Is a basin of attraction necessarily an open set?

Question

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.

The basin of attraction of a fixed point is usually defined for every fixed point, not only for attracting fixed points. Sure about this part of the definition in your question (which should not be called Theorem)? — Did, Apr 23 '16 at 12:40
So, should we understand that there is no basin of attraction around non-attracting fixed points? — Did, Apr 23 '16 at 13:43
A basin of attraction can exists for non-attracting fixed point. As long as the point is a fixed point, a basin of attraction is possible about that point.@Did — Mathematicing, Apr 24 '16 at 01:14
Which is exactly the point I am making and the reason why the definition in your post should be clarified: are you considering all basins of attraction (and then they can be non open), or only basins of attraction around attracting fixed points (and then they are indeed open)? — Did, Apr 24 '16 at 06:56

score 6 · Accepted Answer · answered Apr 03 '18 at 17:55

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.

score 1 · Answer 2 · answered Apr 23 '16 at 05:09

1

It is true. The heuristic argument is that if $x_0$ is in the basin of attraction, then you can find an $\epsilon$ that is very small (dependent on the gradient around $x_0$) such that $x_0+\epsilon$ is also in the basin of attraction because you can pick a small enough $\epsilon$ such that the first iteration for both wind up in "basically the same place" and then you can iterate this idea of being close enough across all $t$

answered Apr 23 '16 at 05:09

Stella Biderman

31,475

When you speak of 'gradient', are you referring to the trajectories of a vector field? – Mathematicing Apr 23 '16 at 05:40
@Mathematicing I mean, this is a differential equation right? So it has a potential function and a gradient? – Stella Biderman Apr 23 '16 at 13:03

score -2 · Answer 3 · answered Apr 24 '16 at 01:34

I'm pretty sure it's an open set and can give a heuristic proof (note: this is excluding basins of attraction for unstable steady states since they are trivial (the point itself) which is closed). In 2D if you apply the Poincure-Bendixon theorem (basically, just that there's no chaos in 2D), then every flow goes to a fixed point or a periodic orbit. But between any of these, you must have an unstable steady state / periodic orbit, and so the boundary between basins of attraction are unstable states/orbits.

In higher dimensions, this is more difficult because of the possibility of chaos. However, it seems like it's always the case that between any region where there is an attractor (either a chaotic attractor, attracting orbit, or attracting steady state) there are unstable fixed points on the boundary. I wouldn't be able to prove it (and it could be wrong in some crazy example).

Is a basin of attraction necessarily an open set?

3 Answers3