If we have $\Delta$ a finite set (For simplicity we can just assume it's $2$) and we are looking at $\Delta^\Bbb N$, we can look at this set as a dynamical system with respect to the action:

$T((a_n))=(a_{n+1})$ (shift left).

Take some sequence $a_n$, how would a minimal dynamical system look in the forward orbit closure of a sequence (where a minimal system is a closed set $Y\subset \bar {O((\alpha_n)_{n=1}^{\infty})} \subset \Delta^\Bbb N$ which is invariant to $T$'s action and any proper closed subset $Z\subset Y$ is not invariant to $T$'s action)?

Is there any way to find such a system for a given function?? Any example would be helpful!

What I've done and thought of so far:

A minimal system in a compact space is a system for which every orbit is dense in, and equivalently every open set $V$ covers the space with a finite number of group elements preimages: $$ A=\bigcup_{g\in G_0} g^{-1}V $$ This gives us that $\bar {O((a_n))}$ is minimal iff the first $k$ elements of $a_n$ appear in any $(b_n)\in\bar {O((a_n))}$ after a bounded number of steps, meaning: $$ \forall n\in\Bbb N,\exists N\in\Bbb N,\exists k\in\Bbb N, k\leq N: $$$$ a_1=b_{k+1},...,a_n=b_{k+n} $$

  • 1
    $\begingroup$ Do you want a complete classification of minimals sets ? or just examples ? $\endgroup$
    – FMB
    Dec 11, 2018 at 14:24
  • $\begingroup$ I would have wanted some kind of a classification, but will also be extremely excited to get some examples! (And non-periodic ones, please) $\endgroup$
    – NL1992
    Dec 11, 2018 at 17:10
  • $\begingroup$ Examples are straightforward if you accept the theorem that any closed invariant set contains a minimal set. For example, choose a preliminary element $(a_n)$ whose closure contains no periodic elements, and so its closure contains a minimal set which is not periodic. $\endgroup$
    – Lee Mosher
    Dec 14, 2018 at 19:25
  • $\begingroup$ @LeeMosher you are right, but what I am looking for is a concrete example, as in a non periodic (nor eventually periodic) function in $2^\Bbb N$ which has a minimal orbit. I'm trying to get rid of the use of choice here and I think that with your assumption it is still used. $\endgroup$
    – NL1992
    Dec 15, 2018 at 2:40
  • $\begingroup$ When you say "get rid of the use of choice", are you referring to the axiom of choice? $\endgroup$
    – Lee Mosher
    Dec 15, 2018 at 15:21

1 Answer 1


I am going to describe an example in $2^{\mathbb Z}$ instead of $2^{\mathbb N}$, which is easier to do because the shift map is invertible in $2^{\mathbb Z}$. It should be straightforward to convert this example into a $2^{\mathbb N}$ example; I'll say a few words about this at the end.

The theory of "unstable foliations", which is part of the theory of hyperbolic dynamical systems, produces many interesting examples of minimal shifts. The idea is to start with a 1-dimensional unstable foliation, and then convert it into a minimal shift map using Markov partitions.

I learned about this from studying William Thurston's theory of pseudo-Anosov surface homeomorphisms in my mathematical youth, and in my mathematical middle age I have also seen it through the lens of the Bestvina-Feighn-Handel theory of attracting laminations and train track maps for automorphisms of free groups.

Here is just about the simplest example I can think of that follows this outline. Consider the following rewriting system (which defines an automorphism of the rank 2 free group $F_2 = F\langle a,b \rangle$, equivalently an isotopy class of homeomorphisms of the once punctured torus): $$a \mapsto ab$$ $$b \mapsto a$$ Write out the third iterate, to get this rewriting system: $$a \mapsto abaab$$ $$b \mapsto aba$$ Now use the fact that $b$ occurs as the middle letter of its image $aba$ to inductively define a sequence of intervals, each of which is the middle subsequence of its image under the rewriting system: $$b$$ $$aba$$ $$abaababaabaab$$ $$abaababaabaababaababaabaababaabaababaababaabaababaababa$$ and so on.

Taking the union of this nested sequence of finite sequences, we obtain a bi-infinite sequence $\underline x = (x_i)_{i=-\infty}^\infty$: the first finite sequence $b$ is $(x_i)_{i=0}^0$; the second finite sequence $aba$ is $(x_i)_{i=-1}^{+1}$; the third finite sequence $abaababaabaab$ is $(x_i)_{i=-6}^{+6}$; and so on.

The sequence $\underline x$ is not periodic, but it has a useful "quasiperiodicity" property: for every $K \ge 0$ there exists $L \ge 0$ such that every length $K$ subsequence $(x_i)_{a \le i \le a+K}$ occurs as a subsequence of every length $L$ subsequence $(x_i)_{b \le i \le b+L}$, meaning that there exists $c$ such that $b \le a+c \le a+c+K \le b+L$ and $x_i = x_{i+c}$ for all $a \le i \le a+K$.

Now, in $2^{\mathbb Z} = \{a,b\}^{\mathbb Z}$, let $C$ be the closure of the orbit of the bi-infinite sequence $\underline x = (x_i)_{i=-\infty}^{+\infty}$. Concretely, this can be described as the set of bi-infinite sequences $\underline y = (y_i)_{i = \infty}^{+\infty}$ such that every finite subsequence $(y_i)_{k \le i \le l}$ occurs as a finite subsequence of $(x_i)_{i=-\infty}^{+\infty}$, meaning that there exists $a$ such that $y_i = x_{i+a}$ for all $k \le i \le l$.

This set $C$ is a closed, minimal, shift-invariant subset of $2^{\mathbb Z}$. This can be proved, without too much trouble, using the quasiperiodicity property. And $C$ is not just a single orbit, because it clearly contains $\underline x$ which is not periodic.

I believe that if you then project $C$ to $2^{\mathbb N}$, composing with the projection $2^{\mathbb Z} \mapsto 2^{\mathbb N}$ which ignores the negative indices, you will get a closed, minimal, shift-invariant subset of $2^{\mathbb N}$.

By the way, the $C$ that I constructed has a "self-similarity map" given by the rewriting system. Using the countably infinite set of possible rewriting systems (and carefully examining them for the required quasiperiodicity property), you get countably many different examples which possess self-similarity maps. However, if you discard self-similarity as a desired property, then you can get uncountably many different examples, but that takes quite a bit more work to describe.


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