The upper and lower bound of the projection of a subshift of finite type I am thinking a problem: given a subshift of finte type of $\{0,1\}^{\mathbb{N}}$ and $2>q>1$, where $q$ is a real number. Then how can we find the largest and smallest numbers of the projection of this subshift of finite type in base $q$. We know that the projection of  the subshift of finite type in base $q$ is a graph-directed self-similar sets. Hence, the problem is that how can we find the extreme points of this set. Here, we only think one-dimensional graph-directed self-similar sets.
The largest and smallest numbers exist as SFT is closed and the projection is compact. 
 A: Let us look at the maximum, for instance.
If $1^k$ is admissible in your SFT for all $k\ge1$, then, clearly, $a=1^\infty$ does the trick.
Otherwise let $k$ be such that $1^k$ is admissible, but $1^{k+1}$ isn't. Then take $a=1^k d_1d_2\dots$, where $(d_n)$ is the greedy expansion of 1 base $q$. This should do it, provided $(d_n)$ is admissible in your SFT. Notice that unless $q$ is a Parry number, $(d_n)$ is not eventually periodic; in particular, it isn't if $q$ is transcendental. (Or $\sqrt2$, for example.)
If $(d_n)$ is not admissible in your SFT, then we might have a problem (and an interesting one, too, it seems). 
A simple example is the SFT whose set of forbidden words is just $\{11\}$ (the ``Fibonacci compactum''). It looks like the maximum is attained at $a=(10)^\infty$ irrespective of whether $q$ is less than the golden ratio or not. But this is a very specific example. 
A: It's a rational point for each $q$. Suppose the SFT has forbidden words of length at most $\ell$. Then given your current symbol and $\ell-1$ previous symbols, you can decide what symbol to put to maximize(minimize) your projection going forward. This means that your point is eventually periodic with period at most $2^\ell$. 
