Let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. $A\subseteq\Omega$ is called stable if for every neighborhood $V$ of $A$, there is a neighborhood $U$ of $A$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag1.$$ $A$ is called attractor if $A$ is
- forward invariant, i.e. $A=\tau(A)$;
- stable; and
- there is a neighborhood $U_0$ of $A$ with $U_0\subseteq E(A)$,
where $E(A)$ is the the basin of attraction for $A$, i.e. the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$.
I've read that if $A_1,A_2\subseteq\Omega$ are different (maybe we need even to assume that they are disjoint) attractors, then $E(A_1)\cap E(A_2)=\emptyset$. How can we show this?
Moreover, I would like to conclude that in that case there cannot be a global attractor $G$, i.e. an attractor with $E(G)=\Omega$. This should immediately follow from the former result, if I'm not missing something.
If the topology on $\Omega$ is induced by a metric $d$, then $$E(A)\subseteq E_0(A):=\left\{x\in\Omega:d(\tau^n(x),A)\xrightarrow{n\to\infty}0\right\}$$ for all $A\subseteq\Omega$ (and "$=$" if $A$ is compact).
So, in that case, the desired claim would clearly follow if even $E_0(A_1)\cap E_0(A_2)=\emptyset$. Now it clearly holds that if $x\in E_0(A_1)\cap E_0(A_2)$, then $$d(A_1,A_2)\le d(\tau^n(x),A_1)+d(\tau^n(x),A_2)\xrightarrow{n\to\infty}0\tag1;$$ hence $d(A_1,A_2)=0$, but that doesn't yield a contradiction to $A_1\ne A_2$ (or $A_1\cap A_2=\emptyset$) in general ...
Your definition of attractor lacks the following property that is usually attributed to attractors: An attractor has no true subset that is an attractor itself. One way to fix this is to require that $\mathop{\text{orb}} x$ in definition basin of attraction is not only eventually in every neighbourhood of $A$ but in every neighbourhood of every point of $A$.
Without that condition, you can have $A_1 \subset A_2$ (e.g., by $A_1$ containing one and $A_2$ containing both fixed points of a bistable system), in which case you certainly don’t have $E(A_1) \cap E(A_2) = \emptyset$.