# Diophantine sets in general are Cantor sets

Fernando Argentieri  Fernando Argentieri

###### Abstract

Let ${\gamma}\in(0;\frac{1}{{2}}),{\tau}\geq 1$ and define the
“${\gamma},{\tau}$ Diophantine set” as:

$D_{\gamma,\tau}:=\\{{\alpha}\in(0;1):||q{\alpha}||\geq\frac{{\gamma}}{q^{{\tau}}}\quad\forall
q\in{N}\\},\qquad||x||:=\inf_{p\in{Z}}|x-p|.$

In this paper we study the topology of these sets and we show that, for large
${\tau}$ and for almost all ${\gamma}>0$, $D_{\gamma,\tau}$ is a Cantor set.

##  1 Introduction

Diophantine sets play an important role in dynamical systems, in particular,
in small divisors problems with applications to KAM theory, Aubry-Mather
theory, conjugation of circle diffeomorphisms, etc. (see, for example, [3],
[5], [9], [12], [13], [14], [16]).

The set $D_{\gamma,\tau}$ is compact and totally disconnected (since
$D_{\gamma,\tau}\cap{Q}=\emptyset$), however, these sets may be not Cantor
sets. In fact, in [17] we have shown various examples in which
$D_{\gamma,\tau}$ have isolated points. In this paper we prove the
following:Theorem Let ${\tau}>\frac{3+\sqrt{17}}{2}$. Then, for almost all
${\gamma}>0$ $D_{\gamma,\tau}$ is a Cantor set.

By [6], for ${\tau}=1$ and $\frac{1}{{3}}<{\gamma}<\frac{1}{{2}}$
$D_{\gamma,\tau}$ is countable (and non empty for ${\gamma}>\frac{1}{{3}}$
small enough). In particular, this result does not holds for ${\tau}=1$. We
expect that ${\tau}>\frac{3+\sqrt{17}}{2}$ can be improved with ${\tau}>3$.
However it is not clear what is the best constant. Following the same proof,
we can prove also that, fixed ${\tau}>\frac{3+\sqrt{17}}{2}$, for almost all
${\gamma}>0$, if ${\alpha}\in D_{\gamma,\tau}$ and $U$ is an open neighborhood
that contains ${\alpha}$, then ${\mu}(D_{\gamma,\tau}\cap U)>0$.

The paper is organized as follows: in the second section we give some basic
definitions and remarks, in the third section we prove our result and, in the
last section are present some natural questions.

##  2 Definitions and remarks

###  2.1 Definitions

  * •

${N}:=\\{1,2,3,...\\}$, ${N}_{0}:=\\{0,1,2,3,...\\}$

  * •

Given $a,b\in{Z}-\\{0\\}$, we indicate with $(a,b)$ the maximum common divisor
of $a$ and $b$.

  * •

Let ${\alpha}$ be a real number. We indicate with $[{\alpha}]$ the integral
part of ${\alpha}$, with $\\{{\alpha}\\}$ the fractional part of ${\alpha}$ .

  * •

Given E$\subseteq{{R}}$, we indicate with $\mathcal{I}$(E) the set of isolated
points of E.

  * •

Given E$\subseteq{{R}}$, we indicate with $\mathcal{A}$(E) the set of
accumulated points of E.

  * •

We say that E$\subseteq{{R}}$ is perfect if $\mathcal{A}$(E)=E.

  * •

Given a Borel set E$\subseteq{{R}}$ we denote with ${\mu}$(E) the Lebesgue
measure of E.

  * •

A topological space X is a totally disconnected space if the points are the
only connected subsets of X.

  * •

$X\subseteq{R}$ is a Cantor set if it is closed, totally disconnected and
perfect.

  * •

For $E\subseteq{R}^{n}$, $\dim_{H}E$ is the Hausdorff dimension of $E$.

  * •

Given ${\alpha}\in{R}$ we define:

$||{\alpha}||:=\min_{p\in{Z}}|{\alpha}-p|$

  * •

Given ${\gamma}>0,{\tau}\geq 1$, we define the $({\gamma},{\tau})$ Diophantine
points in $(0;1)$ as the numbers in the set:

$D_{\gamma,\tau}:=\\{{\alpha}\in(0;1):||q{\alpha}||\geq\frac{{\gamma}}{q^{{\tau}}}\quad\forall
q\in{N}\\}$

  * •

$D^{{R}}_{{\gamma},{\tau}}:=\\{{\alpha}\in{R}:||q{\alpha}||\geq\frac{{\gamma}}{q^{\tau}}\quad\forall
q\in{N}\\},$ $D_{{\tau}}:=\bigcup_{{\gamma}>0}D_{{\gamma},{\tau}},\quad
D:=\bigcup_{{\tau}\geq 1}D_{{\tau}}.$

We call $D$ the set of Diophantine numbers.

  * •

Given ${\tau}\geq 1,{\alpha}\in{R}$, we define:

$\gamma(\alpha,\tau):=\inf_{q\in{N}}q^{{\tau}}||q{\alpha}||$

  * •

Given ${\alpha}\in{R}$ we define:

${\tau}({\alpha}):=\inf\\{{\tau}\geq 1:\gamma(\alpha,\tau)>0\\}$

  * •

Given an irrational number
${\alpha}=[a_{0};a_{1},...]:=a_{0}+\frac{1}{{a_{1}+\frac{1}{{a_{2}+...}}}}$,
we denote with $\\{\frac{p_{n}}{q_{n}}\\}_{n\in{N}_{0}}$ the convergents of
${\alpha}$, ${\alpha}_{n}:=[a_{n};a_{n+1},...]$111for informations about
continued fractions see [4],[8],[15] .

  * •

We indicate with
$[a_{1},a_{2},a_{3},...]:=\frac{1}{{a_{1}+\frac{1}{{a_{2}+\frac{1}{{a_{3}+...}}}}}}$.

  * •

Let ${\alpha}$ be an irrational number. We define:

${\gamma}_{n}({\alpha},{\tau}):=q_{n}^{{\tau}}||q_{n}{\alpha}||=q_{n}^{{\tau}}|q_{n}{\alpha}-p_{n}|$

