# Concentration of dimension in extremal points of left-half lines in the
Lagrange spectrum

Carlos Gustavo Moreira  Carlos Gustavo Moreira: IMPA, Estrada Dona Castorina
110, 22460-320, Rio de Janeiro, Brazil<EMAIL_ADDRESS>and  Christian Villamil
Christian Villamil: IMPA, Estrada Dona Castorina 110, 22460-320, Rio de
Janeiro, Brazil<EMAIL_ADDRESS>
###### Abstract.

We prove that for any $\eta$ that belongs to the closure of the interior of
the Markov and Lagrange spectra, the sets $k^{-1}((-\infty,\eta])$ and
$k^{-1}(\eta)$, which are the sets of irrational numbers with best constant of
Diophantine approximation bounded by $\eta$ and exactly $\eta$ respectively,
have the same Hausdorff dimension. We also show that, as $\eta$ varies in the
interior of the spectra, this Hausdorff dimension is a strictly increasing
function.

###### Key words and phrases:

Hausdorff dimension, horseshoes, Lagrange spectrum, surface diffeomorphisms

The first author was partially supported by CNPq and FAPERJ

The second author was partially supported by CNPq and Viana’s Louis D. prize

###### Contents

  1. 1 Introduction
  2. 2 Preliminares
    1. 2.1 Continued fractions and regular Cantor sets
    2. 2.2 Results on Dynamical Markov and Lagrange spectra
    3. 2.3 The horseshoe $\Lambda(N)$
  3. 3 Proof of the result
    1. 3.1 Connection of subhorseshoes
    2. 3.2 Dimension estimates
    3. 3.3 Putting unstable Cantor sets into $k^{-1}(\eta)$

##  1\. Introduction

The classical Lagrange and Markov spectra are closed subsets of the real line
related to Diophantine approximations. They arise naturally in the study of
rational approximations of irrational numbers and of indefinite binary
quadratic forms, respectively.

Given $\alpha\in\mathbb{R}\setminus\mathbb{Q}$, set

$\displaystyle k(\alpha)$ $\displaystyle=$
$\displaystyle\sup\left\\{k>0:\left|\alpha-\frac{p}{q}\right|<\frac{1}{kq^{2}}\
\mbox{has infinitely many rational solution}\ \frac{p}{q}\right\\}$
$\displaystyle=$
$\displaystyle\limsup_{p\in\mathbb{Z},q\in\mathbb{N},p,q\to\infty}|q(q\alpha-p)|^{-1}\in\mathbb{R}\cup\\{\infty\\}$

for the best constant of Diophantine approximations of $\alpha$.

The classical Lagrange spectrum is the set

$L=\\{k(\alpha):\alpha\in\mathbb{R}\setminus\mathbb{Q},k(\alpha)<\infty\\},$

and the classical Markov spectrum is the set

$M=\left\\{\left(\inf\limits_{(x,y)\in\mathbb{Z}^{2}-\\{(0,0)\\}}|q(x,y)|\right)^{-1}<\infty:q(x,y)=ax^{2}+bxy+cy^{2},b^{2}-4ac=1\right\\}$

that consists of the reciprocal of the minimal values over non-trivial integer
vectors $(x,y)\in\mathbb{Z}^{2}-\\{(0,0)\\}$ of indefinite binary quadratic
forms $q(x,y)$ with unit discriminant.

Perron gave in [19] the following dynamical characterizations of these
classical spectra in terms of symbolic dynamical systems: Given a bi-infinite
sequence
$\theta=(\theta_{n})_{n\in\mathbb{Z}}\in(\mathbb{N}^{*})^{\mathbb{Z}}$, let

$\lambda_{i}(\theta):=[0;a_{i+1},a_{i+2},\dots]+a_{i}+[0;a_{i-1},a_{i-2},\dots].$

If the Markov value $m(\theta)$ of $\theta$ is
$m(\theta):=\sup\limits_{i\in\mathbb{Z}}\lambda_{i}(\theta)$ and the Lagrange
value $\ell(\theta)$ is
$\ell(\theta):=\limsup\limits_{i\to\infty}\lambda_{i}(\theta).$ Then the
Markov spectrum is the set

$M=\\{m(\theta)<\infty:\theta\in(\mathbb{N}^{*})^{\mathbb{Z}}\\}$

and the Lagrange spectrum is the set

$L=\\{\ell(\theta)<\infty:\theta\in(\mathbb{N}^{*})^{\mathbb{Z}}\\}.$

It follows from these characterizations that $M$ and $L$ are closed subsets of
$\mathbb{R}$ and that $L\subset M$.

Markov showed in [11] that

$L\cap(-\infty,3)=M\cap(-\infty,3)=\\{k_{1}=\sqrt{5}<k_{2}=2\sqrt{2}<k_{3}=\frac{\sqrt{221}}{5}<...\\},$

where $k^{2}_{n}\in\mathbb{Q}$ for every $n\in\mathbb{N}$ and $k_{n}\to 3$
when $n\to\infty$.

M. Hall in cf.[6] proved that

$C_{4}+C_{4}=[\sqrt{2}-1,4(\sqrt{2}-1)],$

where for each positive integer $N$, $C_{N}$ is the set of numbers in $[0,1]$
in whose continued fractions the coefficients are bounded by $N$, i.e.,
$C_{N}=\\{x=[0;a_{1},...,a_{n},...]\in[0,1]:a_{i}\leq N,\ \forall i\geq 1\\}.$
Together with Perron characterizations, this implies that $L$ and $M$ contain
the whole half-line $[6,+\infty)$.

Freiman in [5] determined the precise beginning of Hall’s ray (the biggest
half-line contained in $L$), which is

$\frac{2221564096+283748\sqrt{462}}{491993569}=4.52782956616\dots$

The first author in [15] proved several results on the geometry of the Markov
and Lagrange spectra, for example that the map $d:\mathbb{R}\rightarrow[0,1]$,
given by

$d(\eta)=HD(L\cap(-\infty,\eta))=HD(M\cap(-\infty,\eta))$

is continuous, surjective and such that $d(3)=0$ and $d(\sqrt{12})=1$.
Moreover, that

$d(\eta)=\min\\{1,2D(\eta)\\},$

where $D(\eta)=HD(k^{-1}(-\infty,\eta))=HD(k^{-1}(-\infty,\eta])$ is a
continuous surjective function from $\mathbb{R}$ to $[0,1).$ And also, the
limit

$\lim_{\eta\rightarrow\infty}HD(k^{-1}(\eta))=1.$

Recently in [17] was given the estimate

$t^{*}_{1}:=\sup\\{s\in\mathbb{R}:d(s)<1\\}=3.334384...$

In particular, any $t\in\mathbb{R}$ that belongs to the interior of the Markov
and Lagrange spectra, must satisfy $t>t^{*}_{1}.$

Now, let $\varphi:S\rightarrow S$ be a diffeomorphism of a $C^{\infty}$
compact surface $S$ with a mixing horseshoe $\Lambda$ and let
$f:S\rightarrow\mathbb{R}$ be a differentiable function. For $x\in S$,
following the above characterization of the classical spectra, we define the
Lagrange value of $x$ associated to $f$ and $\varphi$ as being the number
$\ell_{\varphi,f}(x)=\limsup_{n\to\infty}f(\varphi^{n}(x))$ and also the
Markov value of $x$ associated to $f$ and $\varphi$ as the number
$m_{\varphi,f}(x)=\sup_{n\in\mathbb{Z}}f(\varphi^{n}(x))$.

The sets

$L_{\varphi,f}(\Lambda)=\\{\ell_{\varphi,f}(x):x\in\Lambda\\}$

and

$M_{\varphi,f}(\Lambda)=\\{m_{\varphi,f}(x):x\in\Lambda\\}$

are called Lagrange Spectrum of $(\varphi,f,\Lambda)$ and Markov Spectrum of
$(\varphi,f,\Lambda)$.

It turns out that dynamical Markov and Lagrange spectra associated to
hyperbolic dynamics are closely related to the classical Markov and Lagrange
spectra. Several results on the Markov and Lagrange dynamical spectra
associated to horseshoes in dimension 2 which are analogous to previously
known results on the classical spectra were obtained recently: in [18] it is
shown that typical dynamical spectra associated to horseshoes with Hausdorff
dimensions larger than one have non-empty interior (as the classical ones). In
[16] it is shown that typical Markov and Lagrange dynamical spectra associated
to horseshoes have the same minimum, which is an isolated point in both
spectra and is the image by the function of a periodic point of the horseshoe.

In [12], in the context of conservative diffeomorphism it is proven (as a
generalization of the results in [4]) that for typical choices of the dynamic
and of the function, the intersections of the corresponding dynamical Markov
and Lagrange spectra with half-lines $(-\infty,t)$ have the same Hausdorff
dimensions, and this defines a continuous function of $t$ whose image is
$[0,\min\\{1,D\\}]$, where $D$ is the Hausdorff dimension of the horseshoe.

For more information and results on classical and dynamical Markov and
Lagrange spectra, we refer to the books [3] and [9].

