Vacuous definition - annoying condition

Question

Can you give me an example of a point of a discrete dynamical system that is neither a fixed point or periodic point, nor in the orbit of a periodic fixed point, but fulfills the definition of either "stable", "attracting" or "asymptotically stable"?
(Pick your favorite way to define these three concepts, as long as you use first-principle definitions and not those where you use eigenvalues, which are actually a consequence of the former, as there are small variations between books.)

How about continuous systems and equilibrium points and periodic solutions?
How about one-dimensional vs. multidimensional discrete/continuous systems?

I'm asking this, because these concepts three usually are only defined in case of discrete systems for fixed points, periodic points and orbits of periodic points - though one could define them actually for any set of points $S\subseteq M$. (And similar considerations hold for continuous systems.)
So the question is, if for any other (set of) point(s), these definitions would always be vacuous (then I'd ask a proof that for points that are neither fixed nor periodc in which case it would make sense to formulated the definition of these concepts only for such restricted classes of points.

Answer 1

Let me summarize my comments in an answer. A pretty simple example of a disrete dynamical system with non-trivial (i.e., not a fixed or periodic point) attractor would be the following: $$ \overline{r} = 1 + \alpha (r - 1), \; \overline{\phi} = \phi + \theta \cdot \frac{1+r^2}{2} \mod 2\pi,$$ where $0 < \alpha <1, \; r \in \mathbb{R}, \; \phi \in \mathbb{S}^1$ and $\frac{\theta}{\pi} \not \in \mathbb{Q}$. The dynamics for $r$ can be analyzed separately from dynamics of $\phi$: it is a linear monotonous mapping which has exactly one fixed point at $r=1$ and no other periodic or fixed points. So, if any periodic points of full mapping do exist, they must be contained in $r=1$. Let's see what happens when $r =1$: $(1, \tilde{\phi}) \mapsto (1, \tilde{\phi} + \theta \mod 2\pi)$, so the set $r =1$ is an invariant set. Because of monotonicity, it is possible to show that the image of any set $\alpha < r < \beta$ with $\alpha < 1 < \beta$ lies in the interior of itself. You can recognize that these properties are enough to call $r = 1$ an attractor in the common sense (see the beginning of Scholarpedia article). However, what happens exactly on the attractor? As we've already shown, the dynamics on this attractor is described by the mapping $\tilde{\phi} \mapsto \tilde{\phi} + \theta \mod 2\pi$. This is known as an irrational rotation of a circle, and for our choice of $\theta$ ($\theta$ and $2\pi$ are rationally inconsummerate, i.e. $\theta / \pi \not \in \mathbb{Q}$) it has no periodic or fixed points.

Answer 2

Let me fix a discrete dynamical system, regarding it as a homeomorphism $f : X \to X$ of a topological space $X$. (One could also have this discussion for continuous self-maps that are not homeomorphisms, but that introduces complications that I would like to ignore).

And I'm going to stick with the concept of "attracting", since I'm not sure that I can guess what you might mean by "stable" or "asymptotically stable".

The concept of "attracting" doesn't make much sense unless you apply it to a subset $C \subset X$ which is invariant under the map $f$, meaning that $f(C)$. This explains, in part, why a fixed point or a periodic orbit is often used when discussing the "attracting" concept: a point is invariant if and only if it is fixed; a finite set is invariant if and only if it is a finite union of periodic orbits.

So now I can define an attracting invariant set:

An invariant set $C \subset X$ is attracting if and only if there exists an open set $V \subset X$ such that $C \subset V$, and such that for any open set $U \subset V$, if $C \subset U$ then there exists an integer $n > 0$ such that $f^n(V) \subset U$. The open set $V$ is called an attracting neighborhood for $C$.

Now, I suppose that you could formulate this definition even in the case that $C$ was not known to be an invariant set.

However, if you added a few more hypotheses then invariance of $C$ would be a consequence of this definition. For example, let's suppose that $X$ is a metric space. If $C \subset X$ is a closed subset then $C$ is equal to the intersection of all open sets $V$ that contain $C$, and from this one can easily show that if $C$ is an attractor then $C$ is an invariant set. In particular, a point (or finite subset) is an attractor only if it is an invariant set.

Furthermore, if $A \subset X$ is an arbitrary subset then one can show that $A$ is an attractor (according to the above definition) if and only if its closure $C = \overline A$ is an attractor, implying that $C$ is invariant, and therefore implying that $A$ is a dense subset of a closed, invariant, attracting set $C$.

So, to study attractors in a metric space, you might as well restrict your attention to invariant sets.