  * •

Let ${\tau}\geq 1$,

${\gamma}_{-}({\alpha},{\tau}):=\inf_{n\in
2{N}_{0}}{\gamma}_{n}({\alpha},{\tau}),$
${\gamma}_{+}({\alpha},{\tau}):=\inf_{n\in
2{N}_{0}+1}{\gamma}_{n}({\alpha},{\tau}),$
${\mathcal{D}_{{\tau}}}:=\\{{\alpha}\in
D_{{\tau}}:{\tau}({\alpha})={\tau}\\},$
${\mathcal{I}}^{1}_{{\gamma},{\tau}}:=\\{{\alpha}\in
D_{{\gamma},{\tau}}:\exists n\not\equiv
m\quad{(\rm{mod}2)},{\gamma}_{n}({\alpha},{\tau})={\gamma}_{m}({\alpha},{\tau})=\gamma(\alpha,\tau)\\},$
${\mathcal{I}}^{2}_{{\gamma},{\tau}}:=\\{{\alpha}\in
D_{{\gamma},{\tau}}:\exists
n\in{N}_{0},{\gamma}_{n}({\alpha},{\tau})={\gamma}({\alpha},{\tau})\\}\cap({\mathcal{I}}^{1}_{{\gamma},{\tau}})^{c},$
${\mathcal{I}}^{3}_{{\gamma},{\tau}}:={\mathcal{I}}(D_{\gamma,\tau})\cap({\mathcal{I}}^{1}_{{\gamma},{\tau}}\cup{\mathcal{I}}^{2}_{{\gamma},{\tau}})^{c},$
${\mathcal{I}}^{1}_{{\tau}}:=\bigcup_{{\gamma}>0}{\mathcal{I}}^{1}_{{\gamma},{\tau}},$
${\mathcal{I}}^{2}_{{\tau}}:=\bigcup_{{\gamma}>0}{\mathcal{I}}^{2}_{{\gamma},{\tau}},$
${\mathcal{I}}^{3}_{{\tau}}:=\bigcup_{{\gamma}>0}{\mathcal{I}}^{3}_{{\gamma},{\tau}}.$

###  2.2 Remarks

We list here some simple remarks. For a proof see [17].

  1. (a)

${\alpha}\in D_{\gamma,\tau}\iff 1-{\alpha}\in D_{\gamma,\tau}$.

  2. (b)

$\gamma(\alpha,\tau)\leq\min\\{{\alpha},1-{\alpha}\\}.$

  3. (c)

Fixed ${\tau}\geq 1$,
${\gamma}(.,{\tau}):D_{{\tau}}\rightarrow(0,\frac{1}{2})$.

  4. (d)

$D_{\gamma,\tau}^{{R}}=\bigcup_{n\in{Z}}(D_{\gamma,\tau}+n)$, thus we can
restrict to study the Diophantine points in $(0,1)$.

  5. (e)

$\left\\{\begin{array}[]{l}{\gamma}_{n}({\alpha},{\tau})=\frac{q_{n}^{{\tau}}}{{\alpha}_{n+1}q_{n}+q_{n-1}},\\\
\frac{1}{{{\gamma}_{n}({\alpha},{\tau})}}=\frac{q_{n+1}}{q_{n}^{{\tau}}}+\frac{1}{{\alpha}_{n+2}q_{n}^{{\tau}-1}}\end{array}\right.$
(1)

  6. (f)

$\gamma(\alpha,\tau)=\inf_{n\in{N}_{0}}{\gamma}_{n}({\alpha},{\tau})$.

  7. (g)

If ${\tau}<{\tau}({\alpha})$, then $\gamma(\alpha,\tau)=0$; if
${\tau}>{\tau}({\alpha})$ then $\gamma(\alpha,\tau)>0$. Moreover, for
${\tau}>{\tau}({\alpha})$ the inf is a minimum.

  8. (h)

${\alpha}\in{\mathcal{D}_{{\tau}}}\iff{\tau}({\alpha})={\tau}$ and
$\gamma(\alpha,\tau)>0$.

  9. (i)

If ${\alpha}\in{\mathcal{I}}^{1}_{{\gamma},{\tau}}$, then ${\alpha}$ is an
isolated point of $D_{\gamma,\tau}$.

  10. (j)

The cardinality of ${{\mathcal{I}}^{1}_{{\tau}}}$ is at most countable.

  11. (k)

${\mu}({\mathcal{D}_{{\tau}}})=0$ for all ${\tau}\geq 1$.

  12. (l)

${\gamma}_{0}({\alpha},{\tau})=\left\\{{\alpha}\right\\}$, in particular
${\gamma}_{0}({\alpha},{\tau})$ does not depend on ${\tau}$.

  13. (m)

Let $\frac{p}{q}$ a rational number.

${\alpha}\in D_{{\tau}}\iff\left\\{{\alpha}+\frac{p}{q}\right\\}\in D_{\tau},$
(2)
${\alpha}\in{\mathcal{D}_{{\tau}}}\iff\left\\{{\alpha}+\frac{p}{q}\right\\}\in{\mathcal{D}_{{\tau}}}.$
(3)

  14. (n)

If ${\tau}>{\tau}({\alpha})$,
${\gamma}_{-}({\alpha},{\tau})={\gamma}_{+}({\alpha},{\tau})$, then
${\alpha}\in{\mathcal{I}_{{\tau}}}$.

  15. (o)

${\alpha}\in D_{\tau}\iff q_{n+1}=O(q_{n}^{{\tau}}).$

##  3 Proof of the Theorem

In the first part of this section, we suppose without loss of generality, that
$n$ is always even. In fact, for $n$ odd it suffices to consider $1-{\alpha}$
(${\alpha}=[a_{0};...,a_{n},...]\in D_{\gamma,\tau}\iff 1-{\alpha}\in
D_{\gamma,\tau}$, and the denominators of the odd convergents to $1-{\alpha}$
are the same of the even convergents to ${\alpha}$, hence, by symmetry, all
that is demonstrated for $n$ even continues to hold if $n$ is odd). Moreover,
in all the section $0<{\gamma}<\frac{1}{{2}}$ (otherwise
$D_{\gamma,\tau}=\emptyset$). We want to prove that, for
${\tau}>\frac{3+\sqrt{17}}{2}$:

${\mu}\left(\left\\{0<{\gamma}<\frac{1}{2}:{\mathcal{I}}(D_{\gamma,\tau})\not=\emptyset\right\\}\right)=0.$

By Remark (j) it is enough to prove it for
${\mathcal{I}}^{2}_{{\gamma},{\tau}}$ and
${\mathcal{I}}^{3}_{{\gamma},{\tau}}$. Observe that the isolated points of
type 2,3 are obtained by infinitely many intersections of intervals centered
in rational numbers $\frac{p}{q}$ with length
$\frac{2{\gamma}}{q^{{\tau}+1}}$. Thus, the first step is to show that, given
${\alpha}\in D_{\gamma,\tau}$, it is enough (up to a set of measure zero and
for ${\tau}$ big enough) to control the intersection of intervals centred in
the convergents. The second step will be to show that, if intervals centred in
the convergents intersects, then the coefficients of the continued fractions
cannot grow too. In the final step we prove that, when intervals centred in
the convergents do not intersect and for big convergents, the interval between
two subsequent convergentes (with the same parity) contains a diophantine sets
with positive mesure.

