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.

Answer 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 $t$

Answer 2

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).