# Transcendence and Normality of Complex Numbers

Felipe García-Ramos  Felipe García-Ramos, Physics Institute, Universidad
Autónoma de San Luis Potosí, México.
Faculty of Mathematics and Computer Science, Uniwersytet Jagielloński, Poland
<EMAIL_ADDRESS>,  Gerardo González Robert  Gerardo González Robert,
Department of Mathematical and Physical Sciences, La Trobe University, Bendigo
3552, Australia<EMAIL_ADDRESS>and  Mumtaz Hussain  Mumtaz Hussain,
Department of Mathematical and Physical Sciences, La Trobe University, Bendigo
3552, Australia<EMAIL_ADDRESS>
###### Abstract.

Hurwitz continued fractions associate an infinite sequence of Gaussian
integers to every complex number which is not a Gaussian rational. However,
the resulting space of sequences of Gaussian integers $\Omega$ is not closed.
In this paper, we show that $\Omega$ contains a natural subset whose closure
$\overline{\mathsf{R}}$ encodes continued fraction expansions of complex non-
Gaussian rationals. Furthermore, by means of an algorithm that we develop, we
exhibit for each complex non-Gaussian rational $z$ a sequence belonging to
$\overline{\mathsf{R}}$ giving a continued fraction expansion of $z$. We also
prove that $(\overline{\mathsf{R}},\sigma)$ is a subshift with a feeble
specification property. As an application, we determine the rank in the Borel
hierarchy of the set of Hurwitz normal numbers with respect to the complex
Gauss measure. More precisely, we prove that it is a $\Pi_{0}^{3}$-complete
set. As a second application, we construct a family of complex transcendental
numbers with bounded partial quotients.

##  1\. Introduction

A regular continued fraction is a finite or infinite expression of the form

(1) $a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{\ddots}}}$

where all the terms $a_{n}$ are integers and $a_{n}\geq 1$ for any $n\geq 1$.
Regular continued fractions are a powerful tool in number theory. They provide
a number system where the expansion of a real number is infinite if and only
if it is irrational; for example,

$\sqrt{2}=1+\cfrac{1}{2+\cfrac{1}{2+\ddots}}.$

From a Diophantine approximation perspective, regular continued fractions give
the best rational approximations to a real number in a systematic way (see,
for example, the classic reference [20]). Regular continued fractions can be
studied from a dynamical perspective with the aid of the Gauss map
$T_{\mathbb{R}}:[0,1)\to[0,1)$ given by $T_{\mathbb{R}}(0)=0$ and
$T_{\mathbb{R}}(x)=x^{-1}-\lfloor x^{-1}\rfloor$ for $x\neq 0$ where
$\lfloor\cdot\rfloor$ is the floor function. This dynamical approach not only
provides a natural framework to explore diversity and complexity but also
establishes connections with other fields such as harmonic analysis,
probability theory, and descriptive set theory [4, 24, 25, 31].

There have been several attempts to obtain multidimensional continued fraction
algorithms (see, for example, [18, 30]) but the higher dimensional theory, in
comparison, is hardly as comprehensive as the one-dimensional theory. One such
example is the continued fraction theory for complex plane that we focus on in
this paper. In his 1887 paper [16], Hurwitz introduced a complex continued
fraction algorithm where all the partial quotients $(a_{n})_{n\geq 1}$ are
Gaussian integers. Despite its age, Hurwitz’s algorithm had not received
extensive attention until recently, when interesting discoveries were made
regarding its Diophantine approximation and dynamical properties [8, 11, 12,
13, 15]. Hurwitz showed that his continued fractions associate to each number
$z\in\mathbb{C}\setminus\mathbb{Q}(i)$ a unique sequence $(a_{n})_{n\geq 0}$
in $\mathbb{Z}[i]$ satisfying

$z=[a_{0};a_{1},a_{2},\ldots]:=\lim_{n\to\infty}a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{\ddots+\cfrac{1}{a_{n}}}}.$

Hurwitz continued fractions can also be approached from a dynamical
perspective. In fact, the role of the unit interval is played by the square

$\mathfrak{F}:=\left\\{z\in\mathbb{C}:-\frac{1}{2}\leq\operatorname{Re}(z)<\frac{1}{2},\;-\frac{1}{2}\leq\operatorname{Im}(z)<\frac{1}{2}\right\\}$

and the definition of the complex Gauss map $T$ is similar to that of
$T_{\mathbb{R}}$ (see Section 3). We say a sequence $(a_{n})_{n\geq 1}$ in
$\mathbb{Z}[i]$ is valid when $z=[0;a_{1},a_{2},\ldots]$ for some
$z\in\mathfrak{F}\setminus\mathbb{Q}(i)$ and we denote by $\Omega$ the space
of valid sequences. A central difference between regular and Hurwitz continued
fractions is that, while $\mathbb{N}^{\mathbb{N}}$ is the space of valid
sequences for regular continued fractions, the set $\Omega$ is not even closed
in $\mathbb{Z}[i]^{\mathbb{N}}$ for Hurwitz continued fractions (see Section
4). Working with closed sets is often crucial for developing a natural
dynamical perspective. One solution might involve taking the closure of
$\Omega$, but this approach can result in several intricate points that prove
challenging to handle. Nevertheless, it turns out that there is a naturally
closed set related to $\Omega$ that still encapsulates its fundamental
characteristics. To be more precise, for each $m\in\mathbb{N}$ and
$(b_{1},\ldots,b_{m})\in\mathbb{Z}[i]^{m}$, define the cylinder of level $m$
based on $\mathbf{b}$ as the set

$\mathcal{C}_{m}(\mathbf{b}):=\\{z\in\mathfrak{F}:a_{1}(z)=b_{1},\ldots,a_{m}(z)=b_{m}\\}$

and denote its interior by $\mathcal{C}^{\circ}_{m}(\mathbf{b})$. Let
$\overline{\mathfrak{F}}$ be the closure of $\mathfrak{F}$. Define the space
of sequences

$\overline{\mathsf{R}}:=\left\\{(a_{n})_{n\geq
1}\in\mathbb{Z}[i]^{\mathbb{N}}:\forall
n\in\mathbb{N}\quad\mathcal{C}^{\circ}_{n}(a_{1},\ldots,a_{n})\neq\varnothing\right\\}$

and let $\overline{\Lambda}:\overline{\mathsf{R}}\to\overline{\mathfrak{F}}$
be the mapping acting on each $\mathbf{a}=(a_{n})_{n\geq
1}\in\overline{\mathsf{R}}$ as follows:

$\overline{\Lambda}(\mathbf{a}):=\lim_{n\to\infty}[0;a_{1},\ldots,a_{n}].$

It will be easy to show that $\overline{\Lambda}$ is well defined (see Section
4). Loosely speaking, our first result tells us that we may replace $\Omega$
with $\overline{\mathsf{R}}$.

######  Theorem 1.1.

$\overline{\Lambda}\left(\overline{\mathsf{R}}\right)=\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)$.

The containment
$\overline{\Lambda}\left(\overline{\mathsf{R}}\right)\subseteq\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)$
will be a consequence of the approximation properties of Hurwitz continued
fractions (Proposition 3.1.v). We will prove
$\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)\subseteq\overline{\Lambda}\left(\overline{\mathsf{R}}\right)$,
using an algorithm whose input is a sequence
$\mathbf{a}\in\Omega\setminus\overline{\mathsf{R}}$ and whose output is a
sequence $\mathbf{b}\in\overline{\mathsf{R}}$ such that
$\overline{\Lambda}(\mathbf{b})=\Lambda(\mathbf{a})$. In [8, Corollary 6.4],
$\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)\subseteq\overline{\Lambda}\left(\overline{\mathsf{R}}\right)$
was obtained by different means but the argument fails to provide for any
$\mathbf{a}\in\Omega\setminus\overline{\mathsf{R}}$ a sequence
$\mathbf{b}\in\overline{\mathsf{R}}$ for which
$\Lambda(\mathbf{a})=\overline{\Lambda}(\mathbf{b})$.

Call $\sigma$ the left-shift map on $\mathbb{Z}[i]^{\mathbb{N}}$. In [33],
Yuri proved that $(\overline{\mathsf{R}},\sigma)$ is a sofic shift realized by
a directed graph whose vertices are open prototype sets (see definition in
Section 3) and with edges labeled by Gaussian integers. Yuri’s paper is
concerned with the ergodic theory of a class of dynamical systems defined in
compact subsets of $\mathbb{R}^{n}$, $n\in\mathbb{N}$. She considered measures
equivalent to the $n$-dimensional Lebesgue measure and, thus, certain
Lebesgue-null sets as well as some subsets of the associated shift space are
discarded. These omissions force us to make a more detailed study of the
relationship between $(\overline{\mathsf{R}},\sigma)$ and the dynamics of
Hurwitz continued fraction when dealing with other problems such as those in
Section 4 and Section 6.

Recall that a real number $x\in[0,1)$ is normal to base $2$ if the asymptotic
frequency of every finite word in $\\{0,1\\}$ of length $k\in\mathbb{N}$ in
the binary expansion of $x$ is exactly $2^{-k}$. In general, given a
numeration system, normal numbers are those for which the asymptotic sequence
of every finite word of digits is typical. For the unit interval and regular
continued fractions, the notion of typical is determined by the Gauss measure
of cylinders (see [3, 4, 5, 21]). In the context of Hurwitz continued
fractions, we consider the unique Borel probability $T$-ergodic measure
$\mu_{\operatorname{H}}$ which is equivalent to the Lebesgue measure on
$\mathfrak{F}$ (see [13, 26]) and we refer to normal numbers as _Hurwitz
normal numbers_ (see Subsection 5.1 for a precise definition). As a first
application of Theorem 1.1 and its proof, we determine how complicated is the
set of Hurwitz normal numbers as a Borel set.

######  Theorem 1.2.

The set of Hurwitz normal numbers cannot be expressed as a countable
intersection of open sets or as a countable union of closed (or even
$G_{\delta}$) sets.

In fact, we will establish a more robust result by categorizing the set of
normal numbers within the framework of the Borel hierarchy. The Borel
hierarchy serves as a natural ordinal rank, quantifying the number of steps
required to construct a Borel set, commencing from open sets. A similar result
was obtained by Dani and Nogueira when they were studying the image of
$\mathbb{Z}[i]^{2}$ under binary quadratic forms [12, Propositions 8.5, 8.6].

Theorem 1.2 is a consequence of the dynamical structure of
$\overline{\mathsf{R}}$ with respect to the left-shift map. We prove a weak
form of specification introduced by Airey, Jackson, Kwietniak, and Mance in
[4].

######  Theorem 1.3.

The subshift $(\overline{\mathsf{R}},\sigma)$ has the right feeble-
specification property.

The subshift $(\overline{\mathsf{R}},\sigma)$ is not a full shift, in other
words, $\overline{\mathsf{R}}$ is not of the form $\mathscr{A}^{\mathbb{N}}$
for some $\mathscr{A}\subseteq\mathbb{Z}[i]$. Nonetheless, Theorem 1.3 shows
that the system $(\overline{\mathsf{R}},\sigma)$ still has a considerable
level of freedom in terms of specification. The specification property was
introduced by Bowen in his seminal paper [6]. Bowen’s Specification Theorem
asserts that all transitive hyperbolic systems possess the “specification
property”, wherein complete independence is established among finite subsets
with bounded gaps. The specification and its variations have proven to be
potent tools unlocking access to various other properties, such as the
uniqueness of measures of maximal entropy [6, 10], universality [27], and the
intricate structure of generic points [4].

The second application of Theorem 1.1 is related to trascendental numbers. In
[2], Adamczewski and Bugeaud solved Cobham’s Conjecture when they proved that,
for any $b\in\mathbb{N}_{\geq 2}$, a real number whose $b$-adic expansion is
non-periodic and generated by finite automaton must be transcendental. They
extended their work to continued fractions [1] and, afterward in [7], Bugeaud
gave a combinatorial condition on non-periodic continued fraction ensuring its
transcendence. This condition can be expressed in an extremely neat manner
using the repetition exponent introduced by Bugeaud and Kim [9]. Given a
finite alphabet $\mathscr{A}$ and an infinite word $\mathbf{a}=(a_{n})_{n\geq
1}$ in $\mathscr{A}$, for each $n\in\mathbb{N}$ define

$r(n,\mathbf{a}):=\min\left\\{m\in\mathbb{N}:\exists
i\in\\{1,\ldots,m-n\\}\quad(a_{i},\ldots,a_{i+n+1})=(a_{m},,\ldots,a_{m+n+1})\right\\}.$

The repetition exponent of $\mathbf{a}$ is

$\operatorname{rep}(\mathbf{a}):=\liminf_{n\to\infty}\frac{r(n,x)}{n}.$

Bugeaud proved that for any bounded non-periodic sequence
$\mathbf{a}=(a_{n})_{n\geq 1}\in\mathbb{N}^{\mathbb{N}}$, if
$\operatorname{rep}(\mathbf{a})<\infty$, the continued fraction (1) is
transcendental for any $a_{0}\in\mathbb{Z}$ [7, Theorem 1.3] (Theorem 6.2). In
[14], a complex version of under the additional assumption
$|a_{n}|\geq\sqrt{8}$ for all $n\in\mathbb{N}$ was proved by the second author
of this manuscript. In this paper we apply the algorithm from the proof of
Theorem 1.1 to construct a family of transcendental numbers whose partial
quotients $(a_{n})_{n\geq 1}$ are bounded and such that $|a_{n}|=2$ for
infinitely many $n\in\mathbb{N}$. Schmidt’s Subspace Theorem, a deep result in
Diophantine approximation, lies at the heart of this construction.

######  Theorem 1.4.

Let $\mathbf{B}:=(B_{n})_{n\geq 1}$ be a bounded sequence in $\mathbb{Z}$
which is not periodic and such that

$\min_{n\in\mathbb{N}}|B_{n}|\geq 3\quad\text{ and
}\quad\operatorname{rep}(\mathbf{B})<\infty.$

Then, $\zeta=[0;-2;1+iB_{1},-2,1+iB_{2},-2,1+iB_{3},-2,1+iB_{4},\ldots]$ is
transcendental.

The method of the proof of Theorem 1.4 allows us to construct other families
of transcendental numbers (Theorem 6.6).

To sum up, these are our applications:

  1. (1)

The set $\overline{\mathsf{R}}:=\left\\{(a_{n})_{n\geq
1}\in\mathbb{Z}[i]^{\mathbb{N}}:\forall
n\in\mathbb{N}\quad(a_{1},\ldots,a_{n})\text{ is regular }\right\\}$ is closed
and dynamically chaotic in the sense that it has a weak type of specification.

  2. (2)

The set of Hurwitz normal numbers is classified in the third level of the
Borel hierarchy.

  3. (3)

We construct a family of transcendental complex numbers with bounded partial
quotients.

The structure of the paper is as follows. First, in Section 2, we establish
notation and conventions. In Section 3, we introduce the basic definitions on
Hurwitz continued fractions, we discuss the classification of cylinders and
prototype sets, and we give some basic results. In Section 4 we prove Theorem
1.1 and state the algorithm without proof. Section 5 starts with some basic
notions on Borel Hierarchy and the feeble-specification property. Afterward,
we show Theorem 1.3 and then Theorem 5.1 (which implies Theorem 1.2). In
Section 6 we show Theorem 1.4. Lastly, Section 7 contains the proof of the
algorithm we use to show Theorem 1.1.

Acknowledgements. Research by F. García-Ramos was funded by the Grant:
U1U/W16/ NO/01.03 of the Strategic Excellence Initiative program of the
Jagiellonian University. Research by G. González Robert was partially funded
by CONAHCyT through the program Estancias posdoctorales por México. Both G.
González Robert and M. Hussain were funded by the ARC Discovery Project
200100994.

##  2\. Notation and conventions

We gather some frequently used notations.

  1. i.

By natural numbers $\mathbb{N}$ we mean the set of positive integers and we
write $\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$.

  2. ii.

The real part of a complex number $z$ is denoted by $\operatorname{Re}(z)$ and
its imaginary part is denoted by $\operatorname{Im}(z)$.

  3. iii.

If $X,Y$ are sets, $A\subseteq X$, $B\subseteq Y$ and $f:X\to Y$ is any
function, we define

$f[A]:=\\{f(a):a\in A\\}\quad\text{and}\quad f^{-1}[B]:=\\{a\in A:f(a)\in
B\\}.$

  4. iv.

If $(X,\tau)$ is a topological space and $A\subseteq X$, we let
$\operatorname{Cl}_{X}(A)$ (resp. $\operatorname{Int}_{X}(A)$ ) be the closure
of $A$ (resp. the interior). We omit the sub-index $X$ if the space is clear.

  5. v.

We use the terms word and sequence interchangeably.

  6. vi.

If $\mathscr{A}$ is a non-empty set and $\mathbf{a}=(a_{n})_{n\geq
1}\in\mathscr{A}^{\mathbb{N}}$, for each $m\in\mathbb{N}$ the prefix of length
$m$ of $\mathbf{a}$ is
$\operatorname{pref}(\mathbf{a};m):=(a_{1},\ldots,a_{m})$.

  7. vii.

If $\mathscr{A}$ is a non-empty set, $\mathbf{a}$ is a word in $\mathscr{A}$
of length $n\in\mathbb{N}$ and $\mathbf{b}=(b_{n})_{n\geq 1}$ is a word in
$\mathscr{A}$, we write

$\mathbf{a}\mathbf{b}:=(a_{1},\ldots,a_{n},b_{1},b_{2},\ldots).$

  8. viii.

For any $a\in\mathbb{Z}[i]$, the function $\tau_{a}:\mathbb{C}\to\mathbb{C}$
is given by $\tau_{a}(z)=z+a$, $z\in\mathbb{C}$. By $\iota$ we mean the
complex inversion: $\iota(z)=z^{-1}$ for $z\in\mathbb{C}\setminus\\{0\\}$.