###### Lemma 1

Let ${\gamma}>0,{\tau}>1,{\alpha}\in D_{\gamma,\tau}$, $\frac{p_{n}}{q_{n}}$
the convergents to ${\alpha}$,

$I_{n}:=\left(\frac{p_{n}}{q_{n}},\frac{p_{n+2}}{q_{n+2}}\right).$

Suppose that $\exists N\in{N}$ such that, for all $n>N$ even:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}.$
(4)

For $n>N$ define

$A_{n}:=\left(\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}},\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\right).$

Moreover, suppose that for every $n$ (even):

${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$ (5)

Then, there exists $N_{1}\in{N}$ such that, for all $n>N_{1}$:

$\frac{p}{q}\not\in I_{n}\ \Longrightarrow\
\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}\not\in
A_{n}.$

Proof Note that it is enough to verify the inequality when
$\frac{p}{q}<{\alpha}$. In fact the inequality is trivial if
$\frac{p}{q}>{\alpha}$ (because of ${\alpha}\in D_{\gamma,\tau}$ implies
$\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}\geq{\alpha}>\frac{p_{n+2}}{q_{n+2}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}$
by (5)). By (4) it follows that $A_{n}\cap A_{m}=\emptyset$ for $n\not=m$,
with $n,m>N$ even. From

${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$

for $n$ even, we get

$\max_{2n\leq
N}\frac{p_{2n}}{q_{2n}}+\frac{{\gamma}}{q_{2n}^{{\tau}+1}}=:C<{\alpha},$

from which it follows that there exists $N_{1}\in{N}$ such that for $n$ even,
$n>N_{1}$:

$\frac{p_{n}}{q_{n}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}}>C.$

If $\frac{p}{q}=\frac{p_{m}}{q_{m}}\not\in I_{n}$ is an even convergent to
${\alpha}$ with $n>N_{2}:=\max\\{N,N_{1}\\}$ then, for $m\leq N$ even:

$\frac{p_{m}}{q_{m}}<\frac{p_{n}}{q_{n}}.$

Moreover, by definition of $N_{1}$ it follows that:

$\frac{p_{m}}{q_{m}}+\frac{{\gamma}}{q_{m}^{{\tau}+1}}\leq
C<\frac{p_{n}}{q_{n}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}},$

from which it follows that the Lemma holds if
$\frac{p}{q}=\frac{p_{m}}{q_{m}}$ is an even convergent to ${\alpha}$ with
$m\leq N$. If $m>N$ and $n>m$ is even:

$\frac{p_{m}}{q_{m}}+\frac{{\gamma}}{q_{m}^{{\tau}+1}}<\frac{p_{m+2}}{q_{m+2}}-\frac{{\gamma}}{q_{m+2}^{{\tau}+1}}\leq\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}$

while, for $n<m$ even:

$\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}>\frac{p_{m-2}}{q_{m-2}}+\frac{{\gamma}}{q_{m-2}^{{\tau}+1}}\geq\frac{p_{n+2}}{q_{n+2}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}.$

So Lemma 1 is true if $\frac{p}{q}$ is an even convergent to ${\alpha}$. Thus,
Lemma 1 remains to be verified when $\frac{p}{q}$ is not a convergent to
${\alpha}$. It is no restrictive to suppose that there exists $m\not=n$ even
for which $\frac{p}{q}\in I_{m}$, otherwise Lemma 1 is trivial. Now we show
that, for $m$ big enough:

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}\in\left(\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}},\frac{p_{m+2}}{q_{m+2}}+\frac{{\gamma}}{q_{m+2}^{{\tau}+1}}\right)$

from which Lemma 1 follows immediately by (5). By the properties of Farey
sequence, for the rationals $\frac{p}{q}\in I_{m}$ we have $q>q_{m}$, so the
inequality:

$\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}$

holds. It remains to show that:

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{m+2}}{q_{m+2}}+\frac{{\gamma}}{q_{m+2}^{{\tau}+1}}.$

This inequality holds for $q\geq\frac{q_{m+2}}{2}$ and $m$ big enough. In
fact, in that case:

$\frac{p_{m+2}}{q_{m+2}}-\frac{p}{q}\geq\frac{1}{qq_{m+2}}>\frac{{\gamma}}{q^{{\tau}+1}}-\frac{{\gamma}}{q_{m+2}^{{\tau}+1}},$

that is true for $m$ big enough (because of ${\tau}>1$). So, we can assume
that $q_{m}<q<\frac{q_{m+2}}{2}$. Because we have assumed that $\frac{p}{q}$
is not a convergent, by Legendre’s Theorem (see [8]), we have:

${\alpha}-\frac{p}{q}>\frac{1}{2q^{2}},$

while, because $\frac{p_{m}}{q_{m}}$ is a convergent, we have:

${\alpha}-\frac{p_{m+2}}{q_{m+2}}<\frac{1}{q_{m+2}^{2}}.$

So, putting together the two inequalities, if $q<\frac{q_{m+2}}{2}$:

$\frac{p_{m+2}}{q_{m+2}}-\frac{p}{q}=\frac{p_{m+2}}{q_{m+2}}-{\alpha}+{\alpha}-\frac{p}{q}>\frac{1}{2q^{2}}-\frac{1}{q_{m+2}^{2}}>-\frac{{\gamma}}{q_{m+2}^{{\tau}+1}}+\frac{{\gamma}}{q^{{\tau}+1}}\iff$
$\frac{1}{2q^{2}}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{1}{q_{m+2}^{2}}-\frac{{\gamma}}{q_{m+2}^{{\tau}+1}},$

that is true for $m$ big enough (it follows by $q_{m}<q<\frac{q_{m+2}}{2}$).
So Lemma 1 is proved.

We know by Farey’s sequence that for $\frac{p}{q}\in I_{n}$, $q>q_{n+1}$. So,
there are a finite numbers of $\frac{p}{q}\in I_{n}$ with $q<q_{n+2}$. In the
next Lemma we want to control the distance between these numbers and
$\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}$.

###### Lemma 2

Let ${\gamma}>0$, ${\tau}>3,{\alpha}\in D_{\gamma,\tau},\frac{p_{n}}{q_{n}}$
the convergents to ${\alpha}$. There exists $N_{1}\in{N}$ such that, for
$n>N_{1}$:

$\frac{p}{q}\in I_{n},q<q_{n+2}\ \Longrightarrow\
\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}.$

Proof Let $n>N$, $\frac{p}{q}\in I_{n}$, so by definition of convergents and
the fact that $\frac{p_{n}}{q_{n}}<\frac{p}{q}<\frac{p_{n+2}}{q_{n+2}}$ we get
that $\frac{p}{q}$ is not a convergent. If $q\geq\frac{q_{n+2}}{2}$ we get:

$\frac{p_{n+2}}{q_{n+2}}-\frac{p}{q}\geq\frac{1}{qq_{n+2}}\geq\frac{1}{q_{n+2}^{2}}>\frac{{\gamma}2^{{\tau}+1}}{q_{n+2}^{{\tau}+1}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}\geq\frac{{\gamma}}{q^{{\tau}+1}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}$

for $n$ big enough (because of ${\tau}>3$). So, for $n$ big enough, the
inequality remain to be proved for $q<\frac{q_{n+2}}{2}$. In that case:

$\frac{p_{n+2}}{q_{n+2}}-\frac{p}{q}=\frac{p_{n+2}}{q_{n+2}}-{\alpha}+{\alpha}-\frac{p}{q}>\frac{1}{2q^{2}}-\frac{1}{q_{n+2}^{2}}>\frac{{\gamma}}{q^{{\tau}+1}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}\iff$
$\frac{1}{2q^{2}}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{1}{q_{n+2}^{2}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}.$

From the fact that

$G(x):=\frac{1}{2x^{2}}-\frac{{\gamma}}{x^{{\tau}+1}}$

is a decreasing function for $x$ big enough, it is enough to show the
inequality for $q=[\frac{q_{n+2}}{2}]$. In this case we get:

$\frac{1}{2q^{2}}-\frac{1}{q_{n+2}^{2}}\geq\frac{2}{q_{n+2}^{2}}-\frac{1}{q_{n+2}^{2}}=\frac{1}{q_{n+2}^{2}}>\frac{{\gamma}}{q^{{\tau}+1}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}$

for $n$ big enough (for ${\tau}>3$), so $\exists N_{1}\in{N}$ such that, when
$n>N_{1}$ is even the inequality is verified.

###### Lemma 3

Let ${\tau}>3$ $,{\alpha}=[a_{1},a_{2},...]\in
D_{\gamma,\tau},\frac{p_{n}}{q_{n}}$ the convergents to ${\alpha}$, then
$\exists N\in{N}$ such that for all $n>N$ even:

${\mu}\left(\bigcup_{\frac{p}{q}\in I_{n},q\geq
q_{n+2}}\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\right)<\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}$

Proof

${\mu}\left(\bigcup_{\frac{p}{q}\in I_{n},q\geq
q_{n+2}}\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\right)$
$<\sum_{q\geq
q_{n+2}}\sum_{q\frac{p_{n}}{q_{n}}<p<q\frac{p_{n+2}}{q_{n+2}}}\frac{2{\gamma}}{q^{{\tau}+1}}<2{\gamma}\left(\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n}}{q_{n}}\right)\sum_{q\geq
q_{n+2}}\frac{1}{{q^{{\tau}}}}$
$<2{\gamma}C\left(\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n}}{q_{n}}\right)\frac{1}{{q_{n+2}^{{\tau}-1}}}=o\left(\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}\right)$

for some constant $C>0$.

###### Lemma 4

Let ${\tau}>1,{\gamma}>0,$ ${\alpha}=[a_{1},a_{2},...]\in
D_{\gamma,\tau},\frac{p_{n}}{q_{n}}$ be the convergents to ${\alpha}$. Then:

$\quad\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\iff$
(6)
$a_{n+2}>\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}}\quad$
(7)

Proof (6) is true if and only if:

$\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n}}{q_{n}}=\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n+1}}{q_{n+1}}+\frac{p_{n+1}}{q_{n+1}}-\frac{p_{n}}{q_{n}}=$
$\frac{1}{q_{n}q_{n+1}}-\frac{1}{q_{n+1}q_{n+2}}>\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\iff$
$\frac{1}{q_{n+2}q_{n+1}}<\frac{1}{q_{n}q_{n+1}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\iff$
$\frac{1}{q_{n+2}q_{n+1}}<\frac{{\gamma}}{q_{n}q_{n+1}}(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\iff$
$\frac{1}{{q_{n+2}}}<\frac{{\gamma}}{q_{n}}(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-q_{n+1}\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\iff$
$\left\\{\begin{array}[]{l}\displaystyle\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}}>\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}},\\\
\\\ \displaystyle
q_{n+2}>\frac{q_{n}}{{\gamma}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}\\\
\end{array}\right.\ \\\ $ (8)

The first inequality is always true because of:

$\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}}>\frac{1}{{\alpha}_{n+2}q_{n}^{{\tau}-1}}>\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}.$

So Lemma 4 follows from the fact that $q_{n+2}=a_{n+2}q_{n+1}+q_{n}$.

###### Lemma 5

Let ${\tau}>1$, for almost all ${\gamma}\in(0,\frac{1}{{2}})$ (for
${\gamma}\geq\frac{1}{{2}}$ $D_{\gamma,\tau}=\emptyset$), given ${\epsilon}>0$
there exists $C=C({\epsilon},{\gamma})>0$ such that:

$\left|\frac{1}{{{\gamma}}}-\frac{p}{q^{{\tau}}}\right|\geq\frac{C}{q^{{\tau}+1+{\epsilon}}}$

for all $\frac{p}{q}\in{Q}$.

Proof Define
$B_{C,k}:=\left\\{{\alpha}:|{\alpha}-\frac{p}{q^{{\tau}}}|\geq\frac{C}{q^{k}}\quad\forall\frac{p}{q}\in{Q}\right\\}$,
so ${\alpha}\in B_{C,k}^{c}\iff$ there exists $\frac{p}{q}$ such that
${\alpha}\in\left(\frac{p}{q}-\frac{C}{q^{k}},\frac{p}{q}+\frac{C}{q^{k}}\right)$.
So, given $N\in{N}$ we get:

${\mu}\left(B_{C,k}^{c}\cap\left(-N,N\right)\right)<\sum_{q>0}\sum_{-Nq^{{\tau}}<p<Nq^{{\tau}}}\frac{2C}{q^{k}}<\sum_{q>0}\frac{4NC}{q^{k-{\tau}}}$

and for $k>{\tau}+1$, $C$ that tends to zero, also

${\mu}\left(B_{C,k}^{c}\cap\left(-N,N\right)\right)$

goes to zero. From the arbitrariness of $N$ we obtain:

${\mu}\left(\bigcap_{C>0}B_{C,k}^{c}\right)=0$

for $k>{\tau}+1$, from which follows Lemma 5.

###### Lemma 6

Let ${\tau}>1$, ${\alpha}=[a_{1},a_{2},...]\in D_{\gamma,\tau}$,
$\frac{p_{n}}{q_{n}}$ the convergents to ${\alpha}$. The inequality:

$\quad\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}$
(9)

is definitively verified if and only if definitively:

${a_{n+2}>\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}-\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}}-\frac{q_{n}}{q_{n+1}}}$
(10)

###### Remark 1

Observe that (10) is definitively true if:

$\limsup\frac{q_{n+1}}{q_{n}^{{\tau}}}<\frac{1}{{\gamma}},$

because in that case:

$\limsup\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}-\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}}-\frac{q_{n}}{q_{n+1}}<1.$

Thus, if for infinitely many $n$ even $(\ref{ci})$ is not verified, for this
$n$, with $n$ big enough:

$\frac{q_{n+1}}{q_{n}^{{\tau}}}\sim\frac{1}{{\gamma}},$

so $q_{n+1}\sim\frac{q_{n}^{{\tau}}}{{\gamma}}$.

