[proofplan] Choose a uniform local product structure radius for the compact hyperbolic set. If two points have orbits that remain sufficiently close for all positive times, then one point lies on the local stable plaque of the other. If their negative iterates also remain close, then it also lies on the local unstable plaque. The uniqueness part of local product structure says that the local stable and unstable plaques through the same base point meet only at that base point, so the two points are equal. [/proofplan]

[step:Choose uniform plaque data] By the [stable manifold theorem](/theorems/2778) and local product structure for compact hyperbolic sets, there are constants $\rho>0$ and $\eta>0$ such that, for every $z\in\Lambda$, the local plaques \begin{align*} W^s_\rho(z),\qquad W^u_\rho(z) \end{align*} are defined, and the following uniqueness property holds: if a point belongs to both $W^s_\rho(z)$ and $W^u_\rho(z)$, then it is $z$. Decrease $\eta$ if necessary so that, whenever $z,w\in\Lambda$ and \begin{align*} d(f^n(z),f^n(w))<\eta \end{align*} for every $n\geq 0$, the point $w$ lies in $W^s_\rho(z)$; similarly, if the same estimate holds for every $n\leq 0$, then $w$ lies in $W^u_\rho(z)$. These implications are the standard local characterization of stable and unstable plaques, made uniform by compactness of $\Lambda$. [/step]

[step:Apply the stable and unstable plaque characterizations]Set \begin{align*} c:=\eta. \end{align*} Suppose $x,y\in\Lambda$ satisfy \begin{align*} d(f^n(x),f^n(y))<c \end{align*} for every $n\in\mathbb Z$. In particular, the inequality holds for every $n\geq 0$, so the forward plaque characterization gives \begin{align*} y\in W^s_\rho(x). \end{align*} The same inequality also holds for every $n\leq 0$, so the backward plaque characterization gives \begin{align*} y\in W^u_\rho(x). \end{align*} Thus \begin{align*} y\in W^s_\rho(x)\cap W^u_\rho(x). \end{align*} By the uniqueness of the local product intersection at the base point $x$, this intersection is $\{x\}$. Hence $y=x$.[/step]

[guided]The uniform plaque data comes from compactness and hyperbolicity of $\Lambda$. The number $\rho$ is a common size for the local stable and unstable plaques, and $\eta$ is chosen so that forward $\eta$-closeness forces membership in the local stable plaque while backward $\eta$-closeness forces membership in the local unstable plaque. Choose \begin{align*} c:=\eta. \end{align*} If $x,y\in\Lambda$ and \begin{align*} d(f^n(x),f^n(y))<c \end{align*} for every $n\in\mathbb Z$, then the forward inequalities imply \begin{align*} y\in W^s_\rho(x), \end{align*} and the backward inequalities imply \begin{align*} y\in W^u_\rho(x). \end{align*} Therefore \begin{align*} y\in W^s_\rho(x)\cap W^u_\rho(x). \end{align*} The local stable and unstable plaques through $x$ meet only at $x$, so $y=x$. This is exactly expansivity for $f|_\Lambda$.[/guided]