  9. ix.

Given finitely many functions $h_{1},\ldots,h_{n}$ that are either $\iota$,
$\tau_{a}$ for some $a\in\mathbb{Z}[i]$ or some element in $\mathsf{Di}_{8}$
(see Section 3), we write $h_{n}\cdots h_{1}:=h_{n}\circ\cdots\circ h_{1}$.

  10. x.

For $z\in\mathbb{C}$ and $\rho>0$, we write

$\displaystyle\mathbb{D}(z;\rho):=\\{w\in\mathbb{C}:|z-w|<\rho\\},\quad$
$\displaystyle\quad\overline{\mathbb{D}}(z;\rho):=\\{w\in\mathbb{C}:|z-w|<\rho\\}$
$\displaystyle\mathbb{D}(z):=\mathbb{D}(z;1),\quad$
$\displaystyle\quad\overline{\mathbb{D}}(z):=\operatorname{Cl}_{\mathbb{C}}\left(\mathbb{D}(z;1)\right),$
$\displaystyle\mathbb{E}(z):=\mathbb{C}\setminus\overline{\mathbb{D}}(z;1),\quad$
$\displaystyle\quad\overline{\mathbb{E}}(z):=\operatorname{Cl}_{\mathbb{C}}\left(\mathbb{E}(z;1)\right),$
$\displaystyle C(z;\rho):=\\{w\in\mathbb{C}:|z-w|=1\\},\quad$
$\displaystyle\quad C(z):=C(z;1).$

  11. xi.

For $z\in\mathbb{C}$,
$\operatorname{Pm}(z):=\min\\{|\operatorname{Re}(z)|,|\operatorname{Im}(z)|\\}$.

We use implicitly the formulas for the inversion of discs and lines in the
complex plane throughout the paper. Namely, for $k\in\mathbb{R}$, if

$L(\operatorname{Re}(z)=k)=\\{z\in\mathbb{C}:\operatorname{Re}(z)=k\\}\quad\text{
and }\quad
L(\operatorname{Im}(z)=k)=\\{z\in\mathbb{C}:\operatorname{Im}(z)=k\\};$

then,

$\iota[L(\operatorname{Re}(z)=k)]=C\left(\frac{1}{2k};\frac{1}{2|k|}\right),\quad\iota[L(\operatorname{Im}(z)=k)]=C\left(-\frac{i}{2k};\frac{1}{2|k|}\right).$

Also, if $z_{0}\in\mathbb{C}$ and $\rho>0$ are such that $\rho\neq|z_{0}|$
then

$\iota\left[C(z_{0};\rho)\right]=C\left(\frac{\overline{z_{0}}}{|z|^{2}-\rho^{2}};\frac{\rho}{|\rho^{2}-|z_{0}|^{2}|}\right).$

##  3\. Hurwitz continued fractions

We start this section by defining Hurwitz continued fractions (other complex
continued fractions algorithms have been discussed in [12, 23, 28]). Let
$\lfloor\,\cdot\,\rfloor:\mathbb{R}\to\mathbb{Z}$ be the usual floor function.
For $z\in\mathbb{C}$, the nearest Gaussian integer $[z]$ to $z$ is

$[z]:=\left\lfloor\operatorname{Re}(z)+\frac{1}{2}\right\rfloor+i\left\lfloor\operatorname{Im}(z)+\frac{1}{2}\right\rfloor.$

We define

$\mathfrak{F}=\\{z\in\mathbb{C}:[z]=0\\}.$

The complex Gauss map, $T:\mathfrak{F}\to\mathfrak{F}$, is given by

$\quad T(z)=\begin{cases}z^{-1}-[z^{-1}],\text{ if }z\neq 0,\\\ 0,\text{ if
}z=0.\end{cases}$

For any $z\in\mathfrak{F}\setminus\\{0\\}$, we define
$a_{1}(z)\colon=[z^{-1}]$ and, if $T^{n-1}(z)\neq 0$, we put
$a_{n}(z)\colon=a_{1}(T^{n-1}(z))$ for $n\in\mathbb{N}$. As it is customary,
the exponent in $T$ denotes iteration. The Hurwitz continued fraction (HCF) of
$z$ is

$[0;a_{1}(z),a_{2}(z),\ldots]\colon=\cfrac{1}{a_{1}(z)+\cfrac{1}{a_{2}(z)+\cfrac{1}{\ddots}}}.$

The terms of the sequence $(a_{n})_{n\geq 1}$ are the partial quotients of
$z$. Define the sequences of Gaussian integers $(p_{n})_{n\geq 0}$,
$(q_{n})_{n\geq 0}$ by:

(2) $\begin{pmatrix}p_{-1}&p_{-2}\\\
q_{-1}&q_{-2}\end{pmatrix}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\qquad\text{
and }\qquad\forall n\in\mathbb{N}_{0}\quad\begin{pmatrix}p_{n}\\\
q_{n}\end{pmatrix}=\begin{pmatrix}p_{n-1}&p_{n-2}\\\
q_{n-1}&q_{n-2}\end{pmatrix}\begin{pmatrix}a_{n}\\\ 1\end{pmatrix}.$

The general theory of continued fractions (see, for example, [20, Section 2])
tells us that

$\forall
n\in\mathbb{N}\quad\frac{p_{n}}{q_{n}}=[0;a_{1},a_{2},\ldots,a_{n}]:=\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{\ddots+\cfrac{1}{a_{n}}}}}.$

There are complex numbers $\xi$ for which there are more than one Gaussian
integer $a_{0}$ solving

(3) $|\xi-a_{0}|=\min\left\\{|\xi-a|:a\in\mathbb{Z}[i]\right\\}.$

For such $\xi$, we have defined $[\xi]$ to be the solution of (3) with the
largest real and imaginary parts. In many applications, the breaking ties
criterion is irrelevant for it applies on a set of Hausdorff dimension $1$,
and hence it is a Lebesgue null set. However, it ensures the uniqueness of the
HCF expansion. If we are willing to loosen slightly the uniqueness condition,
Theorem 1.1 and its proof imply that it makes no difference what solution of
(3) we choose.

We will often need continued fractions which are not necessarily HCFs. In this
case, we write

$\langle
b_{0};b_{1},b_{2},\ldots\rangle:=b_{0}+\cfrac{1}{b_{1}+\cfrac{1}{b_{2}+\ddots}}.$

and reserve the notation $[b_{0};b_{1},b_{2},\ldots]$ to HCFs. We recall some
elementary facts of Hurwitz continued fractions. Define
$\mathscr{D}:=\mathbb{Z}[i]\setminus\\{0,1,i,-1\\}$.

######  Proposition 3.1.

Let $z=[0;a_{1},a_{2},\ldots]\in\mathfrak{F}\setminus\\{0\\}$ and let
$(p_{n})_{n\geq 0}$, $(q_{n})_{n\geq 0}$ be given by Equation (2). Then,

  1. i.

[16, Section I] For every $n$, we have $a_{n}\in\mathscr{D}$.

  2. ii.

[16, Section I] The sequence $(|q_{n}|)_{n\geq 0}$ is strictly increasing.

  3. iii.

[12, Corollary 5.3] If $\psi\colon=\sqrt{\frac{1+\sqrt{5}}{2}}$, then
$|q_{n}|\geq\psi^{n-1}$ for all $n\in\mathbb{N}_{1}$.

  4. iv.

[12, Corollary 3.3] If
$z=[0;a_{1},a_{2},\ldots]\in\mathfrak{F}\setminus\mathbb{Q}(i)$, then

$\forall n\in\mathbb{N}\quad
z=\frac{(a_{n+1}+[0;a_{n+2},a_{n+3},\ldots])p_{n}+p_{n-1}}{(a_{n+1}+[0;a_{n+2},a_{n+3},\ldots])q_{n}+q_{n-1}}.$

  5. v.

[22, Theorem 1] If
$z=[0;a_{1},a_{2},\ldots]\in\mathfrak{F}\setminus\mathbb{Q}(i)$, then

$\forall
n\in\mathbb{N}\quad\left|z-\frac{p_{n}}{q_{n}}\right|<\frac{1}{|q_{n}|^{2}}.$

###  3.1. HCFs Cylinders

Cylinders and prototype sets are the building blocks of the geometric theory
of Hurwitz continued fractions. Given $n\in\mathbb{N}$ and
$\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathscr{D}^{n}$, the prototype set
$\mathfrak{F}_{n}(\mathbf{a})$ is defined by

$\mathfrak{F}_{n}(\mathbf{a})\colon=T^{n}[\mathcal{C}_{n}(\mathbf{a})].$

We denote the interior (resp. closure) of $\mathcal{C}_{n}(\mathbf{a})$ by
$\mathcal{C}^{\circ}_{n}(\mathbf{a})$ (resp.
$\overline{\mathcal{C}}_{n}(\mathbf{a})$) and similar conventions apply on
$\mathfrak{F}$. For brevity, by open cylinders we mean the interior of a
regular cylinder and similarly for prototype sets. For any set $A$, the number
of elements in $A$ is denoted $\\#A$. Cylinders induce a natural
classification of finite words in $\mathscr{D}$. We say that $\mathbf{a}$ is

  1. i.

valid if $\mathcal{C}_{n}(\mathbf{a})\neq\varnothing$;

  2. ii.

regular if $\mathcal{C}^{\circ}_{n}(\mathbf{a})\neq\varnothing$;

  3. iii.

irregular if $\mathcal{C}^{\circ}_{n}(\mathbf{a})\neq\varnothing$; and
$\mathcal{C}^{\circ}_{n}(\mathbf{a})=\varnothing$;

  4. iv.

extremely irregular if $\\#\mathcal{C}_{n}(\mathbf{a})=1$;

  5. v.

full if $T^{n}(\mathcal{C}_{n}(\mathbf{a}))=\mathfrak{F}$;

We denote by $\Omega(n)$ (resp. $\mathsf{R}(n)$, $\mathsf{Ir}(n)$,
$\mathsf{EI}(n)$, $\mathsf{F}(n)$) the set of valid (resp. regular, irregular,
extremely irregular, full) words of length $n$. We extend the notions regular,
irregular, extremely irregular, and full to cylinders and prototype sets in an
obvious way. It is well known that there are only finitely many prototype sets
(see, for example, [13]); in fact, we have
$\\#\\{\mathfrak{F}^{\circ}_{n}(\mathbf{a}):n\in\mathbb{N},\mathbf{a}\in\mathsf{R}(n)\\}=13$
and the interior of every open prototype set is one of the sets

$\mathfrak{F}^{\circ},\quad\mathfrak{F}^{\circ}_{1}(-2)=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(1),$
$\mathfrak{F}^{\circ}_{1}(-2+i)=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(1-i),\quad\mathfrak{F}^{\circ}_{1}(-1+i)=\mathfrak{F}^{\circ}\setminus\left(\overline{\mathbb{D}}(1)\cup\overline{\mathbb{D}}(-i)\right),$

or one of their right-angled rotations. Figure 1 depicts the possible forms of
open prototype sets.

Figure 1.  From left to right: $\mathfrak{F}^{\circ}$,
$\mathfrak{F}^{\circ}_{1}(-2)$, $\mathfrak{F}^{\circ}_{1}(-2+i)$,
$\mathfrak{F}^{\circ}_{1}(-1+i)$.

An infinite word $\mathbf{a}=(a_{n})_{n\geq 1}\in\mathscr{D}^{\mathbb{N}}$ is

  1. i.

valid if it is the HCF of some $z\in\mathfrak{F}$;

  2. ii.

regular if it is valid and $(a_{1},\ldots,a_{n})\in\mathsf{R}(n)$ for all
$n\in\mathbb{N}$;

  3. iii.

irregular if it is valid and $(a_{1},\ldots,a_{n})\in\mathsf{Ir}(n)$ for some
$n\in\mathbb{N}$;

  4. iv.

extremely irregular if it is valid and $(a_{1},\ldots,a_{n})\in\mathsf{EI}(n)$
for some $n\in\mathbb{N}$.

  5. v.

full if it is valid and $(a_{1},\ldots,a_{n})\in\mathsf{F}(n)$ for all
$n\in\mathbb{N}$.

We denote by $\Omega$ (resp. $\mathsf{R}$, $\mathsf{EI}$, $\mathsf{F}$) the
set of valid (resp. regular, extremely irregular, full).

A significant part of our calculations and our arguments is based on an
observation stemming from the definitions of cylinders: if $(a_{n})_{n\geq 1}$
is a sequence in $\mathscr{D}$, then

$\mathcal{C}_{1}(a_{1})=\mathfrak{F}\cap\iota\tau_{a_{1}}[\mathfrak{F}],\quad\mathcal{C}_{1}(a_{1})=\mathfrak{F}\cap\iota\tau_{a_{1}}[\mathfrak{F}]\cap\iota\tau_{a_{2}}\iota\tau_{a_{1}}[\mathfrak{F}];$

in general,

(4) $\forall
n\in\mathbb{N}\quad\mathcal{C}_{n}(a_{1},\ldots,a_{n})=\mathfrak{F}_{1}\cap\iota\tau_{a_{1}}[\mathfrak{F}]\cap\ldots\cap(\iota\tau_{a_{n}}\cdots\iota\tau_{a_{1}})[\mathfrak{F}].$

Since the functions $\iota$ and $\tau_{a}$ are homeomorphisms, we may replace
$\mathfrak{F}$ and $\mathcal{C}$ with $\mathfrak{F}^{\circ}$ and
$\mathcal{C}^{\circ}$, respectively, in (4). Recall that a word finite word
$\mathbf{b}$ on $\mathscr{D}$ is a factor of a word $\mathbf{a}$ on
$\mathscr{D}$ if there are words $\mathbf{c},\mathbf{d}$ in $\mathscr{D}$ such
that $\mathbf{a}=\mathbf{c}\mathbf{b}\mathbf{d}$. Then, an immediate
consequence of (4) is that every factor of a regular word (finite or infinite)
is again regular. Furthermore, letting $\varepsilon$ be the empty word and
$\mathfrak{F}_{0}(\varepsilon):=\mathfrak{F}$, we have the next proposition.

######  Proposition 3.2.

Every sequence $(a_{n})_{n\geq 1}$ in $\mathscr{D}$ satisfies

$\forall
n\in\mathbb{N}\quad\mathfrak{F}_{n}(a_{1},\ldots,a_{n})=-\tau_{a_{n}}\iota\left[\mathfrak{F}_{n-1}(a_{1},\ldots,a_{n-1})\cap\mathcal{C}_{1}(a_{n})\right].$

###  3.2. Regular cylinders

We recall two important properties of regular cylinders. some results on
geometric sequences. The first one tells us that every infinite sequence in
$\mathscr{D}$ with large terms is full. The second proposition reveals the
symmetries of the family of regular cylinders.

######  Proposition 3.3 ([8, Proposition 4.3]).

If $\mathbf{a}=(a_{n})_{n\geq 1}$ is such that $|a_{n}|\geq\sqrt{8}$, then
$\mathbf{a}\in\mathsf{F}$.

Let $\mathsf{rot},\mathsf{mir}_{1}:\mathbb{C}\to\mathbb{C}$ be given by

$\forall z\in\mathbb{C}\quad\mathsf{rot}(z):=iz\quad\text{ and
}\quad\mathsf{mir}_{1}(z):=\overline{z}$

and call $\mathsf{Di}_{8}$ the group of isometries they generate. Obviously,
$\mathsf{Di}_{8}$ is the dihedral group of order $8$. Proposition 4.5 and
Corollary 4.6 in [8] say that any $f\in\mathsf{Di}_{8}$ maps open cylinders
onto open cylinders of the same level. These results are proven by explicitly
determining the image of an open cylinder under $\mathsf{rot}$ and
$\mathsf{mir}_{1}$. In particular, writing
$\mathsf{mir}_{2}:=\mathsf{rot}^{2}\mathsf{mir}_{1}$ and calling
$\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle\subseteq\mathsf{Di}_{8}$ is
the subgroup generated by $\mathsf{mir}_{1}$ and $\mathsf{mir}_{2}$, we have
the next result.

