endgroup$ – Jack Lee Jul 14 at 0:04

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