If $f(x)=x^3-\lambda x$ then $\Big\{ x\in\mathbb R : \Big| \lim_{n\to \infty}f^n(x)\Big|<+\infty \Big\}$ is a Cantor set. 
This problem is from  An introduction to chaotic dynamical systems  Robert L. Devaney 
I've already  prove (a) and (b) but i'm a bit confused about (c).
For (c) we can assume that $\lambda>4$, I have seen a similar post here but it did not helped me at all, i was thinking of a different approach.

$f$ has $3$ fixed points since $f(x)=x\iff x_0=0 $ or $x_1=-\sqrt{1+\lambda}$, or $x_2=\sqrt{1+\lambda}$, if we define
$$ A=\Big\{ x\in\mathbb R : \Big| \lim_{n\to \infty}f^n(x)\Big|<+\infty \Big\}$$
then $A\subseteq [x_1,x_2]\Rightarrow A=\Big\{ x\in[x_1,x_2] : f^n(x)\nrightarrow \pm \infty \Big\}$ , 
But how can i continue from this point on? 
Any suggestions?
 A: Note that for all $x>\sqrt{\lambda+1}, f(x) > x$, and so $\lim_{n\to\infty}f^n(x)=\infty$. Similarly for $x < -\sqrt{\lambda+1}$.
So the interesting thing is the behaviour of $x\in[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$. The critical question is if $x$ starting in $[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$ ever leaves that interval. We first note that the interval contains a local min and a local max, located at $x=\pm\sqrt{\lambda/3}$ the minimum and maximum value reached evaluate to $\pm\lambda^{3/2}\frac{2}{3\sqrt3}$. So $\lambda$ will be large enough to make $x$ jump out of the interval for some value of $x$ if:
$$
\frac{2\lambda^{3/2}}{3\sqrt3} > \sqrt{\lambda+1}
$$
This gives a critical value of $\lambda=3$. For larger values of $\lambda$, subintervals around $x=\pm\sqrt{\lambda/3}$ get mapped outside of the range $[-\sqrt{\lambda+1}, \sqrt{\lambda+1}]$. If we remove these 2 subintervals, since they will eventually go to $\pm\infty$, the 3 remaining subintervals all have $f$ increasing or decreasing monotonically between $-\sqrt{\lambda+1}$, and  $\sqrt{\lambda+1}$. This is just our original interval again, so applying $f$ another time will remove another 2 subintervals from each of our 3 remaining intervals. Applying this process ad infinitum produces a Cantor set.
