Suppose $p$ is an attracting fixed point under a continuous map $f$ and that the basin of attraction of $p$ is the interval $(a,b)$. How do I show that $f(a,b)\subset(a,b)$?
So I was able to show that $f(a,b)\subset(a,b)$ with the help of this: Fixed point, with basin of attraction but I'm trying to figure out if $f(a,b)$ can equal to $(a,b)$. I want to say yes because the basin of attractions are open intervals. My other question is that can all of this apply to discontinuous maps? I want to say yes again because basin of attractions are open intervals, so discontinuous maps are okay. I'm not sure how to go further about this.
It can indeed be equal: for example, the basin of attraction of $0$ for $f(x) = x^3$ is $(-1,1)$, and $f((-1,1)) = (-1,1)$.
I presume the basin of attraction of $p$ for map $f$ is defined as the set of all $x$ such that $\lim_{n \to \infty} f^{n}(x) = p$ (where $f^{n}$ is the $n$-fold iterate of $f$). Then it is certainly true that if $x$ is in the basin of attraction of $p$, so is $f(x)$, and this does not require continuity. That is, $f$ maps the basin of attraction into itself. What does require continuity is that the basin of attraction is an open set.