Is a basin of attraction necessarily an open set? 
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.
 A: 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.  
A: 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 $t$
A: 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).
