The question that I have in mind is the following:

Which kind of closed sets can arise as the $\omega$-limit of a point for a $1$-dimensional dynamical system?

It is probably somewhat naive, but nowhere I looked did I find a hint of what kind of answer I was to expect. Let me specify a bit what got me to formulate my question this way.

Some of the simplest examples of such systems are rotations of the circle. For those, the $\omega$-limits are either a finite orbit or the whole circle and are from the topological point of view very simple. The Denjoy construction takes it to another level, it builds an homeomorphism of the circle for which the orbit of any point accumulates to a Cantor set. The price to pay though is that such a construction cannot be made $\mathcal{C }^2$.

  1. Do those examples cover every possible accumulation sets: periodic orbits, Cantor sets and the whole circle?

  2. Is it possible that the $\omega$-limit has non-empty interior, but is not the whole circle?

  3. Is it possible to build examples for which an $\omega$-limit can be somehow 'hybrid'?

  4. Can a $\omega$-limit be the union of several periodic orbits?

  5. Are there countable $\omega$-limits which are not finite?

My last question is motivated by the existence of 'very complicated' countable closed subsets of $\mathbb{R}$: there exists countable closed subsets of $\mathbb{R}$ whose sequence of derived sets is non empty until any fixed ordinal. Can those appear as the $\omega$-limit of a homeomorphism of $\mathbb{R}$ or $S^1$? If so is there any restriction on the regularity of such an homeomorphism?

All those questions extend to other $1$-dimensional systems such as codimension $1$ foliations on manifolds, interval exchange transformation, groups acting on the circle...

I'd be very pleased if you could share any interesting example/theorem going some way to answering any of the question I asked. I understand some of those might be very stupid, I apologize in advance :)

  • $\begingroup$ Kronecker systems are semisimple - they break down into disjoint union of minimal subsystems, and as every point in a minimal subsystem is dense in the orbit closure and uniformly recurrent, it is easy to classify its w-limits. Thing start to become interesting when you're not inside a minimal system (or semisimple). $\endgroup$ – Asaf Oct 1 '16 at 16:51
  • $\begingroup$ For an example to 4, think about $\times 2$ action on the circle (I think about it as a Bernoulli system). $0$ is a fixed point, and $\{1/3,2/3\}$ is a period, so just track the trajectory along the period, and sometimes just stick a large bunch of $0$'s into your expansion, you'll get a point which for many times close to the period, but for many other times close to $0$, hence you can get such a $\omega$-limit. $\endgroup$ – Asaf Oct 1 '16 at 16:53
  • $\begingroup$ For 2 say, assuming your map is ergodic (wrt to Lebesgue), as you will contain an open set, this open set would contain a generic point, and the orbit of a generic point is dense (and even equidistributed). $\endgroup$ – Asaf Oct 1 '16 at 16:54
  • $\begingroup$ For $5$ I think it's possible to use the strategy of $4$, order somehow the infinitely (countably-)many periodic orbits for $\times 2$ (they are contained inside discrete subgroups of the torus), and then just mimic the construction I gave, you will "see" much of every orbit for many times hence it will be inside your $\omega$-limit. $\endgroup$ – Asaf Oct 1 '16 at 16:57
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    $\begingroup$ @Asaf: for 4 your limit set will also include preperiodic points of the form $00000\ldots 00000\overline{10}$. $\endgroup$ – Ian Morris Oct 4 '16 at 14:19

I'll stick to the context of continuous (but not necessarily invertible) transformations of $S^1$, and with the $\omega$-limit taken in forward time only (if the map is invertible).

$2$. The map $f \colon [-1,1] \to [-1,1]$ given by $f(x):=4x^3-3x$ has a dense orbit and fixes $1$ and $-1$. Take the disjoint union of two copies of $[-1,1]$ with this map defined on each. Now identify $1$ on the first interval with $1$ on the other, and identify $-1$ with the copy of $-1$ too. This creates a map of the circle with two invariant closed intervals. The "dense orbit" in each interval has $\omega$-limit set equal to the whole interval (i.e. a closed semicircle). I think that with a little care one could come up with an analytic example.

$4$. I don't have a complete answer, but here is a quick sketch to show that the limit set cannot consist only of $k$ distinct fixed points $z_1,\ldots,z_k$. I will show that in this case necessarily $k=1$. Choose $\kappa>0$ such that $d(z_i,z_j)>2\kappa$ when $i \neq j$. Now choose $\delta \in (0,\kappa)$ such that $d(f(y_1),f(y_2))<\kappa$ whenever $d(y_1,y_2)<\delta$. Let $X$ denote the set of all points in $S^1$ which are not within an open $\delta$-ball of any $z_i$. Since $X$ is compact but does not intersect the limit set of $x$, there exists $N\geq 1$ such that $f^n(x)\notin X$ when $n \geq N$. In particular, for every $n \geq N$ there exists a unique $i_n \in \{1,\ldots,k\}$ such that $d(f^n(x),z_{i_n})<\delta$. (Uniqueness holds since otherwise $2\kappa<d(z_i,z_j)\leq d(z_i,f^n(x))+d(f^n(x),z_j)<2\delta$, a contradiction.) I claim that $(i_n)_{n=N}^\infty$ is constant, which implies that the $\omega$-limit set is actually a point. Indeed, if this is not the case then $i_{n+1} \neq i_n$ for some $n$, but then $$2\kappa<d(z_{i_n},z_{i_{n+1}}) \leq d(f(z_{i_n}),f(f^n(x)))+ d(f^{n+1}(x),z_{i_{n+1}})< \kappa + \delta$$ contradicting the definition of $\delta$. It follows that the limit set is a point as claimed. Note that no properties of $S^1$ other than compactness and metrisability have been used.

$5$. Asaf's proposed answer to 4 can be adapted to answer 5. Let $f \colon \mathbb{R}/\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$ be the doubling map $f(x)=2x$, and let $x=\sum_{n=1}^\infty 2^{-n^2}$. A point is in the $\omega$-limit set of $x$ if and only if it is either $0$ or $2^{-k}$ for some integer $k$.


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