I am reading James C. Robinson's book "Infinite-dimensional dynamical system -- an introduction to dissipative parabolic PDEs and the theory of global attractors". I have a question regarding the content in chapter 10.

In chapter 10.6, theorem 10.10 says

If X is a compact invariant set, then $$W^u(X) \subset \mathcal{A}$$

Here, $W^u(X)$ is the unstable manifold of set $X$. $\mathcal{A}$ is the global attractor. The proof goes like

Let $u\in W^u(X)$. Then by definition of unstable manifold, $u$ lies on the complete orbit $Y=U_{t\in \mathbb{R}}u(t)$. As $t\to-\infty$ we know that $dist(u(t), X)\to 0$, and as $t\to\infty$ we know that $dist(u(t), \mathcal{A})\to\infty$, so the orbit $u(t)$ is bounded orbit, then by Theorem 10.7, $u\in\mathcal{A}$.

Here, Theorem 10.7 is given as

All complete bounded orbits lie in $\mathcal{A}$. A "complete" orbit $u(t)$ is a solution of the PDE or ODE that is defined for all $t\in \mathbb{R}$.

The proof of theorem 10.10 seems only to use the fact the for $t\to\pm \infty$, the unstable manifold $W^u(X)$ approaches to a bounded set. So according to theorem 10.7, it is contained in the global attractor. What about stable manifolds? This book says nothing about it. I guess not all stable manifold should be contained in the global attractor, otherwise the attractor of an PDE cannot be finite dimensional. But what is the exact reason why stable manifolds are not in the global attractor. Theoretical argument and maybe a simple example are appreciated.