Definition:
The basin of attraction is the defined as the set of all initial conditions $x_{0}$ such that $x(t$) tends to an attracting fixed point $x^{\ast}$ as time $t$ tends to $\infty$.
Is this basin of attraction necessarily an open set?
My text mentioned nothing about the basin of attraction being an open set-Of course this could imply that the audience is meant to think on a deeper level about the said properties of it being an open set. It is in a given example that I concluded that the author implicitly claimed that the basin of attraction is an open set.
I would like to know if it is indeed true that the basin of attraction is an open set and if it is how can it be shown on a heuristic level.
Thanks in advance.
Best Answer
As pointed out by Did, for non-attracting fixed points $x_0$ the set of points $x$ with $\lim_{t\to \infty} x(t)=x_0$ does not have to be open.
For an attractive fixed point, if the dynamical system is continuous then the attraction basin is indeed open (Note: For discontinuous dynamics this is clearly no true; for a counterexample, just pick any set $S$ containing a neighbourhood of $x_0$ and send it to $x_0$ and send the rest to some point $y_0$ not in $S$; the basin of attraction is now $S$).
To see this for continuous dynamics, argue as follows:
Start with $x_1$ in the attraction basin. We will find a neighborhood of it also in the attraction basin.
By definition of attractive fixed point, there exists open $U$ around $x_0$ such that for any $x\in U$ $\lim_{t\to \infty} x(t)=x_0$. The fact that $x_1$ is attracted of $x_0$ means that $\lim_{t\to \infty} x_1(t)=x_0$. In particular, for some large $T$ we have $x_1(T)\in U$. Since the dynamics is continuous, the preimage of $U$ under $\phi_T$ (lets call it $V$) is open, and contains $x_1$.
I claim that all of $V$ is also in the basin of attraction -- which is obvious because any $x\in V$ ends up in $U$ after time $T$ and then attracts to $x_0$; in formulas $\lim_{t\to \infty} x(t)=\lim_{(s=t-T)\to \infty} x(T)(s) =x_0$.
So there you have it: any $x_1$ in the basin has an open neighbourhood $V$ around it which is also entirely in the basin.