This is just a curiosity and the question is really foggy. I'm wondering if there can exist a notion of "minimal smooth compactification" (when I say minimal I think something like adding a finite number of points or at least cells of dimension less than than the dimension of manifold) for a smooth noncompact manifold, in this sense: if the one point compactification of the manifold is smooth and the embedding is smooth, we are done; but what if the one point compactification is singular? Can I embed the manifold in a "minimal" compact manifold of the same dimension?

$\begingroup$ Just a remark, it is interesting to look at $\mathbb{T}^2$ times $(0,1)$ which has two ``boundary components'', but apparently the "minimal" way to embed it in a compact manifold is to include it in $\mathbb{T}^2 \times S^1$ (the other possible compactifications are "singular"). How can we define minimal though? $\endgroup$ – rpotrie Aug 5 '10 at 9:54

4$\begingroup$ If you attach finitely many cells to a CW complex with finite dimensional homology groups, then the resulting CW complex still has finite dimensional homology groups. Thus, if you start with a manifold with a nonfinitely generated homology group, you cannot "complete" it with finitely many cells. Thus, the complement of a closed set of points in the plane with infinitely many components will do. For instance, you can remove the Cantor set in the $x$axis from the plane. $\endgroup$ – damiano Aug 5 '10 at 10:01

$\begingroup$ Maybe the question can be changed to: Is it possible to embed any manifold in a compact manifold in order that the image is open and dense? This seems quite "minimal" to me. $\endgroup$ – rpotrie Aug 5 '10 at 10:09

1$\begingroup$ You should post that as an answer! $\endgroup$ – rpotrie Aug 5 '10 at 10:14

1$\begingroup$ On a slightly different setting, there is a characterization of manifolds that are the interior of a manifold with boundary; This was investigated by Larry Siebenmann and browsing its web page or MathSciNet references should help. $\endgroup$ – Benoît Kloeckner Aug 5 '10 at 18:03
A "surface of infinite genus" $S$ is an example of a manifold that is not an open subset of a compact manifold.
The reason $S$ cannot be embedded in a compact manifold is straightforward: we can find simple closed curves $a_1 , b_1 , \ldots , a_n , b_n , \ldots $ on $S$ such that, for each positive integer $g$, the curves $a_1 , b_1 , \ldots , a_g , b_g$ form a standard basis for a surface of genus $g$. Thus, considering the product in homology of the classes of these curves, we deduce that they are independent. If $S$ were an open subset of a compact manifold $M$, the same argument would imply that the images of the curves constructed above would also be independent in the homology of $M$. This contradicts the fact that the homology of the compact manifold $M$ is finite dimensional.
Observe that this example is not particularly different from the example of the complement in the complex plane of the integers. Indeed, $S$ can be realized as the double cover of $\mathbb{C}$ branched along the integers.
What do you want to do with an open annulus in the plane? Already open subsets of the plane may need infinitely many points added to compactify them in a sensible way.

$\begingroup$ If I understand the question well, the annulus you can compactify it with two points to get the sphere. The annulus is diffeomorphic to the sphere minus two points, so, it embeds there. In the remark I made, there is an example where you need to include cells, but again they are of smaller dimension. $\endgroup$ – rpotrie Aug 5 '10 at 10:00

1$\begingroup$ Yes, but the point made in the comments about infinitelygenerated homology is already in the second sentence, really. The complement o the points (n,0) where n is an integer $can$ be compactified by adding one point for each n, then compactified in the 2sphere. Do yo want that? $\endgroup$ – Charles Matthews Aug 5 '10 at 10:19

$\begingroup$ You are right. This answer gets my vote. I was thinking that possibly it could be saved by asking to compactify with a set with empty interior (this can be done for open subsets of the plane), but the surface of infinite genus also gives a counter example of the more general statement. $\endgroup$ – rpotrie Aug 5 '10 at 10:28