Proof

In a similar way of Lemma 4, (9) is verified if and only if:

$\left\\{\begin{array}[]{l}\displaystyle\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}}>\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}+\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}},\\\
\\\ \displaystyle
q_{n+2}>\frac{q_{n}}{{\gamma}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}-\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}}\\\
\end{array}\right.\ \\\ $ (11)

Because of ${\alpha}\in D_{\gamma,\tau}$, the first of the two conditions is
definitively verified, in fact, for $n$ big enough:

$\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}+\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}<\frac{1}{{\alpha}_{n+2}q_{n}^{{\tau}-1}}<\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}}$

So, from the fact that $q_{n+2}=a_{n+2}q_{n+1}+q_{n}$ we are done.

###### Lemma 7

Let ${\tau}>\frac{3+\sqrt{17}}{2}$. For almost all
${\gamma}\in(0,\frac{1}{2})$, if ${\alpha}=[a_{0},a_{1},...]\in
D_{\gamma,\tau}$ ,for $n$ even big enough: (6) is true if and only if (9) is
true.

Proof If (9) is true, then trivially (6) is true. So we have to show that for
almost all ${\gamma}\in(0,\frac{1}{{2}})$ and for all ${\alpha}\in
D_{\gamma,\tau}$ (with ${\tau}>\frac{3+\sqrt{17}}{2}$) holds the converse. So,
suppose by contradiction that exists $A\subseteq\left(C_{1},C_{2}\right)$,
with $0<C_{1}<C_{2}<\frac{1}{{2}}$, ${\mu}(A)>0$ such that, for all
${\gamma}\in A$ there exists ${\alpha}\in D_{\gamma,\tau}$ that satisfies (6)
but not (9) for infinitely many $n$ even. By Lemma 4 and Lemma 6 it follows
that for all ${\gamma}$ in $A$ there exists ${\alpha}\in D_{\gamma,\tau}$ such
that for infinitely many $n$ even:

$\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}-\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}}-\frac{q_{n}}{q_{n+1}}\geq
a_{n+2}>\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}},$

and by Remark 1 it follows that, for this $n$:

$q_{n+1}\sim\frac{q_{n}^{{\tau}}}{{\gamma}}.$

So, for $n$ big enough such that (6) holds but (9) doesn’t hold we get:

$\frac{q_{n}^{{\tau}}}{C_{2}}<q_{n+1}<\frac{q_{n}^{{\tau}}}{C_{1}}.$

Moreover:

$a_{n+2}>\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}}\iff$
$\frac{a_{n+2}q_{n+1}}{q_{n}}+1=\frac{q_{n+2}}{q_{n}}>\frac{1}{{1-\frac{{\gamma}q_{n+1}}{q_{n}^{{\tau}}}-\frac{{\gamma}q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}\iff$
$1-\frac{{\gamma}q_{n+1}}{q_{n}^{{\tau}}}-\frac{{\gamma}q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}>\frac{q_{n}}{q_{n+2}}\iff$
${\gamma}<\frac{1-\frac{q_{n}}{q_{n+2}}}{\frac{q_{n+1}}{q_{n}^{{\tau}}}+\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}$

In a similar way:

$\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}-\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}}-\frac{q_{n}}{q_{n+1}}\geq
a_{n+2}\iff$
${\gamma}\geq\frac{1-\frac{q_{n}}{q_{n+2}}}{\frac{q_{n+1}}{q_{n}^{{\tau}}}+\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}+\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}.$

Thus:

$\frac{1-\frac{q_{n}}{q_{n+2}}}{\frac{q_{n+1}}{q_{n}^{{\tau}}}+\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}+\frac{2q_{n}q_{n+1}}{q_{n+2}^{{\tau}-1}}}\leq{\gamma}<\frac{1-\frac{q_{n}}{q_{n+2}}}{\frac{q_{n+1}}{q_{n}^{{\tau}}}+\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}$

for infinitely many $n$ even, so for all ${\gamma}\in A$ there exist
infinitely many $q\in{N}$ such that:

$\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}+\frac{2qp}{(Np+q)^{{\tau}-1}}}\leq{\gamma}<\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}}$

for some $N\in{N}$ and some
$\frac{q^{{\tau}}}{C_{2}}<p<\frac{q^{{\tau}}}{C_{1}}$. So for all $M\in{N}$:

$A\subseteq\bigcup_{q>M}\bigcup_{\frac{q^{{\tau}}}{C_{2}}<p<\frac{q^{{\tau}}}{C_{1}}}\bigcup_{N>0}\left(\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}+\frac{2qp}{(Np+q)^{{\tau}-1}}},\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}}\right),$

moreover:

$\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}}-\frac{1-\frac{q}{Np+q}}{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}+\frac{2qp}{(Np+q)^{{\tau}-1}}}<$
$\frac{2qp}{(Np+q)^{{\tau}-1}}\left(\frac{1}{{\frac{p}{q^{{\tau}}}+\frac{qp}{(Np+q)^{{\tau}+1}}}}\right)^{2}<\frac{2qC_{2}^{2}}{N^{{\tau}-1}p^{{\tau}-2}}$

so we obtain:

$m(A)\leq\sum_{q>M}\sum_{\frac{q^{{\tau}}}{C_{2}}<p<\frac{q^{{\tau}}}{C_{1}}}\sum_{N>0}\frac{2qC_{2}^{2}}{N^{{\tau}-1}p^{{\tau}-2}}<$
${\beta}\sum_{q>M}\frac{q^{{\tau}+1}}{q^{{\tau}^{2}-2{\tau}}}={\beta}\sum_{q>M}\frac{1}{{q^{{\tau}^{2}-3{\tau}-1}}}$

for some constant ${\beta}>0$. From the hypothesis
(${\tau}>\frac{3+\sqrt{17}}{2}$) we have that the series converge, so for $M$
that goes to infinity we get that ${\mu}(A)=0$, that contradicts the
hypothesis ${\mu}(A)>0$. Thus, for almost all ${\gamma}\in(C_{1},C_{2})$ we
have that: if (6) holds, then (9) holds, and from the arbitrariness of
$C_{1},C_{2}$ Lemma 7 follows.