######  Proposition 3.4.

If $n\in\mathbb{N}$, $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathsf{R}(n)$, and
$\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$, then

$\mathsf{mir}\left(\mathcal{C}^{\circ}_{n}(\mathbf{a})\right)=\mathcal{C}^{\circ}_{n}(\mathsf{mir}(a_{1}),\ldots,\mathsf{mir}(a_{n})).$

###  3.3. Irregular cylinders

For any pair of complex numbers $z$ and $w$, write

$[z,w):=\\{z+t(w-z):t\in[0,1)\\}.$

Irregular cylinders, and thus prototype sets, exist. Certainly, if

$\alpha\colon=\frac{2-\sqrt{3}}{2}\approx 0.13397,$

for every $m\in\mathbb{Z}$ with $|m|\geq 2$ we have

$\mathfrak{F}_{2}(-2,1+mi)=\begin{cases}\left[-\frac{1}{2}-i\alpha,-\frac{1}{2}+\frac{i}{2}\right),\
\text{\rm if }\ m=2,\\\\[8.61108pt]
\left[-\frac{1}{2}-\frac{i}{2},-\frac{1}{2}+i\alpha\right),\ \text{\rm if }\
m=-2,\\\\[8.61108pt]
\left[-\frac{1}{2}-\frac{i}{2},-\frac{1}{2}+\frac{i}{2}\right),\ \text{\rm if
}\ |m|\geq 3\end{cases}$

(see [8, Section 4]). Additionally, if

$\zeta_{1}\colon=-\frac{1}{2}+i\alpha,\quad\zeta_{2}\colon=-\frac{1}{2}-i\alpha,\quad\zeta_{3}\colon=-\alpha-\frac{i}{2},\quad\zeta_{4}\colon=\alpha-\frac{i}{2},$

straightforward calculations yield $a_{1}(\zeta_{1})=-2$,
$a_{1}(\zeta_{2})=-2+i$, $a_{1}(\zeta_{3})=2i$, $a_{1}(\zeta_{4})=1+2i$ and

(5) $T(\zeta_{1})=\zeta_{4},\quad T(\zeta_{2})=\zeta_{4},\quad
T(\zeta_{3})=\zeta_{2},\quad T(\zeta_{4})=\zeta_{2}.$

As a consequence, we have the HCF expansions

$\displaystyle\zeta_{1}$
$\displaystyle=[0;-2,1+2i,-2+i,1+2i,-2+i,1+2i,-2+i,\ldots],$
$\displaystyle\zeta_{2}$
$\displaystyle=[0;-2+i,1+2i,-2+i,1+2i,-2+i,1+2i,-2+i,\ldots],$
$\displaystyle\zeta_{3}$
$\displaystyle=[0;2i,-2+i,1+2i,-2+i,1+2i,-2+i,1+2i,\ldots],$
$\displaystyle\zeta_{4}$
$\displaystyle=[0;1+2i,-2+i,1+2i,-2+i,1+2i,-2+i,1+2i,\ldots].$

By (4) and the formulas for inverting circles, we may show that $\zeta_{1}$,
$\zeta_{2}$, $\zeta_{3}$, $\zeta_{4}$ are extremely irregular. In fact, by
carefully tracking the image under $T$ of the boundaries of prototype sets, we
may even show that a complex number $\xi\in\mathfrak{F}$ is extremely
irregular if and only if $T^{n}(\xi)=\zeta_{4}$ for some $n\in\mathbb{N}$.

######  Lemma 3.5.

Let $n\in\mathbb{N}$ be arbitrary. If $\mathbf{a}\in\mathsf{R}(n)$ and
$b\in\mathscr{D}$ are such that $\mathbf{a}b\in\mathsf{Ir}(n+1)$, then
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})$ is one of the next sets:

$\displaystyle\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(1),\quad\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(-i)$
$\displaystyle,\quad\mathfrak{F}^{\circ}\setminus\left(\overline{\mathbb{D}}(1)\cup\overline{\mathbb{D}}(-i)\right),$
$\displaystyle\mathfrak{F}^{\circ}\setminus\left(\overline{\mathbb{D}}(-1)\cup\overline{\mathbb{D}}(-i)\right),$
$\displaystyle\quad\mathfrak{F}^{\circ}\setminus\left(\overline{\mathbb{D}}(1)\cup\overline{\mathbb{D}}(i)\right).$

###### Proof.

The open prototype sets left unmentioned in the lemma are:

  1. i.

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}$

  2. ii.

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus
i^{k}\overline{\mathbb{D}}(1+i)$ for $k\in\\{1,2,3,4\\}$,

  3. iii.

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(-1)$
and
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(i)$,

  4. iv.

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus\left(\overline{\mathbb{D}}(i)\cup\overline{\mathbb{D}}(-1)\right)$.

The proof now amounts to check that for each of these options

$\mathbf{a}b\in\Omega(n+1)\quad\text{ implies
}\quad\mathbf{a}b\in\mathsf{R}(n+1),$

which follows from the stronger statement

$\iota\left[\overline{\mathfrak{F}}_{n}(\mathbf{a})\right]\cap\tau_{b}\left[\mathfrak{F}\right]\neq\varnothing\quad\text{
implies
}\quad\iota\left[\mathfrak{F}^{\circ}_{n}(\mathbf{a})\right]\cap\tau_{b}\left[\mathfrak{F}\right]\neq\varnothing.$

For the rest of the proof, we rely on pictures in order to avoid long and
elementary computations.

  1. i.

This is obvious.

  2. ii.

For $k=1$, we have
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(1+i)$
so
$\iota[\overline{\mathfrak{F}}_{n}(\mathbf{a})]=\iota[\overline{\mathfrak{F}}]\cap\overline{\mathbb{E}}(1-i)$
and

$\iota\left[\overline{\mathfrak{F}}_{n}(\mathbf{a})\right]\cap\tau_{-b}\iota[\mathfrak{F}^{\circ}]\;\begin{cases}\neq\varnothing,\text{
if }b\neq 1-i,\\\ =\varnothing,\text{ if }b=1-i.\end{cases}$

Furthermore, as we may see in Figure 2, for $b\neq 1-i$ we have
$\iota\left[\mathfrak{F}^{\circ}_{n}(\mathbf{a})\right]\cap\tau_{b}\left[\mathfrak{F}\right]\neq\varnothing$,
hence $\mathbf{a}b\in\mathsf{R}(n+1)$. The cases $k\in\\{2,3,4\\}$ are
similar.

  3. iii.

If
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(-1)$,
then

$\iota\left[\overline{\mathfrak{F}}_{n}(\mathbf{a})\right]=\iota\left[\overline{\mathfrak{F}}\right]\cap\left\\{z\in\mathbb{C}:\operatorname{Re}(z)\geq-\frac{1}{2}\right\\}$

and, by Figure 3,

$\iota\left[\overline{\mathfrak{F}}_{n}(\mathbf{a})\right]\cap\tau_{b}[\mathfrak{F}]\begin{cases}\neq\varnothing,\text{
if }\operatorname{Re}(b)\geq 0,\\\ =\varnothing,\text{ if
}\operatorname{Re}(b)<0.\end{cases}$

Figure 3 also tells us that $\operatorname{Re}(b)\geq 0$ yields
$\iota[\mathfrak{F}^{\circ}_{n}(\mathbf{a})]=\iota[\overline{\mathfrak{F}}]\cap\tau_{b}[\mathfrak{F}]$
and, thus, $\mathbf{a}b\in\mathsf{R}(n+1)$.

  4. iv.

When
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\setminus(\overline{\mathbb{D}}(i)\cup\overline{\mathbb{D}}(-1))$,
by Figure 4, we have
$\tau_{b}\left[\mathfrak{F}^{\circ}_{n}(\mathbf{a})\right]\cap\iota[\overline{\mathfrak{F}}]\neq\varnothing$
if and only if $\operatorname{Re}(b)\geq 0$ and $\operatorname{Im}(b)\geq 0$.
In this case,
$\iota[\mathfrak{F}^{\circ}_{n}(\mathbf{a})]\cap\tau_{b}[\mathfrak{F}^{\circ}]\neq\varnothing$
and $\mathbf{a}b\in\mathsf{R}(n+1)$.

∎

Figure 2.  The sets $\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(1+i)$
(left) and $\iota[\overline{\mathfrak{F}}\setminus\mathbb{D}(1+i)]$ (right).

Figure 3.  The sets $\mathfrak{F}^{\circ}\setminus\overline{\mathbb{D}}(-1)$
(left) and $\iota[\overline{\mathfrak{F}}\setminus\mathbb{D}(-1)]$ (right).

Figure 4.  The sets
$\mathfrak{F}^{\circ}\setminus(\overline{\mathbb{D}}(i)\cup\overline{\mathbb{D}}(-1))$
(left) and
$\iota[(\overline{\mathfrak{F}}\setminus(\mathbb{D}(i)\cup\mathbb{D}(-1)))]$
(right).

##  4\. Proof of Theorem 1.1

###  4.1. The shift space

Endow $\mathscr{D}^{\mathbb{N}}$ with the product topology assuming each
factor has a discrete topology. Then, $\mathscr{D}^{\mathbb{N}}$ can be
metricized with the metric $\operatorname{dist}$ given by

$\forall\mathbf{a},\mathbf{b}\in\mathscr{D}^{\mathbb{N}}\quad\operatorname{dist}(\mathbf{a},\mathbf{b}):=\begin{cases}0,\text{
if }\mathbf{a}=\mathbf{b},\\\
2^{-\min\\{n\in\mathbb{N}:\operatorname{pref}(\mathbf{a};n)\neq\operatorname{pref}(\mathbf{b};m)\\}},\text{
if }\mathbf{a}\neq\mathbf{b}.\end{cases}$

The shift map
$\sigma\colon\mathscr{D}^{\mathbb{N}}\to\mathscr{D}^{\mathbb{N}}$ is the map
that satisfies $\sigma(x)(n)=x(n+1)$ for all $n\in\mathbb{N}$. We link the
dynamical systems $(\Omega,\sigma)$ and
$(\mathfrak{F}\setminus\mathbb{Q}(i),T)$ through the map
$\Lambda:\Omega\to\mathfrak{F}\setminus\mathbb{Q}(i)$ given by

$\forall\mathbf{a}=(a_{n})_{n\geq
1}\in\Omega\quad\Lambda(\mathbf{a}):=[0;a_{1},a_{2},a_{3},\ldots].$

Let $\mathbf{a}\in\Omega$ and $\varepsilon>0$ be arbitrary. Take
$m\in\mathbb{N}$ such that $\frac{2}{\psi^{m-1}}<\varepsilon$ and denote by
$p_{m}/q_{m}$ the $m$-th convergent of $\Lambda(\mathbf{a})$. If
$\mathbf{b}\in\Omega$ is such that
$\operatorname{dist}(\mathbf{a},\mathbf{b})<2^{-m}$, then
$\operatorname{pref}(\mathbf{a};m)=\operatorname{pref}(\mathbf{b};m)$ and, by
Proposition 3.1.v,

$\left|\Lambda(\mathbf{a})-\Lambda(\mathbf{b})\right|\leq\left|\Lambda(\mathbf{a})-\frac{p_{m}}{q_{m}}\right|+\left|\Lambda(\mathbf{b})-\frac{p_{m}}{q_{m}}\right|\leq\frac{2}{|q_{m}|^{2}}\leq\frac{2}{\psi^{m-1}}<\varepsilon.$

In other words, $\Lambda$ is uniformly continuous. Call
$\Lambda|_{\mathsf{R}}$ the restriction of $\Lambda$ to $\mathsf{R}$ and let
$\overline{\Lambda}$ be the unique continuous extension of
$\Lambda|_{\mathsf{R}}$ to $\overline{\mathsf{R}}$.

The next lemma contains the core of Theorem 1.1. We will prove it by
constructing algorithmically the sequence $\mathbf{b}$ from $\mathbf{a}$.

######  Lemma 4.1.

If $\mathbf{a}\in\mathsf{Ir}$, then there is some
$\mathbf{b}\in\overline{\mathsf{R}}$ such that
$\overline{\Lambda}(\mathbf{b})=\Lambda(\mathbf{a})$.

We are now in the position of proving Theorem 1.1.

###### Proof of Theorem 1.1.

The continuity of $\overline{\Lambda}$ implies
$\overline{\Lambda}\left(\overline{\mathsf{R}}\right)\subseteq\overline{\mathfrak{F}}$.
Let $\mathbf{a}\in\overline{\mathsf{R}}$ be arbitrary and put
$z=\overline{\Lambda}(\mathbf{a})$. By Proposition 3.1.v, we have

$\forall
n\in\mathbb{N}\quad\left|z-\frac{p_{n}(a_{1},\ldots,a_{n})}{q_{n}(a_{1},\ldots,a_{n})}\right|\leq\frac{1}{|q_{n}(a_{1},\ldots,a_{n})|^{2}}$

and, since $(|q_{n}|)_{n\geq 1}$ is strictly increasing,
$z\in\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)$.

Now take any $z\in\overline{\mathfrak{F}}\setminus\mathbb{Q}(i)$. Assume that
$z=[0;a_{1},a_{2},\ldots]\in\mathfrak{F}$. If $\mathbf{a}\in\mathsf{R}$, then
$z=\overline{\Lambda}(\mathbf{a})$. When $\mathbf{a}\in\mathsf{Ir}$, the
sequence $\mathbf{b}$ provided by Lemma 4.1 solves the problem. Assume now
that $z\not\in\mathfrak{F}$, hence

$\operatorname{Re}(z)=\tfrac{1}{2}\quad\text{ or
}\quad\operatorname{Im}(z)=\tfrac{1}{2}.$

If $\operatorname{Im}(z)=\frac{1}{2}$, then
$w=\mathsf{mir}_{1}(z)\in\mathfrak{F}$ and we can obtain some
$\mathbf{b}=(b_{n})_{n\geq 1}\in\overline{\mathsf{R}}$ such that
$\Lambda(\mathbf{b})=w$. Thus, the sequence
$\mathbf{c}:=(\mathsf{mir}_{1}(b_{j}))_{j\geq 1}$ also belongs to
$\overline{\mathsf{R}}$, by Proposition 3.4, and satisfies
$\Lambda(\mathbf{c})=z$. A similar argument replacing $\mathsf{mir}_{1}$ with
$\mathsf{mir}_{2}$ gives the proof when $\operatorname{Re}(z)=\frac{1}{2}$. ∎

###  4.2. The algorithm

For each $a\in\mathscr{D}$ such that

$a\in\\{1+im,-1+im,m+i,m-i\\}\text{ for some }m\in\mathbb{Z},\;|m|\geq 2,$

define

$S(a)\colon=\begin{cases}im,\text{ if }a\in\\{1+im,-1+im\\},\\\ m,\text{ if
}a\in\\{m+i,m-i\\}.\end{cases}$

#### Algorithm

  * •

`Input.` A sequence $\mathbf{a}=(a_{n})_{n\geq 1}\in\mathsf{Ir}$.

  * •

`Output.` A sequence $\mathbf{b}=(b_{j})_{j\geq 1}\in\mathscr{D}^{\mathbb{N}}$
satisfying:

    * •

$(b_{1},\ldots,b_{j})\in\mathsf{R}(j)$ for $j\in\mathbb{N}$,

    * •

$\Lambda(\mathbf{a})\in\overline{\mathcal{C}}_{j}(a_{1},\ldots,a_{j})$ for all
$j\in\mathbb{N}$,

    * •

$\Lambda(\mathbf{a})=\langle
0;b_{1},b_{2},\ldots\rangle=\overline{\Lambda}(\mathbf{b})$.

  1. 1.

Put $N=0$ and define $\mathbf{b}_{N}\colon=(b_{N}(j))_{j\geq 1}$ by

$\forall j\in\mathbb{N}\quad b_{N}(j):=a_{j}.$

  2. 2.

If possible, pick $j_{N}\in\mathbb{N}$ such that

(6) $(b_{N}(1),\ldots,b_{N}(j_{N}))\in\mathsf{R}(j_{N})\quad\text{ and
}\quad(b_{N}(1),\ldots,b_{N}(j_{N}),b_{N}(j_{N}+1))\not\in\mathsf{R}(j_{N}+1).$

Define the sequence $\mathbf{b}_{N+1}=(b_{N+1}(j))_{j\geq 1}$ as follows:

    1. i.

For $l\in\\{1,\ldots,j_{N}\\}$, write

$b_{N+1}(l):=b_{N}(l).$

    2. ii.

For $l=j_{N}+1$, write

$b_{N+1}(j_{N}+1):=S\left(b_{N}(j_{N}+1)\right).$

    3. iii.

For $l\in\\{j_{N}+1+n:n\in\mathbb{N}\\}$, let $m\in\mathbb{Z}$ be such that
$b_{N}(j_{N}+1)\in\\{1+im,-1+im,m+i,m-i\\}$.

       * •

If $b_{N}(j_{N}+1)\in\\{m+i,m-i\\}$ for some $m\in\mathbb{Z}$, put

$\forall n\in\mathbb{N}\quad
b_{N+1}(j_{N}+1+n):=\mathsf{mir}_{1}\left(b_{N}(j_{N}+1+n)\right).$

       * •

If $b_{N}(j_{N}+1)\in\\{1+im,-1+im\\}$, then

$\forall n\in\mathbb{N}\quad
b_{N+1}(j_{N}+1+n):=\mathsf{mir}_{2}\left(b_{N}(j_{N}+1+n)\right).$

If there is no $j_{N}\in\mathbb{N}$ such that (6) holds, put
$\mathbf{b}_{N+1}\colon=\mathbf{b}_{N}$. Put $N:=N+1$.

  3. 3.

Repeat step 2.

  4. 4.

Take $\mathbf{b}:=\displaystyle\lim_{N\to\infty}\mathbf{b}_{N}$.
$\blacksquare$

###  4.3. Examples

For the sake of clarity, we illustrate the algorithm with a couple of
examples.

