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Behaviour of two distinct, arbitrary points of the Thue-Morse Subshift


Nice corollaries to Poincaré-Bendixson theoremBi-infinite sequences of $0$'s and $1$'sOrbit of shift operator in symbolic dynamicsShowing that a family of metrics induce all the same topology on special sequence spaceElementary properties of gradient systemsErgodic measure for the left shift supported on an orbitA group action on an infinite set with the cofinite topology$omega(x_0)=mathcal{O}(x_0)$ for periodic points $x_0$ergodic system has dense orbits a.e., continuity of TMinimal dynamical systems in $2^{mathbb N}$













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$begingroup$


Let $x^+$ denote the infinite Thue-Morse word, $$x^+ = 0110100110010110ldots,$$ which is defined as being the only fixed point starting with $0$ of the morphism $S$ (defined over the words on the alphabet ${0,1}$) given by $S(0) = 01$ and $S(1) = 10$ - that is, $$x^+ = S^{omega}(0) = limlimits_{n geqslant 1} S^n(0);$$ this is only one among many different, equivalent definitions of such word (for instance, pick $A_0 = 0$ and let $A_{n+1} = A_n,^frownoverline{A_{n}}$ - where $^frown$ denotes concatenation and $overline{A_n}$ denotes the bitwise complement of $A_n$. Then $x^+ = limlimits_{n in mathbb{N}} A_n$). From now on, let $x^+ = (t_n)_{n in mathbb{N}}$.



It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword $w$ of $x^+$ there is some $n in mathbb{N}$ such that, for every $i in mathbb{N}$, the block $t_{i + 1}t_{i + 2}ldots t_{i + n}$ contains some ocurrence of $w$) but it is $mathbf{textrm{not}}$ ultimately periodic (meaning, it is not true that there are $p geqslant 1$ and $N geqslant 0$ such that $t_i = t_{i + p}$ for all $i geqslant N$).



The two-sided Thue Morse sequence $x = (x_i)_i in mathbb{Z}$ is defined as $x = x^{-},^frown x^+$, meaning that $x_n = t_n$ for all $n geqslant 0$ and $x_{-n} = t_{n - 1}$ for all $n geqslant 1$. So, $$x = ldots 100101100.1101001 ldots$$



Let $sigma$ denote (as usual) the left shift map on $^mathbb{Z}{0,1}$. The Thue-Morse subshift $X$ is the closure (in the Tychonoff topology) of the orbit ${sigma^n(x): n geqslant 0}$. Equivalently, $X$ is the family of all doubly infinite sequences $z$ such that every finite subword of $z$ is a subword of $x^+$. It is known that $X$ is a infinite, minimal subshift (that is, every point of $X$ has a dense orbit in $X$).



My question is:



Let $s = (s_i)_{i in mathbb{Z}},$ and $u = (u_i)_{i in mathbb{Z}},$ be two distinct points of the Thue-Morse subshift $X$ - that is, there is some integer $i in mathbb{Z}$ such that $s_i neq u_i$. Consider the set of integers given by $${i in mathbb{Z}: s_i neq u_i }.$$ Is it true that such set is, necessarily, an $mathbf{textrm{infinite}}$ set of integers ?



My conjecture is that the answer is "yes", but I wasn't able to produce a proof.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Let $x^+$ denote the infinite Thue-Morse word, $$x^+ = 0110100110010110ldots,$$ which is defined as being the only fixed point starting with $0$ of the morphism $S$ (defined over the words on the alphabet ${0,1}$) given by $S(0) = 01$ and $S(1) = 10$ - that is, $$x^+ = S^{omega}(0) = limlimits_{n geqslant 1} S^n(0);$$ this is only one among many different, equivalent definitions of such word (for instance, pick $A_0 = 0$ and let $A_{n+1} = A_n,^frownoverline{A_{n}}$ - where $^frown$ denotes concatenation and $overline{A_n}$ denotes the bitwise complement of $A_n$. Then $x^+ = limlimits_{n in mathbb{N}} A_n$). From now on, let $x^+ = (t_n)_{n in mathbb{N}}$.



    It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword $w$ of $x^+$ there is some $n in mathbb{N}$ such that, for every $i in mathbb{N}$, the block $t_{i + 1}t_{i + 2}ldots t_{i + n}$ contains some ocurrence of $w$) but it is $mathbf{textrm{not}}$ ultimately periodic (meaning, it is not true that there are $p geqslant 1$ and $N geqslant 0$ such that $t_i = t_{i + p}$ for all $i geqslant N$).



    The two-sided Thue Morse sequence $x = (x_i)_i in mathbb{Z}$ is defined as $x = x^{-},^frown x^+$, meaning that $x_n = t_n$ for all $n geqslant 0$ and $x_{-n} = t_{n - 1}$ for all $n geqslant 1$. So, $$x = ldots 100101100.1101001 ldots$$



    Let $sigma$ denote (as usual) the left shift map on $^mathbb{Z}{0,1}$. The Thue-Morse subshift $X$ is the closure (in the Tychonoff topology) of the orbit ${sigma^n(x): n geqslant 0}$. Equivalently, $X$ is the family of all doubly infinite sequences $z$ such that every finite subword of $z$ is a subword of $x^+$. It is known that $X$ is a infinite, minimal subshift (that is, every point of $X$ has a dense orbit in $X$).



    My question is:



    Let $s = (s_i)_{i in mathbb{Z}},$ and $u = (u_i)_{i in mathbb{Z}},$ be two distinct points of the Thue-Morse subshift $X$ - that is, there is some integer $i in mathbb{Z}$ such that $s_i neq u_i$. Consider the set of integers given by $${i in mathbb{Z}: s_i neq u_i }.$$ Is it true that such set is, necessarily, an $mathbf{textrm{infinite}}$ set of integers ?



