Smale horseshoe

Example

Let $Q = [0,1]^2$, and let $H_0 = [0,1] \times [0, \tfrac13]$ and $H_1 = [0,1] \times [\tfrac23, 1]$ be two horizontal strips. Consider a diffeomorphism $f$ of the plane that bends $Q$ into a horseshoe shape and is affine on the strips: $$f(x, y) = \left(\tfrac{x}{3}, \; 3y\right) \text{ on } H_0, \qquad f(x, y) = \left(1 - \tfrac{x}{3}, \; 3 - 3y\right) \text{ on } H_1.$$ Then $f(H_0) = [0, \tfrac13] \times [0,1]$ and $f(H_1) = [\tfrac23, 1] \times [0, 1]$, and the middle strip of $Q$ is mapped outside $Q$. Let $\Lambda = \bigcap_{n \in \Z} f^n(Q)$ be the set of points whose whole orbit stays in $Q$.

Check of the structure. Consider the maps $y \mapsto 3y$ on $[0, \tfrac13]$ and $y \mapsto 3 - 3y$ on $[\tfrac23, 1]$. A point stays in $Q$ for all forward times exactly when its $y$-coordinate never leaves $[0, \tfrac13] \cup [\tfrac23, 1]$ under these maps. This means that $y$ lies in the middle-thirds Cantor set $C$. By the same argument for $f^{-1}$, a point stays in $Q$ for all backward times exactly when $x \in C$. So $\Lambda = C \times C$, a Cantor set. The map $h \colon \Lambda \to \set{0,1}^{\Z}$, with $h(p)_n = i$ when $f^n(p) \in H_i$, is a homeomorphism with $h \circ f = \sigma \circ h$, so $f|_\Lambda$ is topologically conjugate to the full shift on two symbols. On $\Lambda$, $Df$ equals $\operatorname{diag}(\tfrac13, 3)$ or $\operatorname{diag}(-\tfrac13, -3)$. So the horizontal direction is contracted and the vertical one expanded by the factor $3$, and $\Lambda$ is a hyperbolic set with $C = 1$ and $\lambda = \tfrac13$.