#### Example 1.

Let $(B_{n})_{n\geq 1}$ be a sequence in $\mathbb{Z}$ such that $|B_{n}|\geq
3$ for all $n\in\mathbb{N}$ and $\mathbf{a}=(a_{n})_{n\geq 1}$ by

$\mathbf{a}=(-2,1+iB_{1},-2,1+iB_{2},-2,1+iB_{3},-2,1+iB_{4},\ldots).$

First, let us show that $\mathbf{a}\in\mathsf{Ir}$. Since $|B_{1}|\geq 3$, we
have

$\mathfrak{F}_{2}(-2,1+iB_{1})=\left[-\frac{1}{2}-\frac{i}{2},-\frac{1}{2}+\frac{i}{2}\right),$

so $(-2,1+iB_{1})\in\mathsf{Ir}(2)$ and
$\overline{\mathcal{C}}_{1}(-2)\cap\overline{\mathfrak{F}}_{2}(-2,1+iB_{1})\subseteq\mathfrak{F}_{2}(-2,1+iB_{1})$,
which implies

$\overline{\mathcal{C}}_{3}(-2,1+iB_{1},-2)\subseteq\mathcal{C}_{2}(-2,1+iB_{1}).$

Moreover, we have $\mathfrak{F}_{3}(-2,1+iB_{1},-2)=\mathfrak{F}\cap C(1;1)$,
so

$\iota\left[\mathfrak{F}_{3}(-2,1+iB_{1},-2)\cap\tau_{1+iB_{2}}[\overline{\mathfrak{F}}]\right]=\left[1+i\left(B_{2}-\frac{1}{2}\right),1+i\left(B_{2}+\frac{1}{2}\right)\right],$

and hence

$\overline{\mathcal{C}}_{4}(-2,1+iB_{1},-2,1+iB_{2})\subseteq\mathcal{C}_{3}(-2,1+iB_{1},-2)$

In fact, we may continue this way to show

$\forall
n\in\mathbb{N}\quad\overline{\mathcal{C}}_{n+1}(a_{1},\ldots,a_{n+1})\subseteq\mathcal{C}_{n}(a_{1},\ldots,a_{n}).$

Since $|\mathcal{C}_{n}(a_{1},\ldots,a_{n})|\to 0$ as $n\to\infty$, we can
single out a complex number $z$ satisfying

$\\{z\\}=\bigcap_{n\in\mathbb{N}}\mathcal{C}_{n}(a_{1},\ldots,a_{n}).$

As a consequence, $z=[0;a_{1},a_{2},\ldots]$ and $\mathbf{a}\in\mathsf{Ir}$,
because $(-2,1+iB_{1})\in\mathsf{Ir}(2)$. The algorithm gives us the sequence

$\mathbf{b}=(b_{j})_{j\geq
1}=(-2,iB_{1},2,iB_{2},-2,iB_{3},2,iB_{4},-2,iB_{5},2,iB_{6},\ldots),$

To see it, we worked out the first two iterations.

  1. i.

Take $N=0$ and $\mathbf{b}_{0}\colon=\mathbf{a}$. Since $j_{0}=1$, we have

$\displaystyle b_{1}(1)$ $\displaystyle:=a_{1}=-2,$ $\displaystyle b_{1}(2)$
$\displaystyle:=S(a_{2})=iB_{1},$ $\displaystyle\forall l\in\mathbb{N}_{\geq
3}\quad b_{1}(l)$ $\displaystyle:=\mathsf{mir}_{2}(a_{l}),$

which means

$\mathbf{b}_{1}=(-2,iB_{1},2,-1+iB_{2},2,-1+iB_{3},2,-1+iB_{4},\ldots).$

Add $1$ to $N$.

  2. ii.

For $N=1$, we have $j_{1}=3$ because

$(-2,iB_{1},2)\in\mathsf{R}(3)\quad\text{ and
}\quad(-2,iB_{1},2,-1+iB_{2})\not\in\mathsf{R}(3).$

Note that $(-2,iB_{1},2,-1+iB_{2})\not\in\Omega(3)$. We have

$\displaystyle\forall l\in\\{1,2,3\\}\quad b_{2}(l)$
$\displaystyle:=b_{1}(l),$ $\displaystyle b_{2}(4)$
$\displaystyle:=S(b_{1}(3))=iB_{2},$ $\displaystyle\forall
l\in\mathbb{N}_{\geq 5}\quad b_{2}(l)$
$\displaystyle:=\mathsf{mir}_{2}(b_{1}(l))=\mathsf{mir}_{2}^{2}(a_{l})=a_{l}.$

which is

$\mathbf{b}_{2}=(-2,iB_{1},2,iB_{2},-2,1+iB_{3},-2,1+iB_{4},\ldots).$

#### Example 2.

Similarly, from the extremely irregular sequence

$\mathbf{a}=(-2,1+2i,-2+i,1+2i,-2+i,1+2i,-2+i,\ldots),$

where the block $1+2i,-2+i,1+2i,-2+i$ is repeated, we get

$\mathbf{b}=(-2,2i,2,-2i,-2,2i,2,-2i,\ldots).$

Observe that the sequences

$\displaystyle(-2+i,\,-2i,\,2,\,2i,-2,\,-2i,\,2,\,2i,-2,\,-2i,\,2,\,2i,\ldots),$
$\displaystyle(-2+i,\,1-2i,\,-2,\,2i,\,2,\,-2i,\,-2,\,2i,\,2,\,-2i,\ldots)$

also belong to $\overline{\mathsf{R}}$ and are mapped to $\zeta_{1}$ under
$\overline{\Lambda}$ (cfr. [8, Proposition 5.5]).

##  5\. First application: Borel hierarchy of the set of Hurwitz normal
numbers

The main aim of this section is to determine how complicated are Hurwitz
normal numbers as Borel sets. First, we give a rigorous definition of Hurwitz
normal numbers. Then, we recall the construction of the Borel Hierarchy.
Afterward, we give some definitions and results from symbolic dynamics. Then,
we show Theorem 1.3. We finish this section proving Theorem 5.1 below, which
implies Theorem 1.2.

######  Theorem 5.1.

The set $\operatorname{Norm}(\mu_{\operatorname{H}})$ is
$\mathbf{\Pi}_{3}^{0}(\mathbb{C})$-complete.

###  5.1. Hurwitz normal numbers

In [26], Nakada proved the existence and uniqueness of a $T$-ergodic Borel
probability measure on $\mathfrak{F}$, $\mu_{\operatorname{H}}$, equivalent to
the Lebesgue measure. For any $A\subseteq\mathfrak{F}$ let $\chi_{A}$ be its
indicator function; that is, $\chi_{A}(x)=1$ if $x\in A$ and $\chi_{A}(x)=0$
if $x\in\mathfrak{F}\setminus A$. By virtue Birkhoff’s Ergodic Theorem [32,
Theorem 1.14], for every $n\in\mathbb{N}$ and every
$\mathbf{b}\in\mathsf{R}(n)$, almost every $z\in\mathfrak{F}$ with respect to
$\mu_{\operatorname{H}}$ (or, equivalently, to Lebesgue measure) satisfies

(7)
$\lim_{n\to\infty}\frac{\\#\\{j\in\\{0,\ldots,n-1\\}:(a_{j}(z),\ldots,a_{j+n-1}(z))=\mathbf{b}\\}}{n}=\mu_{\operatorname{H}}(\mathcal{C}_{n}(\mathbf{b})).$

In fact, for almost every $z\in\mathfrak{F}$ the equality (7) holds for every
finite regular word $\mathbf{b}$. We refer to such numbers $z$ as Hurwitz
normal numbers and we denote them by
$\operatorname{Norm}(\mu_{\operatorname{H}})$.

As for regular continued fractions or integer base expansions, normality is
closely related to equidistribution. Given a sequence $(z_{n})_{n\geq 1}$ in
$\mathfrak{F}$, let $\delta_{z_{n}}$ be the point mass measure on $z_{n}$ for
each $n\in\mathbb{N}$. We call $(z_{n})_{n\geq 1}$ uniformly
distributed$\pmod{\mu_{\operatorname{H}}}$ if the sequence of measures
$\frac{1}{n}\sum_{j=1}^{n}\delta_{z_{n}}$ converges weakly to
$\mu_{\operatorname{H}}$. From the Portmanteau Theorem [17, Theorem 5.25], we
can derive the next equality:

$\operatorname{Norm}(\mu_{\operatorname{H}})=\left\\{z\in\mathfrak{F}:(T^{n}(z))_{n\geq
1}\text{ is uniformly distributed}\pmod{\mu_{\operatorname{H}}}\right\\}.$

In particular, for every $z\in\operatorname{Norm}(\mu_{\operatorname{H}})$ the
set $\mathcal{O}^{T}_{+}(z):=\\{T^{n}(z):n\in\mathbb{N}_{0}\\}$ is dense in
$\mathfrak{F}$.

###  5.2. Borel hierarchy

We recall some definitions on the Borel hierarchy. The proofs for the
statements in this subsection can be found in [19, Chapter 11]. Let $(X,\tau)$
be a Polish space, that is, a separable completely metrizable topological
space. Denote the first uncountable ordinal by $\omega_{1}$. Write
$\Sigma_{1}^{0}(X):=\tau$ and for each ordinal $\xi$ with
$1\leq\xi<\omega_{1}$ put

$\displaystyle\mathbf{\Pi}_{\xi}^{0}(X)$
$\displaystyle\colon=\left\\{X\setminus A\colon
A\in\mathbf{\Sigma}_{\xi}^{0}(X)\right\\}$
$\displaystyle\mathbf{\Sigma}_{\xi}^{0}(X)$
$\displaystyle\colon=\left\\{\bigcup_{n\in\mathbb{N}}A_{n}:n\in\mathbb{N},\,\xi_{n}<\xi,A_{n}\in\mathbf{\Pi}_{\xi_{n}}^{0}(X)\right\\}.$

Then, for any two ordinals $\xi,\xi^{\prime}$ satisfying
$1\leq\xi<\xi^{\prime}<\omega_{1}$ we have

$\mathbf{\Pi}_{\xi}^{0}(X)\cup\mathbf{\Sigma}_{\xi}^{0}(X)\subseteq\mathbf{\Pi}_{\xi^{\prime}}^{0}(X)\quad\text{
and
}\quad\mathbf{\Pi}_{\xi}^{0}(X)\cup\mathbf{\Sigma}_{\xi}^{0}(X)\subseteq\mathbf{\Sigma}_{\xi^{\prime}}^{0}(X).$

Let $\mathscr{B}(X)$ be the Borel $\sigma$-algebra on $X$. It is a well
established fact that

$\mathscr{B}(X)=\bigcup_{\xi<\omega_{1}}\mathbf{\Pi}_{\xi}^{0}(X)=\bigcup_{\xi<\omega_{1}}\mathbf{\Sigma}_{\xi}^{0}(X).$

and that, whenever $X$ is also uncountable, all the families
$\mathbf{\Pi}_{\xi}^{0}(X)$ and $\mathbf{\Sigma}_{\xi}^{0}(X)$ are different.
This allows us to stratify $\mathscr{B}(X)$ in terms of their complexity in
the so-called Borel hierarchy.

For any ordinal $\xi<\omega_{1}$, we say that a set $A\subseteq X$ is
$\mathbf{\Pi}_{\xi}^{0}(X)$-hard (resp. $\mathbf{\Sigma}_{\xi}^{0}(X)$-hard)
if $A\not\in\mathbf{\Sigma}_{\xi}^{0}(X)$ (resp.
$A\not\in\mathbf{\Pi}_{\xi}^{0}(X)$). We say that $A$ is
$\mathbf{\Sigma}_{\xi}^{0}(X)$-complete (resp.
$\mathbf{\Pi}_{\xi}^{0}(X)$-complete) if it belongs to
$\mathbf{\Sigma}_{\xi}^{0}(X)$ and $\mathbf{\Sigma}_{\xi}^{0}(X)$-hard (resp.
$\mathbf{\Pi}_{\xi}^{0}(X)$ and $\mathbf{\Pi}_{\xi}^{0}(X)$-hard). In simple
terms, a Borel set is $\mathbf{\Pi}_{\xi}^{0}(X)$-complete when its first
appearance in the Borel hierarchy is precisely $\mathbf{\Pi}_{\xi}^{0}(X)$.

###  5.3. Symbolic dynamics

Let $\mathscr{A}$ be non-empty and at most countable set. We denote non-empty
words $\mathbf{u}$ on $\mathscr{A}$ as $\mathbf{u}=(u_{1},u_{2},\ldots)$. In
symbolic dynamics, it is customary to denote words without parenthesis, i.e.
$\mathbf{u}=u_{1}u_{2}\cdots$. He has decided to keep the parentheses for the
notation’s consistency. Let $\mathscr{A}^{<\omega}$ denote the set of finite
words in $\mathscr{A}$.

The length of a non-empty finite word $\mathbf{u}$ is denoted by
$|\mathbf{u}|$; for example, for $\mathbf{u}=u_{1}\cdots u_{n}$ we have
$|\mathbf{u}|=n$ and if $\epsilon$ is the empty word, we define
$|\epsilon|=0$. We remind the reader that given two finite words
$\mathbf{u}=(u_{1},\ldots,u_{n})$ and $\mathbf{v}=(v_{1},\ldots,v_{m})$, the
concatenation of $\mathbf{u}$ and $\mathbf{v}$ is

$\mathbf{u}\mathbf{v}\colon=(u_{1},\ldots,u_{n},v_{1},\ldots v_{m}).$

For any two non-empty finite words $\mathbf{v}=v_{1}\cdots v_{n}$,
$\mathbf{w}=w_{1}\cdots w_{n}$ of the same length $n\in\mathbb{N}$, the
normalized Hamming distance between $\mathbf{v}$ and $\mathbf{w}$ is

$d_{H}(\mathbf{v},\mathbf{w})=\frac{\\#\\{j\in\\{1,\ldots,n\\}:v_{j}\neq
w_{j}\\}}{n}.$

Endow $\mathscr{A}^{\mathbb{N}}$ with the product topology. A subshift $X$ of
$\mathscr{A}^{\mathbb{N}}$ is a closed and strongly shift invariant subset of
$\mathscr{A}^{\mathbb{N}}$, that is $\sigma(X)=X$. For any
$\mathbf{b}\in\mathscr{A}^{<\omega}$, $\mathbf{b}\neq\varepsilon$, the
symbolic cylinder based on $\mathbf{b}$ is the set

$[\mathbf{b}]_{X}\colon=\left\\{\mathbf{a}\in
X:\operatorname{pref}(\mathbf{a};|\mathbf{b}|)=\mathbf{b}\right\\}.$

We denote by $\mathscr{L}(X)$ the set of all finite words that appear as
factors of some words in $X$. The subshift $X$ has the right feeble
specification property if there is some $\mathscr{G}\subseteq\mathscr{L}(X)$
with the following properties:

  1. i.

If $\mathbf{u},\mathbf{v}\in\mathscr{G}$, then
$\mathbf{u}\mathbf{v}\in\mathscr{G}$;

  2. ii.

For every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for any
$\mathbf{u}\in\mathscr{G}$ and any $\mathbf{v}\in\mathscr{L}(X)$ satisfying
$|\mathbf{v}|\geq N$ there are some
$\mathbf{s}^{\prime},\mathbf{v}^{\prime}\in\mathscr{A}^{<\omega}$ satisfying

$|\mathbf{v}^{\prime}|=|\mathbf{v}|,\quad
0\leq|\mathbf{s}|\leq\varepsilon|\mathbf{v}|,\quad
d_{H}(\mathbf{v},\mathbf{v}^{\prime})<\varepsilon,\quad\mathbf{u}\mathbf{s}\mathbf{v}^{\prime}\in\mathscr{G}.$

For any $\mathbf{x}=(x_{1},x_{2},x_{3},\ldots)\in\mathscr{A}^{\mathbb{N}}$ and
$\mathbf{w}\in\mathscr{A}^{<\omega}$ define

$e(\mathbf{w},\mathbf{x},N)\colon=\\#\left\\{j\in\\{1,\ldots,N\\}:(x_{j},x_{j+1},\ldots,x_{j+n-1})=\mathbf{w}\right\\}.$

The quasi-regular set $Q(X)$ of $X$ is the set of words for which the
frequency of every finite word exist:

$Q(X)\>=\left\\{\mathbf{x}\in
X:\forall\mathbf{w}\in\mathscr{L}(X)\quad\lim_{N\to\infty}\frac{e(\mathbf{w},\mathbf{x},N)}{N}\text{
exists }\right\\}.$

For any $\sigma$-invariant Borel measure $\nu$ on $X$, we say that a word
$\mathbf{x}\in X$ is $\nu$-generic if

$\forall\mathbf{w}\in\mathscr{A}^{<\omega}\quad\lim_{N\to\infty}\frac{e(\mathbf{w},\mathbf{x},N)}{N}=\nu\left([\mathbf{w}]_{X}\right).$

We denote by $G_{\nu}$ the set of $\nu$-generic points. Clearly,
$Q(X)\subseteq G_{\nu}$.

In [4], Airey, Jackson, Kwietniak, and Mance determined the place in the Borel
Hierarchy of certain Borel sets under the right feeble specification property
and some additional mild conditions.

######  Theorem 5.2 ([4, Theorem 6]).

Let $X$ be a closed subshift with the right feeble specification property and
at least two shift-invariant Borel probability measures. Let $\nu$ be a shift-
invariant Borel probability measure. If $B$ is a Borel set such that
$G_{\nu}\subseteq B\subseteq Q(X)$, then $B$ is
$\mathbf{\Pi}_{3}^{0}(X)$-hard. In particular, $G_{\nu}$ and $Q(X)$ are
$\mathbf{\Pi}_{3}^{0}(X)$-complete.

###  5.4. Proof of Theorem 1.3

The following lemma is essentially [12, Proposition 8.4].

######  Lemma 5.3.

Let $n\in\mathbb{N}$ and $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathsf{R}(n)$ be
arbitrary.

  1. i.

If $\operatorname{Pm}(a_{n})\geq 3$ and $\mathbf{b}\in\mathsf{R}(m)$ for some
$m\in\mathbb{N}$, then $\mathbf{a}\mathbf{b}\in\mathsf{R}(m+n)$.

  2. ii.

There exists some $b\in\mathbb{Z}[i]$ such that $\operatorname{Pm}(b)\geq 3$
and $\mathbf{a}b\in\mathsf{F}(n+1)$.

###### Proof of Theorem 1.3.

By definition, $\overline{\mathsf{R}}$ is closed in
$\mathscr{D}^{\mathbb{N}}$. Moreover, $\overline{\mathsf{R}}$ is strong
$\sigma$-invariant because factors of regular words are regular by (4). Define
the set

$\mathscr{G}\colon=\left\\{\mathbf{c}=(c_{1},\ldots,c_{m})\in\bigcup_{n\in\mathbb{N}}\mathsf{R}(n):\operatorname{Pm}(c_{m})\geq
3\right\\}.$

Let us verify that $\mathscr{G}$ has the conditions in the definition of the
right feeble specification property.