    My conjecture is that the answer is "yes", but I wasn't able to produce a proof.










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Let $x^+$ denote the infinite Thue-Morse word, $$x^+ = 0110100110010110ldots,$$ which is defined as being the only fixed point starting with $0$ of the morphism $S$ (defined over the words on the alphabet ${0,1}$) given by $S(0) = 01$ and $S(1) = 10$ - that is, $$x^+ = S^{omega}(0) = limlimits_{n geqslant 1} S^n(0);$$ this is only one among many different, equivalent definitions of such word (for instance, pick $A_0 = 0$ and let $A_{n+1} = A_n,^frownoverline{A_{n}}$ - where $^frown$ denotes concatenation and $overline{A_n}$ denotes the bitwise complement of $A_n$. Then $x^+ = limlimits_{n in mathbb{N}} A_n$). From now on, let $x^+ = (t_n)_{n in mathbb{N}}$.



      It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword $w$ of $x^+$ there is some $n in mathbb{N}$ such that, for every $i in mathbb{N}$, the block $t_{i + 1}t_{i + 2}ldots t_{i + n}$ contains some ocurrence of $w$) but it is $mathbf{textrm{not}}$ ultimately periodic (meaning, it is not true that there are $p geqslant 1$ and $N geqslant 0$ such that $t_i = t_{i + p}$ for all $i geqslant N$).



      The two-sided Thue Morse sequence $x = (x_i)_i in mathbb{Z}$ is defined as $x = x^{-},^frown x^+$, meaning that $x_n = t_n$ for all $n geqslant 0$ and $x_{-n} = t_{n - 1}$ for all $n geqslant 1$. So, $$x = ldots 100101100.1101001 ldots$$



      Let $sigma$ denote (as usual) the left shift map on $^mathbb{Z}{0,1}$. The Thue-Morse subshift $X$ is the closure (in the Tychonoff topology) of the orbit ${sigma^n(x): n geqslant 0}$. Equivalently, $X$ is the family of all doubly infinite sequences $z$ such that every finite subword of $z$ is a subword of $x^+$. It is known that $X$ is a infinite, minimal subshift (that is, every point of $X$ has a dense orbit in $X$).



      My question is:



      Let $s = (s_i)_{i in mathbb{Z}},$ and $u = (u_i)_{i in mathbb{Z}},$ be two distinct points of the Thue-Morse subshift $X$ - that is, there is some integer $i in mathbb{Z}$ such that $s_i neq u_i$. Consider the set of integers given by $${i in mathbb{Z}: s_i neq u_i }.$$ Is it true that such set is, necessarily, an $mathbf{textrm{infinite}}$ set of integers ?



      My conjecture is that the answer is "yes", but I wasn't able to produce a proof.










      share|cite|improve this question











      $endgroup$




      Let $x^+$ denote the infinite Thue-Morse word, $$x^+ = 0110100110010110ldots,$$ which is defined as being the only fixed point starting with $0$ of the morphism $S$ (defined over the words on the alphabet ${0,1}$) given by $S(0) = 01$ and $S(1) = 10$ - that is, $$x^+ = S^{omega}(0) = limlimits_{n geqslant 1} S^n(0);$$ this is only one among many different, equivalent definitions of such word (for instance, pick $A_0 = 0$ and let $A_{n+1} = A_n,^frownoverline{A_{n}}$ - where $^frown$ denotes concatenation and $overline{A_n}$ denotes the bitwise complement of $A_n$. Then $x^+ = limlimits_{n in mathbb{N}} A_n$). From now on, let $x^+ = (t_n)_{n in mathbb{N}}$.



      It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword $w$ of $x^+$ there is some $n in mathbb{N}$ such that, for every $i in mathbb{N}$, the block $t_{i + 1}t_{i + 2}ldots t_{i + n}$ contains some ocurrence of $w$) but it is $mathbf{textrm{not}}$ ultimately periodic (meaning, it is not true that there are $p geqslant 1$ and $N geqslant 0$ such that $t_i = t_{i + p}$ for all $i geqslant N$).



      The two-sided Thue Morse sequence $x = (x_i)_i in mathbb{Z}$ is defined as $x = x^{-},^frown x^+$, meaning that $x_n = t_n$ for all $n geqslant 0$ and $x_{-n} = t_{n - 1}$ for all $n geqslant 1$. So, $$x = ldots 100101100.1101001 ldots$$



      Let $sigma$ denote (as usual) the left shift map on $^mathbb{Z}{0,1}$. The Thue-Morse subshift $X$ is the closure (in the Tychonoff topology) of the orbit ${sigma^n(x): n geqslant 0}$. Equivalently, $X$ is the family of all doubly infinite sequences $z$ such that every finite subword of $z$ is a subword of $x^+$. It is known that $X$ is a infinite, minimal subshift (that is, every point of $X$ has a dense orbit in $X$).



      My question is:



      Let $s = (s_i)_{i in mathbb{Z}},$ and $u = (u_i)_{i in mathbb{Z}},$ be two distinct points of the Thue-Morse subshift $X$ - that is, there is some integer $i in mathbb{Z}$ such that $s_i neq u_i$. Consider the set of integers given by $${i in mathbb{Z}: s_i neq u_i }.$$ Is it true that such set is, necessarily, an $mathbf{textrm{infinite}}$ set of integers ?



      My conjecture is that the answer is "yes", but I wasn't able to produce a proof.







      dynamical-systems






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday







      Samuel G. Silva

















      asked Sep 19 '17 at 17:54









      Samuel G. SilvaSamuel G. Silva

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