Does global Lyapunov stability imply unique equilibrium?

Recall that given a time-invariant dynamical system

$$\dot x = f(x)$$

We say that an equilibrium point at the origin, $x_e \in \mathbb{R}^n$, of the above system is stable (in the sense of Lyapunov) if:

$$\forall \epsilon > 0, \exists \delta > 0, \text{ s.t. } \|x(t_0) - x_e\| < \delta \implies \|x(t) - x_e\|< \epsilon, \forall t \geq t_0$$

Suppose that the above condition holds for all $x_0 = x(t_0) \in \mathbb{R}^n$, then we can say that the equilibrium point is globally stable.

Suppose that $x_e$ is globally stable, then is it the unique equilibrium of $\dot x = f(x)$?

Note that global asymptotic stability implies uniqueness because every trajectory has to converge to that point.

• There's no $x_0$ in the condition, what do you mean when you say that it holds for all $x_0\in\mathbb{R}^n$? Jul 22 '17 at 0:07

Consider $\dot x =0$: every point is a globally stable equilibrium.
For a less ridiculous example, $\dot x = (-x_1, 0)$ has a line of globally stable equilibria.