Let $X\_1,\dots,X\_n$ be complete vector fields on $\mathbb R^n$ and suppose that $(X\_1(p),\dots,X\_n(p))$ is a basis for all $p \in \mathbb R^n$.


Question: Is it possible to choose a cube $C$ around the origin of $\mathbb R^n$ such that there is for every $p \in C$ a piecewise smooth curve $\alpha \subset C$ which connects $p$ with $0$ where the smooth parts of the curve are given by the flows of the vector fields $\pm X\_1,\dots,\pm X\_n$? (With other words: is it possible to travel from $0$ to $p$ following only the integral curves of the given vector fields in a bounded domain?)


For $n=2$ this is pretty clear; w.l.o.g. $X\_1=\partial/\partial x\_1$ and the image of $x\_1=0$ under the flow $X\_2$ fills all of $\mathbb R^2$, since the flow lines of $X\_2$ intersect $x\_1=0$ transversally. Now it is easy to find a $C$ and $\alpha$ for a $p \in C$. But I can not generalize this for $n$ arbitrary.


**Edit:** Instead of demanding $C$ to be a cube, one could also ask if there is a open neighbourhood of the origin with the desired properties.



Linearly independent vector fields
For a suitable nbd of the origin $U$, yes, even following the $n$ flows in a prescribed order. Assuming $X\_1,\dots,X\_n$ locally Lipschitz continuous, the corresponding flows $\phi\_1(t,x),\dots, \phi\_n(t,x)$ are $C^1$ maps (in the pair) so the map $\Phi$
$$(t \_ 1,\dots,t\_ n)\mapsto \phi \_ n (t \_ n, \phi\_{n-1}( t \_ {n-1}, \dots \phi \_1 (t\_1, 0))\dots)$$
is of class $C^1$ on a nbd of $0\in\mathbb{R}^n$. Since for $(t,x)\to 0$ in $ \mathbb{R} \times\mathbb{R}^n$,
$$ \phi\_j(t,x)= x + tX\_j(0) + o\big((t,x) \big) $$
it follows that the differential of $\Phi$ at $0\in\mathbb{R}^n$ is the linear map
$$(t \_ 1,\dots,t\_ n)\mapsto t\_1X\_1(0)+\dots t\_n X\_n(0)$$
which is invertible because by assumption $X\_1(0),\dots, X\_n(0)$ is a base. By the implicit function theorem, $\Phi$ is a local diffeo between a cube $Q$ and a nbd $U$ of $0$. The images of the broken lines in $Q$ with edges parallel to the axes is what you want and reach any point of $U$.


**(edit)** Also, given any connected open subset $\Omega\subset \mathbb{R}^n$, any two points $p$ and $q$ of $\Omega$ can be joined by a continuous curve made by finitely many flow-lines of the fields $X\_1,\dots,X\_n$, just because the set of points that may be reached this way starting from $p$ is a non-empty open and closed subset of $\Omega$, thanks to the preceding local fact. 


However, there could be problems to reach some point if $\Omega$ is not open, e.g a closed cube. Take e.g. in $n=2$ and two constant fields $X\_1:=(2,1)$, $X\_2:=(1,2)$. Then, the vertex $(r,r)$ of the square $Q:=[−r,r]\times[−r,r]$ is easily joined to the origin by a two-edges broken flow-line, but the vertex $(r,−r)$ can't be reached by any such broken line in $Q,$ in any finite number of steps.