In this paper, we use that dynamical Markov and Lagrange spectra associated to
conservative horseshoes in surfaces are natural generalizations of the
classical Markov and Lagrange spectra. In fact, classical Markov and Lagrange
spectra are not compact sets, so they cannot be dynamical spectra associated
to horseshoes. However, in [7] is showed that, for any $N\geq 2$ with $N\neq
3$, the initial segments of the classical spectra until $\sqrt{N^{2}+4N}$
(i.e., $M\cap(-\infty,\sqrt{N^{2}+4N}]$ and $L\cap(-\infty,\sqrt{N^{2}+4N}]$)
coincide with the sets $M(N)$ and $L(N)$, given, in the notation we used in
Perron’s characterization of $M$ and $L$ by

$M(N)=m(\Sigma(N))=\\{m(\theta):\theta\in\Sigma(N)\\}$

and

$L(N)=\ell(\Sigma(N))=\\{\ell(\theta):\theta\in\Sigma(N)\\}$

where $\Sigma(N):=\\{1,2,\dots,N\\}^{\mathbb{Z}}.$

It is proved also that $M(N)$ and $L(N)$ are dynamical Markov and Lagrange
spectra associated to a smooth real function $f$ and to a horseshoe
$\Lambda(N)$ defined by a smooth conservative diffeomorphism $\varphi$, and
also that they are naturally associated to continued fractions with
coefficients bounded by N.

Here we use this relation between classical and dynamical spectra in order to
understand better the fractal geometry (Hausdorff dimension) of the preimage
of half-lines by the function $k$. We can state our main result as:

######  Theorem 1.1.

Define $T:=int(L)=int(M)$. For any $\eta\in\overline{T}$,
$D(\eta)=HD(k^{-1}(\eta))$ i.e.

$HD(k^{-1}((-\infty,\eta)))=HD(k^{-1}((-\infty,\eta]))=HD(k^{-1}(\eta)).$

Even more,

  * •

if $\eta$ is accumulated from the left by points of $T$, then

$D(\eta)>D(t),\ \ \forall t<\eta$

  * •

if $\eta$ is accumulated from the right by points of $T$, then

$D(\eta)<D(t),\ \ \forall t>\eta.$

In particular, $D|_{X}$ is strictly increasing, where $X$ is $T$ or any
interval contained in $\overline{T}$.

##  2\. Preliminares

###  2.1. Continued fractions and regular Cantor sets

The continued fraction expansion of an irrational number $\alpha$ is denoted
by

$\alpha=[a_{0};a_{1},a_{2},\dots]=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\ddots}}},$

so that the Gauss map $G:(0,1)\to[0,1)$,
$G(x)=\dfrac{1}{x}-\left\lfloor\dfrac{1}{x}\right\rfloor$ acts on continued
fraction expansions by

$G([0;a_{1},a_{2},\dots])=[0;a_{2},\dots].$

For an irrational number $\alpha=\alpha_{0}\in(0,1)$, the continued fraction
expansion $\alpha=[0;a_{1},\dots]$ is recursively obtained by setting
$a_{n}=\lfloor\alpha_{n}\rfloor$ and
$\alpha_{n+1}=\frac{1}{\alpha_{n}-a_{n}}=\frac{1}{G^{n}(\alpha_{0})}$. The
rational approximations

$\frac{p_{n}}{q_{n}}:=[0;a_{1},\dots,a_{n}]\in\mathbb{Q}$

of $\alpha$ satisfy the recurrence relations

(2.1) $p_{n}=a_{n}p_{n-1}+p_{n-2}\ \mbox{and}\ q_{n}=a_{n}q_{n-1}+q_{n-2},\ \
n\geq 0$

with the convention that $p_{-2}=q_{-1}=0$ and $p_{-1}=q_{-2}=1$. If
$0<a_{j}\leq N$ for all $j$, it follows that

$\frac{p_{n}}{N+1}\leq p_{n-1}\leq p_{n}\ \mbox{and}\ \frac{q_{n}}{N+1}\leq
q_{n-1}\leq q_{n}\ \ n\geq 1.$

Given a finite sequence $(a_{1},a_{2},\dots,a_{n})\in(\mathbb{N}^{*})^{n}$, we
define

$I(a_{1},a_{2},\dots,a_{n})=\\{x\in[0,1]:x=[0;a_{1},a_{2},\dots,a_{n},\alpha_{n+1}],\
\alpha_{n+1}\geq 1\\}$

then by 2.1, $I(a_{1},a_{2},\dots,a_{n})$ is the interval with extremities
$[0;a_{1},a_{2},\dots,a_{n}]=\frac{p_{n}}{q_{n}}$ and
$[0;a_{1},a_{2},\dots,a_{n}+1]=\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}$ and so

$\lvert
I(a_{1},a_{2},\dots,a_{n})\rvert=\left|\frac{p_{n}}{q_{n}}-\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}\right|=\frac{1}{q_{n}(q_{n}+q_{n-1})},$

because $p_{n}q_{n-1}-p_{n-1}q_{n}=(-1)^{n-1}.$

Also, for $(a_{0},a_{1},\dots,a_{n})\in(\mathbb{N}^{*})^{n+1}$ we set

$I(a_{0};a_{1},\dots,a_{n})=\\{x\in[0,1]:x=[a_{0};a_{1},a_{2},\dots,a_{n},\alpha_{n+1}],\
\alpha_{n+1}\geq 1\\},$

clearly as $I(a_{0};a_{1},\dots,a_{n})=a_{0}+I(a_{1},a_{2},\dots,a_{n})$, we
have

(2.2) $\lvert I(a_{0};a_{1},\dots,a_{n})\rvert=\lvert
I(a_{1},a_{2},\dots,a_{n})\rvert.$

An elementary result for comparing continued fractions is the following lemma

######  Lemma 2.1.

Let $\alpha=[a_{0};a_{1},\dots,a_{n},a_{n+1},\dots]$ and
$\tilde{\alpha}=[a_{0};a_{1},\dots,a_{n},b_{n+1},\dots]$, then:

  * •

$\lvert\alpha-\tilde{\alpha}\rvert<1/2^{n-1},$

  * •

If $a_{n+1}\neq b_{n+1}$, $\alpha>\tilde{\alpha}$ if and only if
$(-1)^{n+1}(a_{n+1}-b_{n+1})>0.$

The next lemma is from [15] (see lemma A.1)

######  Lemma 2.2.

If $a_{0},a_{1},a_{2}\dots,a_{n},a_{n+1},\dots$ and $b_{n+1},b_{n+2},\dots$
are positive integers bounded by $N\in\mathbb{N}$ and $a_{n+1}\neq b_{n+1}$
then

$\displaystyle\lvert[a_{0};a_{1},a_{2}\dots,a_{n},a_{n+1},\dots]-[a_{0};a_{1},a_{2}\dots,a_{n},b_{n+1},\dots]\rvert$
$\displaystyle>$ $\displaystyle c(N)/q^{2}_{n-1}$ $\displaystyle>$
$\displaystyle c(N)\lvert I(a_{1},a_{2},\dots,a_{n})\rvert$

for some positive constant $c(N).$

For the sequel, the following application of lemma 2.1 will also be useful

######  Lemma 2.3.

Given $R,N\in\mathbb{N}$, let
$\beta^{1},\beta^{2},\beta^{3}\in\Sigma(N)^{+}:=\\{1,2,\dots,N\\}^{\mathbb{N}}$
such that $[0;\beta^{1}]<[0;\beta^{2}]<[0;\beta^{3}]$. If for two sequences
$\alpha=(\alpha_{n})_{n\in\mathbb{Z}}\ \mbox{and}\
\tilde{\alpha}=(\tilde{\alpha}_{n})_{n\in\mathbb{Z}}\ \mbox{in}\ \Sigma(N)$ it
is true that
$\alpha_{0},\dots,\alpha_{2R+1}=\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1}$.
Then for all $j\leq 2R+1$ we have

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\alpha_{-2},\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{2}))<\max\\{m(\dots,\alpha_{-2},\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}),$
$\displaystyle
m(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1},\beta^{3})\\}+1/2^{R-1}.$

###### Proof.

It is just an application of lemma 2.1. Indeed, for $j\leq R+1$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{2}))<\lambda_{0}(\sigma^{j}(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}))+1/2^{R-1}$
$\displaystyle\leq\max\\{m(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}),m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1},\beta^{3})\\}+1/2^{R-1}.$

For $R+1<j\leq 2R+1$, if
$[\alpha_{j};\dots,\alpha_{2R+1},\beta^{2}]<[\tilde{\alpha}_{j};\dots,\tilde{\alpha}_{2R+1},\beta^{3}]$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\alpha_{-2},\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{2}))<\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1},\beta^{3}))+1/2^{R}$
$\displaystyle\leq\max\\{m(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}),m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1},\beta^{3})\\}+1/2^{R}.$

