Question

Show that orbit of the Thue-Morse sequence is dense in the shift space.

Answer 1

Let x+x+ denote the infinite Thue-Morse word,

x+=0110100110010110…,x+=0110100110010110…,

which is defined as being the only fixed point starting with 00 of the morphism SS (defined over the words on the alphabet {0,1}{0,1}) given by S(0)=01S(0)=01 and S(1)=10S(1)=10 - that is,

x+=Sω(0)=limn⩾1Sn(0);x+=Sω(0)=limn⩾1Sn(0);

this is only one among many different, equivalent definitions of such word (for instance, pick A0=0A0=0 and let An+1=An⌢An¯¯¯¯¯¯An+1=An⌢An¯ - where ⌢⌢ denotes concatenation and An¯¯¯¯¯¯An¯ denotes the bitwise complement of AnAn. Then x+=limn∈NAnx+=limn∈NAn). From now on, let x+=(tn)n∈Nx+=(tn)n∈N.

It is known that the Thue-Morse word is uniformly recorrent (meaning, for every finite subword ww of x+x+ there is some n∈Nn∈N such that, for every i∈Ni∈N, the block ti+1ti+2…ti+nti+1ti+2…ti+n contains some ocurrence of ww) but it is notnot ultimately periodic (meaning, it is not true that there are p⩾1p⩾1 and N⩾0N⩾0 such that ti=ti+pti=ti+p for all i⩾Ni⩾N).

The two-sided Thue Morse sequence x=(xi)i∈Zx=(xi)i∈Z is defined as x=x−⌢x+x=x−⌢x+, meaning that xn=tnxn=tn for all n⩾0n⩾0 and x−n=tn−1x−n=tn−1 for all n⩾1n⩾1. So,

x=…100101100.1101001…x=…100101100.1101001…

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

My question is:

Let s=(si)i∈Zs=(si)i∈Z and u=(ui)i∈Zu=(ui)i∈Z be two distinct points of the Thue-Morse subshift XX - that is, there is some integer i∈Zi∈Z such that si≠uisi≠ui. Consider the set of integers given by

{i∈Z:si≠ui}.