  1. i.

If $\mathbf{u},\mathbf{v}\in\mathscr{G}$, then
$\mathbf{u}\mathbf{v}\in\mathscr{G}$ by Lemma 5.3.(i).

  2. ii.

Let $\varepsilon>0$ be arbitrary and pick $N\in\mathbb{N}$ such that
$N^{-1}<\varepsilon$. Take
$\mathbf{v}=(v_{1},\ldots,v_{m})\in\mathscr{L}(\overline{\mathsf{R}})$ with
$m\geq N$. Because of $\mathbf{v}\in\mathsf{R}(m)$, we have
$(v_{1},\ldots,v_{m-1})\in\mathsf{R}(m-1)$ and, by Lemma 5.3.(ii), we may
choose $a\in\mathscr{D}$ satisfying

$\operatorname{Pm}(a)\geq 3\quad\text{ and
}\quad\mathbf{v}^{\prime}\colon=(v_{1},\ldots,v_{m-1},a)\in\mathscr{G}.$

Letting $\mathbf{s}$ be the empty word, for every $\mathbf{u}\in\mathscr{G}$
we have

$0\leq|\mathbf{s}|\leq\varepsilon|\mathbf{v}|,\quad|\mathbf{v}^{\prime}|=|\mathbf{v}|,\quad
d_{H}(\mathbf{v},\mathbf{v}^{\prime})<\varepsilon,\quad\mathbf{u}\mathbf{d}\mathbf{v}^{\prime}\in\mathscr{G}.\;\qed$

###  5.5. Proof of Theorem 5.1

We split the proof of Theorem 5.1 into Lemma 5.4 and Lemma 5.6.

######  Lemma 5.4.

The set $\operatorname{Norm}(\mu_{\operatorname{H}})$ is
$\mathbf{\Pi}_{3}^{0}(\mathbb{C})$-hard.

###### Proof.

The space $(\overline{\mathsf{R}},\sigma)$ has infinitely many
$\sigma$-invariant measures. Certainly, by Proposition 3.3, any constant
sequence $\mathbf{a}=(a,a,\ldots)\in\mathscr{D}^{\mathbb{N}}$ with
$|a|\geq\sqrt{8}$ belongs to $\overline{\mathsf{R}}$ and the point mass
measure based on $\mathbf{a}$ is $\sigma$-invariant.

Let us construct the $\sigma$-invariant Borel probability measure $\nu$ on
$\overline{\mathsf{R}}$ on which we apply Theorem 5.2. Let
$\mathscr{I}\subseteq\overline{\mathsf{R}}$ be the set

$\mathscr{I}\colon=\bigcap_{n\in\mathbb{N}}\bigcup_{\mathbf{a}\in\mathsf{R}(n)}\mathcal{C}^{\circ}_{n}(\mathbf{a})$

and define the map $\Phi:\Lambda^{-1}[\mathscr{I}]\to\mathscr{I}$ by

$\forall\mathbf{a}\in\Lambda^{-1}[\mathscr{I}]\quad\Phi(\mathbf{a}):=\overline{\Lambda}(\mathbf{a}).$

It is easy to show that $\Lambda^{-1}[\mathscr{I}]$ is a Borel subset of
$\overline{\mathsf{R}}$ and that $\Phi$ is a homeomorphism. Then, the
following conditions define a unique Borel probability measure $\nu$ on
$\overline{\mathsf{R}}$ given by:

  1. i.

For every Borel subset $B$ of $\Lambda^{-1}[\mathscr{I}]$ we have
$\nu(B):=\mu_{\operatorname{H}}(\Phi[B])$.

  2. ii.

$\nu\left(\overline{\mathsf{R}}\setminus\overline{\Lambda}\,^{-1}[\mathscr{I}]\right)=0$.

The measure $\nu$ is $\sigma$-invariant, by virtue of the $T$-invariance of
$\mu_{\operatorname{H}}$ and the equality $\Lambda\circ\sigma=T\circ\Lambda$
on $\mathsf{R}$. Therefore, by Theorem 5.2, the set
$G_{\nu}\subseteq\overline{\mathsf{R}}$ is
$\Pi_{3}^{0}(\overline{\mathsf{R}})$-complete. To finish the proof, we must
translate this conclusion into the complex plane. First, note that if $z$
belongs to the boundary of some prototype set, then $T(z)$ also belongs to the
boundary of some prototype set. The proof of this assertion amounts to
applying the complex inversion to a finite collection of line segments and
arches. As a consequence, since for each
$z\in\operatorname{Norm}(\mu_{\operatorname{H}})$ the set
$\mathcal{O}^{T}_{+}(z)$ is dense in $\mathfrak{F}$, we have
$\operatorname{Norm}(\mu_{\operatorname{H}})\subseteq\mathscr{I}$. Moreover,
the definitions of $G_{\nu}$ and $\operatorname{Norm}(\mu_{\operatorname{H}})$
yield

$G_{\nu}=\Phi^{-1}[\operatorname{Norm}(\mu_{\operatorname{H}})].$

Therefore, the set $\operatorname{Norm}(\mu_{\operatorname{H}})$ is
$\mathbf{\Pi}_{3}^{0}(\mathscr{I})$-complete and, thus, it is
$\mathbf{\Pi}_{3}^{0}(\mathbb{C})$-hard (see Remark 5.5 below). ∎

######  Remark 5.5.

If $X,Y$ are Polish spaces, $A\subseteq X$, $B\subseteq Y$, a Wadge reduction
of $A$ to $B$ is a continuous function $f:X\to Y$ such that $f^{-1}[B]=A$. It
is easy to verify that if, for some ordinal $\xi<\omega_{1}$, the set $B$ is
$\mathbf{\Pi}_{\xi}^{0}$-hard and there is Wadge reduction of $A$ to $B$, then
$A$ is $\mathbf{\Pi}_{\xi}^{0}$-hard. For more on Wadge reductions, see [19,
Section 21.E].

######  Lemma 5.6.

The set $\operatorname{Norm}(\mu_{\operatorname{H}})$ belongs to
$\mathbf{\Pi}_{3}^{0}(\mathbb{C})$.

###### Proof.

Let $(\varepsilon_{k})_{k\geq 1}$ be a strictly decreasing sequence of
positive numbers converging to $0$. For $M,n,k\in\mathbb{N}$ and
$\mathbf{b}\in\mathsf{R}(n)$, define the sets

$\displaystyle E_{n}(\mathbf{b},M,k)$
$\displaystyle\colon=\left\\{z\colon=[0;a_{1},a_{2},\ldots]:\left|\frac{1}{M}\sum_{j=0}^{M-1}\chi_{\mathcal{C}_{n}(\mathbf{b})}\left(T^{j}(z)\right)-\mu_{\operatorname{H}}(\mathcal{C}_{n}(\mathbf{a}))\right|\leq\varepsilon_{k}\right\\},$
$\displaystyle F_{n}(\mathbf{b},M,k)$
$\displaystyle\colon=\operatorname{Cl}_{\mathbb{C}}(E_{n}(\mathbf{b},M,k)),$
$\displaystyle G_{n}(\mathbf{b},M,k)$
$\displaystyle\colon=\operatorname{Int}_{\mathbb{C}}(E_{n}(\mathbf{b},M,k)).$

Then, the definition of $\operatorname{Norm}(\mu_{\operatorname{H}})$ gives

$\operatorname{Norm}(\mu_{\operatorname{H}})=\bigcap_{n\in\mathbb{N}}\bigcap_{\mathbf{b}\in\mathsf{R}(n)}\bigcap_{k\in\mathbb{N}}\bigcup_{N\in\mathbb{N}}\bigcap_{M\in\mathbb{N}_{\geq
N}}E_{n}(\mathbf{b},M,k).\\\ $

Note that each $E_{n}(\mathbf{b},M,k)$ is a finite union of cylinders of level
$n+M$. Put

$\displaystyle\mathfrak{L}$
$\displaystyle\colon=\bigcap_{n\in\mathbb{N}}\bigcap_{\mathbf{b}\in\mathsf{R}(n)}\bigcap_{k\in\mathbb{N}}\bigcup_{N\in\mathbb{N}}\bigcap_{M\in\mathbb{N}_{\geq
N}}G_{n}(\mathbf{b},M,k),$ $\displaystyle\mathfrak{M}$
$\displaystyle\colon=\bigcap_{n\in\mathbb{N}}\bigcap_{\mathbf{b}\in\mathsf{R}(n)}\bigcap_{k\in\mathbb{N}}\bigcup_{N\in\mathbb{N}}\bigcap_{M\in\mathbb{N}_{\geq
N}}F_{n}(\mathbf{b},M,k),$

so
$\mathfrak{L}\subseteq\operatorname{Norm}(\mu_{\operatorname{H}})\subseteq\mathfrak{M}$.
Now we show $\mathfrak{M}\subseteq\mathfrak{L}$ in order to conclude the
lemma. Let $z\in\mathfrak{M}$ be arbitrary and assume that

(8)
$z\in\bigcap_{n\in\mathbb{N}}\mathcal{C}^{\circ}_{n}\left(a_{1}(z),\ldots,a_{n}(z)\right).$

Let $n,k\in\mathbb{N}$ and $\mathbf{b}\in\mathsf{R}(n)$ be arbitrary and
consider $N\in\mathbb{N}$ such that

$\forall M\in\mathbb{N}_{\geq N}\quad z\in F_{n}(\mathbf{b},M,k).$

Take any $M\in\mathbb{N}_{\geq N}$ and let $\mathbf{d}\in\mathsf{R}(M+n)$ be
such that

$\mathcal{C}_{M+n}(\mathbf{d})\subseteq E_{n}(\mathbf{b},M,k)\quad\text{ and
}\quad z\in\overline{\mathcal{C}}_{M+n}(\mathbf{d}).$

Since different regular cylinders of the same level have disjoint interior,
from (8) we get $\mathbf{d}=\left(a_{1}(z),\ldots,a_{M+n}(z)\right)$, so $z\in
G_{n}(\mathbf{b},M,k)$. This shows $\mathfrak{M}\subseteq\mathfrak{L}$ and,
hence, $\operatorname{Norm}(\mu_{\operatorname{H}})=\mathfrak{M}$.

Lastly, we now show that (8) is indeed true. To this end, it suffices to prove
that $\mathcal{O}_{T}^{+}(z)$ is dense in $\mathfrak{F}$, which in turns
follows from

(9) $\forall
n\in\mathbb{N}\quad\forall\mathbf{b}\in\mathsf{R}(n)\quad\left\\{T^{j}(z):j\in\mathbb{N}_{0}\right\\}\cap\mathcal{C}^{\circ}_{n}(\mathbf{b})\neq\varnothing.$

Take any $n\in\mathbb{N}$ and $\mathbf{b}\in\mathsf{R}(n)$. Let
$c\in\mathscr{D}$ be such that

$\mathbf{b}c\in\mathsf{R}(n+1)\quad\text{ and
}\quad\overline{\mathcal{C}}_{n+1}(\mathbf{b}c)\subseteq\mathcal{C}^{\circ}_{n}(\mathbf{b}).$

Choose $k\in\mathbb{N}$ so large that

$\varepsilon_{k}<\frac{\mu_{\operatorname{H}}(\mathcal{C}_{n+1}(\mathbf{b}c))}{4}$

and let $N\in\mathbb{N}$ be such that

$\forall M\in\mathbb{N}_{\geq N}\quad z\in F_{n+1}(\mathbf{b}c,M,k).$

Note that $E_{n+1}(\mathbf{b}c,M,k)$ is a finite union of cylinders of level
$M+n$. Take $\mathbf{d}\in\mathsf{R}(M+n)$ satisfying

$\mathcal{C}_{M+n}(\mathbf{d})\subseteq E_{n+1}(\mathbf{b}c,M,k)\quad\text{
and }\quad z\in\overline{\mathcal{C}}_{M+n}(\mathbf{d}).$

The choice of $k$ and the definition of $E_{n+1}(\mathbf{b}c,M,k)$ guaranty
the existence of some $r\in\\{n+1,\ldots,M+n\\}$ satisfying

$(d_{r-n},\ldots,d_{r-1},d_{r})=(b_{1},\ldots,b_{n},c).$

Then, by choice of $c$, we have

$\overline{\mathcal{C}}_{M+n-1}(\mathbf{d})\subseteq\overline{\mathcal{C}}_{r}(d_{1},\ldots,d_{r})\subseteq\mathcal{C}^{\circ}_{r-1}(d_{1},\ldots,d_{r-1});$

therefore, applying $T^{r-n-1}$ we conclude
$T^{r-n-1}(z)\in\mathcal{C}^{\circ}_{n}(\mathbf{b})$ and, hence, (9). ∎

##  6\. Second application: construction of transcendental numbers

Recall from the introduction that, given a finite alphabet $\mathscr{A}$ and
an infinite word $\mathbf{a}=(a_{n})_{n\geq 1}$ on $\mathscr{A}$, for each
$n\in\mathbb{N}$ we define

$r(n,\mathbf{a}):=\min\left\\{m\in\mathbb{N}:\exists
i\in\\{1,\ldots,m-n\\}\quad(a_{i},\ldots,a_{i+n+1})=(a_{m},,\ldots,a_{m+n+1})\right\\}$

and that the repetition exponent of $\mathbf{a}$ is

$\operatorname{rep}(\mathbf{a}):=\liminf_{n\to\infty}\frac{r(n,x)}{n}.$

Bugeaud and Kim [9] showed that the finiteness of
$\operatorname{rep}(\mathbf{x})$ for a given word $\mathbf{x}$ is equivalent
to a combinatorial condition used in [1, 7] to construct real transcendental
numbers.

######  Lemma 6.1 ([9, Section 10]).

Let $\mathscr{A}$ be a finite alphabet and $\mathbf{a}=(a_{n})_{n\geq
1}\in\mathscr{A}^{\mathbb{N}}$. If $\operatorname{rep}(\mathbf{a})<\infty$,
then there are three sequences $(W_{n})_{n\geq 1}$, $(U_{n})_{n\geq 1}$,
$(V_{n})_{n\geq 1}$ of finite words in $\mathscr{A}$ such that

  1. 1.

For all $n\in\mathbb{N}$, the word $W_{n}U_{n}V_{n}U_{n}$ is a prefix of
$\mathbf{a}$;

  2. 2.

The sequence $\left((|W_{n}|+|V_{n}|)/|U_{n}|\right)_{n\geq 1}$ is bounded
above;

  3. 3.

$|W_{n}|\to\infty$ as $n\to\infty$.

Lemma 6.1 allows us to state a well-known result by Bugeaud [7] as follows.

######  Theorem 6.2 ([7, Theorem 1.3]).

Let $\mathbf{a}=(a_{n})_{n\geq 1}\in\mathbb{N}^{\mathbb{N}}$ be a bounded and
non-periodic sequence such that $\operatorname{rep}(\mathbf{a})<\infty$. Then,
$\langle 0;a_{1},a_{2},a_{3},\ldots\rangle$ is transcendental.

Theorem 6.3 is the HCF version of Theorem 6.2.

######  Theorem 6.3.

[14, Theorem 5.1] Let $\mathbf{a}\in\Omega$ be a bounded and non-periodic
sequence such that $\operatorname{rep}(\mathbf{a})<\infty$. Let
$(W_{n})_{n\geq 1}$, $(U_{n})_{n\geq 1}$, and $(V_{n})_{n\geq 1}$ be sequences
as in Lemma 6.1.

  1. 1.

If $\displaystyle\liminf_{n\to\infty}|W_{n}|<\infty$, then
$\Lambda(\mathbf{a})$ is transcendental.

  2. 2.

If $\displaystyle\liminf_{n\to\infty}|W_{n}|=\infty$ and $|a_{n}|\geq\sqrt{8}$
for all $n\in\mathbb{N}$, then $\Lambda(\mathbf{a})$ is transcendental.

We conjecture that the condition $|a_{n}|\geq\sqrt{8}$ for all
$n\in\mathbb{N}$ can be removed from the second point of Theorem 6.3. The
second application of our work, Theorem 1.4, is a construction of a family of
transcendental numbers overlooked by Theorem 6.3. This provides some evidence
supporting our conjecture. In [29], Schmidt extended his celebrated Subspace
Theorem to number fields [29]. Theorem 6.4 below is the corresponding
statement for $\mathbb{Q}(i)$. For $m\in\mathbb{N}$ and
$\mathbf{z}=(z_{1},\ldots,z_{m})\in\mathbb{C}^{m}$, we write
$\|\mathbf{z}\|=\max\\{|z_{1}|,\ldots,|z_{m}|\\}$ and
$\overline{\mathbf{z}}=(\overline{z_{1}},\ldots,\overline{z_{m}})$, where the
bar means complex conjugation.

######  Theorem 6.4 ([29, Theorem 3]).

Let $m\in\mathbb{N}$. Assume that we have two collections of linearly
independent linear forms $\mathscr{L}_{1},\ldots,\mathscr{L}_{m}$ and
$\mathscr{M}_{1},\ldots,\mathscr{M}_{m}$ in $m$ variables with real or complex
algebraic coefficients. For each $\varepsilon>0$ there is a finite collection
$T_{1},\ldots,T_{k}$ of proper subspaces of $\mathbb{C}^{m}$ such that every
solution $\boldsymbol{\beta}=(\beta_{1},\ldots,\beta_{m})\in\mathbb{Z}[i]^{m}$
of

$\left|\mathscr{L}_{1}(\boldsymbol{\beta})\cdots\mathscr{L}_{m}(\boldsymbol{\beta})\right|\left|\mathscr{M}_{1}(\boldsymbol{\beta})\cdots\mathscr{M}_{m}(\boldsymbol{\beta})\right|<\|\boldsymbol{\beta}\|^{-\varepsilon}$

lies in $T_{1}\cup\cdots\cup T_{k}$.