And for $R+1<j\leq 2R+1$, if
$[\alpha_{j};\dots,\alpha_{2R+1},\beta^{2}]<[\alpha_{j};\dots,\alpha_{2R+1},\beta^{1}]$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{2}))<\lambda_{0}(\sigma^{j}(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}))$
$\displaystyle\leq\max\\{m(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{2R+1},\beta^{1}),m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{2R+1},\beta^{1})\\}.$

Then we have proved the result. ∎

We end this subsection with one definition

######  Definition 2.4.

A set $K\subset\mathbb{R}$ is called a $C^{1+\alpha}$-regular Cantor set,
$\alpha>0$, if there exists a collection
$\mathcal{P}=\\{I_{1},I_{2},...,I_{r}\\}$ of compacts intervals and a
$C^{1+\alpha}$-expanding map $\psi$, defined in a neighbourhood of
$\displaystyle\cup_{1\leq j\leq r}I_{j}$ such that

  1. (i)

$K\subset\cup_{1\leq j\leq r}I_{j}$ and $\cup_{1\leq j\leq r}\partial
I_{j}\subset K$,

  2. (ii)

For every $1\leq j\leq r$ we have that $\psi(I_{j})$ is the convex hull of a
union of $I_{r}$’s, for $l$ sufficiently large $\psi^{l}(K\cap I_{j})=K$ and

$K=\bigcap_{n\geq 0}\psi^{-n}(\bigcup_{1\leq j\leq r}I_{j}).$

More precisely, we also say that the triple $(K,\mathcal{P},\psi)$ is a
$C^{1+\alpha}$-regular Cantor set.

For example, in our context of sets of continued fractions. Let, as before,
$G$ be the Gauss map and $C_{N}=\\{x=[0;a_{1},a_{2},...]:a_{i}\leq N,\forall
i\geq 1\\}$. Then,

$C_{N}=\bigcap_{n\geq 0}G^{-n}(I_{N}\cup...\cup I_{1}),$

where $I_{j}=[a_{j},b_{j}]$ and $a_{j}=[0;j,\overline{1,N}]$ and
$b_{j}=[0;j,\overline{N,1}].$ That is, $C_{N}$ is a regular Cantor set.

###  2.2. Results on Dynamical Markov and Lagrange spectra

Let $\varphi:S\rightarrow S$ be a diffeomorphism of a $C^{\infty}$ compact
surface $S$ with a mixing horseshoe $\Lambda$ and let
$f:S\rightarrow\mathbb{R}$ be a differentiable function. Fix a Markov
partition $\\{R_{a}\\}_{a\in\mathcal{A}}$ with sufficiently small diameter
consisting of rectangles $R_{a}\sim I_{a}^{s}\times I_{a}^{u}$ delimited by
compact pieces $I_{a}^{s}$, $I_{a}^{u}$, of stable and unstable manifolds of
certain points of $\Lambda$, see [1] theorem 2, page 172. Recall that the
stable and unstable manifolds of $\Lambda$ can be extended to locally
invariant $C^{1+\alpha}$ foliations in a neighborhood of $\Lambda$ for some
$\alpha>0$. Using these foliations it is possible define projections
$\pi^{u}_{a}:R_{a}\rightarrow I^{s}_{a}\times\\{i^{u}_{a}\\}$ and
$\pi^{s}_{a}:R_{a}\rightarrow\\{i^{s}_{a}\\}\times I^{u}_{a}$ of the
rectangles into the connected components $I^{s}_{a}\times\\{i^{u}_{a}\\}$ and
$\\{i^{s}_{a}\\}\times I^{u}_{a}$ of the stable and unstable boundaries of
$R_{a}$, where $i^{u}_{a}\in\partial I^{u}_{a}$ and $i^{s}_{a}\in\partial
I^{s}_{a}$ are fixed arbitrarily. In this way, we have the unstable and stable
Cantor sets

$K^{u}=\bigcup_{a\in\mathcal{A}}\pi^{s}_{a}(\Lambda\cap R_{a})\ \mbox{and}\
K^{s}=\bigcup_{a\in\mathcal{A}}\pi^{u}_{a}(\Lambda\cap R_{a}).$

In fact $K^{u}$ and $K^{s}$ are $C^{1+\alpha}$ dynamically defined, associated
to some expanding maps $\psi_{s}$ and $\psi_{u}$. The stable and unstable
Cantor sets, $K^{s}$ and $K^{u}$, respectively, are closely related to the
fractal geometry of the horseshoe $\Lambda$. For instance, it is well-known
that,

$HD(\Lambda)=HD(K^{s})+HD(K^{u})$

and that in the conservative case

$HD(K^{s})=HD(K^{u}).$

The study of the intersection of the spectra with half-lines is related to the
study of fractal dimensions of the set

$\Lambda_{t}=\bigcap\limits_{n\in\mathbb{Z}}\varphi^{-n}(\\{y\in\Lambda:f(y)\leq
t\\})=\\{x\in\Lambda:m_{\varphi,f}(x)=\sup\limits_{n\in\mathbb{Z}}f(\varphi^{n}(x))\leq
t\\}$

for $t\in\mathbb{R}$. Following this, we also consider the subsets
$\Lambda_{t}$ through its projections on the stable and unstable Cantor sets
of $\Lambda$

$K^{u}_{t}=\bigcup_{a\in\mathcal{A}}\pi^{s}_{a}(\Lambda_{t}\cap R_{a})\
\mbox{and}\ K^{s}_{t}=\bigcup_{a\in\mathcal{A}}\pi^{u}_{a}(\Lambda_{t}\cap
R_{a}).$

In [12] is showed the following result

######  Theorem 2.5.

Let $r\geq 2$ and $\varphi\in\text{Diff}^{2}(S)$ a conservative diffeomorphism
preserving a smooth form $\omega$ and take $\Lambda$ a mixing horseshoe of
$\varphi$. If $f\in C^{r}(S,\mathbb{R})$ satisfies that $\forall\
z\in\Lambda,\ \nabla f(z)\neq 0$, then the functions

$t\mapsto HD(K^{u}_{t})\ \mbox{and}\ t\mapsto HD(K^{s}_{t})$

are equal and continuous. Even more, one has

$HD(\Lambda_{t})=2HD(K^{u}_{t}).$

###  2.3. The horseshoe $\Lambda(N)$

Given an integer $N\geq 2$, write $\tilde{C}_{N}=\\{1,2,...,N\\}+C_{N}$ and
define

$\Lambda(N)=C_{N}\times\tilde{C}_{N}.$

If $x=[0;a_{1},a_{2},...]$ and $y=[a_{0};a_{-1},a_{-2},...]$ then we take
$\varphi:\Lambda(N)\rightarrow\Lambda(N)$ given by

$\displaystyle\varphi(x,y)$ $\displaystyle=$ $\displaystyle(G(x),a_{1}+1/y)$
$\displaystyle=$
$\displaystyle([0;a_{2},a_{3},...],a_{1}+[0;a_{0},a_{-1},...]).$

Also, equip $\Lambda(N)$ with the real map $f(x,y)=x+y$. We note that
$\varphi$ can be extended to a $C^{\infty}$-diffeomorphism on a diffeomorphic
copy of the 2-dimensional sphere $\mathbb{S}^{2}$.

Notice also that $\varphi$ is conjugated to the restriction to $C_{N}\times
C_{N}$ of the map $\psi:(0,1)\times(0,1)\to[0,1)\times(0,1)$ given by

$\psi(x,y)=\left(G(x),\frac{1}{y+\lfloor 1/x\rfloor}\right)$

and following [2] and [20] we know that $\psi$ has an invariant measure
equivalent to the Lebesgue measure. In particular, $\varphi$ also has an
invariant measure equivalent to the Lebesgue measure and then $\varphi$ is
conservative.

Indeed, if $\mathcal{S}=\\{(x,y)\in{\mathbb{R}}^{2}|0<x<1,0<y<1/(1+x)\\}$ and
$T:\mathcal{S}\to\mathcal{S}$ is given by

$T(x,y)=(G(x),x-x^{2}y),$

then $T$ preserves the Lebesgue measure in the plane. If
$h:\mathcal{S}\to[0,1)\times(0,1)$ is given by $h(x,y)=(x,y/(1-xy))$ then $h$
is a conjugation between $T$ and $\psi$ (and thus $\psi$ preserves the smooth
measure $h_{*}$(Leb).

