Is every basin of attraction completely invariant? I can't seem to find a definitive answer in the literature.  I believe the answer is yes, but my focus has been on the rational maps on the Riemann sphere.  At the very least I'm confident that if the rational map is conjugate to a polynomial, then the answer ought to be yes.  If it isn't true, can someone provide a counter example, and also explain what is flawed with my reasoning?  
The relevant definitions and my reasoning follow.
Given a holomorphic map $R$, define the basin of attraction of $R$ containing the fixed point $c$ by 
$B_{R,c}=\{z\in\mathbb{C}:\lim_{n\rightarrow\infty}R^{\circ n}(z) = c\}$, 
where $R^{\circ n}$ denotes $R$ composed with itself $n$ times.  
A set $U$ is completely invariant with respect to $R$ if 
  $$R(U)\subset U$$ 
and 
$$R^{-1}(U)\subset U$$
If $R$ is a polynomial with a basin of attraction $B_{R,c}$, then $B_{R,c}$ is completely invariant
Proof: That $R(B_{R,c})\subset B_{R,c}$ is clear.  This can be easily found in the literature.  To see that $R^{-1}(B_{R,c})\subset B_{R,c}$, suppose $z\in R^{-1}(B_{R,c})$, then there is some $w\in B_{R,c}$ such that $R(z) = w$, but then $$\lim_{n\rightarrow\infty}R^{\circ n}(z) = \lim_{n\rightarrow\infty}R^{\circ n-1}(R(z)) = \lim_{n\rightarrow\infty}R^{\circ n-1}(w) = c$$
Thus $z\in B_{R,c}$ and so $R^{-1}(B_{R,c})\subset B_{R,c}$, concluding the proof.
The result is extended quite easily to rational maps conjugate to a  polynomial via Conjugacy theory.  I would expect this to be true in greater generality, but again, because I have not seen such a result in the literature, it seems that it might not be true.  Any thoughts and suggestions are greatly appreciated!
 A: Yes, every basin of attraction is completely invariant even for any function.
This result can be generalised to metric spaces ( even to topological, but we stick with metric).
Let $(X,d)$ be a metric space. Let $R:X\to X$ be a function and $c\in X$ a point. Denote $$B_{R,c}=\{x\in X\:\lim_{x\to\infty}R^n(x)=c\}.$$
We will show that $R(B_{R,c})\subset B_{R,c}$ and $R^{-1}(B_{R,c})\subset B_{R,c}$.
$R(B_{R,c})\subset B_{R,c}:$ Let $R(z)\in R(B_{R,c}).$ Then
$$\lim_{n\to\infty}R^n(R(z))=\lim_{n\to\infty}R^{n+1}(z)=c.$$
Hence $R(z)\in B_{R,c}.$
$R^{-1}(B_{R,c})\subset B_{R,c}:$ Suppose $z\in R^{−1}(B_{R,c})$, then there is some $w\in B_{R,c}$ such that $R(z)=w$, but then 
$$\lim_{n\rightarrow\infty}R^{ n}(z) = \lim_{n\rightarrow\infty}R^{n-1}(R(z)) = \lim_{n\rightarrow\infty}R^{ n-1}(w) = c.$$
In your case we have $(X,d)=(\mathbb{C},||\cdot||)$ or $(S^2,d)$ in Riemann sphere case.

If you want to find this type of results in literature, you should look in books about discrete dynamical systems.
