A shift space $(X, \sigma)$ is a coded system if there exist a countable collection of finite words $(\omega^n)_{n \in \mathbb{N}}$, called generators, such that $X$ is the closure of the set of sequences obtained by freely concatenating the generators.

In Lind and Marcus book An introduction to symbolic dynamics and coding, page 451 there is stated a nice topological equivalence: $(X,\sigma)$ is a coded system if and only if $X$ contain an increasing sequence of transitive subshifts of finite type whose union is dense in $X$.

It is mentioned that this equivalence is due to Krieger and it is cited on Blanchard and Hansel paper Sofic constant-to-one extensions of subshifts of finite type, Proc. Amer.Math. Soc. 112 (1991), 259-265. The given reference is a personal communication between W. Krieger and the authors.

I would like to know if there exist a reference where I can find the proof of this statement.

  • $\begingroup$ I think you could find a proof in the book by Blanchard and Maass, "Topics in Symbolic Dynamics and Applications, but I don't have the book handy so can't check for sure. $\endgroup$ – Douglas Lind Feb 15 '12 at 17:53
  • $\begingroup$ Thank you very much! I'm going to the library right know. I'll let you know if the proof is there. $\endgroup$ – Rafael Alcaraz Barrera Feb 15 '12 at 18:28

In case the book referred to in Doug Lind's comment didn't have what you're looking for, a proof of this statement can be found in Section 2 of this paper:

Doris Fiebig and Ulf-Rainer Fiebig, Invariants for subshifts via nested sequences of shifts of finite type, Ergodic Theory and Dynamical Systems 21 (2001), pp 1731-1758.

There are also some other references there, including an article by Krieger with the proof, and another paper by Fiebig and Fiebig with lots more information on coded systems.

| cite | improve this answer | |
  • $\begingroup$ Thanks Vaughn, a direct explicit reference is exactly what we needed here! -Doug $\endgroup$ – Douglas Lind Feb 16 '12 at 6:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.