For $\Lambda(N)$ we have the Markov partition $\\{R_{a}\\}_{a\in\mathcal{A}}$
where $\mathcal{A}=\\{1,2,\dots,N\\}$ and $R_{a}$ is such that
$R_{a}\cap\Lambda(N)=C_{N}\times(C_{N}+a)=C_{N}\times C_{N}+(0,a)$. It is
clear then that $\varphi|_{\Lambda_{N}}$ is topologically conjugated to
$\sigma:\Sigma(N)\rightarrow\Sigma(N)$ (via a map
$\Pi:\Lambda(N)\to\Sigma(N)$), and that in sequences, $f$ becomes
$\tilde{f}:\Sigma(N)\rightarrow\mathbb{R}$ given by

$\tilde{f}(\theta)=[0;a_{1}(\theta),a_{2}(\theta),...]+a_{0}(\theta)+[0;a_{-1}(\theta),a_{-2}(\theta),...]=\lambda_{0}(\theta),$

where $\theta=(a_{i}(\theta))_{i\in\mathbb{Z}},$ and so

$L_{\varphi,f}(\Lambda(N))=L(N)\ \ \mbox{and}\ \
M_{\varphi,f}(\Lambda(N))=M(N).$

In this context, let
$\alpha=(a_{s_{1}},a_{s_{1}+1},...,a_{s_{2}})\in\mathcal{A}^{s_{2}-s_{1}+1}$
any word where $s_{1},s_{2}\in\mathbb{Z},\ s_{1}<s_{2}$ and fix $s_{1}\leq
s\leq s_{2}$. Define then

$R(\alpha;s):=\bigcap_{m=s_{1}-s}^{s_{2}-s}\varphi^{-m}(R_{a_{m+s}}).$

Note that if $x\in R(\alpha;s)\cap\Lambda(N)$ then the symbolic representation
of $x$ is in the way $(...a_{s_{1}}...a_{s-1};a_{s},a_{s+1}...a_{s_{2}}...)$
where on the right of the ; is the $0$-th position.

Finally, let us consider $A_{N}=[0;\overline{N,1}]$ and
$B_{N}=[0;\overline{1,N}]$. As

$NA_{N}+A_{N}B_{N}=1\ \mbox{and}\ B_{N}+B_{N}A_{N}=1,$

we have $A_{N}=\dfrac{B_{N}}{N}.$ Thus $B_{N}=\frac{-N+\sqrt{N^{2}+4N}}{2}\
\mbox{,}\ A_{N}=\frac{-N+\sqrt{N^{2}+4N}}{2N}$ and then

$\max f|_{\Lambda(N)}=2B_{N}+N=\sqrt{N^{2}+4N},\ \min
f|_{\Lambda(N)}=2A_{N}+1=\frac{\sqrt{N^{2}+4N}}{N}.$

##  3\. Proof of the result

###  3.1. Connection of subhorseshoes

For the next, it will be useful to give the following definition

######  Definition 3.1.

Given $\Lambda^{1}$ and $\Lambda^{2}$ subhorseshoes of a horseshoe $\Lambda$
and $t\in\mathbb{R}$, we said that $\Lambda^{1}$ _connects_ with $\Lambda^{2}$
or that $\Lambda^{1}$ and $\Lambda^{2}$ _connect_ before $t$ if there exist a
subhorseshoe $\tilde{\Lambda}\subset\Lambda$ and some $q<t$ with
$\Lambda^{1}\cup\Lambda^{2}\subset\tilde{\Lambda}\subset\Lambda_{q}$.

######  Lemma 3.2.

Suppose $\Lambda^{1}$ and $\Lambda^{2}$ are subhorseshoes of $\Lambda$ and for
some $x,y\in\Lambda$ we have $x\in W^{u}(\Lambda^{1})\cap W^{s}(\Lambda^{2})$
and $y\in W^{u}(\Lambda^{2})\cap W^{s}(\Lambda^{1})$. If for some
$t\in\mathbb{R}$, it is true that

$\Lambda^{1}\cup\Lambda^{2}\cup\mathcal{O}(x)\cup\mathcal{O}(y)\subset\Lambda_{t},$

then for every $\epsilon>0$, $\Lambda^{1}$ and $\Lambda^{2}$ connect before
$t+\epsilon$.

###### Proof.

Take a Markov partition $\mathcal{P}$ for $\Lambda$ with diameter small enough
such that $\max f|_{\bigcup\limits_{P\in\mathcal{R}}P}<t+\epsilon,$ where
$\mathcal{R}=\\{P\in\mathcal{P}:P\cap(\Lambda^{1}\cup\Lambda^{2}\cup\mathcal{O}(x)\cup\mathcal{O}(y))\neq\emptyset\\}$
and consider

$\Lambda_{\mathcal{R}}=\bigcap\limits_{n\in\mathbb{Z}}\varphi^{-n}(\bigcup\limits_{P\in\mathcal{R}}P).$

Evidently
$\Lambda^{1}\cup\Lambda^{2}\cup\mathcal{O}(x)\cup\mathcal{O}(y)\subset\Lambda_{\mathcal{R}}\subset\Lambda_{t+\epsilon}$,
then the lemma will be proved if we show that $\Lambda^{1}$ and $\Lambda^{2}$
form part of the same transitive component of $\Lambda_{\mathcal{R}}$.

Let $x_{1}\in\Lambda^{1}$, $x_{2}\in\Lambda^{2}$ and $\rho_{1},\rho_{2}>0$.
Take

$\eta=\frac{1}{2}\min\\{\rho_{1},\rho_{2},\min\\{d(P,Q):P,Q\in\mathcal{R}\
\mbox{and}\ P\neq Q\\}\\}.$

By the shadowing lemma there exist $0<\delta\leq\eta$ such that every
$\delta$-pseudo orbit of $\Lambda$ is $\eta$-shadowed by the orbit of some
element of $\Lambda$.

On the other hand, as $\varphi|_{\Lambda^{1}}$ is transitive and $x\in
W^{u}(\Lambda^{1})$ there exist $y_{1}\in\Lambda^{1}\cap B(x_{1},\delta)$ and
$N_{1},M_{1}\in\mathbb{N}$ such that
$d(\varphi^{M_{1}}(y_{1}),\varphi^{-N_{1}}(x))<\delta$ and analogously as
$\varphi|_{\Lambda^{2}}$ is transitive and $x\in W^{s}(\Lambda^{2})$ there
exist $y_{2}\in\Lambda^{2}$ and $N_{2},M_{2}\in\mathbb{N}$ such that
$d(\varphi^{N_{2}}(x),y_{2})<\delta$ and
$d(x_{2},\varphi^{M_{2}}(y_{2}))<\delta$. Define then the $\delta$-pseudo
orbit:

$\dots,\varphi^{-1}(y_{1});y_{1},\varphi(y_{1}),\dots,\varphi^{M_{1}-1}(y_{1}),\varphi^{-N_{1}}(x),\dots,\varphi^{N_{2}-1}(x),y_{2},\varphi(y_{2}),\dots$

Then there exists $w\in\Lambda$ that $\eta$-shadowed that orbit. Moreover as
the $\delta$-pseudo orbit have all its terms in
$\bigcup\limits_{P\in\mathcal{R}}P$ and
$\eta\leq\frac{1}{2}\min\\{d(P,Q):P,Q\in\mathcal{R}\ \mbox{and}\ P\neq Q\\}$
we have also $\mathcal{O}(w)\subset\bigcup\limits_{P\in\mathcal{R}}P\ ;$ that
is, $w\in\Lambda_{\mathcal{R}}$ and furthermore

$w\in
B(x_{1},\rho_{1})\quad\mbox{and}\quad\varphi^{M_{1}+N_{1}-1+N_{2}+M_{2}}(w)\in
B(x_{2},\rho_{2}).$

The proof that there exists $w\in B(x_{2},\rho_{2})$ and $M\in\mathbb{N}$ such
that $\varphi^{M}(w)\in B(x_{1},\rho_{1})$ is analog. ∎

######  Corollary 3.3.

Suppose $\Lambda^{1}$ and $\Lambda^{2}$ are subhorseshoes of $\Lambda$ with
$\Lambda^{1}\cup\Lambda^{2}\subset\Lambda_{t}$ for some $t\in\mathbb{R}$. If
$\Lambda^{1}\cap\Lambda^{2}\neq\emptyset$, then for every $\epsilon>0$,
$\Lambda^{1}$ and $\Lambda^{2}$ connects before $t+\epsilon$.

###### Proof.

If $\Lambda^{1}\cap\Lambda^{2}\neq\emptyset$, then every
$w\in\Lambda^{1}\cap\Lambda^{2}$ satisfies $w\in W^{u}(\Lambda^{1})\cap
W^{s}(\Lambda^{2})$ and $w\in W^{u}(\Lambda^{2})\cap W^{s}(\Lambda^{1})$ and
then we have the conclusion. ∎

######  Corollary 3.4.

Let $\Lambda^{1}$, $\Lambda^{2}$ and $\Lambda^{3}$ subhorseshoes of $\Lambda$
and $t\in\mathbb{R}$. If $\Lambda^{1}$ connects with $\Lambda^{2}$ before $t$
and $\Lambda^{2}$ connects with $\Lambda^{3}$ before $t$. Then also
$\Lambda^{1}$ connects with $\Lambda^{3}$ before $t$.

