I'm looking to understand how to approximate certain countable alphabet subshifts by Markov shifts, and realised that I don't know how to do it even in the finite alphabet case. My guess is that the answer to the following question is well known, but I couldn't work it out this morning. 


Let $X\subset\{0,1\}^{\mathbb Z}$ be closed and shift invariant. It is a classical result that there exist Markov shifts $X^n\supset X$ such that $$\lim\_{n\to\infty} h\_{top}(X^n)= h\_{top}(X),$$ where


$$h\_{top}(X):=\lim\_{n\to\infty}\frac{1}{n}\log(\text{ no. of words of length $n$ in $X$}).
$$


This is easy to see, one just lets $X^n$ be the space made by freely concatenating all words of length $n$ in $X$, which is $n$-step Markov.


Question: Can we do such an approximation from below? i.e. Can we find Markov shifts $X\_n\subset X$ with $\lim\_{n\to\infty} h\_{top} X\_n= h\_{top}(X)?$



Approximating Subshifts From Below
If $X$ is minimal and not a periodic orbit then it cannot contain a periodic orbit and hence in particular cannot contain a Markov shift. A classical construction by Grillenberger shows that one can construct uniquely ergodic (hence in particular minimal) subshifts with arbitrary entropy. (This result also follows from the symbolic version of the Jewett-Krieger Theorem.)