###### Proposition 1

Let ${\tau}>\frac{3+\sqrt{17}}{2}$. For almost every
$0<{\gamma}<\frac{1}{{2}}$: if ${\alpha}\in D_{\gamma,\tau}$,
$\frac{p_{n}}{q_{n}}$ are the convergents to ${\alpha}$,
${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$, and
definitively:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},$

then ${\alpha}$ is an accumulation point of $D_{\gamma,\tau}$ and in
particular, for $n$ even big enough:

${\mu}\left(D_{\gamma,\tau}\cap\left(\frac{p_{n}}{q_{n}},\frac{p_{n+2}}{q_{n+2}}\right)\right)>0$

Proof By Lemma 1 it follows that $\exists N_{1}\in{N}$ such that for $n>N_{1}$
even:

$\frac{p}{q}\not\in I_{n}\ \Longrightarrow\
\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}\not\in
A_{n},$

and by Lemma 7 for almost all ${\gamma}\in(0,\frac{1}{{2}})$:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\
\Longrightarrow\
\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}},$

therefore, up to a set of measure zero we can suppose that ${\gamma}$
satisfies this property. Moreover, by Lemma 2, for $n$ even big enough, if
$\frac{p}{q}\in I_{n},$ $q<q_{n+2}$ then:

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}.$

So, if we define:

$c_{n}:=\max_{\frac{p}{q}\in[\frac{p_{n}}{q_{n}},\frac{p_{n+2}}{q_{n+2}}),q<q_{n+2}}\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}},$

we obtain:

$c_{n}<\frac{p_{n+2}}{q_{n+2}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}.$

By Lemma 1, if $n>N_{1}$ is even and $\frac{p}{q}\not\in I_{n}$, then

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}\not\in
A_{n},$

so, if

$\frac{p}{q}<\frac{p_{n}}{q_{n}}\ \Longrightarrow\
\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\leq
c_{n},$

while for $\frac{p}{q}>\frac{p_{n+2}}{q_{n+2}}$ we get $q>q_{n+2}$, so:

$\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},$

but from:

${\beta}\in
D_{\gamma,\tau}^{c}\iff\exists\frac{p}{q}\in(0,1):{\beta}\in\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)$

we get that for $n>N_{1}$ even, holds:

${\mu}\left(D_{\gamma,\tau}^{c}\cap
I_{n}\right)\leq{\mu}\left(\bigcup_{\frac{p}{q}\in[\frac{p_{n}}{q_{n}},\frac{p_{n+2}}{q_{n+2}}),q<q_{n+2}}\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\cap
I_{n}\right)$ $+{\mu}\left(\bigcup_{\frac{p}{q}\in I_{n},q\geq
q_{n+2}}\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\right)+{\mu}\left(\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},\frac{p_{n+2}}{q_{n+2}}\right).$

So by Lemma 3:

${\mu}(D_{\gamma,\tau}^{c}\cap I_{n})\leq
c_{n}-\frac{p_{n}}{q_{n}}+\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}<{\mu}(I_{n})=\frac{p_{n+2}}{q_{n+2}}-\frac{p_{n}}{q_{n}}\iff$
$c_{n}<\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}-\frac{2{\gamma}}{q_{n+2}^{{\tau}-1}},$

that follows from the definition of $c_{n}$.

So, given ${\tau}>3$, for almost all ${\gamma}>0$: if ${\alpha}\in
D_{\gamma,\tau}$ is not an isolated point of the first type and definitively
the intervals centered in the convergents have an empty intersection, then
${\alpha}$ is an accumulation point in $D_{\gamma,\tau}$. The second step is
to show that: if ${\tau}>3$, ${\gamma}>0$, ${\alpha}\in D_{\gamma,\tau}$ but
${\alpha}$ is not an isolated point of the first type and
${\tau}>{\tau}({\alpha})$, then ${\alpha}$ is an accumulation point in
$D_{\gamma,\tau}$.

###### Lemma 8

Let ${\tau}>3$. For almost all ${\gamma}\in(0,\frac{1}{{2}})$: given
${\alpha}\in D_{\gamma,\tau}$, if for infinitely many $n$ even:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}>\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},$

then there exists $C>0$ such that for this $n$:

$a_{n+2}\leq Cq_{n}^{2+{\epsilon}},$

with ${\epsilon}>0$ arbitrarily small.

Proof By Lemma 4 it follows that, given ${\alpha}\in D_{\gamma,\tau}$ that
satisfies the hypothesis of Lemma 8, for $n$ even big enough:

$a_{n+2}\leq\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}},$

so, up to a set of measure zero, by Lemma 5 we can suppose that there exist
${\epsilon}>0,C>0$ such that $\frac{1}{{{\gamma}}}\in
B_{C,{\tau}+1+{\epsilon}}$ with ${\tau}+1+{\epsilon}<{\tau}^{2}-1$, from which
it follows that:

$\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{(\frac{1}{{\gamma}}-\frac{q_{n+1}}{q_{n}^{{\tau}}})-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}}\leq\frac{q_{n}}{{\gamma}q_{n+1}}\frac{1}{{\frac{C}{q_{n}^{{\tau}+1+{\epsilon}}}-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}}}-\frac{q_{n}}{q_{n+1}},$

moreover, by Remark 1 it follows that
$q_{n+1}\sim\frac{q_{n}^{{\tau}}}{{\gamma}}$, from which we obtain:

$\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}<\frac{q_{n}}{q_{n+1}^{{\tau}}}\sim\frac{{\gamma}^{{\tau}}}{q_{n}^{{\tau}^{2}-1}},$

so, if $n$ is big enough, by ${\tau}+1+{\epsilon}<{\tau}^{2}-1$ we have:

$\frac{C}{q_{n}^{{\tau}+1+{\epsilon}}}-\frac{q_{n}q_{n+1}}{q_{n+2}^{{\tau}+1}}>\frac{C}{2q_{n}^{{\tau}+1+{\epsilon}}}.$

So we obtain:

$a_{n+2}<\frac{q_{n}}{q_{n+1}}\frac{2q_{n}^{{\tau}+1+{\epsilon}}}{C}\sim\frac{2{\gamma}}{C}q_{n}^{2+{\epsilon}}<\frac{4{\gamma}}{C}q_{n}^{2+{\epsilon}}=C^{\prime}q_{n}^{2+{\epsilon}}$

definitively, from which we get Lemma 8.

###### Lemma 9

Let ${\tau}>\frac{3+\sqrt{17}}{2},{\gamma}>0$, ${\alpha}\in D_{\gamma,\tau}$.
If for infinitely many $m$ even, for $n<m$ even holds:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}<\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}},$
(12)

and ${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$ for all
$n$ even, then ${\alpha}$ is in ${\mathcal{A}}(D_{\gamma,\tau})$.

