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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...)

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  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.

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  • $\begingroup$ 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? $\endgroup$ – Rhjg Feb 15 '18 at 19:25
  • $\begingroup$ 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 $x_0 \in B$ $\endgroup$ – Glougloubarbaki Feb 15 '18 at 19:34
  • $\begingroup$ Doesn't $f-$invariant mean that $f(B)\subseteq B$? $\endgroup$ – Rhjg Feb 15 '18 at 19:36
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Glougloubarbaki Feb 15 '18 at 19:44
  • $\begingroup$ $\Omega(f)$ 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 $x.$ $\endgroup$ – Dap Feb 17 '18 at 6:41
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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 $N$.

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