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In "Introduction to Modern Theory of Dynamical System" by Katok and Hasselblatt, the following definition of attractor is given:

Definition 3.3.1 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 $A=\bigcap_{n\in\mathbb{N}}f^n(V)$.

Afterwards the following remark is given:

Remark. Considering $V'=\bigcap_{n=0}^{N-1}f^n(V)$ we may take $N=1$ in the definition.

I do not understand this remark. I neither see that $V'$ is a neighborhood of $A$, nor that $A=\bigcap_{n\in\mathbb{N}}f^n(V')$.

The only thing that I see is that $f(V')\subset V'$.

Maybe you can help.

Alp Uzman
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    $\def\N{\mathbb{N}}$For the second thing we have $\bigcap_{n ∈ \N} f^n(V) ⊆ V' ⊆ V$, and hence $\bigcap_{n ∈ \N} f^n(V) ⊆ f^n(V') ⊆ f^n(V)$ for every $n$ since $f(\bigcap_{n ∈ \N} f^n(V)) ⊆ \bigcap_{n ∈ \N} f^n(V)$. Hence, $\bigcap_{n ∈ \N} f^n(V) = \bigcap_{n ∈ \N} f^n(V')$. – Adam Bartoš Sep 24 '15 at 18:23
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    Just an idea: what is $f$? If $f$ is a homeomorphism and $V$ is open, then the sets $f^n(V)$ are clearly open. – Adam Bartoš Sep 24 '15 at 18:30
  • @user87690 Your first comment seems right to me. Thanks. Your second comment: I do not know exactly what f is. The book I took this from is so big that I cannot read out what is supposed for f at this place. - It may be that f is a homeomorphism but I dont think that V is supposed to be open since they only say V is a neighborhood (not open neighborhood). –  Sep 24 '15 at 18:42
  • A homeomorphism is an open function, isn't it? Then V does not need to be open, since it contains an open neighborhood of A. Then $f^n(V)$ is a neighborhood of $A$. –  Sep 24 '15 at 18:45
  • I had the idea with homeomorphism since sometimes dynamical systems are defined using a homeomorphism. If we have $A ⊆ U ⊆ V$ where $U$ is open, how do you know that $A ⊆ f(U)$? We know that $f(A) ⊆ A$, but I think that we don't know that $f(A) = A$. – Adam Bartoš Sep 24 '15 at 18:48
  • $A=\bigcap_{n=1}^{\infty}f^n(V)\subset f^N(V)\subset V$, so imho $A=\bigcap_{n=0}^{\infty}f^n(V)$. Then $f(A)=\bigcap_{n=0}^{\infty}f^{n+1}(V)=\bigcap_{n=1}^{\infty}f^n(V)=A$. Or do I missing sth? –  Sep 24 '15 at 18:53
  • It is not true that $f(X ∩ Y) = f(X) ∩ f(Y)$ in general. But you are right that it is true if $f$ is bijection. So if we again assume that $f$ is a homeomorphism, it works. – Adam Bartoš Sep 24 '15 at 18:55
  • True, I mixed it up with the $\bigcup$-case where this is the case. –  Sep 24 '15 at 19:05
  • In fact, isn't it enough to assume that $f$ is injective? –  Sep 24 '15 at 19:07
  • Yes, it is enough. Note that if $f$ is injective and open, then $f: X \to f(X)$ is a homeomorphism onto a clopen subset of $X$. – Adam Bartoš Sep 24 '15 at 19:12
  • Puh, how do you see that? . . I do not. –  Sep 24 '15 at 19:19
  • No, injectivity itself is not enough. If $f$ is injective and $U$ is open, then $f(U)$ is open in $f(X)$ since $f$ is closed, but we don't know that $f(X)$ is open. But if $f(X)$ is open, then $f$ is open (in our case). – Adam Bartoš Sep 24 '15 at 19:21
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    Well, an open continuous injective map is an open embedding, which is always a homeomorphism onto its image, which is open subset of the codomain. In our case, $f$ is also closed, so the image is clopen. – Adam Bartoš Sep 24 '15 at 19:23
  • Ok, but it surely is enough to assume that f is a homeomorphism... maybe this is the solution.... back to the question where this one arose from, we then have that $V'$ is a neighborhood of A, don't we? –  Sep 24 '15 at 19:27
  • Yes, the stronger property of being a homeomorphism is definitely enough. – Adam Bartoš Sep 24 '15 at 19:29

1 Answers1

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$\def\N{\mathbb{N}}$To sum up the discussion, we have both things if we assume that $f: X \to X$ is a homeomorphism. In that case $f(A) = f(\bigcap_{n ∈ \N} f^n(V)) = \bigcap_{n ∈ \N} f^{n + 1}(V) = A$ (since $f^N(V) ⊆ V$).

$\bigcap_{n ∈ \N} f^n(V') = A$. We have $A ⊆ V' ⊆ V$, and hence $A ⊆ f^n(V') ⊆ f^n(V)$ for every $n$, and hence $A ⊆ \bigcap_{n ∈ \N} f^n(V') ⊆ \bigcap_{n ∈ \N} f^n(V) = A$.

$V'$ is a neigborhood of $A$. There is an open set $U$ such that $A ⊆ U ⊆ V$, and hence $A ⊆ f^n(U) ⊆ f^n(V)$ and $f^n(U)$ is open for every $n$. Hence, $A ⊆ \bigcap_{n < N} f^n(U) ⊆ V'$.

Adam Bartoš
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