Proof Let $\frac{p_{n}}{q_{n}}<\frac{p}{q}<\frac{p_{n+2}}{q_{n+2}}$ with $n$
even and $n<m-2$, for $\frac{q_{n+2}}{2}\leq q$:

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{n+2}}{q_{n+2}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}$

is definitively true, while for $q<\frac{q_{n+2}}{2}$:

$\frac{p_{n+2}}{q_{n+2}}-\frac{p}{q}=\frac{p_{n+2}}{q_{n+2}}-{\alpha}+{\alpha}-\frac{p}{q}>\frac{1}{2q^{2}}-\frac{1}{q_{n+2}^{2}}>\frac{{\gamma}}{q^{{\tau}+1}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}}\iff$
$\frac{1}{2q^{2}}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{1}{q_{n+2}^{2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},$

that is true for $q$ big enough, so $\exists T\in{N}$ such that the inequality
is verified for $q\geq T$ (from the fact that
$G(x):=\frac{1}{2x^{2}}-\frac{{\gamma}}{x^{{\tau}+1}}$ is definitively
decreasing and ${\tau}>3>1$). From the hypothesis that
${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$ for all $n$
even:

$v:=\max_{\frac{p}{q}<{\alpha},q\leq
T}\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<{\alpha},$

so there exists $T_{1}\in{N}$ such that for $n>T_{1}$:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}>v.$

By Lemma 2, for $m$ big enough, $\frac{p}{q}\in I_{n}$, with $n<m-2$ even:

$\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\leq\max\left\\{\frac{p_{n+2}}{q_{n+2}}+\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},v\right\\}\leq\frac{p_{m-2}}{q_{m-2}}+\frac{{\gamma}}{q_{m-2}^{{\tau}+1}}<\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}+1}},$

while by Lemma 2, for $m$ big enough:

$\frac{p}{q}\in I_{m-2},q<q_{m-2}\ \Longrightarrow\
\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}<\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}},$

so if we define:

$c_{m}:=\max\left\\{\max_{\frac{p}{q}\in
I_{m-2},q<q_{m}}\left(\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right),\max_{\frac{p}{q}\leq\frac{p_{m-2}}{q_{m-2}}}\left(\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\right\\},$

for $m$ even big enough:

$c_{m}<\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}}.$

Moreover, by Lemma 3, from ${\tau}>3>2$, for $m$ even big enough:

${\mu}\left(\bigcup_{\frac{p}{q}\in I_{m-2},q\geq
q_{m}}\left(\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}},\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)\right)<\frac{2{\gamma}}{q_{m}^{{\tau}-1}}.$

Finally, if $\frac{p}{q}>\frac{p_{m}}{q_{m}}$, by the properties of continued
fractions we obtain $q>q_{m}$, so
$\frac{p}{q}-\frac{{\gamma}}{q^{{\tau}+1}}>\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}$.
Thus:

${\mu}\left(D_{\gamma,\tau}^{c}\cap\left(\frac{p_{m-2}}{q_{m-2}},\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}\right)\right)<c_{m}-\frac{p_{m-2}}{q_{m-2}}+\frac{2{\gamma}}{q_{m}^{{\tau}-1}}$
$<\frac{p_{m}}{q_{m}}-\frac{p_{m-2}}{q_{m-2}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}={\mu}(\frac{p_{m-2}}{q_{m-2}},\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}),$

then

$D_{\gamma,\tau}\cap\left(\frac{p_{m-2}}{q_{m-2}},\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}\right)\not=\emptyset,$

and from the fact that this holds for infinitely many $m$ even, then
${\alpha}$ is an accumulation point of $D_{\gamma,\tau}$.

###### Remark 2

Let ${\tau}>\frac{\sqrt{17}+3}{2},{\gamma}>0$, ${\alpha}\in D_{\gamma,\tau}$,
if ${\alpha}\in{\mathcal{I}}^{2}_{{\gamma},{\tau}}$ or
${\mathcal{I}}^{3}_{{\gamma},{\tau}}$, then ${\tau}({\alpha})={\tau}$. In fact
if this doesn’t hold, from
${\alpha}\not\in{\mathcal{I}}^{1}_{{\gamma},{\tau}}$ we get that for all $n$
even or for all $n$ odd:

$\left|{\alpha}-\frac{p_{n}}{q_{n}}\right|>\frac{{\gamma}}{q_{n}^{{\tau}+1}}.$

Suppose for example that this property holds for all $n$ even. If on the
contrary ${\tau}({\alpha})<{\tau}$, by Remark 1, the hypothesis of Proposition
1 are satisfied, so ${\alpha}\in{\mathcal{A}}(D_{\gamma,\tau})$,
contradiction.

###### Corollary 1

If ${\tau}>\frac{3+\sqrt{17}}{2}$:

${\mu}\left(\left\\{{\gamma}>0:{\mathcal{I}}^{2}_{{\gamma},{\tau}}\not=\emptyset\right\\}\right)=0.$

Proof Observe that, if ${\alpha}\in{\mathcal{I}}^{2}_{{\gamma},{\tau}}$, then
there exists $n\in{N}$ such that

$\left|{\alpha}-\frac{p_{n}}{q_{n}}\right|=\frac{{\gamma}}{q_{n}^{{\tau}+1}}.$

Suppose for example that $n$ is even, thus:

${\alpha}=\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}.$

Moreover, for almost all ${\gamma}\in(0,\frac{1}{{2}})$:

${\tau}\left(\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)={\tau}\left(\frac{{\gamma}}{q^{{\tau}+1}}\right)=1.$

Taking the union on all the $\frac{p}{q}$ we obtain that for almost all
${\gamma}\in(0,\frac{1}{{2}})$ and for all $\frac{p}{q}\in{Q}$,

${\tau}\left(\frac{p}{q}+\frac{{\gamma}}{q^{{\tau}+1}}\right)=1.$

So Corollary 1 follows by Remark 2.

It remains the last one step, in which we get the Theorem.

###### Lemma 10

Let ${\tau}>3$. For almost all ${\gamma}>0$, if
${\alpha}\in{\mathcal{I}}(D_{\gamma,\tau})$, there exists $N\in{N}$ such that,
for all $m>N$ even there is some $n<m$ even with:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}}$

.

Proof By Corollary 1 and Remark (j) it follow that, up to a set of measure
zero, we can suppose that
${\mathcal{I}}^{1}_{{\gamma},{\tau}}={\mathcal{I}}^{2}_{{\gamma},{\tau}}=\emptyset$,
so observe that if the Lemma were not true, it would exist
${\alpha}\in{\mathcal{I}}^{2}_{{\gamma},{\tau}}$ with the even convergents
that satisfy the hypothesis of Lemma 9, that implies
${\alpha}\in{\mathcal{A}}(D_{\gamma,\tau})$, contradiction.

Theorem Let ${\tau}>\frac{3+\sqrt{17}}{2}$. Then, for almost all ${\gamma}>0$
$D_{\gamma,\tau}$ is a Cantor set.