Recall that, in general, if $A_{0},A_{1},\ldots,A_{n}$ are variables and
$q_{0},q_{1},\ldots,q_{n}$ are the denominators of the continued fraction
$\langle A_{0};A_{1},A_{2},A_{3},\ldots,A_{n}\rangle$, then

(10) $\frac{q_{n-1}}{q_{n}}=\langle
0;A_{n},A_{n-1},\ldots,A_{2},A_{1}\rangle.$

Equation (10) is known as mirror formula [20, Theorem 6].

######  Lemma 6.5.

Under the assumptions of Theorem 1.4, there exists an $\varepsilon>0$ such
that for all $n\in\mathbb{N}$ we have
$\psi^{2u_{n}}\geq|q_{2w_{n}}{q_{2w_{n}+2u_{n}+2v_{n}}}|^{\varepsilon}$.

###### Proof.

This is essentially [14, Lemma 5.2]. ∎

The shuffle of two words of the same length $A=(a_{1},\ldots,a_{n})$ and
$B=(b_{1},\ldots,b_{n})$ is the word $s(A,B)=(c_{1},\ldots,c_{2n})$ given by
$c_{2j-1}=a_{j}$ and $c_{2j}=b_{j}$ for all $j\in\\{1,\ldots,n\\}$; that is,

$s(A,B)=(a_{1},\,b_{1},\,a_{2},\,b_{2},\ldots,\,a_{n},\,b_{n}).$

###### Proof of Theorem 1.4.

We omit several lengthy but direct computations. Similar arguments have
appeared in [1, 7] for real numbers and in [14, Section 5.2] for complex
numbers.

Assume that $\zeta$ is an algebraic number, then
$[\mathbb{Q}(i,\zeta):\mathbb{Q}(i)]\geq 3$ because its HCF is not periodic
(see, for example, [12, Theorem 4.2]). Let $(W_{n})_{n\geq 1}$,
$(U_{n})_{n\geq 1}$, and $(V_{n})_{n\geq 1}$ be the sequences associated to
$\mathbf{B}$ as in Lemma 6.1 and define the sequences of non-negative integers
$(w_{n})_{n\geq 1}$, $(u_{n})_{n\geq 1}$, and $(v_{n})_{n\geq 1}$ by

$\forall n\in\mathbb{N}\quad
w_{n}:=|W_{n}|,\;u_{n}:=|U_{n}|,\;v_{n}:=|V_{n}|,\;t_{n}:=w_{n}+u_{n}+v_{n}.$

We may assume that every $u_{n}$ is even. Indeed, suppose that $u_{n}$ is odd
for every large $n$. For such $n$, call $x$ be the last letter of $U_{n}$ and
$\widehat{U}:=\operatorname{pref}(U_{n},u_{n}-1)$, so

$W_{n}U_{n}V_{n}U_{n}=W_{n}\,\widehat{U}_{n}x\,V_{n}\widehat{U}_{n}x$

and we can replace $U_{n}$ with $\widehat{U}_{n}$ and $V_{n}$ with $x\,V_{n}$.

When $\displaystyle\liminf_{n\to\infty}w_{n}<\infty$, the sequence
$(-2,1+iB_{1},-2,1+iB_{2},\ldots)$ also has finite repetition exponent and its
corresponding sequence $(W_{n})_{n\geq 1}$ has a constant subsequence. The
transcendence of $\zeta$ then follows from Theorem 6.3. Assume that
$w_{n}\to\infty$ when $n\to\infty$. It was observed in [7, p. 1013] that we do
not lose any generality by assuming $B_{w_{n}}\neq B_{t_{n}}$ for all
$n\in\mathbb{N}$. Define $\mathbf{a}=(a_{n})_{n\geq 1}$ by

$\forall n\in\mathbb{N}\quad a_{n}:=\begin{cases}-2,\text{ if }n\text{ is
odd},\\\ 1+iB_{\frac{n}{2}},\text{ if }n\text{ is even}.\\\ \end{cases}$

By Example 1, we have $\mathbf{a}\in\mathsf{Ir}$ and the algorithm produces
the sequence

$\mathbf{b}=(b_{1},b_{2},\ldots)=(-2,iB_{1},2,iB_{2},-2,iB_{3},2,iB_{4},\ldots).$

Note that

(11) $\forall
n\in\mathbb{N}\quad(b_{1},\ldots,b_{n}),\;(b_{n},\ldots,b_{1})\in\mathsf{R}(n).$

For each $n\in\mathbb{N}$, let $A_{n}$, $B_{n}$, $C_{n}$ be the alternating
words

$\displaystyle A_{n}$
$\displaystyle:=\left(-2,\,2,\,\ldots,\,(-2)^{w_{n}}\right),$ $\displaystyle
B_{n}$
$\displaystyle:=\left((-2)^{w_{n}+1},\,(-2)^{w_{n}+2},\ldots,\,(-2)^{w_{n}+u_{n}}\right),$
$\displaystyle C_{n}$
$\displaystyle:=\left((-2)^{w_{n}+1},\,(-2)^{w_{n}+2},\ldots,\,(-2)^{w_{n}+v_{n}}\right),$

and define

$W_{n}^{\prime}:=s(A_{n},W_{n}),\quad U_{n}^{\prime}:=s(B_{n},U_{n}),\quad
V_{n}^{\prime}:=s(C_{n},V_{n}).$

Therefore, since for all $n\in\mathbb{N}$ we have
$\operatorname{pref}(\mathbf{b},2n)\in\mathsf{F}(n)$ and $u_{n}$ is even, the
periodic sequence
$W_{n}^{\prime}\,U_{n}^{\prime}\,V_{n}^{\prime}\,U_{n}^{\prime}\,V_{n}^{\prime}\,U_{n}^{\prime}\cdots$
belongs to $\overline{\mathsf{R}}$ and the number

$\zeta_{n}:=\overline{\Lambda}(W_{n}^{\prime}\,U_{n}^{\prime}\,V_{n}^{\prime}\,U_{n}^{\prime}\,V_{n}^{\prime}\,U_{n}^{\prime}\,V_{n}^{\prime}\,\cdots)$

is a quadratic irrational over $\mathbb{Q}(i)$ solving the polynomial

$P_{n}(X):=\begin{vmatrix}q_{2w_{n}-1}&q_{2t_{n}-1}\\\
q_{2w_{n}}&q_{2t_{n}}\end{vmatrix}X^{2}-\left(\begin{vmatrix}q_{2w_{n}-1}&p_{2t_{n}-1}\\\
q_{2w_{n}}&p_{2t_{n}}\end{vmatrix}+\begin{vmatrix}p_{2w_{n}-1}&q_{2t_{n}-1}\\\
p_{2w_{n}}&q_{2t_{n}}\end{vmatrix}\right)X+\begin{vmatrix}p_{2w_{n}-1}&p_{2t_{n}-1}\\\
p_{2w_{n}}&p_{2t_{n}}\end{vmatrix}$

(see the proof of [12, Theorem 4.2]). We can rely on the basic properties of
determinants and part iii of Lemma 3.1 (cfr. [14, p. 21]) to show that, for
some constant $\kappa_{1}=\kappa_{1}(\zeta)>0$, we have

$\forall
n\in\mathbb{N}\quad|P_{n}(\zeta)|\leq\kappa_{1}\frac{|q_{2t_{n}}|}{|q_{2w_{n}}q^{2}_{2t_{n}+2u_{n}}|}.$

Let $(\mathbf{x}_{n})_{n\geq 1}$ be the sequence in $\mathbb{Z}[i]^{4}$ given
by

$\mathbf{x}_{n}:=\left(\begin{vmatrix}q_{2w_{n}-1}&q_{2t_{n}-1}\\\
q_{2w_{n}}&q_{2t_{n}}\\\
\end{vmatrix},\;\begin{vmatrix}q_{2w_{n}-1}&p_{2t_{n}-1}\\\
q_{2w_{n}}&p_{2t_{n}}\\\
\end{vmatrix},\;\begin{vmatrix}p_{2w_{n}-1}&q_{2t_{n}-1}\\\
p_{2w_{n}}&q_{2t_{n}}\\\
\end{vmatrix},\;\begin{vmatrix}p_{2w_{n}-1}&p_{2t_{n}-1}\\\
p_{2w_{n}}&p_{2t_{n}}\\\ \end{vmatrix}\right),$

so $\|\mathbf{x}_{n}\|\leq 2|q_{2w_{n}}q_{2t_{n}}|$. Note that, if
$M\in\mathbb{N}$ is such that $|B_{n}|\leq M$ for all $n\in\mathbb{N}$, we
have

$\forall n\in\mathbb{N}\quad\left|\zeta x_{n,1}-x_{n,2}\right|\leq
2M\left|\frac{|q_{2w_{n}}}{q_{2t_{n}}}\right|.$

Certainly, for any $n\in\mathbb{N}$ we have

$\displaystyle\left|\zeta x_{n,1}-x_{n,2}\right|$
$\displaystyle=\left|q_{2w_{n}-1}(\zeta
q_{2t_{n}}-p_{2t_{n}})-q_{2w_{n}}(\zeta q_{2t_{n}-1}-p_{2t_{n}-1})\right|$
$\displaystyle\leq\left|\frac{q_{2w_{n}-1}}{q_{2t_{n}}}\right|+\left|\frac{q_{2w_{n}}}{q_{2t_{n}-1}}\right|\leq
2(M+1)\left|\frac{q_{2w_{n}}}{q_{2t_{n}-1}}\right|.$

We can obtain a similar bound for $|\zeta x_{n,1}-x_{n,3}|$. Consider the
linear forms $\mathscr{L}_{1,1}$, $\mathscr{L}_{1,2}$, $\mathscr{L}_{1,3}$,
$\mathscr{L}_{1,4}$ in the variable $\mathbf{X}=(X_{1},X_{2},X_{3},X_{4})$ and
the linear forms $\mathscr{M}_{1,1}$, $\mathscr{M}_{1,2}$,
$\mathscr{M}_{1,3}$, $\mathscr{M}_{1,4}$ in the variable
$\mathbf{Y}=(Y_{1},Y_{2},Y_{3},Y_{4})$ given by

$\displaystyle\mathscr{L}_{1,1}(\mathbf{X})$
$\displaystyle=\zeta^{2}X_{1}-\zeta(X_{2}+X_{3})+X_{4},$
$\displaystyle\quad\mathscr{M}_{1,1}(\mathbf{X})=\overline{\mathscr{L}_{1,1}(\overline{\mathbf{Y}})},$
$\displaystyle\mathscr{L}_{1,2}(\mathbf{X})$ $\displaystyle=\zeta
X_{1}-X_{2},$
$\displaystyle\quad\mathscr{M}_{1,2}(\mathbf{Y})=\overline{\mathscr{L}_{1,2}(\overline{\mathbf{Y}})}$
$\displaystyle\mathscr{L}_{1,3}(\mathbf{X})$ $\displaystyle=\zeta
X_{1}-X_{3},$
$\displaystyle\quad\mathscr{M}_{1,3}(\mathbf{Y})=\overline{\mathscr{L}_{1,3}(\overline{\mathbf{Y}})}$
$\displaystyle\mathscr{L}_{1,4}(\mathbf{X})$ $\displaystyle=X_{1},$
$\displaystyle\quad\mathscr{M}_{1,4}(\mathbf{Y})=\overline{\mathscr{L}_{1,4}(\overline{\mathbf{Y}})}.$

Take $\varepsilon>0$ as in Lemma 6.5. Then, for some positive constants
$\kappa_{2},\kappa_{3}$ and all $n\in\mathbb{N}$ we have

$\displaystyle\left|\mathscr{L}_{1,1}(\mathbf{x}_{n})\mathscr{L}_{1,2}(\mathbf{x}_{n})\mathscr{L}_{1,3}(\mathbf{x}_{n})\mathscr{L}_{1,4}(\mathbf{x}_{n})\right|$
$\displaystyle=\left|P_{n}(\zeta)\right|\left|\zeta
x_{n,1}-x_{n,2}\right|\left|\zeta x_{n,1}-x_{n,3}\right|\left|x_{n,1}\right|$
$\displaystyle\leq\kappa_{2}\left|\frac{q_{2t_{n}}}{q_{2w_{n}}q^{2}_{2t_{n}+2u_{n}}}\frac{q_{2w_{n}}}{q_{2t_{n}}}\frac{q_{2t_{n}}}{q_{2w_{n}}}q_{2w_{n}}q_{2t_{n}}\right|$
$\displaystyle=\kappa_{2}\left|\frac{q_{2t_{n}}^{2}}{q_{2t_{n}+2u_{n}}^{2}}\right|$
$\displaystyle\leq\frac{\kappa_{3}}{\psi^{4u_{n}}}\leq\frac{\kappa_{3}}{|q_{2w_{n}}q_{2t_{n}}|^{\varepsilon}}=\frac{\kappa_{3}}{\|\mathbf{x}_{n}\|^{\varepsilon}}.$

By the definition of $\mathscr{M}_{1,1}$, $\mathscr{M}_{1,2}$,
$\mathscr{M}_{1,3}$, $\mathscr{M}_{1,4}$, for each $n\in\mathbb{N}$ we have

$\left|\mathscr{M}_{1,1}(\mathbf{x}_{n})\mathscr{M}_{1,2}(\mathbf{x}_{n})\mathscr{M}_{1,3}(\mathbf{x}_{n})\mathscr{M}_{1,4}(\mathbf{x}_{n})\right|=\left|\mathscr{L}_{1,1}(\mathbf{x}_{n})\mathscr{L}_{1,2}(\mathbf{x}_{n})\mathscr{L}_{1,3}(\mathbf{x}_{n})\mathscr{L}_{1,4}(\mathbf{x}_{n})\right|.$

Therefore, by Theorem 6.4 there is some non-zero
$\mathbf{x}=(x_{1},x_{2},x_{3},x_{4})\in\mathbb{Z}[i]^{4}$ and an infinite set
$\mathcal{N}_{1}\subseteq\mathbb{N}$ such that

(12) $\forall n\in\mathcal{N}_{1}\quad
x_{1}x_{n,1}+x_{2}x_{n,2}+x_{3}x_{n,3}+x_{4}x_{n,4}=0.$

Let $(Q_{n})_{n\geq 1}$ and $(R_{n})_{n\geq 1}$ be the sequences given by

$\forall n\in\mathbb{N}\qquad
Q_{n}:=\frac{q_{2w_{n}-1}q_{2t_{n}}}{q_{2w_{n}}q_{2t_{n}-1}},\quad
R_{n}:=\zeta-\frac{p_{2n}}{q_{2n}}.$

Then, $(Q_{n})_{n\geq 1}$ is then bounded away from $0$ since both sequences
$(q_{2w_{n}-1}/q_{2w_{n}})_{n\geq 1}$ and $(q_{2t_{n}-1}/q_{2t_{n}})_{n\geq
1}$ are bounded and bounded away from $0$. For each $n\in\mathcal{N}_{1}$, we
divide (12) by $q_{2w_{n}}q_{2t_{n}-1}$ to obtain

$\displaystyle 0$ $\displaystyle=x_{1}(Q_{n}-1)+x_{2}\left(Q_{n}(\zeta-
R_{2t_{n}})-(\zeta-R_{2t_{n}-1})\right)+$
$\displaystyle\;+x_{3}\left(Q_{n}(\zeta-R_{2w_{n}-1})-(\zeta-
R_{2w_{n}})\right)+$ $\displaystyle\;+x_{4}\left(Q_{n}(\zeta-
R_{2w_{n}-1})(\zeta-R_{2t_{n}})-(\zeta-R_{2w_{n}})(\zeta-R_{2t_{n}-1})\right)$

and, since $R_{n}\to 0$ as $n\to\infty$, we conclude

$\lim_{\begin{subarray}{c}n\to\infty\\\
n\in\mathcal{N}_{1}\end{subarray}}(Q_{n}-1)\left(x_{1}+(x_{2}+x_{3})\zeta+\zeta^{2}x_{4}\right)=0.$

In order to show

(13) $x_{1}+(x_{2}+x_{3})\zeta+\zeta^{2}x_{4}=0,$

take an infinite subset $\mathcal{N}_{1}^{\prime}$ of $\mathcal{N}_{1}$ and
two different constants $a,b\in\mathbb{Z}$ such that $a=B_{w_{n}}$ and
$b=B_{t_{n}}$ for all $n\in\mathcal{N}_{1}^{\prime}$ and such that the next
limits exist:

$\gamma:=\lim_{\begin{subarray}{c}n\to\infty\\\
n\in\mathcal{N}_{1}^{\prime}\end{subarray}}\frac{q_{2w_{n}}}{q_{2w_{n}-1}},\quad\eta:=\lim_{\begin{subarray}{c}n\to\infty\\\
n\in\mathcal{N}_{1}^{\prime}\end{subarray}}\frac{q_{2t_{n}}}{q_{2t_{n}-1}}.$

Since the compact sets

$\tau_{ia}\left(\overline{\mathcal{C}}_{1}(-2)\cup\overline{\mathcal{C}}_{1}(2)\right)\quad\text{
and
}\quad\tau_{ib}\left(\overline{\mathcal{C}}_{1}(-2)\cup\overline{\mathcal{C}}_{1}(2)\right)$

are disjoint, the sequence $(Q_{n})_{n\geq 1}$ is bounded away from $1$ and,
therefore, (13) is true. In this step we have used the mirror formula and
(11). Therefore, by $[\mathbb{Q}(i,\zeta),\mathbb{Q}(i)]\geq 3$, we must have

$x_{1}=x_{2}+x_{3}=x_{4}=0.$

As a consequence, for $n\in\mathcal{N}_{1}$, the quadratic irrational
$\zeta_{n}$ is a root of