###### Proof.

By hypothesis we have two subhorseshoes $\Lambda^{1,2}$ and $\Lambda^{2,3}$
and $q_{1},q_{2}<t$ with

$\Lambda^{1}\cup\Lambda^{2}\subset\Lambda^{1,2}\subset\Lambda_{q_{1}}\
\mbox{and }\
\Lambda^{2}\cup\Lambda^{3}\subset\Lambda^{2,3}\subset\Lambda_{q_{2}}.$

Applying corollary 3.3 to $\Lambda^{1,2}$ and $\Lambda^{2,3}$, with
$\tilde{t}=\max\\{q_{1},q_{2}\\}$ and $\epsilon=t-\tilde{t}$ we have the
result. ∎

###  3.2. Dimension estimates

Fix an integer $m\geq 1$ and consider the horseshoe

$\Lambda:=\Lambda(m+3)=C(m+3)\times\tilde{C}(m+3)$

equipped with the diffeomorphism $\varphi$ and the map $f$ given in the
previous section. Also, consider

$\eta\in(m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}],m+4)\cap\overline{T}$

which is accumulated from the left by points of $T$.

Given $t\in(m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}],\eta)\cap T$ and
$0<\epsilon<\eta-t$, take $\ell(t,\epsilon)\in\mathbb{N}$ sufficiently large
such that for the set

$C(t,\epsilon)=\\{\alpha=(a_{0},a_{1}\cdots,a_{2\ell(t,\epsilon)})\in\\{1,2,\cdots,m+3\\}^{2\ell(t,\epsilon)+1}:R(\alpha;\ell(t,\epsilon))\cap\Lambda_{t+\epsilon/4}\neq\emptyset\\}$

if $\alpha\in C(t,\epsilon)\ \mbox{and}\ z,y\in R(\alpha;\ell(t,\epsilon))\
\mbox{then}\ \lvert f(x)-f(y)\rvert<\epsilon/4.$

Define

$P(t,\epsilon):=\bigcap\limits_{n\in\mathbb{Z}}\varphi^{-n}(\bigcup\limits_{\alpha\in
C(t,\epsilon)}R(\alpha;\ell(t,\epsilon))).$

Note that by construction, $\Lambda_{t+\epsilon/4}\subset
P(t,\epsilon)\subset\Lambda_{t+\epsilon/2}$ and being $P(t,\epsilon)$ a
hyperbolic set of finite type, it admits a decomposition

$P(t,\epsilon)=\bigcup\limits_{x\in\mathcal{X}}\tilde{\Lambda}_{x}$

where $\mathcal{X}$ is a finite index set and for $x\in\mathcal{X}$,
$\tilde{\Lambda}_{i}$ is a subhorseshoe or a transient set i.e a set of the
form $\tau=\\{x\in M:\alpha(x)\subset\tilde{\Lambda}_{i_{1}}\ \mbox{and }\
\omega(x)\subset\tilde{\Lambda}_{i_{2}}\\}$ where $\tilde{\Lambda}_{i_{1}}$
and $\tilde{\Lambda}_{i_{2}}$ with $i_{1},i_{2}\in\mathcal{X}$ are
subhorseshoes.

As for every transient $\tau$ set as before, we have

$HD(\tau)=HD(K^{s}(\tilde{\Lambda}_{i_{1}}))+HD(K^{u}(\tilde{\Lambda}_{i_{2}}))$

and for every subhorseshoe $\tilde{\Lambda}_{i}$, being $\varphi$
conservative, one has

$HD(\tilde{\Lambda}_{i})=HD(K^{s}(\tilde{\Lambda}_{i}))+HD(K^{u}(\tilde{\Lambda}_{i}))=2HD(K^{u}(\tilde{\Lambda}_{i}))$

therefore

(3.1)
$HD(P(t,\epsilon))=\max\limits_{x\in\mathcal{X}}HD(\tilde{\Lambda}_{x})=\max\limits_{\begin{subarray}{c}x\in\mathcal{X}:\
\tilde{\Lambda}_{x}\ is\\\
subhorseshoe\end{subarray}}HD(\tilde{\Lambda}_{x}).$

Now, in [15] was proved for $s\leq\max f|_{\Lambda}$ that

$D(s)=HD(k^{-1}(-\infty,s])=HD(K^{u}_{s})$

and by theorem 2.5, we have

$HD(K^{u}_{s})=\frac{1}{2}HD(\Lambda_{s}).$

Then, for some $x\in\mathcal{X}$, $HD(\tilde{\Lambda}_{x})\geq 1$ because
$\Lambda_{t}\subset P(t,\epsilon)$ and

$t^{*}_{1}=\sup\\{s\in\mathbb{R}:\min\\{1,HD(\Lambda_{s})\\}<1\\}=\sup\\{s\in\mathbb{R}:HD(\Lambda_{s})<1\\}<t.$

We will show that any subhorseshoe contained in $P(t,\epsilon)$ with Hausdorff
dimension greater or equal than $1$ connects with the periodic orbit $\xi$,
given by the kneading sequence $(1)_{i\in\mathbb{Z}}$, before any time bigger
than $t+\epsilon/2$. To do that, take any $\delta>0$ and write

$\tilde{P}(t,\epsilon)=\bigcup\limits_{\begin{subarray}{c}x\in\mathcal{X}:\
\tilde{\Lambda}_{x}\ is\\\
subhorseshoe\end{subarray}}\tilde{\Lambda}_{x}=\bigcup\limits_{i\in\mathcal{I}}\tilde{\Lambda}_{i}\cup\bigcup\limits_{i\in\mathcal{J}}\tilde{\Lambda}_{j}$

where

$\mathcal{I}=\\{i\in\mathcal{X}:\tilde{\Lambda}_{i}\ \mbox{is a subhorseshoe
and it connects with}\ \xi\ \mbox{before}\ t+\epsilon/2+\delta\\}$

and

$\mathcal{J}=\\{j\in\mathcal{X}:\tilde{\Lambda}_{j}\ \mbox{is a subhorseshoe
and it doesn't connect with}\ \xi\ \mbox{before}\ t+\epsilon/2+\delta\\}.$

By proposition 3.2, given $j\in\mathcal{J}$ as
$\tilde{\Lambda}_{j}\cup\xi\subset\Lambda_{t+\epsilon/2}$ we cannot have at
the same time the existence of two points $x\in W^{u}(\tilde{\Lambda}_{j})\cap
W^{s}(\xi)$ and $y\in W^{u}(\xi)\cap W^{s}(\tilde{\Lambda}_{j})$ such that
$\mathcal{O}(x)\cup\mathcal{O}(y)\subset\Lambda_{t+\epsilon/2+\delta/2}$.
Without loss of generality suppose that there is no $x\in
W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)$ with $m_{\varphi,f}(x)\leq
t+\epsilon/2+\delta/2$ (the argument for the other case is similar). We will
show that this condition forces the possible letters that may appear in the
sequences that determine the unstable Cantor set of $\tilde{\Lambda}_{j}$.

Let us begin fixing $R\in\mathbb{N}$ large enough such that
$1/2^{R-1}<\delta/2$ and consider the set
$\mathcal{C}_{2R+1}=\\{I(a_{0};a_{1},\dots,a_{2R+1}):I(a_{0};a_{1},\dots,a_{2R+1})\cap
K^{u}(\tilde{\Lambda}_{j})\neq\emptyset\\}$, clearly $\mathcal{C}_{2R+1}$ is a
covering of $K^{u}(\tilde{\Lambda}_{j})$. We will give a mechanism to
construct coverings $\mathcal{C}_{k}$ with $k\geq 2R+1$ that can be used to
_efficiently_ cover $K^{u}(\tilde{\Lambda}_{j})$ as $k$ goes to infinity.

Indeed, if for some $k\geq 2R+1$, and
$I(a_{0};a_{1},\dots,a_{k})\in\mathcal{C}_{k}$, $(a_{0},a_{1},\dots,a_{k})$
has continuations with forced first letter. That is, for every
$\alpha=(\alpha_{n})_{n\in\mathbb{Z}}\in\Pi(\tilde{\Lambda}_{j})$ with
$\alpha_{0},\alpha_{1},\dots,\alpha_{k}=a_{0},a_{1},\dots,a_{k}$ one has
$\alpha_{k+1}=a_{k+1}$ for some fixed $a_{k+1}$, then we can refine the
original cover $\mathcal{C}_{k}$, by replacing the interval
$I(a_{0};a_{1},\dots,a_{k})$ with the interval
$I(a_{0};a_{1},\dots,a_{k},a_{k+1})$.

