# Box dimension of the set of Pisot numbers?

A Pisot number is an algebraic integer bigger than $1$ with all of its Galois conjugates having modulus less than $1$. The set of Pisot numbers is known to be countably infinite and is not dense in $(1,\infty)$. What is the box dimension of the set of Pisot numbers? I have been unable to find anything in the literature that would answer this, but I wouldn't be surprised if it follows directly from something that's already known.

• Presumably by "box dimension" you mean the Minkowski–Bouligand dimension? Oct 30, 2016 at 23:37
• Yes this is what I'm asking for. Oct 30, 2016 at 23:40

The set $\mathcal P$ of all Pisot numbers is known to be closed (Salem). Its limit points $\mathcal P'$ are also known in $(1,2)$ (Talmoudi). The smallest element of $\mathcal P'$ is the golden ratio $G=\frac{1+\sqrt5}2$, and $\mathcal P\cap (1,G)$ is also known (Dufresnoy-Pisot).
In particular, the roots $\eta_n\in(1,2)$ of the polynomials $x^n(x^2-x-1)+1$ are all Pisot numbers, and $\eta_n\uparrow G$ as $n\to\infty$. The question is how fast this convergence is.
We have $$\eta_n^n(\eta_n^2-\eta_n-1)+1=0,$$ and $$\eta_n^n(G^2-G-1)=0.$$ Subtracting, we get that $\eta_n^{-n}\asymp G-\eta_n$, whence $$G-\eta_n\asymp G^{-n}.$$ Thus, the convergence is exponential. Now, if we recall the toy example $\Omega=\{2^{-n}\mid n\ge1\}\cup \{0\}$, we have $\dim_B(\Omega)=0$, so my conjecture is $\dim_B(\mathcal P)=0$ as well.
To prove this rigorously for $\mathcal P\cap(1,2)$, one probably needs to look at all limit points and all sequences of Pisot numbers converging to them (F. L. Talmoudi, ‘Sur les nombres de S ∩ [1, 2[’, C. R. Acad. Sci. Paris S´er. A–B 287 (1978) A739–A741.)
For $\mathcal P'\cap (2,\infty)$, to my best knowledge, not much is known. But I still expect exponential convergence and, consequently, zero box dimension.
Update. As I expected, outside $(1,2)$ this is a very tough problem. The main issue is to deal with $\mathcal P''$, i.e., the set of limit points for the limit points. Even at $\theta=2$ (which is its minimum), it is unclear how to show that $\mathcal P\cap (2, 2+\delta)$ has zero box dimension -- simply because not much is known about $\mathcal P'\cap (2,2+\delta)$.