**Definition**: Let be $f:M\to M$ a diffeomorphism of a compact manifold. We say that $A\subset M$ is an attractor when there exists a neighborhood $U\supset A$ such that $f( \overline{U})\subset int (U)$ and 
$$
A=\bigcap\_{n\geq 0}f^n(U)
$$
$U$ is called an basin of attraction of $f$.


**Theorem**: Let $M$ be a compact manifold and let $f:M→M$ a diffeomorphism. If 
$\overline{Per f}$ has hyperbolic structure, then 
can be partitioned into a finite number of compact, invariant and topologically transitive sets, called basic sets:
 $$\overline{Per(f)}=⋃\_{i=1}^{m}Λ\_i$$ 


**Definition**: Le be $\Lambda\_i$ a basic set of $f$, then we define $$W^s(\Lambda\_i)=\{x: d(f^n(x), \Lambda\_i)\to 0,~n\to\infty \}$$ 


**Question**: Supose that the chain recurrent set of $f$, $\mathcal{R}(f)$ has hyperbolic structure, I would like to see that if $int (W^s(\Lambda\_i))\neq \varnothing$ then the basic set $\Lambda\_i$ is an attractor.



Relationship between basic sets and attractors
Ok, here is the answer assuming that the basic set $\Lambda$ has local product structure.


Step 1. We will prove that $\forall x\in\Lambda$ $W^u(x)\subset \Lambda$. Then any point $y$ sufficiently close to $\Lambda$ belongs to a local stable manifold of some point in $\Lambda$, hence $\Lambda$ is an attractor.


Step 2. So we need to prove that local unstable of every point in $\Lambda$ is in it. Basic set is transitive and closed; local unstable manifolds vary continuously. Therefore it sufficient to show that local unstable manifold of a point $z\in\Lambda$ is in $\Lambda$, where $z$ is a point whose tranjectory is dense.


Step 3. It is easy to show that 
$$
W^s(\Lambda)=\bigcup\_{x\in\Lambda}W^s(x)
$$
See for example Proposition 9.1 of Shub's book. (Here local product structure is essential.) 


Step 4. Therefore if $p$ is in the interior of the above set. Then $\exists q\in \Lambda$ such that $p\in W^s(q)$. Take a transvesral to $W^s(q)$ that passes through $p$ and apply the $\lambda$-lemma to it to conclude that $W^u(q)$ is in $W^s(\Lambda)$. Hence $W^u(q)\subset W^s(\Lambda)$. Note that the same reasoning applies to points sufficiently close to $q$. In particular to some iterate of $z$. Hence we are done.