On the other hand, if $(a_{0},a_{1},\dots,a_{k})$ has two continuations with
different initial letter, said $\gamma_{k+1}=(a_{k+1},a_{k+2},\dots)$ and
$\beta_{k+1}=(a^{*}_{k+1},a^{*}_{k+2},\dots)$ with $a_{k+1}\neq a^{*}_{k+1}$.
Take $\alpha=(\alpha_{n})_{n\in\mathbb{Z}}\in\Pi(\tilde{\Lambda}_{j})$ and
$\tilde{\alpha}=(\tilde{\alpha}_{n})_{n\in\mathbb{Z}}\in\Pi(\tilde{\Lambda}_{j})$,
such that
$\alpha=(\dots,\alpha_{-2},\alpha_{-1};a_{0},a_{1},\dots,a_{k},\gamma_{k+1})$
and
$\tilde{\alpha}=(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};a_{0},a_{1},\dots,a_{k},\beta_{k+1})$.
If
$a_{k+1}=i$ then, necessarily either $a^{*}_{k+1}=i+1$ or $a^{*}_{k+1}=i-1$
because if for example $a_{k+1}+1<a^{*}_{k+1}$ we can set $s=a_{k+1}+1$ and
therefore by lemma 3.5 as
$[0;\beta_{k+1}]<[0;s,\overline{1}]<[0;\gamma_{k+1}]$, we would have for all
$j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},s,\overline{1}))$
$\displaystyle\leq$
$\displaystyle\max\\{m(\dots,\alpha_{-1};\alpha_{0},\dots,\alpha_{k},\gamma_{k+1}),$
$\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})\\}+1/2^{R-1}$
$\displaystyle<$ $\displaystyle t+\epsilon/2+\delta/2.$

For $j=k+1$,

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},s,\overline{1}))$
$\displaystyle=$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+s+[0;\overline{1}]$
$\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+s+1$
$\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+a^{*}_{k+1}$
$\displaystyle+[0;a^{*}_{k+2},a^{*}_{k+3},\dots]$ $\displaystyle=$
$\displaystyle\lambda_{0}(\sigma^{k+1}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1}))$
$\displaystyle\leq$ $\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})$
$\displaystyle\leq$ $\displaystyle t+\epsilon/2$

and for $j>k+1$, clearly

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},s,\overline{1}))<3<t+\epsilon/2.$

Then taking
$x=\Pi^{-1}((\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},s,\overline{1}))$
one would have

$x\in W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)\ \mbox{and}\
m_{\varphi,f}(x)\leq t+\epsilon/2+\delta/2$

that is a contradiction. The case $a_{k+1}-1>a^{*}_{k+1}$ is quite similar.

Now, suppose $a_{k+1}=i$ and $a^{*}_{k+1}=i+1$. We affirm that $a_{k+2}=1$
because in other case by lemma 3.5, as
$[0;\beta_{k+1}]<[0;i,\overline{1}]<[0;\gamma_{k+1}]$, we would have again for
all $j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,\overline{1}))<t+\epsilon/2+\delta/2.$

For $j>k+1$, one more time

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,\overline{1}))<t+\epsilon/2$

and for $j=k+1$,

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,\overline{1}))$
$\displaystyle=$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+i+[0;\overline{1}]$
$\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+i+1$
$\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+a^{*}_{k+1}$
$\displaystyle+[0;a^{*}_{k+2},a^{*}_{k+3},\dots]$ $\displaystyle=$
$\displaystyle\lambda_{0}(\sigma^{k+1}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1}))$
$\displaystyle\leq$ $\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})$
$\displaystyle\leq$ $\displaystyle t+\epsilon/2.$

Then for
$x=\Pi^{-1}((\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,\overline{1}))$
one would get the contradiction

$x\in W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)\ \mbox{and}\
m_{\varphi,f}(x)\leq t+\epsilon/2+\delta/2.$

Even more, we have $a_{k+3}\in\\{m+1,m+2,m+3\\}$ because if $a_{k+3}=\ell\leq
m$, then $[0;\beta_{k+1}]<[0;i,1,\ell+1,\overline{1}]<[0;\gamma_{k+1}]$ and by
lemma 3.5 we would have for all $j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,1,\ell+1,\overline{1}))<t+\epsilon/2+\delta/2.$

For $j=k+1$,

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,1,\ell+1,\overline{1}))$
$\displaystyle=$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+i+$
$\displaystyle[0;1,\ell+1,\overline{1}]$ $\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+a^{*}_{k+1}$
$\displaystyle+[0;a^{*}_{k+2},a^{*}_{k+3},\dots]$ $\displaystyle=$
$\displaystyle\lambda_{0}(\sigma^{k+1}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1}))$
$\displaystyle\leq$ $\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})$
$\displaystyle\leq$ $\displaystyle t+\epsilon/2$

and for $j>k+1$,

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,1,\ell+1,\overline{1}))<m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}]<t+\epsilon/2$

then taking
$x=\Pi^{-1}((\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,1,\ell+1,\overline{1}))$
one would have

$x\in W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)\ \mbox{and}\
m_{\varphi,f}(x)\leq t+\epsilon/2+\delta/2$

that is again a contradiction.

In a similar way, we must have $a^{*}_{k+2}\in\\{m+1,m+2,m+3\\}$ because if
$a^{*}_{k+2}=\ell\leq m$, then
$[0;\beta_{k+1}]<[0;i+1,\ell+1,\overline{1}]<[0;\gamma_{k+1}]$ and as before
we would have for all $j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i+1,\ell+1,\overline{1}))<t+\epsilon/2+\delta/2,$

for $j=k+1$,

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i+1,\ell+1,\overline{1}))$
$\displaystyle=$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+i+1+$
$\displaystyle[0;\ell+1,\overline{1}]$ $\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+a^{*}_{k+1}$
$\displaystyle+[0;a^{*}_{k+2},a^{*}_{k+3},\dots]$ $\displaystyle=$
$\displaystyle\lambda_{0}(\sigma^{k+1}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1}))$
$\displaystyle\leq$ $\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})$
$\displaystyle\leq$ $\displaystyle t+\epsilon/2$

and for $j>k+1$,

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i+1,\ell+1,\overline{1}))<m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}]<t+\epsilon/2$

that let us get a contradiction again.

In particular, in this case, we can refine the cover $\mathcal{C}_{k}$ by
replacing the interval $I(a_{0};a_{1},\dots,a_{k})$ with the six intervals
$I(a_{0};a_{1},\dots,a_{k},i,1,m+1),I(a_{0};a_{1},\dots,a_{k},i,1,m+2),I(a_{0};a_{1},\dots,a_{k},i,1,m+3),I(a_{0};a_{1},\dots,a_{k},i+1,m+1),I(a_{0};a_{1},\dots,a_{k},i+1,m+2)\
\mbox{and}\ I(a_{0};a_{1},\dots,a_{k},i+1,m+3)$ for one and only one
$i=1,\dots,m+2.$

Observe that, in fact, some of the intervals considered in the last paragraph,
may not be possible. For example, if $\eta=m+3$ then $t+\epsilon/2<m+3$;
therefore the letter $m+3$ cannot appear in the kneading sequence of any point
of $\tilde{\Lambda}_{j}$. But this will not affect our argument. Indeed, we
affirm that this procedure doesn’t increase the 0.49-sum,
$H_{0.49}(\mathcal{C}_{k})=\sum\limits_{I\in\mathcal{C}_{k}}\lvert
I\rvert^{0.49}$ of the cover $\mathcal{C}_{k}$ of
$K^{u}(\tilde{\Lambda}_{j})$. That is, by 2.2 we need to prove that

$\sum\limits_{j=m+1}^{m+3}\lvert
I(a_{1},\dots,a_{k},i,1,j)\rvert^{0.49}+\sum\limits_{j=m+1}^{m+3}\lvert
I(a_{1},\dots,a_{k},i+1,j)\rvert^{0.49}<\lvert
I(a_{1},\dots,a_{k})\rvert^{0.49}$

or

(3.2) $\sum\limits_{j=m+1}^{m+3}\left(\frac{\lvert
I(a_{1},\dots,a_{k},i,1,j)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}+\sum\limits_{j=m+1}^{m+3}\left(\frac{\lvert
I(a_{1},\dots,a_{k},i+1,j)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}<1$

where $i=1,\dots,m+2.$

In this direction, we have the following lemmas

######  Lemma 3.5.

Given $a_{0},a_{1},\dots,a_{n},a,b,c\in\\{1,\dots,m+3\\}$ we have

$\frac{\lvert I(a_{1},\dots,a_{n},a,b)\rvert}{\lvert
I(a_{1},\dots,a_{n})\rvert}=\frac{1+r}{(ab+1+br)(ab+a+1+(b+1)r)}$

and

$\frac{\lvert I(a_{1},\dots,a_{n},a,b,c)\rvert}{\lvert
I(a_{1},\dots,a_{n})\rvert}=\frac{1+r}{(abc+c+a+(bc+1)r)(abc+c+a+ab+1+(bc+b+1)r)}$

where $r\in(0,1).$

###### Proof.

Recall that the length of $I(b_{1},\dots,b_{m})$ is

$\lvert I(b_{1},\dots,b_{m})\rvert=\frac{1}{q_{m}(q_{m}+q_{m-1})},$

where $q_{s}$ is the denominator of $[0;b_{1},\dots,b_{s}]$. And that we also
have the recurrence formula

