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Let $X$ be a metrisable (compact) topological space and $f\colon X\to X$ continuous.

Show: If $A$ is an attractor for $f$, then there exists an open neighborhood $V$ of $A$ such that $\omega(x)\subset A$ for all $x\in V$.

Here are the involved definitions:

Attractor A compact set $A\subset X$ is called an attractor for $f$, if there exists a neighborhood $V$ of $A$ and $N\in\mathbb{N}$ such that $f^N(V)\subset V$ and $A=\bigcap_{n\in\mathbb{N}}f^n(V)$. $V$ is called the basin of attraction of $A$.

<p>A point $y\in X$ is called a $\omega$-limit point for $x\in X$ with respect to $f$, if there exists a sequence $(n_k)_{k\in\mathbb{N}}$ of moments of time going to $+\infty$ such that the images of $x$ converge to $y$, i.e. $f^{n_k}(x)\to y$ as $n_k\to\infty$. The set of all $\omega$-limit points for $x$ with respect to $f$ is denoted by $\omega(x)$.</p>

Now back to the statement that is to show.

I do not know how to choose the open neighborhood $V$. My first intuition was to let $V$ be the basin of attraction of $A$.

Let $x\in V$, V being the basin of attraction of $A$. Let $y\in\omega(x)$. Then $y=\lim_{k\to\infty}f^{n_k}(x)$ for some sequence $(n_k)$. But why should $y$ be in $A=\bigcap_{n\in\mathbb{N}}f^n(V)$? I do not see that.

Rhjg
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2 Answers2

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An idea: $\def\cl #1{\overline{#1}}$Suppose that $N = 1$. Put $B_n(x) := \{f^{n_k}(x): n_k ≥ n\}$. We have $B_n(x) ⊆ f^n(V)$ and $ω(x) ⊆ \cl{B_n} ⊆ \cl{f^n(V)} = f^n(\cl{V})$. Hence, $ω(x) ⊆ \bigcap_{n ∈ ω} f^n(\cl{V})$.

Adam Bartoš
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  • Why do you suppose that $N=1$. I do not see if this says that the basin of attraction is a right idea or not. – Rhjg Sep 22 '15 at 13:44
  • @Rhjg: If you want to prove it, then you have to be able to prove it for the easier case that $N = 1$. An in fact this is enough. To your second question, there are many basins for particular attractor, but you can choose one with some nicer properties. – Adam Bartoš Sep 22 '15 at 13:47
  • If I have a basin $V$ in the definition where $N$ is arbitrary why does this fulfill the desired property that $\omega(x)\subset A$ for $x\in V$? – Rhjg Sep 22 '15 at 13:49
  • @Rhjg: See the second paragraph in the edited answer. – Adam Bartoš Sep 22 '15 at 13:51
  • I see that, if $N=1$, then $\omega(x)\subset\bigcap_{n\in\mathbb{N}}f^n(\overline{V})$. But I do not see (1.) why the latter equals $A$ and (2.) the whole statement if $N\neq 1$. Moreover I do not see why this should be an open neighborhood. – Rhjg Sep 22 '15 at 14:57
  • @Rhjg: For (2) try to prove my edited claim. For (1) I'm not sure, I though that I have a solution, now I don't see it. – Adam Bartoš Sep 22 '15 at 15:15
  • @Rhjg: Sorry, now I see nothing. I'll think about that. – Adam Bartoš Sep 22 '15 at 15:22
  • If $f^N(V)\subset V$, is this equivalent with $f(V)\subset V$? – Rhjg Sep 22 '15 at 15:34
  • @Rhjg: No, it is not. Consider $f(x) = -x$ on ${-1, 1}$. – Adam Bartoš Sep 22 '15 at 15:39
  • How do you know that $\overline{f^n(V)}=f^n(\overline{V})$? In general, this is not the case for continuous functions. – Rhjg Sep 24 '15 at 18:00
  • @Rhjg: But here we are in a compact space, so $f^n(\overline{V})$ is closed. – Adam Bartoš Sep 24 '15 at 18:14
  • Could you please explain that? By continuity, we have $f^n(\overline{V})\subset \overline{f^n(V)}$, right? But why the other implication? – Rhjg Sep 24 '15 at 18:16
  • @Rhjg: By compactness $f^n(\overline{V})$ is closed, and it contains $f^n(V)$. Hence, $f^n(\overline{V}) ⊇ \overline{f^n(V)}$. – Adam Bartoš Sep 24 '15 at 18:25
  • Do you suppose we have a compact Hausdorff metric space? Or how do you know that $f^n(\overline{V})$ is closed by compactness? – Rhjg Sep 24 '15 at 18:27
  • @Rhjg: I suppose it, because you suppose it. We have that $X$ is Hausdorff compact, and hence $f$ is a closed mapping. – Adam Bartoš Sep 24 '15 at 18:32
  • Ah yes, I supposed that... then $\overline{V}$ is a closed subset of a compact set, hence compact (hence X is metrisable) and $f(\overline{V})$ is compact as image of a compact set (since f is continuous) and hence $f(\overline{V})$ is closed, since Hausdorff. Yes? – Rhjg Sep 24 '15 at 18:36
  • @Rhjg: Yes. Note that you don't need metrizability. – Adam Bartoš Sep 24 '15 at 18:37
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Suggestion: Suppose that $f$ is a homeomorphism.

