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$\newcommand{\O}{\mathcal{O}}\newcommand{\G}{\mathcal{G}}\newcommand{\T}{\mathcal{T}}$TLDR; skip to the end of the preamble - we know that the tent map system is topologically transitive. However, do we know explicit descriptions of points which are transitive? We know they exist, but can we find the needle in the haystack?

This question is about topological dynamical systems, in particular about the transitivity of the tent map. Some definitions:

A topological dynamical system is a pair $(K;\varphi)$ where $K$ is a nonempty compact Hausdorff space and $\varphi:K\to K$ is continuous.

The system is (topologically) transitive if there exists at least one $x_0\in K$ such that the orbit of $x_0$ is dense: $$\overline{\{\varphi^n(x_0):n\in\Bbb N_0\}}=K$$

It is a theorem that if the following holds:

($\ast$) Let $(K;\varphi)$ be a topological dynamical system. For all nonempty open $U,V$ in $K$, there exists an $n\in\Bbb N$ such that $\varphi^n(U)\cap V\neq\emptyset$.

And if the system is metrisable, then it the system is also topologically transitive. Proof:

Let ($\ast$) hold in a topological dynamical system $(K;\varphi)$. $(\ast)$ is equivalent to the assertion that $\varphi^{-n}(U)\cap V\neq\emptyset$ for some $n$ as well, which we use below. If $K$ is metrisable, its compactness implies it is also second countable and complete. Let $\{\O_m\}_{m\in\Bbb N}$ be a countable base for the topology on $K$. By $(\ast)$, each of the following sets is dense in $K$: $$\G_n:=\bigcup_{m\in\Bbb N}\varphi^{-m}(\O_n)$$And by continuity of $\varphi$ each $\G_n$ is also open. By the Baire Category Theorem, using metrisability and completeness, we have that: $$\T:=\bigcap_{n\in\Bbb N}\G_n$$Is dense in $K$. Fixing any $x_0\in\T$, if $\mathcal{O}\subseteq K$ is an arbitrary nonempty set we have that there exists at least one $n\in\Bbb N$, $\O_n\subseteq\O$, by basis, and by definition of $\T$ there must exist $k_n\in\Bbb N$, $\varphi^{k_n}(x_0)\in\O_n\subseteq\O$, so the orbit of $x_0$ is dense for each $x_0\in\T$.

Consider now the "tent map":

The topological dynamical system $([0,1];T)$ is the Euclidean subspace $[0,1]$ and the map $T:[0,1]\to[0,1]$ which maps $x\mapsto1-|2x-1|$.

I recently showed that the tent map system is topologically transitive by showing it satisfies $(\ast)$ (the OP uses $(\ast)$ as their base definition, I however find it a less interesting definition). So, we know that there are infinitely many, dense, transitive points. How can we pin down the mystery set $\T$? I could not find any research online pertaining to this, so I am asking here as I am not equipped with the tools to tackle this myself.

Intuitively, any point $x_0$ for which $T^n(x_0)$ is eventually in any open set must be irrational. It certainly cannot be a rational with a power-of-two denominator. Calculating the preimages in the $\G_n$ is nightmarish, and computing their intersection would be even worse. As far as I know, that approach is intractable. Does anyone know any technique for identifying the elements of $\T$? Many thanks. I think it would be really interesting if we could find examples, e.g. it would be nice if, say, $1/e$ were a transitive point.

Alp Uzman
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FShrike
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  • Rational numbers are indeed not transitive points: if $x_0$ has denominator $d$ then every iterate $T^n(x_0)$ also has denominator (dividing) $d$, and so the orbit will not intersect $(0,\frac1d)$ for example. – Greg Martin May 16 '22 at 17:27
  • There are irrational numbers that are not transitive also: The orbit ${T^n(x_0)}$ is extremely closely related to the sequence of fractional parts ${ 2^nx_0 - \lfloor 2^nx_0 \rfloor}$, and the latter sequence will be dense for base-2 normal numbers but not otherwise; so for example, the Prouhet–Thue–Morse constant is not a transitive point. – Greg Martin May 16 '22 at 17:27
  • @GregMartin Great. The Prouhet-Thue-Morse constant is very strange; all I know about Thue-Morse is that it is uniformly recurrent in the two-letter shift system. How do you show it is not transitive? I mean, I am inferring that you are strongly suggesting transitive iff. base-2 normal but don't know for sure – FShrike May 16 '22 at 17:33
  • One has $T^n(x) = T({2^{n-1}x})$ where ${\cdot}$ denotes the fractional-part function. Since the P-T-M constant never has three $0$s or three $1$s in a row in its binary expansion, ${2^{n-1}x}$ never lies in $(0,\frac18)$ or $(\frac78,1)$, and so $T^n(x_0)$ never lies in $(0,\frac14)$. – Greg Martin May 16 '22 at 17:58
  • @GregMartin Nice! Thank you for that counterexample. – FShrike May 16 '22 at 18:04
  • Somewhat related: https://math.stackexchange.com/q/654042/169085 – Alp Uzman May 16 '22 at 23:51

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