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Let $x \in R^n$ and $f : R^n \to R^n$, $f\in C^1$

$$ \frac{\mathrm{d}}{\mathrm{d}t} x(t) = f(x(t)) $$

be such that $f(0) = 0$ is asymptotically stable. The domain of attraction is the set of initial conditions $x(0) \in R^n$ such that the solutions $x(t)$ remain bounded and converge to $0$.

I am reading the book Stability of motion by Hahn. There it is stated that a subset of the domain of attraction can be computed. Quote:

Let $v(x)$ be positive definite in a domain $B$. Let there exist a closed hypersurface $F$ entirely contained in the interior of $B$ with the following properties:

  1. The origin is on the inside of $F$.
  2. $\dot{v}(x) = 0$ for $x \in F$
  3. for $x \neq 0, \dot{v}(x) < 0$ if $x$ lies on the inside of $F$, $\dot{v}(x) > 0$ if $x$ lies outside of $F$.
  4. The hypersurface $v(x) = c_1$ (closed by hypothesis) lies entirely inside of $F$.

Then the domain $v(x) \leq c_1$ is a subset of the domain of attraction.

Note: $\dot{v}(x) = \nabla v(x) f(x)$

Question: What if $v(x) = c_1$ leads to two disjoint closed sets, both inside of $F$ but only one of them contains $0$? As far as I can see, this possibility is not ruled out. Is this impossible or am I missing something? If not, isn't this a problem because there might be no guarantee that solutions end up in the set containing the origin?

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