Why do we need the Hartman-Grobman theorem & the Stable Manifold Theorem to prove that any sink is asymptotically stable & source/saddle is unstable? I am reading Perko's book on Differential Equations and Dynamical Systems (3e) and I have the following question:
Why do we need the Hartman-Grobman theorem and the Stable Manifold Theorem to prove that any sink is asymptotically stable and source/saddle is unstable? Why is Hartman-Grobman not enough?
The passage in question is on p.130:

The following is said earlier, which I think makes implicitly use of Hartman-Grobman:

Many thanks in advance!
Attempt of an answer:
Sink:
If we have a sink, then the eigenvalues of $Df(x_{0})$ are all less than zero. To show asymptotical stability I would apply the diffeomorphism from H-G-thm, i.e.
$$\lim_{t\to\infty}||\Phi_{t}(x)-x_{0}||=\lim_{t\to\infty}||H^{-1}\circ e^{At}H(x)-x_{0}||.$$
Then (hopefully) by continuity we have
$$\lim_{t\to\infty}||\Phi_{t}(x)-x_{0}||=\lim_{t\to\infty}||H^{-1}\circ e^{At}H(x)-x_{0}||=0,$$
since $\lim_{t\to\infty}e^{At}H(x)=H(x_{0})$.
Saddle/Source: If we have a saddle then there are eigenvalues with positive and negative real part. Hence, $E^{s},E^{u}\neq\emptyset$. By the Stable Manifold Theorem it follows that $W^{s},W^{u}\neq\emptyset$, too, since they are of the same dimension as $E^{s},E^{u}$ respectively.  Hence, the stable manifold guarantees the existence of a trajectory that leaves any $B_{\epsilon}(x_{0})$. But then we can not have stability.
I hope it makes sense.

Theorems and Definitions:






 A: It seems to me that in fact the Stable Manifold Theorem is enough, instead of the other way around. In order to discuss if an equilibrium point is stable/asymptotically stable/unstable we would need unbounded times, but a priori the Hartman-Grobman Theorem gives a conjugacy local in time. On the other hand the Stable Manifold Theorem also gives information, albeit conditional, about the time intervals for which solutions are defined.
If $x_0$ is a sink, then by the Stable Manifold Theorem $x_0$ has an $n$-dimensional ball $B$ contained in the local stable manifold $\mathcal{S}_{x_0,\text{loc}}$ at $x_0$. By the forward invariance of stable manifolds $x_0$ is stable and by the limit property for any $y\in B$, $\lim_{t\to\infty} \phi_t(y)=x_0$.
If $x_0$ is a source or saddle, then again by the Stable Manifold Theorem the local unstable manifold $\mathcal{U}_{x_0,\text{loc}}$ at $x_0$ is not $0$-dimensional (i.e. the local unstable manifold contains points other than $x_0$). To show that $x_0$ is unstable we need to verify the following:
$$\exists \epsilon_0\in\mathbb{R}_{>0},\forall \delta\in\mathbb{R}_{>0}, \exists y_\delta\in E,\exists t_\delta \in\mathbb{R}_{\geq0}: d(y_\delta,x_0)<\delta, d(\phi_{t_\delta}(y_\delta),x_0)\geq\epsilon_0,$$
where $d$ is a distance on the open set $E\subseteq \mathbb{R}^d$ (say the Euclidean distance).
Putting $z_\delta=\phi_{t_\delta}(y_\delta)$, we have $y_\delta=\phi_{-t_\delta}(z_\delta)$, this is equivalent to showing that
$$\exists \epsilon_0\in\mathbb{R}_{>0},\forall \delta\in\mathbb{R}_{>0}, \exists z_\delta\in E,\exists t_\delta \in\mathbb{R}_{\geq0}: d(\phi_{-t_\delta}(z_\delta),x_0)<\delta, d(z_\delta,x_0)\geq\epsilon_0.$$
Let $\epsilon_0$ be small enough so that there is a $z\in \mathcal{U}_{x_0,\text{loc}}$ such that $d(z,x_0)\geq\epsilon$. Then by the limiting property $\lim_{t\to\infty}d(\phi_{-t}(z),x_0)=0$. Then for any positive $\delta$, there is a time $t_\delta>0$ such that $d(\phi_{-t_\delta}(z),x_0)<\delta$. This shows that $x_0$ is unstable (the $z_\delta$ ends up not being dependent on $\delta$.)
