Let ${\scr F} = \{w_1, w_2,\dots\}$ be a countable collection of words in some finite alphabet, and let $X$ denote the subshift of bi-infinite strings that avoid the elements of $\scr{F}$.
For each $n\in\mathbb{N}$, the collection of strings that just avoid $w_1,\dots,w_n$ is a subshift of finite type and $X$ is the countable nested intersection of the $X_n$.
Is it true that the entropy of these subshifts satisfy $$h(X_n)\longrightarrow h(X)?$$
If so, a reference would be most welcome. This is not exactly my neck of the mathematical woods...
This is true and quite obvious. Coincidently, we are using this very fact in our current paper (soon to be posetd on arXiv) and we decided to not even bother with a reference. It doesn't matter that the subshifts are of finite type. Whenever $X$ is a nested intersection of some subshifts $X_n$, then $$h(X) = \lim h(X_n) = \inf h(X_n).$$ Here is why. Obviously $h(X)\le h(X_n)$ for each $n$, hence $h(X)\le \lim h(X_n).$ On the other hand, each of $X_n$ has a measure of maximal entropy, say $\mu_n$, i.e., such that $h(\mu_n)=h(X_n).$ Any weak-$*$ accumulation point $\mu$ of the sequence $\mu_n$ is supported by $X$ (this takes a few lines to prove, but is very easy), and by upper semicountuity of entropy in subshifts, $$h(X)\ge h(\mu)\ge \limsup h(\mu_n) = \lim_k h(X_n). $$ Cheers!