A subshift is a subset $X$ of $A^\mathbb{N}$ or $A^\mathbb{Z}$ (with $A$ finite), such that $X$ is topologically closed and closed under the shift operation. The shift operation is defined by $\sigma(\{x_i\}_{i\in G}) = \{x_{i+1}\}_{i\in G}$ (where $G = \mathbb{N}$ or $\mathbb{Z}$).

The entropy of a subshift is $\lim_{n\rightarrow\infty} \frac{\log(|B_n|)}{n}$ where $B_n\subseteq A^n$ is the set of length-$n$ strings which appear in some element of the subshift.

Is there a one-dimensional subshift $X$ of positive entropy s, all of whose sub-subshifts $Y\subseteq X$ also have entropy s?


1 Answer 1


Yes. If $X$ is minimal (every orbit is dense) then the only subshift $Y\subset X$ is $X$ itself. The Jewett-Krieger theorem allows the construction of minimal subshifts with positive entropy, which therefore have the property you desire (albeit somewhat vacuously). Googling "minimal subshifts with positive entropy" brings up this paper by Henk Bruin, which includes such a construction and also gives two references to earlier constructions:

  1. Hahn and Katznelson, "On entropy of uniquely ergodic transformations", Trans. AMS 126 (1967) 335-360
  2. C. Grillenberger, "Construction of strictly ergodic systems I. Given entropy", Z. Wahrscheinlichkeitstheorie 25 (1972/1973) 323-334
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    $\begingroup$ Is it false for every non-minimal SFT? $\endgroup$ Apr 28, 2016 at 16:42
  • $\begingroup$ If X is a positive entropy SFT, then there is always a subshift Y contained in X such that 0 < h(Y) < h(X). If X is transitive, construct Y by adding a word to the list of forbidden words. If the word you add is long enough, Y will still be a positive entropy SFT. But its entropy is smaller than that of X since the unique MME of X gives full weight to every open set. For the general case when X is not transitive, first replace X by one of its transitive components. $\endgroup$ Apr 28, 2016 at 16:58

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