Consider the set
$$ N = S^1 \times \{ (x,y)\in \mathbb{R} : x^2+y^2 \leq 1 \} $$
and the $\operatorname{map} f: N \rightarrow N$ defined by
$$ f(\theta, x, y)=\left(2 \theta, \lambda x+\frac{1}{2} \cos (2 \pi \theta), \mu y+\frac{1}{2} \sin (2 \pi \theta)\right) $$
for some constants $\lambda, \mu \in(0,1 / 2)$.
Define $ \text { the solenoid } \Lambda=\bigcap_{n \in \mathbb{N}} f^n(N)$ which is an f-invariant compact set.
To prove hyperbolicity of $\Lambda$ I need to construct the unstable space $E^u(x)$ or stable space $E^s(x)$ of $\Lambda$.
Does anyone know what is the unstable space or stable space of $\Lambda$ ?
Edit - Regarding the importance of the question This question is relevant because the solenoid has great importance in dynamical systems and this importance stems from the fact that the solenoid is a hyperbolic set.
First of all, for each $p = (\theta,x,y) \in N$ we have $$D_p f = \begin{pmatrix} \partial f_1 / \partial \theta & \partial f_1 / \partial x& \partial f_1 / \partial y\\ \partial f_2 / \partial \theta & \partial f_2 / \partial x & \partial f_2 / \partial y \\ \partial f_3 / \partial \theta & \partial f_3 / \partial x & \partial f_3 / \partial y \end{pmatrix} = \begin{pmatrix} 2 & 0 & 0 \\ -\pi \sin(2\pi\theta) & \lambda & 0 \\ \pi \cos(\pi\theta) & 0 & \mu \end{pmatrix} $$ where $f = (f_1,f_2,f_3)$ as in your post.
It follows that when we decompose $T_p N = \underbrace{\langle \partial / \partial \theta \rangle}_{T^\theta_p N} \oplus \underbrace{\langle \partial / \partial x, \partial / \partial y \rangle}_{T^{x,y}_p N}$, then second direct factor $T^{xy}_p N$ defines a $Df$-invariant 2-dimensional subbundle of $T N$ that I shall denote $T^{xy} N$. Furthemore, the action of $Df$ on this bundle shrinks every vector of $T^{xy} N$ by a factor greater than $2$, because of the fact that $\lambda,\mu \in (0,1/2)$. So we can now define $$E^s \Lambda = T^{xy} N \, \big| \, \Lambda $$
Defining $E^u\Lambda$ is a bit trickier, although I can cheat a bit by incorporating some knowledge about the topological structure of $\Lambda$. For each $p = (\theta,x,y) \in \Lambda$ there is a unique 1-dimensional subspace $T_p\Lambda \subset T_p N$ having the property that $\Lambda$ contains a smooth arc through $x$ that is tangent to $T_p\Lambda$. (Even more is true: these smooth arcs piece can be pieced together to give a neighborhood of $p$ homeomorphic to the Cartesian product of an open interval and a Cantor set.) Also, the tangent line $T_p\Lambda$ is never horizontal, meaning that it is never tangent to the $(x,y)$ plane of the $(\theta,x,y)$ coordinate system. Also, these 1-d subspaces $T_p \Lambda$ fit together continuously, to form a 1-dimensional vector space $T\Lambda$ over $\Lambda$. Finally, $T\Lambda$ is invariant under the action of $f$, meaning that $D_p f(T_p \Lambda) = T_{f(p)}\Lambda$.
The final fact we need is that $T_p \Lambda = E^u_p \Lambda$. To prove this, consider a unit vector $\vec u_p \in T_p \Lambda$, and write it as a sum $\vec u_p = \vec u^\theta_p + \vec u^{xy}_p$ of a vector in $T^\theta_p N$ and $T^{xy}_p N$. Since $\vec u_p$ is not horizontal, the angle it makes with the $T^{xy}_p N$ is positive. Since this angle varies continuously as a function of $p \in \Lambda$ and since $\Lambda$ is compact, there is a positive lower bound to this angle, independent of $p$. It follows that the length of $\vec u^\theta_p$ has a positive lower bound $c$ independent of $p$. Putting this together we have $0 < c \le |\vec u^\theta_p| \le |\vec u_p|$. But now, as we iterate, the lengths of the vectors $D_p f^i \vec u^\theta_p$ grow by powers of $2$. It follows that the lengths of the vectors $D_p f^i \vec u_p$ grow at least as fast as $c2^i$, proving that $E^u_p \Lambda$ contains vector $\vec u_p$.