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Let $M$ be a $n$-dimensional smooth non-compact manifold such that the singular cohomology groups $H^{k}(M,\mathbb{Z})$ are finitely generated for $k\geq 0$. Can we find a sufficiently big integer $N$, a sufficiently small $\varepsilon>0$ and a smooth embedding $$i_{\varepsilon}:M\hookrightarrow \mathbb{R}^{N}$$ with the following features:

  1. $i_{\varepsilon}(M)$ is closed in $\mathbb{R}^{N}$

  2. for every $x\in i_{\varepsilon}(M)$ and for every $0<\delta<\varepsilon$ the intersection $B_{\delta}(x)\cap i_{\varepsilon}(M)$ is connected. With $B_{\delta}(x)$ I mean the euclidean ball of $\mathbb{R}^{N}$ centered at the point $x$ and of radius $\delta$.

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I think you can always find such embedding, for all manifolds, without restrictions. Take any proper embedding $i:M \to \Bbb R^N$ such that the coordinate functions restricted to $i(M)$, $x_i : i(M) \to \Bbb R$, are Morse. Up to isotopy,we can assume that the distance between any two singular values of $x_i$, is bounded below by $10\epsilon > 0$. Then this embedding satisfies the required properties. – Daniele Zuddas May 25 '13 at 18:50
@Daniele: Interesting idea. However, it does not seem to suffice. For example, you can embed a line properly into the plane in a U-shape with the U getting increasingly thinner at the top. It seems this issue can be avoided by first rescaling the embedding (non-linearly in general; this would "spread out" the embedding and separate the two ends of the U), and then applying your idea. Nevertheless, I still do not understand how the result would necessarily verify the required conditions. Is that written somewhere? – Ricardo Andrade May 25 '13 at 21:11
The case of the U is exactly the type of embedding i want to rule out by asking condition 2. Maybe i'm wrong but i think that in the case of the U with ends getting closer and closer at infinity to the $y$-axis is that the $x$ coordinate has "critical points" at $\pm\infty$ with the same value. Can it be? – student May 26 '13 at 1:26
Now I'm thinking that my previous comment was incorrect, thanks Ricardo. The issue is probably due to the fact that open manifolds can have very complicated ends (homeomorphic to a Cantor set, for instance, so you cannot separate points outside a very big ball by a constant). So, in that case, I think the best you can do is to take for $\epsilon$ a continuous function, not a constant. However, if the space of ends is finite the statement should be true, by assuming that the intersection of $i(M)$ with a sphere of radius $r$ has connected components of distance bounded below by a constant. – Daniele Zuddas May 26 '13 at 6:49
@Daniele, I think your idea works perfectly. Take the embedding as you describe. Note that in addition you can assume that between any pair of critical values there is an interval where the embedding is radial; i.e., it is swapped by segments in the radial direction. Now choose a function $f:\mathbb R_+\to \mathbb R_+$ such that it is locally constant away of the radial segments and "rescale by this function"; i.e., in the spherical coordinates you new embedding looks like $(f(\rho)\cdot \rho,\theta)$ if the old one was $(\rho,\theta)$. For right choice of $f$ you get the needed embedding. – Anton Petrunin May 26 '13 at 10:04

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