Is this statement about connected components of attractors true or known?

Question

I have a conjecture, which belongs in the field of chaos theory, as far as I can tell.

First some definitions:

• Let $B(0,1)$ be a ball in a finite-dimensional space of dimension greater than 1.

• Let $f: {B(0,1)} \times {\mathbb{R}} \rightarrow {B(0,1)}$ be a continuous surjective function such that:

  • $f(f(x,t_1),t_2) = f(x,t_1+t_2)$ for all x,$\text{ }t_1$ and $t_2$
  • for all $x_1, x_2: f(x_1, \mathbb{R}) \cap f(x_2, \mathbb{R}) = \emptyset $ or $f(x_1, \mathbb{R}) = f(x_2, \mathbb{R})$

• Regularity on $f$

  • $\frac{df}{dt}$ is define on a dense, connected area, is continue and always of norme 1
  • at x fixed either $f(x, \mathbb{R})$ is a point either for all t: $\frac{df}{dt}(x,t)$ is define

• Let $A$ be the attractor of $f$ which I define as follows: $$ u\in A \iff \exists x, (t_n)\rightarrow +\infty: f(x,t_n)\rightarrow u.$$

• What I call inside in the following, I can’t define, but this is an example: The inside of a circle in 2D space is the disk with its border. In 3D space, the inside of the circle is the circle himself.

• $P\subset A$ is a maximally connected attractive component of $A$, iff:

  • there is no connected component of $A$ such that $P$ is in the inside of this subset, with the exception of $P$ itself,
  • there is an $x$ in the outside of $P$ and $(t_n)\rightarrow +\infty$ such that: $f(x,t_n)\rightarrow u \in P$

Now the conjecture: The attractor has one maximally connected attractive component.

This result would be incredibly important for me. Is it true? Does it already exist? Any idea of a proof? I encourage you to make drawing to convince you or find a counter example.

Answer 1

Usually, attractors are required to be minimal by definition, i.e., an attractor¹ must not have a true subset that is also an attractor¹.

Your definition of attractor is missing this and hence the union of all attractors (usually called the attracting set) is also an attractor² in your sense. If you can find two or more distinct attractors¹ for your system, i.e., you have a multistable system, each of them would be a maximally connected attractive component in your sense. However, the union of all attractors would also be an attractor² and consist of two or more maximally connected attractive components.

To give a specific example, for $\dot{x} =-x(x-1)(x+1)$, both $\{-1\}$ and $\{1\}$ are maximally connected attractive components and $\{-1,1\}$ is an attractor². While this example is one-dimensional, it works in an arbitrary number of dimensions. You can just add $\dot{y} = -y$ and so on as additional components, or you can build arbitrary multistable examples like this (scaled down to fit inside your ball):

If, on the other hand, you additionally require that an attractor¹ must not have a true subset that is also an attractor¹, you immediately get that an attractor¹ consists of exactly one connected component. If it consisted of two connected components, each of them would be an attractor¹, which the requirement forbids. Thus your conjecture would hold even without considering maximality and attractiveness: An attractor¹ has one connected component. It is maximal (as per your definition) through uniqueness and attractive per definition.

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^{¹ usual definition
² your definition}