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Let $M$ be a non-compact contractible bounded submanifold of the Euclidean space. Is it possible to find a compact submanifold $S$ of the Euclidean space such that $M$ is a submanifold of $S$?

As a positive example, let $M = \{(x,y) \in \mathbb{R}^2: (x,y) \in S^1, x>0, y>0\}$ be the portion of the 1-sphere $S^1$ lying in the positive orthant of $R^2$, and $S = S^1$.

Edit: the following example is wrong, as pointed out by a counter-example in a comment.

More generally, let $M$ be the image of an embedding $\gamma: (0,1) \to \mathbb{R}^2$ with $\gamma(0) \neq \gamma(1)$, assuming $M$ be bounded. Then, roughly speaking, $S$ can be obtained built joining the endpoints of $M$ by a "well-behaved" smooth curve $\gamma'$ that does not intersect $M$.

I posted a similar version of this question on mathoverflow, and I was redirected here.

DavideL
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    Hint: Consider the spiral $e^{-t}(\sin(t), \cos(t))$, $t>0$. What can you say about its length? Can it be a subset of a compact smooth planar curve? – Moishe Kohan Mar 18 '24 at 18:35
  • @Moishe That curve has finite length. But multiply by $1/(t+1)$ instead. – Ted Shifrin Mar 18 '24 at 18:51
  • @TedShifrin: Per their comments, OP appears to be interested in metrics of finite diameter on $M$. – Moishe Kohan Mar 18 '24 at 18:56
  • @Moishe Oh, sorry. I completely missed your point. But, in fairness, I don't see why his description constrains $M$ to have finite diameter. Oh, you mean the comment to John's response. I think OP needs to revise his post significantly. – Ted Shifrin Mar 18 '24 at 19:01
  • @TedShifrin: agree.... – Moishe Kohan Mar 18 '24 at 19:03
  • Thanks. I will accept John's post, which does answer the original question, and reformulate my question more carefully. – DavideL Mar 19 '24 at 09:38

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I think the answer is "no" in general.

Suppose that $M$ is the interior of the Alexander horned sphere. How would you "complete" $M$ to a closed manifold $K$? This completion would need to contain the limit point of the 'arms', but that point would have no disk-like (or half-disk-like) neighborhood in $N$.

John Hughes
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  • can you think of some additional assumption on $M$ to exclude this kind of pathological objects? In particular I'm interested in $M = F(X)$ being the image under a Nash embedding $F$ of a convex non-compact bounded Riemannian manifold $(X,g)$. The convexity of $X$ is not preserved by isometries, but there might be a way to consider only a special class of metrics $g$ such that the Nash embedding is "nice enough" and $F(X)$ does not have this kind of pathological behaviour. – DavideL Mar 18 '24 at 16:51
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    Nope...I got nuttin'. But what you're asking seems like a separate question, so perhaps you should ask that. – John Hughes Mar 18 '24 at 17:18