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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.

Rodrigo de Azevedo
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MelHA
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  • If the map is not continuous, the existence of a fixed point is not sure, even if the map is contracting the interval. – Lutz Lehmann Oct 30 '17 at 06:51
  • Okay, I guess I'm a little confused as to how I would show that. I'm assuming it's where the discontinuity is at that we wouldn't be able to take it to $p$ or anywhere in the Basin of Attraction. Right? I'm confused because below, they're saying we can. – MelHA Nov 01 '17 at 05:04

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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.

Robert Israel
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  • First and foremost thank you! Secondly, okay, so we do need to have a continuous $f$ in order to have basin of attraction and have a fixed point, but can $f(a,b) = f(a,b)$? I'm having a hard time understanding if it can or can't. – MelHA Oct 31 '17 at 07:13
  • Of course it is possible. For example, from a continuous example take two generic points $c, d \in (a,b)$ and interchange $f(c)$ with $f(d)$. You get a discontinuous function, but $f(a,b)$ is still $f(a,b)$. – Robert Israel Oct 31 '17 at 07:34
  • Okay, I think I get it. What about including the end points of the basin of attraction? I'm trying to see if I can show $f({a,b}) \subset {a,b}$. If they can't equal each other, then what happens? – MelHA Nov 01 '17 at 05:05