(14) $P_{n}(X)=\begin{vmatrix}q_{2w_{n}-1}&q_{2t_{n}-1}\\\
q_{2w_{n}}&q_{2t_{n}}\end{vmatrix}X^{2}-2\begin{vmatrix}q_{2w_{n}-1}&p_{2t_{n}-1}\\\
q_{2w_{n}}&p_{2t_{n}}\end{vmatrix}X+\begin{vmatrix}p_{2w_{n}-1}&p_{2t_{n}-1}\\\
p_{2w_{n}}&p_{2t_{n}}\end{vmatrix}.$

Let $(\mathbf{v}_{n})_{n\geq 1}$,
$\mathbf{v}_{n}=(v_{n,1},v_{n,2},v_{n,3})\in\mathbb{Z}[i]^{3}$, be given by

$\forall
n\in\mathbb{N}\quad\mathbf{v}_{n}:=\left(\begin{vmatrix}q_{2w_{n}-1}&q_{2t_{n}-1}\\\
q_{2w_{n}}&q_{2t_{n}}\end{vmatrix},\begin{vmatrix}q_{2w_{n}-1}&p_{2t_{n}-1}\\\
q_{2w_{n}}&p_{2t_{n}}\end{vmatrix},\begin{vmatrix}p_{2w_{n}-1}&p_{2t_{n}-1}\\\
p_{2w_{n}}&p_{2t_{n}}\end{vmatrix}\right).$

hence, $\|\mathbf{v}_{n}\|_{\infty}\leq 2|q_{2w_{n}}q_{2t_{n}}|$. Consider the
linear forms $\mathscr{L}_{2,1}$, $\mathscr{L}_{2,2}$, $\mathscr{L}_{2,3}$ in
the variable $\mathbf{X}=(X_{1},X_{2},X_{3})$ and the linear forms
$\mathscr{M}_{2,1}$, $\mathscr{M}_{2,2}$, $\mathscr{M}_{2,3}$ in the variable
$\mathbf{Y}=(Y_{1},Y_{2},Y_{3})$ given by

$\displaystyle\mathscr{L}_{2,1}(\mathbf{X})$
$\displaystyle=\zeta^{2}X_{1}-2\zeta X_{2}+X_{3},$
$\displaystyle\quad\mathscr{M}_{2,1}(\mathbf{X})=\overline{\mathscr{L}_{2,1}(\overline{\mathbf{Y}})},$
$\displaystyle\mathscr{L}_{2,2}(\mathbf{X})$ $\displaystyle=\zeta
X_{1}-X_{2},$
$\displaystyle\quad\mathscr{M}_{2,2}(\mathbf{Y})=\overline{\mathscr{L}_{1,2}(\overline{\mathbf{Y}})}$
$\displaystyle\mathscr{L}_{2,3}(\mathbf{X})$ $\displaystyle=X_{1},$
$\displaystyle\quad\mathscr{M}_{2,3}(\mathbf{Y})=\overline{\mathscr{L}_{2,3}(\overline{\mathbf{Y}})}$

Using the new formula for $P_{n}(X)$, we may see that for some positive
constants $\kappa_{4}=\kappa_{4}(\zeta)$, $\kappa_{5}=\kappa_{5}(\zeta)$, and
$\kappa_{6}=\kappa_{6}(\zeta)$, for each $n\in\mathcal{N}_{1}$ we have

$\displaystyle\left|\mathscr{L}_{2,1}(\mathbf{v}_{n})\mathscr{L}_{2,2}(\mathbf{v}_{n})\mathscr{L}_{2,3}(\mathbf{v}_{n})\right|$
$\displaystyle=|P_{n}(\zeta)(\zeta v_{n,1}-v_{n,2})v_{n,1}|$
$\displaystyle\leq\kappa_{4}\frac{|q_{2w_{n}}q_{2t_{n}}|}{|q_{2t_{n}+2u_{n}}^{2}|}\leq\kappa_{4}\frac{|q_{2t_{n}}|}{|q_{2t_{n}+2u_{n}}|}\leq\frac{\kappa_{5}}{|q_{2w_{n}}q_{2t_{n}}|^{\varepsilon}}=\frac{\kappa_{6}}{\|\mathbf{v}_{n}\|^{\varepsilon}}$

as well as

$\left|\mathscr{M}_{2,1}(\mathbf{v}_{n})\mathscr{M}_{2,2}(\mathbf{v}_{n})\mathscr{M}_{2,3}(\mathbf{v}_{n})\right|=\left|\mathscr{L}_{2,1}(\mathbf{v}_{n})\mathscr{L}_{2,2}(\mathbf{v}_{n})\mathscr{L}_{2,3}(\mathbf{v}_{n})\right|.$

Hence, there is an infinite set $\mathcal{N}_{2}\subseteq\mathcal{N}_{1}$ and
a non-zero $\mathbf{r}=(r_{1},r_{2},r_{3})\in\mathbb{Z}[i]^{3}$ such that

$\forall n\in\mathcal{N}_{2}\quad r_{1}v_{n,1}+r_{2}v_{n,2}+r_{3}v_{n,3}=0.$

Dividing by $q_{2w_{n}}q_{2t_{n}-1}$, we obtain

$\lim_{\begin{subarray}{c}n\to\infty\\\
n\in\mathcal{N}_{2}\end{subarray}}(Q_{n}-1)(r_{3}\zeta^{2}+r_{2}\zeta+r_{1})=0.$

We may show as before that $(Q_{n})_{n\in\mathcal{N}_{2}}$ has a limit point
different from $1$, so

$r_{3}\zeta^{2}+r_{2}\zeta+r_{1}=0,$

which implies $[\mathbb{Q}(i,\zeta):\mathbb{Q}(i)]\leq 2$, contradicting our
hypothesis. ∎

The proof of Theorem 1.4 is robust enough for us to obtain similar results not
covered by Theorem 6.3.

######  Theorem 6.6.

Let $\mathbf{B}=(B_{n})_{n\geq 1}$ be a sequence in $\mathbb{Z}$.

  1. i.

If $\mathbf{B}$ is non-periodic, $\min\\{|B_{n}|:n\in\mathbb{N}\\}\geq 2$,
$\mathbf{A}=(-2,-2,\ldots)$ is a constant sequence, and
$s(\mathbf{A},\mathbf{B})\in\Omega$, then

$[0;-2,1+iB_{1},-2,1+iB_{2},-2,1+iB_{3},-2,1+iB_{4},\ldots]$

is transcendental.

  2. ii.

Let $\mathbf{A}=(a_{n})_{n\geq 1}$ be a sequence in $\\{2,2i,-2,-2i\\}$ and
let $\mathbf{C}=s(\mathbf{A},\mathbf{B})$. If $\mathbf{C}$ is not periodic,
valid, and $\operatorname{rep}(\mathbf{C})<\infty$, then
$[0;a_{1},b_{1},a_{2},b_{2},a_{3},b_{3},\ldots]$ is transcendental.

##  7\. Proof of the algorithm

###  7.1. Preliminary lemmas

We omit the proofs of some of the next lemmas. They amount to check directly
all the possible forms of open prototype sets and to argue recursively.

######  Lemma 7.1.

If $n\in\mathbb{N}$ and $(a_{1},\ldots,a_{n-1},-2+i)\in\mathsf{R}(n)$, then
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}_{1}(-2)$ or
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}_{1}(-2+i)$. As a
consequence, $(a_{1},\ldots,a_{n-1},-2+i,b)\not\in\mathsf{R}(n+1)$ implies
$\operatorname{Re}(b)\geq 1$.

######  Lemma 7.2.

If $a,b\in\mathscr{D}$ satisfy
$\mathfrak{F}^{\circ}_{1}(a)\cap\mathcal{C}^{\circ}_{1}(b)\neq\varnothing$ and
$\mathcal{C}^{\circ}_{1}(b)\setminus\mathfrak{F}^{\circ}_{1}(a)\neq\varnothing$,
then $|a|=\sqrt{5}$.

###### Proof.

This can be proved by considering the next cases: $|a|=\sqrt{2}$, $|a|=2$,
$|a|=\sqrt{5}$, $|a|\geq\sqrt{8}$. ∎

We can show inductively that for $n\in\mathbb{N}$ and
$\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathsf{R}(n)$ with $|a_{n}|=\sqrt{5}$
there is some $j\in\\{0,1,2,3\\}$ such that

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}_{1}(2i^{j})\quad\text{
or }\quad\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\left(i^{j}(1+2i)\right).$

Also, we may show that for $n\in\mathbb{N}$ and
$\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathsf{R}(n)$ with $|a_{n}|=\sqrt{8}$
there is some $j\in\\{0,1,2,3\\}$ for which

$\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\mathfrak{F}^{\circ}\quad\text{ or
}\quad\mathfrak{F}^{\circ}_{n}(\mathbf{a})=\left(i^{j}(1+2i)\right).$

This gives the next proposition.

######  Lemma 7.3.

If
$\mathfrak{F}^{\circ}_{n}(\mathbf{a})\cap\mathcal{C}^{\circ}_{1}(b)\neq\varnothing$
and
$\mathcal{C}^{\circ}_{1}(b)\setminus\mathfrak{F}^{\circ}_{n}(\mathbf{a})\neq\varnothing$,
then $|a_{n}|\in\\{\sqrt{5},\sqrt{8}\\}$.

######  Lemma 7.4.

Take $n,k\in\mathbb{N}$, $(x_{1},\ldots,x_{n})\in\mathscr{D}^{n}$,
$(d_{1},\ldots,d_{k})\in\mathscr{D}^{k}$, $m\in\mathbb{Z}$ such that $|m|\geq
2$ and $c\in\\{m,im\\}$. If
$(x_{1},\ldots,x_{n},c,d_{1},\ldots,d_{k})\in\mathsf{R}(n+k+1)$, then

$\mathfrak{F}^{\circ}_{n+k+1}(x_{1},\ldots,x_{n},c,d_{1},\ldots,d_{k})=\mathfrak{F}^{\circ}_{k}(d_{1},\ldots,d_{k}).$

###### Proof.

If $|c|\geq 3$, then
$\mathfrak{F}^{\circ}_{n+1}(x_{1},\ldots,x_{n},c)=\mathfrak{F}^{\circ}$ and,
by (4) and the definition of $T$,

$\displaystyle\mathfrak{F}^{\circ}_{n+k+1}(x_{1},\ldots,x_{n},c,d_{1},\ldots,d_{k})$
$\displaystyle=T^{n+k+1}\left(\mathcal{C}^{\circ}_{n+k+1}(x_{1},\ldots,x_{n},c,d_{1},\ldots,d_{k})\right)$
$\displaystyle=T^{k}\left(\mathfrak{F}^{\circ}_{n+1}(x_{1},\ldots,x_{n},c)\cap\mathcal{C}^{\circ}_{k}(d_{1},d_{2},\ldots,d_{k})\right)$
$\displaystyle=T^{k}\left(\mathfrak{F}^{\circ}\cap\mathcal{C}^{\circ}_{k}(d_{1},\ldots,d_{k})\right)$
$\displaystyle=\mathfrak{F}^{\circ}_{k}(d_{1},\ldots,d_{k}).$

When $|c|=2$, we check the $4$ different forms that
$\mathfrak{F}^{\circ}_{n+1}(x_{1},\ldots,x_{n},c)\neq\varnothing$ may take to
conclude

$\mathfrak{F}^{\circ}_{n+2}(x_{1},\ldots,x_{n},c,d_{1})=\mathfrak{F}^{\circ}_{1}(d_{1}).$

For instance, if $c=-2$, then

$\iota\left(\mathfrak{F}^{\circ}_{n}(-2)\right)=\iota[\mathfrak{F}^{\circ}]\cap\\{z\in\mathbb{C}:\operatorname{Re}(z)<1\\}$

and for all $d\in\mathscr{D}$ we have either
$\tau_{d}[\mathfrak{F}^{\circ}]\subseteq\iota[\mathfrak{F}^{\circ}]$ or
$\tau_{d}[\mathfrak{F}^{\circ}]\cap\iota[\mathfrak{F}^{\circ}]=\varnothing$.
From here we argue as in the case $|d|\geq 3$. ∎

######  Lemma 7.5.

If $m\in\mathbb{Z}$ is such that $|m|\geq 2$ and $a\in\\{1+im,m+i\\}$, then

$(\mathsf{mir}_{1}\circ S)(a)=(S\circ\mathsf{mir}_{1})(a)\quad\text{ and
}\quad(\mathsf{mir}_{2}\circ S)(a)=(S\circ\mathsf{mir}_{2})(a).$

######  Lemma 7.6.

Take $\mathbf{b}=(b_{n})_{n\geq 1}\in\Omega$. If
$(b_{1},b_{2})\in\mathsf{Ir}(2)$, then

$\operatorname{Re}([b_{1};b_{2},b_{3},\ldots])=\frac{1}{2}\quad\text{ or
}\quad\operatorname{Im}([b_{1};b_{2},b_{3},\ldots])=\frac{1}{2}.$

######  Lemma 7.7.

If $\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$, then
$\overline{\Lambda}\circ\mathsf{mir}=\mathsf{mir}\circ\overline{\Lambda}$.

###### Proof.

The lemma follows the continuity of $\mathsf{mir}_{1}$ and $\mathsf{mir}_{2}$
from the fact that for every
$\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$ we have

$\mathsf{mir}\left(\langle 0;z_{1},z_{2}\rangle\right)=\langle
0;\mathsf{mir}(z_{1}),\mathsf{mir}(z_{2})\rangle$

for any $z_{1},z_{2}\in\mathbb{C}$ such that both sides of the equality are
defined. ∎

###  7.2. Proof of the algorithm

Let $\mathbf{a}=(a_{n})_{n\geq 1}$ be an irregular sequence. Since every
cylinder of level $1$ is regular, there is some $j_{0}\in\mathbb{N}$ such that

$(a_{1},\ldots,a_{j_{0}})\in\mathsf{R}(j_{0})\text{ and
}(a_{1},\ldots,a_{j_{0}},a_{j_{0}+1})\in\mathsf{R}(j_{0}+1).$

Note that
$\mathfrak{F}^{\circ}_{j_{0}}(a_{1},\ldots,a_{j_{0}})\neq\mathfrak{F}^{\circ}$,
for otherwise $(a_{1},\ldots,a_{j_{0}},a_{j_{0}+1})$ would be regular. The
open prototype set $\mathfrak{F}^{\circ}_{j_{0}}(a_{1},\ldots,a_{j_{0}})$ can
assume twelve forms and, by checking all of them, we conclude the existence of
some $m\in\mathbb{Z}$ such that

$|m|\geq 2\quad\text{ and }\quad a_{j_{0}+1}\in\left\\{1+im,m+i\right\\}.$

Define $\mathbf{b}_{1}$ as in the algorithm; that is, letting $t=1$ if
$a_{j_{0}+1}=m+i$ and $t=2$ if $a_{j_{0}+1}=1+m$,

$\mathbf{b}_{1}:=(a_{1},\ldots,a_{j_{0}},S(a_{j_{0}+1}),\mathsf{mir}_{t}(a_{j_{0}+2}),\mathsf{mir}_{t}(a_{j_{0}+3}),\mathsf{mir}_{t}(a_{j_{0}+4}),\ldots).$

Assume that $|m|\geq 3$, then

$\mathfrak{F}^{\circ}_{j_{0}+1}\left(a_{1},\ldots,a_{j_{0}},S(a_{j_{0}+1})\right)=\mathfrak{F}^{\circ}\quad\text{
and }\quad(b_{1}(1),\ldots,b_{1}(j_{0}+1))\in\mathsf{R}(j_{0}+1).$

If $a_{j_{0}+1}=1+im$, then

(15)
$\operatorname{Re}\left([a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]\right)=\frac{1}{2}$

and, since $z=1-\overline{z}$ holds whenever
$\operatorname{Re}(z)=\frac{1}{2}$,

$\displaystyle[a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]$
$\displaystyle=1-\overline{[a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]}$
$\displaystyle=\left\langle
1-\overline{a_{j_{0}+1}};\overline{a_{j_{0}+2}},\overline{a_{j_{0}+3}},\ldots\right\rangle$
$\displaystyle=\left\langle
S(a_{j_{0}+1});\mathsf{mir}_{2}(a_{j_{0}+1});\mathsf{mir}_{2}(a_{j_{0}+2}),\mathsf{mir}_{2}(a_{j_{0}+3}),\ldots\right\rangle.$

Therefore,

$\displaystyle\Lambda(\mathbf{a})$
$\displaystyle=[0;a_{1},\ldots,a_{j_{0}},a_{j_{0}+1},a_{j_{0}+2},\ldots]$
$\displaystyle=\langle
0;a_{1},a_{2},\ldots,a_{j_{0}},S(a_{j_{0}+1}),\mathsf{mir}_{2}(a_{j_{0}+2}),\mathsf{mir}_{2}(a_{j_{0}+3})\ldots\rangle$
$\displaystyle=\langle
0;b_{1}(1),b_{1}(2),\ldots,b_{1}(j_{0}),b_{1}(j_{0}+1),b_{1}(j_{0}+2),b_{1}(j_{0}+3),\ldots\rangle$
$\displaystyle=\overline{\Lambda}(\mathbf{b}_{1}).$

When $a_{j_{0}+1}=m+i$, we have

(16)
$\operatorname{Im}\left([a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]\right)=\frac{1}{2}.$

Then, since $z=\overline{z}+i$ whenever $\operatorname{Im}(z)=\frac{1}{2}$, we
have

$\displaystyle[a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]$
$\displaystyle=i+\overline{[a_{j_{0}+1};a_{j_{0}+2},a_{j_{0}+3},\ldots]}$
$\displaystyle=\left\langle
i+\overline{a_{j_{0}+1}};\overline{a_{j_{0}+2}},\overline{a_{j_{0}+3}},\ldots\right\rangle$
$\displaystyle=\langle
S\left(a_{j_{0}+1}\right);\mathsf{mir}_{1}(a_{j_{0}+1});\mathsf{mir}_{1}(a_{j_{0}+2}),\mathsf{mir}_{1}(a_{j_{0}+3}),\ldots\rangle$

and we also conclude $\Lambda(\mathbf{a})=\overline{\Lambda}(\mathbf{b}_{1})$.

