Can we foliate the punctured space by tori? Is it possible to have  a 2  dimensional foliation of $\mathbb{R}^{3}-\{0\}$ such that each leaf is homeomorphic to the torus? what algebraic topological obstruction exist?
Another question: is there a foliation as above with the following additional property :
The foliation is stable at the origin. that is for every neighborhood V of origin there is a smaller neighborhood W such that the  saturation of W is contained in V? 
the motivation for this question is the concept of "Blow up" of  singularities of vector field. when we blow up a singularity in R^3, we replace the singularity by  a S^2. The  blow up processes is  based on  the fact that R^3-{0} is  foliated by a familly of S^2.  Now if the answer of the above question is positive we can obtain a new type of torus- blow up.
However the blowing up was my main motivation for this question, but ,by this question, I never mean "is R^3-{0} homeomorphic to R x tori?"
 A: I would like to offer another explanation of the impossibility of foliating $R^3-0$ by tori (or by higher genus closed surfaces), at least in the $C^\infty$ case.  
Previously I commented that "foliations are rather far from fibrations". Closer to the truth, foliations are `submersions' onto their (potentially very weird!) leaf spaces. So the distance between foliations and fibrations is, in some respects, comparable with the distance between submersions and fibrations. In the case of smoothly foliating open manifolds by compact submanifolds, the distinction however is very small -- in fact, I want to claim below that foliations in this setting are exactly fibrations.  
The starting point is Ehresmann's fibration theorem: if $f:V\to M$ is a proper submersion of smooth manifolds, then $f$ is a locally trivial fibration. A proof can be found in Brocker & Janich's "Introduction to differential topology", section 8.12. 
Hence if we have a proper smooth function $f$ on $R^3-0$ having no critical points, then the fibres $f^{-1}(pt)$ foliate $R^3-0$ by compact embedded submanifolds. Ehresmann's theorem tells us $f$ is a locally trivial fibration, and of course as $R$ is contractible, $f$ actually defines a globally trivial fibration. In otherwords, all the fibres are diffeomorphic and we have $R^3-0 \simeq f^{-1}(pt) \times R$. From here we can determine that any fibre must be $\simeq S^2$. 
An important point which needs some further justification is this: given a smooth foliation $\mathscr{F}$ of $R^3-0$ by, say, compact tori, how do I know that the quotient map from $R^3-0$ to the leaf space is a smooth submersion of $R^3-0$ onto a smooth $1$-manifold? I see how compactness ensures distinct leafs remain separated (and so the quotient is hausdorff), however i am not clear on pinning down the smooth structure of the quotient map. 
Granting the above point, we would know that the quotient is necessarily a noncompact smooth 1-manifold, i.e. the real line $R$, and hence the total space must be a product $T\times R$ -- which we know it isn't.
These arguments, however, suffer the shortcoming of not being able to establish whether or not $R^3-0$ can be foliated by punctured surfaces, a question which seems interesting itself.
A: The answer is "No": it is impossible.
If $\mathbb{R}^3 - \{0\}$ is foliated by tori, then it has to homeomorphic to a torus bundle over a 1-manifold. It is obviously this is impossible.
To explain why it is a torus bundle over a 1-manifold, we only need to prove for each leaf
$T$ of the foliation, there exists a neighborhood $U(T)$ such that the foliation restricted in $U(T)$ is trivial ($U(T)= T^2 \times \mathbb{R}$). Essentially, this is the direct consequence of the holonomy group of $T$ is trivial based on two things:


*

*the assumption that each leaf is compact;

*each embedded torus in $\mathbb{R}^3 - \{0\}$ is separating.



More precisely, we can understand the local structure as follows.


*

*choosing two simple closed curves $m$ and $l$ in $T$ such that they inersect at one point $P$. - choosing a small interval $l_0$ transverse to the foliation passing through $P$ ($l_0$ is called a transversal).

*choosing a small neighborhood $U(T)$ of $T$ and a leave $F$ close to $T$ and $F\cap l_0 \neq \emptyset$.  

*along the orientation of $m$, we can lift $(m,P)$  to $(m_0,(P_0,P_1))$. Here $m_0\subset F$,   $P_0,P_1 \in l_0$ and $P_0$ and $P_1$ are the starting point and ending point of $m_0$ respectively.

*we claim that $P_0 =P_1$, otherwise there are two possibilites:

*(1) $P_0$ and $P_1$ are in two sides of $T$. This impossible since $T$ is separating in $\mathbb{R}^3 -\{0\}$.

*(2) $P_0$ and $P_1$ are in one side of $T$. This will provide a path $L$ in $F$   approaching to $T$. This conflicts to the fact $F$ is compact ($F$ is a torus). 
(the path $L$ is the union of many lift $m_0$, $m_1$, $m_2$, ...  of $m$: the starting point of $m_{i+1}$ is the ending point of $m_i$ in the transversal $l_0$)

*$T - l \cup m$ is a disk (contractable). 

*Finally, "The foliation structure restricted in $U(T)$ is trivial" can be followed by  the above statements.  