$q_{s+2}=b_{s+2}q_{s+1}+q_{s}.$

Using this two and three times respectively, we have

$\lvert
I(a_{1},\dots,a_{n},a,b)\rvert=\frac{1}{((ab+1)q_{n}+bq_{n-1})((ab+a+1)q_{n}+(b+1)q_{n-1})}$

and

$\lvert
I(a_{1},\dots,a_{n},a,b,c)\rvert=\frac{1}{((abc+c+a)q_{n}+(bc+1)q_{n-1})((abc+c+a+ab+1)q_{n}+(bc+b+1)q_{n-1})}$

so, we conclude

$\displaystyle\frac{\lvert I(a_{1},\dots,a_{n},a,b)\rvert}{\lvert
I(a_{1},\dots,a_{n})\rvert}$ $\displaystyle=$
$\displaystyle\frac{q_{n}(q_{n}+q_{n-1})}{((ab+1)q_{n}+bq_{n-1})((ab+a+1)q_{n}+(b+1)q_{n-1})}$
$\displaystyle=$ $\displaystyle\frac{1+r}{(ab+1+br)(ab+a+1+(b+1)r)}$

and

$\displaystyle\frac{\lvert I(a_{1},\dots,a_{n},a,b,c)\rvert}{\lvert
I(a_{1},\dots,a_{n})\rvert}$ $\displaystyle=$
$\displaystyle\frac{q_{n}(q_{n}+q_{n-1})}{((abc+c+a)q_{n}+(bc+1)q_{n-1})((abc+c+a+ab+1)q_{n}+(bc+b+1)q_{n-1})}$
$\displaystyle=$
$\displaystyle\frac{1+r}{(abc+c+a+(bc+1)r)(abc+c+a+ab+1+(bc+b+1)r)}$

with $r=\frac{q_{n-1}}{q_{n}}\in(0,1)$. ∎

######  Lemma 3.6.

Fix $x,y,z,w>0$, then

$\frac{d}{dr}\left(\frac{1+r}{(x+yr)(z+wr)}\right)=\frac{(x-y)(z-w)-yw(r+1)^{2}}{(ywr^{2}+(xw+yz)r+xz)^{2}}<\frac{(x-y)(z-w)-yw}{(ywr^{2}+(xw+yz)r+xz)^{2}}$

for $r\geq 0$.

###### Proof.

It’s a straightforward computation. ∎

Using the previous lemmas, that $i\geq 1$, $m\geq 1$, $r\in(0,1)$ and for
$j\in\\{m+1,m+2,m+3\\}$

$(2j+1-(j+1))(2j+3-(j+2))-(j+1)(j+2)=j(j-1)-(j+1)(j+2)<0,$

for the first sum one has

$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{\lvert
I(a_{1},\dots,a_{k},i,1,j)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}$ $\displaystyle=$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{1+r}{(ij+j+i+(j+1)r)(ij+j+2i+1+(j+2)r)}\right)^{0.49}$
$\displaystyle\leq$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{1+r}{(2j+1+(j+1)r)(2j+3+(j+2)r)}\right)^{0.49}$
$\displaystyle<$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{1}{(2j+1)(2j+3)}\right)^{0.49}$
$\displaystyle\leq$ $\displaystyle\left(\frac{1}{5\times
7}\right)^{0.49}+\left(\frac{1}{7\times
9}\right)^{0.49}+\left(\frac{1}{9\times 11}\right)^{0.49}$ $\displaystyle<$
$\displaystyle 0.412$

and for the second sum

$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{\lvert
I(a_{1},\dots,a_{k},i+1,j)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}$ $\displaystyle=$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{1+r}{((i+1)j+1+jr)((i+1)j+i+2+(j+1)r)}\right)^{0.49}$
$\displaystyle\leq$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{1+r}{(2j+1+jr)(2j+3+(j+1)r)}\right)^{0.49}$
$\displaystyle<$
$\displaystyle\sum\limits_{j=m+1}^{m+3}\left(\frac{2}{(2j+1)(2j+3)}\right)^{0.49}$
$\displaystyle\leq$ $\displaystyle\left(\frac{2}{5\times
7}\right)^{0.49}+\left(\frac{2}{7\times
9}\right)^{0.49}+\left(\frac{2}{9\times 11}\right)^{0.49}$ $\displaystyle<$
$\displaystyle 0.579$

that proves 3.2 and so let us conclude that
$HD(K^{u}(\tilde{\Lambda}_{j}))\leq 0.49$. Finally, as we are in the
conservative setting

$HD(\tilde{\Lambda}_{j})=2HD(K^{u}(\tilde{\Lambda}_{j}))<0.99.$

Fix $\delta=\epsilon/2$. By definition, for $i\in\mathcal{I}$,
$\tilde{\Lambda}_{i}$ connects with $\xi$ before $t+\epsilon$, then we can
apply proposition 4 at most $\lvert\mathcal{I}\rvert-1$ times to see that
there exists a subhorseshoe $\tilde{\Lambda}(t,\epsilon)\subset\Lambda$ and
some $q(t,\epsilon)<t+\epsilon$ such that

$\bigcup\limits_{i\in\mathcal{I}}\tilde{\Lambda}_{i}\subset\tilde{\Lambda}(t,\epsilon)\subset\Lambda_{q(t,\epsilon)}.$

Now, remember that for any subhorseshoe $\tilde{\Lambda}\subset\Lambda$ being
locally maximal we have

$W^{s}(\tilde{\Lambda})=\bigcup\limits_{y\in\tilde{\Lambda}}W^{s}(y)\ \
\mbox{and}\ \
W^{u}(\tilde{\Lambda})=\bigcup\limits_{y\in\tilde{\Lambda}}W^{u}(y).$

Then, there exists an $y\in\tilde{\Lambda}$ with
$\lim\limits_{n\to\infty}d(f(\varphi^{n}(x)),f(\varphi^{n}(y)))=0$ for every
$x\in\Lambda$, such that $\omega(x)\subset\tilde{\Lambda}$ , and so
$\ell_{\varphi,f}(x)=\ell_{\varphi,f}(y)$. Using this, we have

$\ell_{\varphi,f}(P(t,\epsilon))=\ell_{\varphi,f}(\tilde{P}(t,\epsilon))=\bigcup\limits_{i\in\mathcal{I}}\ell_{\varphi,f}(\tilde{\Lambda}_{i})\cup\bigcup\limits_{j\in\mathcal{J}}\ell_{\varphi,f}(\tilde{\Lambda}_{j}).$

On the other hand

$HD(\bigcup\limits_{j\in\mathcal{J}}\ell_{\varphi,f}(\tilde{\Lambda}_{j}))=\max\limits_{j\in\mathcal{J}}HD(\ell_{\varphi,f}(\tilde{\Lambda}_{j}))\leq\max\limits_{j\in\mathcal{J}}HD(f(\tilde{\Lambda}_{j}))\leq\max\limits_{j\in\mathcal{J}}HD(\tilde{\Lambda}_{j})<1$

so
$int(\bigcup\limits_{j\in\mathcal{J}}\ell_{\varphi,f}(\tilde{\Lambda}_{j}))=\emptyset.$

Also, as was proved in lemma $5.2$ of [12], for $\tilde{t}\leq\max
f|_{\Lambda}$

$L\cap(-\infty,\tilde{t})=\bigcup\limits_{s<\tilde{t}}\ell_{\varphi,f}(\Lambda_{s}),$

therefore

$\displaystyle t\in int(m_{\varphi,f}(\Lambda_{t+\epsilon/4}))$
$\displaystyle=$ $\displaystyle
int(M\cap(-\infty,t+\epsilon/4))=int(L\cap(-\infty,t+\epsilon/4))$
$\displaystyle=$ $\displaystyle
int(\bigcup\limits_{s<t+\epsilon/4}\ell_{\varphi,f}(\Lambda_{s}))=int(\ell_{\varphi,f}(\Lambda_{t+\epsilon/4}))$
$\displaystyle\subset$ $\displaystyle int(\ell_{\varphi,f}(P(t,\epsilon)))$

and then, we must have

$t<\sup(\bigcup\limits_{i\in\mathcal{I}}\ell_{\varphi,f}(\tilde{\Lambda}_{i}))\leq\sup(\ell_{\varphi,f}(\tilde{\Lambda}(t,\epsilon)))\leq\sup
f(\tilde{\Lambda}(t,\epsilon))=\max f|_{\tilde{\Lambda}(t,\epsilon)}.$

We have then proved the following result

######  Proposition 3.7.

Given $t\in(m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}],\eta)\cap T$ and
$\epsilon<\eta-t$ there exist some $q(t,\epsilon)<t+\epsilon$ and a
subhorseshoe $\tilde{\Lambda}(t,\epsilon)\subset\Lambda_{q(t,\epsilon)}$ with
$HD(\tilde{\Lambda}(t,\epsilon))\geq 1$ such that

  1. (1)

$HD(\Lambda_{t})\leq HD(\tilde{\Lambda}(t,\epsilon))$

  2. (2)

for every subhorseshoe $\tilde{\Lambda}\subset\Lambda_{t}$ with
$HD(\tilde{\Lambda})\geq 0.99$ one has
$\tilde{\Lambda}\subset\tilde{\Lambda}(t,\epsilon)$

  3. (3)

$t<\max f|_{\tilde{\Lambda}(t,\epsilon)}.$

###  3.3. Putting unstable Cantor sets into $k^{-1}(\eta)$

Let
$\eta\in(m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}],m+4)\cap\overline{T}$
accumulated from the left by points of $T$ and $\epsilon>0$ such that
$m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}]<\eta-\epsilon$. Take any
strictly increasing sequence $\\{t_{n}\\}_{n\geq 0}$ of points of $T$ such
that $t_{0}>\eta-\epsilon$ and $\lim\limits_{n\rightarrow\infty}t_{n}=\eta$.
Proposition 3.7 let us find a sequence of subhorsehoes
$\\{\Lambda^{n}\\}_{n\geq
0}=\\{\tilde{\Lambda}(t_{n},(t_{n+1}-t_{n})/2)\\}_{n\geq 0}$ with the
following properties

  1. (1)

$HD(\Lambda_{t_{n}})\leq HD(\Lambda^{n})$

  2. (2)

$\Lambda^{n}\subset\Lambda^{n+1}$

  3. (3)

$t_{n}<\max f|_{\Lambda^{n}}<t_{n+1}.$

Now, we will construct a local homeomorphism
$\theta:K^{u}(\Lambda^{0})\rightarrow k^{-1}(\eta)$ with local Hölder inverse
and exponent arbitrarily close to one.

