Complex continued fractions with given digits

Question

For which sets of Gaussian integers $A\subseteq\mathbb{Z}[i]$ is the set of continued fractions with digits in $A$ $$ C(A)=\{a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cdots}}\mid a_i\in A\} $$ totally disconnected? Using dimension theory I get a partial result, but I like to know if there is a topological argument.

Answer 1

There exists a constant $c$, such that if $|\alpha| \ge c$ for all $\alpha \in A$, then $C(A)$ is totally disconnected.

To prove this, first let $a_0=0$, and consider $K= \{x+i y \mid x,y \in [-1/2,1/2)\}$. Let $T:K\to K$ be given by $$ Tz = \begin{cases} \frac{1}{z} - \left[ \frac{1}{z} \right] & z \neq 0\\ 0 & z = 0. \end{cases}, $$ where $[z]$ is defined as the point in $\mathbb{Z}[i]$ such that $z-[z] \in K$.

For each $z$, we can define a continued fraction expansion for it by $a_n(z) = [T^{n-1}z]$. Note that if we had picked $K$ differently, we may have obtained a different sequence of continued fraction digits.

The cylinder set $C[a_1]$ can be defined as the set of all $z \in K$ such that $a_1(z)= a_1$. We can define rank $n$ cylinder sets $C[a_1, a_2, \dots, a_n]$ as the set of points $z\in K$ such that $a_i(z) = a_i$ for $1 \le i \le n$. The cylinder sets of rank $n$ are necessarily disjoint from one another. Note that $T$ acts continuously on any cylinder set.

For all $\alpha$ with sufficiently large norm, the cylinder set $C[\alpha]$ is full - that is, $TC[\alpha]= K$. Let $\mathcal{F}$ denote the set of all $\alpha \in \mathbb{Z}[i]$ such that $C[\alpha]$ is full.

The fullness of cylinder sets implies that if you have a string $\{a_1, a_2, a_3, \dots \}$ with each $a_n \in \mathcal{F}$, then there exists a unique $z \in K$ such that $a_n(z) = a_n$ for all $n$. This implies that if $A \subset \mathcal{F}$, then for any $$ z = a_0+ \cfrac{1}{a_1+\dots} \in C(A) $$ we have $a_n(z-a_0)= a_n$, i.e., the continued fraction expansion given by $C(A)$ is the same as the expansion given by $T$.

Now, suppose that - in addition to $A \subset \mathcal{F}$ - all the $\alpha$ have norm sufficiently large so that $C[\alpha]$ is bounded away from the boundary of $K$. In particular, there exists an open set $U$ such that $C[\alpha] \subset U \subset K$, for all $\alpha \in A$.

To show that $C(A)$ is totally disconnected, it suffices to show that no two points $z, z' \in C(A)$ can exist in the same connected component. Without loss of generality we may assume $z, z' \in K \cap C(A)$. There exists a minimal $n$ such that $a_n(z) \neq a_n(z')$. We also have that $T^n z$ and $T^n z'$ are both in $U$. But then $$ z \in T|_{C[a_1(z)]}^{-1} T|_{C[a_2(z)]}^{-1} \dots T|_{C[a_n(z)]}^{-1} U \subset C[a_1(z), \dots, a_n(z)]$$ and $$ z' \in T|_{C[a_1(z')]}^{-1} T|_{C[a_2(z')]}^{-1} \dots T|_{C[a_n(z')]}^{-1} U \subset C[a_1(z'), \dots, a_n(z')]. $$ These preimages of $U$ are open, contain all points $C(A)$ that are in the corresponding rank $n$ cylinder set, and are necessarily disjoint since $a_n(z) \neq a_n(z')$, so $z, z'$ are in separate connected components.

I hope that helps.