Show that $\omega(x)\subset A$ for all $x\in V$, $A$ attractor

Question

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 $x\in V$.

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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 $A$.

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 $\omega(x)$.

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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.

Answer 1

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, $ω(x) ⊆ \bigcap_{n ∈ ω} f^n(\cl{V})$.

Answer 2

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.

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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?