Let $(X\_i)$ be a sequence of compact metric spaces and $(f\_i)$ a sequence of transitive transformations $f\_i:X\_i \to X\_i$ with $0 < h\_{top}(f\_i) < \infty$.


The sequence of dynamical systems satifies:


* $X\_i \subset X\_{i+1}$, $h\_{top}(f\_i) < h\_{top}(f\_{i+1}) $;
* $X\_i$ converges to a compact metric space $X$;
* $f\_{i+1}\mid\_{X\_i} = f\_i$ for every $i$;
* Besides, there is a transformation $f:X \to X$ such that f is transitive, $0 < h\_{top}(f) < \infty$ and $f\mid\_{X\_i} = f\_i$.
* $h\_{top}(f\_i)$ converges to $h\_{top}(f)$


Assume now that the system $(X\_i, f\_i)$ is intrinsically ergodic for all $i\ge0$, i.e., it has a unique measure of maximal entropy. 


QUESTION. Is $(X,f)$ intrinsically ergodic? 


(If it helps, each $(X\_i,f\_i)$ in my set-up is a transitive subshift of finite type (SFT), but $(X,f)$ is not an SFT.) 


If the answer is yes, does there exist a natural way to project the (unique) measure of maximal entropy $\mu$ on $X$ onto $X\_i$ so that the projection of $\mu$ is the measure of maximal entropy $\mu\_i$ on $X\_i$?



Limits of intrinsically ergodic systems
The answer is no. It's based on a (un?)published example of Crannell, Rudolph and Weiss.


The example is the following shift: $X$ is the subset of $\lbrace 0,\pm 1\rbrace ^{\mathbb Z}$ with the property that $x\_k\cdot x\_{k+2^n}$ is not allowed to be $-1$ for any values of $k$ and $n$.


What they prove is that there are 2 measures of maximal entropy for $X$: one the Bernoulli (1/2,1/2) measure living on sequences of 0's and 1's; the other the Bernoulli (1/2,1/2) measure living on sequences of 0's and $-1$'s. In fact I showed with Ayse Şahin that these are the unique measures of maximal entropy.


Now if you let $X\_i$ be the subset of $X$ where you can't have $i$ consecutive $-1$'s, then $X\_i$ is intrinsically ergodic, but $X$ is not.