Let's first assume that $V$ is closed.

Let $f^N(V)\subset V$, $N$ arbitrary and consider $$ V'=\bigcap_{n=0}^{N-1}f^n(V). $$ Then $V'$ is closed, $A\subset V'$, $f(V')\subset V'$, $A=\bigcap_{n\in\mathbb{N}}f^n(V')$ and $V'$ is a neighborhood of $A$. To see the latter, let $A\subset O\subset V$ with $O$ open neighborhood contained in $V$. Since $f^n(A)=A\subset f^n(O)=\subset f^n(V)$ for all $0\le n\le N-1$, and $f$ is an open map, we have that $A\subset\underbrace{\bigcap_{n=0}^{N-1}f^n(O)}_{=:O'}\subset\bigcap_{n=0}^{N-1}f^n(V)=V'$. As a finite intersection of open sets, $O'$ is open.

Now, let $x\in V', y\in\omega(x)$. By definition of $\omega(x)$, $$ y\in\overline{\left\{f^k(x): k>n\right\}}\subset\overline{f^n(V')}~\forall n\in\mathbb{N}. $$ Hence, we have $$ y\in\bigcap_{n\in\mathbb{N}}\overline{f^n(V')}. $$ Here, we have $\overline{f^n(V')}=f^n(\overline{V'})$. Since $V'$ is supposed to be closed, we get $$ y\in\bigcap_{n\in\mathbb{N}}\overline{f^n(V')}=\bigcap_{n\in\mathbb{N}}f^n(V')=A. $$ Thus, $\omega(x)\subset A$ for all $x\in V'$. The same holds for all $x\in O'$ since each $x\in O'$ is in $V'$. Hence we have found an open neighborhood of $A$ fulfilling the statement.

And one notation thing:

I would suggest to call $V'$ and $O'$ basins of attraction. In case $N=1$, i.e. $f(V)\subset V$, it is $V'=V$ and your $V$ is a basin of attraction. But if $N>1$, I would not call $V$ a basin of attraction but $V'$ and $O'$. Do you agree?

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    That's what I thought, but how do you know that $V'$ is a neighborhood of $A$ and that $V'$ is closed? – Adam Bartoš Sep 22 '15 at 16:46
  • Isn't $V'$ a neighborhood of $A$ because $A\subset V=f^0(V)$ and $A=\bigcap_{n\in\mathbb{N}}f^n(V)\subset V'$? Concerning the closedness, I am not sure right now. –  Sep 22 '15 at 16:51
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    How do you know that the sets $f^n(V)$ are open? – Adam Bartoš Sep 22 '15 at 16:52
  • Where is it used that $f^n(V)$ are open? Sorry, I am confused now. :-) –  Sep 22 '15 at 16:55
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    I don't understand why $V'$ is a neighborhood of $A$. Since $\bigcap_{n ∈ ω} f^n(V)$ doesn't have to be open, I don't understand your previous comment. – Adam Bartoš Sep 22 '15 at 16:57
  • I am not sure if this is correct. but in topology each neighborhood contains an open neighborhood. Since $V$ is a neighborhood (it is not said that it is open) of $A$, there is some open neighborhood $O$ of $A$ contained in $V$, i.e. $A\subset O\subset V$. And since $V\subset V'$, $V'$ should be a neighborhood of $A$, too, since $A\subset O\subset V'$. –  Sep 22 '15 at 17:00
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    How do you know that $V ⊆ V'$? It's the other way round. – Adam Bartoš Sep 22 '15 at 17:02
  • Yes, you are right. Sorry. –  Sep 22 '15 at 17:03
  • @user87690 have asked this here http://math.stackexchange.com/questions/1446853/katok-hasselblatt-attractor-of-dynamical-system-how-to-understand-this-remark since i really have no idea –  Sep 22 '15 at 20:06
  • @user87690 I edited my answer, making the assumption that $f$ is a homeomorphism. If you do not mind, please have a look on it. Thanks. –  Sep 24 '15 at 20:36
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    I don't get how we deal with the situation when $V$ is not closed. $V'$ is a neigborhood of $A$ such that $ω(x) ⊆ \bigcap_{n ∈ \mathbb{N}} f^n(\overline{V})$ for every $x ∈ V'$. But if $V$ is not closed, we don't know that the intersection is $A$. – Adam Bartoš Sep 24 '15 at 20:55
  • You are right. I was wrong with it. Dont see it either. –  Sep 24 '15 at 22:11
  • @user87690 it is really strange to me that the statement ist formulated for a neighborhood V of A (and not for an open neighborhood of $A$) since usually such statements are formulated for open neighborhoods. so I think it may be indeed very important that V is closed - and that the statement is not correct for open neighborhoods? –  Sep 25 '15 at 16:01
  • You can't be sure unless you find a couterexample. – Adam Bartoš Sep 25 '15 at 18:21
  • Maybe it is possible to argue that $A=\bigcap_{n\in\mathbb{N}}\overline{f^n(V)}$ even if V is open... –  Sep 25 '15 at 21:59
  • How? And what if $V$ is neither open nor closed? – Adam Bartoš Sep 26 '15 at 08:03