I want to know if there exists a characterization of k-foliations of $\mathbb{R}^n$ which have all the leaves closed. 


Do exists a $k$-foliation of $\mathbb{R}^n$ with a non-closed leaf?


In general, is there any characterization of manifolds in which all the foliations have only closed leaves?


Thanks in advance.



Closed leaves on foliations of $\mathbb{R}^n$
For all $n\ge 3$ and $k\in\{1,\dots, n-1\}$, there is a dimension $k$ foliation of $\mathbb{R}^n$ with non-closed leaves.


I first describe a relatively obvious construction which only works in codimension $2$ and higher. In $\mathbb{R}^3$ (thus $k=1$), consider the Hopf fibration on the sphere minus one point. It is a foliation by circles and one line which is a fibration around any circle. You can thus find a solid torus $T$ foliated as $D\times S^1$ where $D$ is the unit disc and $S^1$ the circle. Take any diffeomorphism $h:D\to D$ which is the identity near the boundary and has an attracting fixed point in the interior and consider $T'$ the suspension of $h$. It is a foliated solid torus with a circle to which infinitely many leaves accumulate. Cut out $T$ from $\mathbb{R}^3$ and replace it by $T'$, and you get the desired foliation.For higher $n$ and $k$, just take product foliations (taking a partial product for the leaves to get them to be of the wanted dimension).


Now for $n=2$ (thus $k=1$), I thought it could not be done but I changed my mind. Let the $y$ axis be one leaf, and the graph $L$ of $x\mapsto \sin(1/x)/x$ (for $x>0$) be another. Take as other leafs the vertical translate of $L$ and their symmetric with respect to the $y$ axis. Then you get a foliation of $\mathbb{R}^2$ with only one closed leaf. I do not know if one can get all leaves to be non-closed. By taking a product, you get the codimension $1$ case for all $n$. This construction seems more difficult to adapt to other manifold, which is why I kept the other one.


