# Global attractor

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...)

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.

• You write that $\Omega(f)\subseteq E$ is always the case. (1) For any $X$ and $f$ or under the assumption that $X$ is compact and $f$ continuous?? (2) Why is any invariant set contained in E? – Rhjg Feb 15 '18 at 19:25
• For any $f$ and $X$. Also, if $f(B) =B$ (ie $B$ is $f$-invariant) then for every $x_{-n} \in B$ there is $x_{-n-1}$ such that $f(x_{-n-1})=x_{-n}$. by induction, this proves that $x_0 \in f^n(X)$ for all $n$, and therefore $x_0 \in E$ for any $x_0 \in B$ – Glougloubarbaki Feb 15 '18 at 19:34
• Doesn't $f-$invariant mean that $f(B)\subseteq B$? – Rhjg Feb 15 '18 at 19:36
• – Glougloubarbaki Feb 15 '18 at 19:44
• $\Omega(f)$ is $f$-stable but not necessarily $f$-invariant even for compact subsets of $\mathbb R^2.$ See the example in this answer (Continuous Maps of the Interval with Finite Nonwandering Set Example D). It has $\Omega(T)=\{x,y\}$ but both $x$ and $y$ are sent to $x.$ – Dap Feb 17 '18 at 6:41

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 $$N$$.