\vdots \\ x_N  \end{pmatrix} =
    B
 \begin{pmatrix}u_{-n} \\ u_{-n+1} \\ \vdots \\ u_{-1} \end{pmatrix}.$$ So long as some $N\times N$ sub-matrix of $B$ has non-zero determinant, we have that for all $x_i$ sufficiently close to $0$, there is a solution of (\[eq:un\]) with small $u_j$. Specifically, we can choose $\delta$ such that if $|x_i| < \delta$, then there is a solution with $|u_j| \leq 1$.

Fix a vector $(x_1, x_2, \dots, x_N)$ in the neighbourhood of $(0,\dots,0)$, where each $|x_i| < \delta$. We will construct a sequence $(a_j)$ with $a_j \in \{-1, 1\}$ such that $$\pi (a_1 a_2 a_3 \dots) = (x_1, x_2, \dots, x_n),$$ which will prove the result. To do this, we first solve equation  for $u_{-n}, \dots, u_{-1}$ with $|u_i| \leq 1$. We will then choose the $u_j$ and $a_j$ for $j = 0, 1, 2, 3, \dots$ by induction, such that $$u_j := a_j - \sum_{k=0}^{n-1} b_{k} u_{j+k-n}$$ and such that $u_j \in [-1,1]$ and $a_j \in \{-1, +1\}$. We see that this is possible, as, by induction, all $u_j$ with $j < 0$ are such that $|u_j| \leq 1$. Furthermore, $$\begin{aligned}
\left|\sum_{k=0}^{n-1} b_j u_{j+k-n}\right| & \leq & \sum_{k=0}^{n-1} |b_k u_{j+k-n}| \\
                                 & \leq & \sum_{k=0}^{n-1} 