# Does there exist a compact smooth embedding submanifold $N\subset M$ with or without boundary such that $N\supset A$?

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 $$N\supset A$$?

• Certainly you cannot expect such an $N$ to not have boundary. – Eric Wofsey Jul 13 at 4:51
• @EricWofsey $\Bbb S^2\subset \Bbb R^3$ – Born to be proud Jul 13 at 8:39

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.

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