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.
$\def\N{\mathbb{N}}$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, $A ⊆ \bigcap_{n < N} f^n(U) ⊆ V'$.