Assume that $|m|=2$. Since $\mathfrak{F}$ contains the left and the bottom
sides, we must have $a_{j_{0}+1}\in\\{1+2i,1-2i,2+i,-2+i\\}$. If
$a_{j_{0}+1}\in\\{1+2i,1-2i\\}$, then (15) holds, we have
$S(a_{j_{0}+1})\in\\{2i,-2i\\}$ and

$(b_{1}(1),\ldots,b_{1}(j_{N}),b_{1}(j_{N}+1))=\left(a_{1},\ldots,S(a_{j_{0}}),\mathsf{mir}_{2}(a_{j_{0}+1})\right)\in\mathsf{R}(j_{1}+1).$

The last statement is verified case by case. For instance, when
$a_{j_{0}+1}=1+2i$, then $[0;a_{j+2},a_{j+3},\ldots]$ belongs to the line
segment determined by $\zeta_{1}$ (included) and $\frac{-1+i}{2})$ (excluded),
so $a_{j_{0}+2}\in\\{-2,-2-i,-1-i\\}$ and
$\mathsf{mir}_{2}(a_{j_{0}+2})\in\\{2,2-i,1-i\\}$ and the claim holds because

$\mathfrak{F}^{\circ}_{2}(2i,2)\neq\varnothing,\quad\mathfrak{F}^{\circ}_{2}(2i,2-i)\neq\varnothing,\quad\mathfrak{F}^{\circ}_{2}(2i,1-i)\neq\varnothing.$

Similarly, when $a_{j_{0}+1}\in\\{2+i,-2+i\\}$, equation (16) holds,
$S(a_{j_{0}+1})\in\\{2,-2\\}$ and

$(b_{1}(1),\ldots,b_{1}(j_{N}),b_{1}(j_{N}+1))=\left(a_{1},\ldots,S(a_{j_{0}}),\mathsf{mir}_{1}(a_{j_{0}+1})\right)\in\mathsf{R}(j_{1}+1).$

Irrespective to what case occurs when $|m|=2$, we may conclude
$\Lambda(\mathbf{a})=\overline{\Lambda}(\mathbf{b}_{1})$ as when $|m|\geq 3$.
To sum up, the sequence $\mathbf{b}_{1}$ has the next properties:

  1. i.

$(b_{1}(1),\ldots,b_{1}(j_{0}),b_{1}(j_{0}+1))\in\mathsf{R}(j_{N}+1)$,

  2. ii.

$\Lambda(\mathbf{a})=\langle
b_{1}(1),b_{1}(2),b_{1}(3),\ldots\rangle=\overline{\Lambda}(\mathbf{b}_{1})$.

Assume that for some $N\in\mathbb{N}$ we have the following:

  * •

We are able to perform $N$ iterations of the procedure and the resulting
sequence $\mathbf{b}_{N}$ satisfies:

    1. i.

$(b_{N}(1),\ldots,b_{N}(j_{N-1}+1))\in\mathsf{R}(j_{N-1}+1)$,

    2. ii.

$\Lambda(\mathbf{a})=\langle
0;b_{N}(1),b_{N}(2),b_{N}(3),\ldots\rangle=\overline{\Lambda}(\mathbf{b}_{N})$.

  * •

Either

(17)
$\operatorname{Re}\left([a_{j_{N-1}+1};a_{j_{N-1}+2},\ldots]\right)=\frac{1}{2}\;\text{
or
}\;\operatorname{Im}\left([a_{j_{N-1}+1};a_{j_{N-1}+2},\ldots]\right)=\frac{1}{2}.$

######  Remark 7.8.

By the definition of $\mathbf{b}_{N}$, for each $M\in\\{1,\ldots,N\\}$ there
exists some $\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$
such that

$\forall n\in\mathbb{N}\quad
b_{M}(j_{M-1}+1+n)=\mathsf{mir}\left(b_{M-1}(j_{M-1}+1+n)\right).$

As a consequence, there is some
$\mathsf{mir}^{\prime}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$ for
wchich

$\forall n\in\mathbb{N}\quad
b_{M}(j_{M-1}+1+n)=\mathsf{mir}^{\prime}\left(a_{j_{M-1}+1+n}\right).$

Suppose there is some $j_{N}\in\mathbb{N}$ satisfying

(18) $\left(b_{N}(1),\ldots,b_{N}(j_{N})\right)\in\mathsf{R}(j_{N})\quad\text{
and
}\quad\left(b_{N}(1),\ldots,b_{N}(j_{N}+1)\right)\not\in\mathsf{R}(j_{N}+1).$

Note that $j_{N}-j_{N-1}\in\mathbb{N}$ by the induction hypothesis. We
consider the cases $j_{N}-j_{N-1}=1$ and $j_{N}-j_{N-1}\geq 2$ in order to
prove

(19)
$\left(b_{N}(1),\ldots,b_{N}(j_{N}),S(b_{N}(j_{N}+1))\right)\in\mathsf{R}(j_{N}+1),$
(20)
$\operatorname{Re}\left([a_{j_{N}+1};a_{j_{N}+2},\ldots]\right)=\frac{1}{2}\quad\text{
or
}\quad\operatorname{Im}\left([a_{j_{N}+1};a_{j_{N}+2},\ldots]\right)=\frac{1}{2},$
(21) $\Lambda(\mathbf{a})=\overline{\Lambda}(\mathbf{b}_{N+1}).$

Case $j_{N}=j_{N-1}+1$. Since
$b_{N}(j_{N})=b_{N}(j_{N-1}+1)=S(b_{N-1}(j_{N-1}+1))$, there is some
$m\in\mathbb{Z}$, $|m|\geq 2$, such that $b_{N}(j_{N})\in\\{m,im,-m,-im\\}$.
If we had $|m|\geq 3$, we would have

$\mathfrak{F}^{\circ}_{j_{N}}(b_{N}(1),\ldots,b_{N}(j_{N}))=\mathfrak{F}^{\circ}_{j_{N-1}+1}(b_{N-1}(1),\ldots,S(b_{N-1}(j_{N-1}+1))=\mathfrak{F}^{\circ},$

which implies
$(b_{N}(1),\ldots,b_{N}(j_{N}),b_{N}(j_{N}+1))\in\mathsf{R}(j_{N}+1)$,
contradicting (18). Therefore, $|m|=2$ and $b_{N}(j_{N})\in\\{-2,-2i,2,2i\\}$.
Let us further assume that $b_{N}(j_{N})=-2$, the cases are treated similarly.
Then, we have

$b_{N-1}(j_{N})=b_{N-1}(j_{N-1}+1)\in\\{-2+i,-2-i\\}$

We work under the assumption $b_{N-1}(j_{N})=-2+i$, an analogous argument
works for $-2-i$. Under these new assumptions, (18) give

(22)
$\mathfrak{F}^{\circ}_{j_{N}}\left(b_{N}(1),\ldots,b_{N}(j_{N})\right)=\mathfrak{F}^{\circ}_{1}(-2)\quad\text{
and }\quad\operatorname{Re}\left(b_{N}(j_{N}+1)\right)\geq 1.$

Note that for some
$\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$ have

(23) $b_{N-1}(j_{N-1}+1)=\mathsf{mir}(a_{j_{N}})\quad\text{ and }\quad
b_{N-1}(j_{N}+1)=\mathsf{mir}(a_{j_{N}+1}).$

Certainly, when $N=1$, we may consider $\mathsf{mir}$ to be the identity map
on $\mathbb{C}$. For $N>1$, the induction hypothesis tells us that
$j_{N-1}+j_{N-2}\in\mathbb{N}$, so

$j_{N-1}+1=j_{N-2}+(j_{N-1}+j_{N-2})+1\geq j_{N-2}+2$

and our claim follows from Remark 7.8. We now consider four possibilities for
$\mathsf{mir}$; in fact, we will show that $\mathsf{mir}$ is the identity map.

  * •

When $\mathsf{mir}$ is the identity map, we have $a_{j_{N}}=-2+i$ and
$\operatorname{Re}(a_{j_{N}+1})\geq 1$. Hence, the set
$\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}})\in\mathsf{Ir}(j_{N})$ is of one
of the forms depicted on Figure 7. We consider each case separately.

    1. (a)

In this case, we have
$[0;a_{j_{N}+1},a_{j_{N}+2},a_{j_{N}+3},\ldots]=\zeta_{4}$ and (20) holds.
This equality also implies $a_{{j_{N}+1}}=1+2i$ and, by (23),

$\left(b_{N-1}(j_{N}),b_{N-1}(j_{N}+1)\right)=(-2+i,1-2i).$

Then, by $b_{N}(j_{N})=S(b_{N-1}(j_{N}+1))$ and
$b_{N}(j_{N}+1)=\mathsf{mir}_{1}(b_{N-1}(j_{N}+1))$, we have

$\left(b_{N}(j_{N}),b_{N}(j_{N}+1)\right)=(-2,1+2i)$

and $S(b_{N}(j_{N}+1))=2i$. The first condition in (22) yields

$\mathfrak{F}^{\circ}_{j_{N}+1}\left(b_{N}(1),\ldots,b_{N}(j_{N}),S(b_{N}(j_{N}+1))\right)=\mathfrak{F}_{1}(2i)\neq\varnothing$

and (19) follows.

    2. (b)

In this case, from $a_{j_{N}}=-2+i$ and (17), we get
$\operatorname{Im}\left([a_{j_{N-1}+1};a_{j_{N-1}+2},\ldots]\right)=\tfrac{1}{2}$.
And, in view of

$\left\\{w\in\mathbb{C}:\operatorname{Im}(w)=\frac{1}{2}\right\\}\cap\tau_{a_{j_{N}}}\left(\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}})\right)=\left\\{\tau_{a_{j_{N}}}(\zeta_{4})\right\\}$

we obtain $[0;a_{j_{N}+1},a_{j_{N}+2},\ldots]=\zeta_{4}$. The argument in (a)
ensures (19) and (20).

    3. (c)

As in case (b), we must have $[0;a_{j_{N}+1},a_{j_{N}+2},\ldots]=\zeta_{4}$
and we now argue as in (a).

    4. (d)

We apply the complex inversion on
$\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}})$ to conclude
$a_{j_{N}+1}\in\\{-1+i,1-+2i,2i,1+2i\\}$ and, since
$\operatorname{Re}(a_{j_{N}+1})\geq 1$, we have $a_{j_{N}+1}=1+2i$. Because of

$\iota\left[\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}}\right]\cap\tau_{a_{j_{N}+1}}[\mathfrak{F}]\cap\tau_{a_{j_{N}}}\left(\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}})\right)=\left\\{\tau_{a_{j_{N}}}(\zeta_{2})\right\\},$

we may proceed as in case (b) to conclude (19) and the equality for the real
part in (20).

The discussion leading (22) actually tells us that
$b_{N-1}(j_{N-1}+n)=a_{j_{N-1}+n}$ for all $n\in\mathbb{N}$; hence, by the
basis of induction,

$\overline{\Lambda}\left(b_{N}(j_{N-1}+n)_{n\geq
1}\right)=\Lambda\left((a_{j_{N-1}+n})_{n\geq 1}\right).$

Lastly, we conclude

$\overline{\Lambda}\left(b_{N}(j_{N-1}+n)_{n\geq
1}\right)=\overline{\Lambda}\left(b_{N+1}(j_{N-1}+n)_{n\geq 1}\right)$

arguing as in the case $N=0$ and applying Lemma 7.7.

  * •

If $\mathsf{mir}=\mathsf{mir}_{1}$, then $a_{j_{N}}=-2-i$ but (17) cannot
hold.

  * •

When $\mathsf{mir}=\mathsf{mir}_{2}$, then we have $a_{j_{N}}=2+i$ and
$\operatorname{Re}(a_{j_{N}+1})\leq-1$. Also, in (20), we have the equality
for the imaginary part. We consider the cases shown in Figure 7.

    1. (a)

This cannot happen, by virtue of (20).

    2. (b)

In this configuration, (20). would imply
$[0;a_{j_{N}+1},a_{j_{N}+2},\ldots]=\zeta_{3}$, so we would have
$a_{j_{N}+1}=2i$, contradicting $\operatorname{Re}(a_{j_{N}+1})\leq-1$. We
have

    3. (c)

Same as (b).

    4. (d)

In this case, we would have $a_{j_{N}+1}\in\\{2i,1+2i,1+i\\}$ which is
incompatible with $\operatorname{Re}(a_{j_{N}+1})\leq-1$.

  * •

When $\mathsf{mir}=\mathsf{mir}_{1}\mathsf{mir}_{2}$, then $a_{j_{N}}=2-i$ and
$\operatorname{Re}(a_{j_{N}}+1)\leq-1$. We only have two possibilities for
$\mathfrak{F}_{j_{N}}(a_{1},\ldots,a_{j_{N}})$ (see Figure 7).

    1. (a)

We have

$[0;a_{j_{N}+1},a_{j_{N}+2},a_{j_{N}+3},\ldots]=\zeta_{1}\quad\text{ and
}\quad\operatorname{Im}([a_{j_{N}+1};a_{j_{N}+2},a_{j_{N}+3},\ldots])=\frac{1}{2}.$

    2. (b)

This configuration is incompatible with
$\operatorname{Re}(a_{j_{N}}+1)\leq-1$.

Figure 5.  Possible forms of $\mathfrak{F}(a_{1},\ldots,a_{j_{N}+1})$ when
$\mathsf{mir}$ is the identity map.

Figure 6.  Possible forms of $\mathfrak{F}(a_{1},\ldots,a_{j_{N}+1})$ when
$\mathsf{mir}=\mathsf{mir}_{2}$.

Figure 7.  Possible forms of $\mathfrak{F}(a_{1},\ldots,a_{j_{N}+1})$ when
$\mathsf{mir}=\mathsf{mir}_{1}\mathsf{mir}_{2}$.

Case $j_{N}\geq j_{N-1}+2$. By construction, for some $m\in\mathbb{Z}$ with
$|m|\geq 2$ we have

$b_{N}(j_{N-1}+1)=S\left(b_{N-1}(j_{N-1}+1)\right)\in\left\\{m,im\right\\}.$

Therefore, by Lemma 7.4 and $j_{N-1}+2=j_{N}$,

$\displaystyle\mathfrak{F}^{\circ}_{j_{N}}\left(b_{N}(1),\ldots,b_{N}(j_{N})\right)$
$\displaystyle=\mathfrak{F}^{\circ}_{j_{N}-j_{N}-1}(b_{N}(j_{N-1}+2),\ldots,b_{N-1}(j_{N}))\neq\varnothing,$
$\displaystyle\mathfrak{F}^{\circ}_{j_{N}+1}\left(b_{N}(1),\ldots,b_{N}(j_{N}+1)\right)$
$\displaystyle=\mathfrak{F}^{\circ}_{j_{N}-j_{N}}\left(b_{N}(j_{N-1}+2),\ldots,b_{N}(j_{N}),b_{N}(j_{N}+1)\right)=\varnothing.$

Moreover, by Remark 7.8, we may pick
$\mathsf{mir}\in\langle\mathsf{mir}_{1},\mathsf{mir}_{2}\rangle$ such that

$\forall n\in\mathbb{N}\quad
b_{N}(j_{N-1}+1+n)=\mathsf{mir}\left(a_{j_{N-1}+1+n}\right).$

As a consequence, recalling the definition of $j_{N}$ and Proposition 3.4, we
have

$\left(a_{j_{N-1}+2},\ldots,a_{j_{N}}\right)\in\mathsf{R}(j_{N}-j_{N-1}-1)\;\text{
and
}\;\left(a_{j_{N-1}+2},\ldots,a_{j_{N}+1}\right)\in\mathsf{Ir}(j_{N}-j_{N-1}-1).$

Applying the case $N=0$ we get

$\left(a_{j_{N-1}+2},\ldots,a_{j_{N}},S(a_{j_{N}+1})\right)\in\mathsf{R}(j_{N}-j_{N-1}-1)$

and, by Lemma 7.6, we conclude (19). The case $N=0$ also implies (20). Lastly,
let $\mathbf{d}$ be the sequence in $\mathscr{D}$ obtained from the first
iteration of the algorithm applied on $\mathbf{a}^{\prime}:=(a_{n})_{n\geq
j_{N-1}+2}$. Then,
$\Lambda(\mathbf{a}^{\prime})=\overline{\Lambda}(\mathbf{d})$. If
$\mathbf{b}^{\prime}:=(b_{N}(n))_{n\geq j_{N-1}+2}$ and
$\mathbf{e}^{\prime}:=(b_{N}(n))_{n\geq j_{N-1}+2}$, then
$\mathsf{mir}(\mathbf{a}^{\prime})=\mathbf{b}^{\prime}$,
$\mathsf{mir}(\mathbf{d})=\mathbf{e}$, and an application of Lemma 7.7 yields

$\Lambda(\mathsf{mir}(\mathbf{a}^{\prime}))=\mathsf{mir}\left(\Lambda(\mathbf{a}^{\prime})\right)=\mathsf{mir}\left(\overline{\Lambda}(\mathbf{d})\right)=\overline{\Lambda}(\mathsf{mir}(\mathbf{d}))=\overline{\Lambda}(\mathbf{e})$

Therefore, (21) is true.

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