Proof By Corollary 1 and Remark (j) it follows that, up to a set of measure
zero, we can suppose that
${\mathcal{I}}^{1}_{{\gamma},{\tau}}={\mathcal{I}}^{2}_{{\gamma},{\tau}}=\emptyset$.
Suppose by contradiction that the statement doesn’t hold, and take
$0<C_{1}<C_{2}$ such that:

${\mu}\left(\left\\{C_{1}<{\gamma}<C_{2}:{\mathcal{I}}(D_{\gamma,\tau})\not=\emptyset\right\\}\right)>0,$

and define
$A:=\\{C_{1}<{\gamma}<C_{2}:{\mathcal{I}}(D_{\gamma,\tau})\not=\emptyset\\}$.
By Lemma 10, for almost all ${\gamma}>0$ there exists
${\alpha}\in{\mathcal{I}}(D_{\gamma,\tau})$ and there exists $N\in{N}$ such
that for all $m>N$ even, there is some $n<m$ even, with:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}}.$

Now we want to show that, for almost all chosen of ${\gamma}\in A$ we have:

$\limsup\frac{q_{2k+2}}{q_{2k+1}^{{\tau}}}<\frac{1}{{\gamma}}.$

In fact if it doesn’t hold, by Remark 1 we get that for infinitely many $m$
even:

$q_{m}\sim\frac{q_{m-1}^{{\tau}}}{{\gamma}},$

and for $m>N$ exists $n<m$ even, with:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}}$

By Lemma 7, up to a set of measure zero in $A$:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}}\iff\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}.$

By the properties of convergents:

${\alpha}-\frac{p_{m}}{q_{m}}<\frac{1}{q_{m}^{2}},$

from which we get:

$\frac{1}{q_{m}^{2}}>{\alpha}-\frac{p_{n}}{q_{n}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}.$

Moreover:

${\alpha}-\frac{p_{n}}{q_{n}}=\frac{1}{{q_{n}(q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}})}},$

so:

$\frac{1}{q_{m}^{2}}>\frac{1}{{q_{n}(q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}})}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}$

For $m$ big enough:

$\frac{1}{q_{m}^{2}}+\frac{{\gamma}}{q_{m}^{{\tau}+1}}<\frac{2}{q_{m}^{2}},$

so:

$\frac{2}{q_{m}^{2}}>\frac{1}{{q_{n}(q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}})}}-\frac{{\gamma}}{q_{n}^{{\tau}+1}}\iff$
${\gamma}>\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}-\frac{2q_{n}^{{\tau}+1}}{q_{m}^{2}},$

moreover:

${\gamma}\leq\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}.$

So we obtain:

$\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}-\frac{2q_{n}^{{\tau}+1}}{q_{m}^{2}}<{\gamma}\leq\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}$

From

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{m}}{q_{m}}-\frac{{\gamma}}{q_{m}^{{\tau}+1}}-\frac{2{\gamma}}{q_{m}^{{\tau}-1}},$

we get:

$\frac{p_{n}}{q_{n}}+\frac{{\gamma}}{q_{n}^{{\tau}+1}}\geq\frac{p_{n+2}}{q_{n+2}}-\frac{{\gamma}}{q_{n+2}^{{\tau}+1}},$

moreover, from
${\alpha}-\frac{p_{n}}{q_{n}}>\frac{{\gamma}}{q_{n}^{{\tau}+1}}$ for all $n$
even, when $m$ increase, also $n$ increase, and by the last inequality and
Remark 1 we get that $q_{n+1}\sim\frac{q_{n}^{{\tau}}}{{\gamma}}$. So

$q_{m}\sim\frac{q_{m-1}^{{\tau}}}{{\gamma}}\geq\frac{q_{n+1}^{{\tau}}}{{\gamma}}\sim\frac{q_{n}^{{\tau}^{2}}}{{\gamma}^{{\tau}}}\geq\frac{q_{n}^{{\tau}^{2}}}{C_{2}^{{\tau}}}.$

So we obtain:

$\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}-\frac{C}{q_{n}^{2{\tau}^{2}-{\tau}-1}}<{\gamma}\leq\frac{q_{n}^{{\tau}}}{q_{n+1}+\frac{{\alpha}_{n+2}}{q_{n}}}$

with a constant $C>0$. By Lemma 8, up to a set of measure zero, we can suppose
that there exists ${\epsilon}>0$ arbitrarily small such that, for $n$ big
enough:

$a_{n+2}<q_{n}^{2+{\epsilon}}.$

So, up to a set of measure zero, we can suppose that for all ${\gamma}\in A$,
there exists infinitely many $q>0$,
$\frac{q^{{\tau}}}{2C_{2}}<p<\frac{2}{C_{1}q^{{\tau}}}$, $N<q^{2+{\epsilon}}$
such that:

$\frac{q^{{\tau}}}{p+\frac{N}{q}}-\frac{C}{q^{2{\tau}^{2}-{\tau}-1}}<{\gamma}\leq\frac{q^{{\tau}}}{q+\frac{N}{q}}.$

So, for all $M\in{N}$:

$A\subseteq\bigcup_{q>M}\bigcup_{\frac{q^{{\tau}}}{2C_{2}}<p<\frac{2q^{{\tau}}}{C_{1}}}\bigcup_{N<q^{2+{\epsilon}}}\left(\frac{q^{{\tau}}}{p+\frac{N}{q}}-\frac{C}{q^{2{\tau}^{2}-{\tau}-1}},\frac{q^{{\tau}}}{q+\frac{N}{q}}\right),$

Thus:

${\mu}(A)<\sum_{q>M}\sum_{\frac{q^{{\tau}}}{2C_{2}}<p<\frac{2q^{{\tau}}}{C_{1}}}\sum_{N<q^{2+{\epsilon}}}\frac{C}{q^{2{\tau}^{2}-{\tau}-1}}$
$<{\beta}\sum_{q>M}\frac{1}{{q^{2{\tau}^{2}-2{\tau}-3-{\epsilon}}}}$

with some constant ${\beta}>0$. Because of ${\tau}>\frac{3+\sqrt{17}}{2}$, for
${\epsilon}$ small enough the series converge, so for $M$ that tends to
infinity we obtain ${\mu}(A)=0$, contradiction. So we have proved that:

$\limsup\frac{q_{2k+2}}{q_{2k+1}^{{\tau}}}<\frac{1}{{\gamma}}.$

But, by Remark 1 and Proposition 1 (used with $n$ odd) we have that
${\alpha}\in{\mathcal{A}}(D_{\gamma,\tau})$, contradiction. So ${\mu}(A)=0$.

The estimate ${\tau}>\frac{3+\sqrt{17}}{2}$ can be improved putting a better
inequality in Lemma 5. Probably the Proposition holds also with ${\tau}>3$.

##  4 Questions

  * •

By [17] we konw that, for some choice of ${\gamma},{\tau}$,
${\mathcal{I}}^{1}_{{\gamma},{\tau}}\not=\emptyset$. What about
${\mathcal{I}}^{2}_{{\gamma},{\tau}},{\mathcal{I}}^{3}_{{\gamma},{\tau}}$?

  * •

What is the best ${\tau}>1$ such that the result holds?

  * •

Is it true that, for all ${\tau}\geq 1$ there exists
${\gamma}_{\tau}\in(0,\frac{1}{{2}})$ such that $D_{\gamma,\tau}$ is a Cantor
set for almost all ${\gamma}\in(0,{\gamma}_{\tau})$?

#### Acknowledgement

I am very grateful to Prof. Luigi Chierchia for his suggestions, remarks, for
his special support and for encouraging me to complete this work.

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