Global attractor

Question

I have three little questions.

1. Let $X$ be a topological space and $f\colon X\to X$.

   How is a global attractor defined?

   In Katok and Hasselblatt’s Modern Theory of Dynamical Systems the following definition of an attractor is given:

   A compact subset $A\subset X$ is called an attractor for $f$ if there exist a neighbourhood $V$ of $A$ and $N\in\mathbb{N}$, such that $f^N(V)\subset V$ and $A=\bigcap_{n\in\mathbb{N}}f^n(V)$.

   There is no definition of a global attractor in this book but I guess this is an attractor for which $V=X$?

2. Assume $X$ is compact and $f$ is continuous. Am I right that $X$ itself as well as $$ E:=\bigcap_{n\in\mathbb{N}}f^n(X) $$ are global attractors? (Using the definition above, I have that $E$ is a compact subset of $X$ and I can choose $V=X$ and $N=1$.)

3. Assume again that $X$ is compact and $f\colon X\to X$ continuous. Consider the so-called non-wandering set $$ \Omega(f)=\left\{x\in X: \text{for each neighbourhoood U of x}\exists~N\geq 1: f^N(U)\cap U\neq\emptyset\right\}. $$ It is known that $\Omega(f)$ is $f$-invariant.

   Am I right that a global attractor contains every $f$-invariant set and therefore $$\Omega(f)\subseteq E=\bigcap_{n\in\mathbb{N}}f^n(X)?$$

   (If $E$ is a global attractor...)

Answer 1

1. Yes.

2. Yes: $X$ is a trivial attractor; some authors exclude $X$ in their definition of attractor (it is not interesting in any case).

   Also yes, $E$ is an attractor (which is of course equal to $X$ whenever $f$ is surjective, which is often the case).

3. Am I right that a global attractor contains every f-invariant set […]?

   No, this is incorrect. If $f(x)=2x$ on $X=\mathbb{RP}^1 = \mathbb R \cup \{ \infty\}$, then $f(X)=X$ but $\Omega(f)=\{0,\infty\}$ is strictly contained in $X$. Moreover, the only non-trivial attractor (i.e., an attractor different from $X$) is $\{\infty\}$ in this case, so $0 \in \Omega(f)$ is not in any non-trivial attractor.

   However, it is indeed always true that $\Omega(f) \subset E$. Indeed, $\Omega(f)$ is $f$-invariant, and every invariant set is contained in $E$.

   Addendum: some authors require that $f^N(A)$ is compactly contained in $A$ for $\bigcap_{n \in \mathbb N} f^n(A)$ to be called an attractor.

Answer 2

Note that the quoted definition of attractor is omits the property that the attractor should be minimal, i.e., it should not contain other attractors. (Every other definition of attractor I know captures this and you end up with something completely different if you omit this.)

1. There is no definition of a global attractor in this book but I guess this is an attractor for which $V=X$?

   I don’t know about this book, but going by what other definitions I can find, yes. Also, being pedantic, we are talking about the maximal $V$ fulfilling the definition here (which is typically called the basin of attraction).

2. $X$ cannot be an attractor because if it is compact, it cannot have a neighbourhood. (Note that if $X$ is not compact it fails the requirement that attractors should be compact.)

   $E$ would be an attractor according the quoted definition.

3. No, repelling fixed points are $f$-invariant, but not attractors. For example, for $f(x)=2x$, $0$ is a fixed point and thus in $Ω(f)$. However, there is no neighbourhood $V$ of $0$ such that $f^N(V) \subset V$ for any $N$.