Given $n\geq 0$, being $\Lambda^{n}$ a mixing horseshoe (because
$\xi\subset\Lambda^{n}$), we can find some $c(n)\in\mathbb{N}$ such that given
two letters $a$ and $b$ in the alphabet $\mathcal{A}(\Lambda^{n})$ of
$\Lambda^{n}$ there exists some finite word of size $c(n)$:
$(a_{1},\dots,a_{c(n)})$ (in the letters of $\mathcal{A}(\Lambda^{n})$) such
that $(a,a_{1},\dots,a_{c(n)},b)$ is admissible; given $a$ and $b$ consider
always a fixed $(a_{1},\dots,a_{c(n)})$ as before. Also, as $\Lambda^{n}$ is a
subhorseshoe of $\Lambda$, it is the invariant set in some rectangles
determined for a set of words of size $2p(n)+1$ for some $p(n)\in\mathbb{N}$.

Now, take $n\geq 1$ and consider the kneading sequence
$\\{x^{n}_{r}\\}_{r\in\mathbb{Z}}$ of some point $x_{n}\in\Lambda^{n}$ such
that $f(x_{n})=\max f|_{\Lambda^{n}}$. Also take $r(n)>p(n+1)+p(n)+p(n-1)$ big
enough such that for any
$\alpha=(a_{0},a_{1}\cdots,a_{2r(n)})\in\\{1,2,\cdots,m+3\\}^{2r(n)+1}$ and
$z,y\in R(\alpha;r(n))$ we have $\lvert f(x)-f(y)\rvert<\min\\{(t_{n+1}-\max
f|_{\Lambda^{n}})/2,(\max f|_{\Lambda^{n}}-t_{n})/2\\}.$ Finally, set
$s(n)=\sum_{k=1}^{n}2r(k)+2c(k)+1$.

Given $a=[a_{0};a_{1},a_{2},\dots]\in K^{u}(\Lambda^{0})$ for $n\geq 1$ set
$a^{(n)}:=(a_{s(n)!+1},\dots,a_{s(n+1)!})$, so one has

$a=[a_{0};a_{1},a_{2},\dots]=[a_{0};a_{1},\dots,a_{s(1)!},a^{(1)},a^{(2)},\dots,a^{(n)},\dots].$

Define then

$\theta(a):=[a_{0};a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},h_{2},a^{(2)},\dots,h_{n},a^{(n)},h_{n+1},\dots]$

where

$h_{n}=(c_{1}^{n},x^{n}_{-r(n)},\dots,x^{n}_{-1},x^{n}_{0},x^{n}_{1},\dots,x^{n}_{r(n)},c_{2}^{n})$

and $c_{1}^{n}$ and $c_{2}^{n}$ are words in the original alphabet
$\mathcal{A}=\\{1,\dots,m+3\\}$ with $\lvert c_{1}^{n}\rvert=\lvert
c_{2}^{n}\rvert=c(n)$ such that
$(a_{0},a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},h_{2},\dots,h_{n},a^{(n)})$
appears in the kneading sequence of some point of $\Lambda^{n}$.

It is easy to see using the construction of $\theta$ that for every $a\in
K^{u}(\Lambda^{0})$, $k(\theta(a))=\eta$, so we have defined the map

$\displaystyle\theta:K^{u}(\Lambda^{0})$ $\displaystyle\rightarrow$
$\displaystyle k^{-1}(\eta)$ $\displaystyle a$ $\displaystyle\rightarrow$
$\displaystyle\theta(a)$

that is clearly continuous and injective.

On the other hand, given any small $\rho>0$ because of the growth of the
factorial map, we have
$\lvert\tilde{a}_{1}-\tilde{a}_{2}\rvert=O(\lvert\theta(\tilde{a}_{1})-\theta(\tilde{a}_{2})\rvert^{1-\rho})$
for any $\tilde{a}_{1},\tilde{a}_{2}\in K^{u}(\tilde{\Lambda}_{i})$ and
$\lvert\tilde{a}_{1}-\tilde{a}_{2}\rvert$ small. Indeed, if $\tilde{a}_{1}$
and $\tilde{a}_{2}$ are such that the letters in their continued fraction
expressions are equal up to the $s$-nth letter and $n\in\mathbb{N}$ is maximal
such that $s(n)!<s$ then because $\lvert h_{k}\rvert=2r(k)+2c(k)+1$;
$\theta(\tilde{a}_{1})$ and $\theta(\tilde{a}_{2})$ coincide exactly in their
first

$s+\sum_{k=1}^{n}2r(k)+2c(k)+1=s+s(n)$

letters.

Now, if $\alpha$ and $\beta$ are finite words of positive integers bounded by
$N\in\mathbb{N}$ and $I_{N}(\alpha)$ is the convex hull of $I(\alpha)\cap
C_{N}$. The so-called bounded distortion property let us conclude that for
some constant $C_{N}>1$

$C_{N}^{-1}\lvert I_{N}(\alpha)\rvert\lvert I_{N}(\beta)\rvert\leq\lvert
I_{N}(\alpha\beta)\rvert\leq C_{N}\lvert I_{N}(\alpha)\rvert\lvert
I_{N}(\beta)\rvert$

also, for some positive constants $\lambda_{1},\lambda_{2}<1$, one has

$C_{N}^{-1}\lambda_{1}^{\lvert\alpha\rvert}\leq\lvert I_{N}(\alpha)\rvert\leq
C_{N}\lambda_{2}^{\lvert\alpha\rvert}$

So, if $s$ is big such that
$s(n)/(s+s(n))<\frac{\rho\log\lambda_{2}}{\log\lambda_{1}-4\log C_{m+3}}$,
using lemma 2.2, we have for some constant $\tilde{C}(m+3)$

$\displaystyle\lvert\theta(\tilde{a}_{1})-\theta(\tilde{a}_{2})\rvert^{1-\rho}$
$\displaystyle\geq$ $\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{1-\rho}$
$\displaystyle\geq$ $\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{1-\rho}$
$\displaystyle=$ $\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert$
$\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$
$\displaystyle\geq$
$\displaystyle\frac{1}{C_{m+3}^{2n}}\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!})\rvert\lvert I_{m+3}(a^{(1)})\rvert\dots\lvert
I_{m+3}(a^{(n-1)})\rvert$ $\displaystyle\lvert
I_{m+3}(a_{s(n)!+1},\dots,a_{s})\rvert\lvert I_{m+3}(h_{1})\rvert\dots\lvert
I_{m+3}(h_{n})\rvert$ $\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$
$\displaystyle\geq$
$\displaystyle\frac{1}{C_{m+3}^{3n}}\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},a_{2},\dots,a_{s})\rvert\lvert I_{m+3}(h_{1})\rvert\dots\lvert
I_{m+3}(h_{n})\rvert$ $\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots
a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$ $\displaystyle\geq$
$\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},a_{2},\dots,a_{s})\rvert e^{-4n\log
C_{m+3}}e^{s(n).\log\lambda_{1}}$ $\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots
a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$ $\displaystyle\geq$
$\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},a_{2},\dots,a_{s})\rvert e^{(\log\lambda_{1}-4\log
C_{m+3})s(n)}$ $\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots
a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$

$\displaystyle\geq$ $\displaystyle\tilde{C}(m+3)^{1-\rho}\lvert
I_{m+3}(a_{1},a_{2},\dots,a_{s})\rvert e^{\rho(s+s(n))\log\lambda_{2}}$
$\displaystyle\lvert I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots
a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$ $\displaystyle\geq$
$\displaystyle\frac{\tilde{C}(m+3)^{1-\rho}}{C_{m+3}^{\rho}}\lvert
I_{m+3}(a_{1},a_{2},\dots,a_{s})\rvert$ $\displaystyle\lvert
I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots,a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{\rho}$
$\displaystyle\lvert I_{m+3}(a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},\dots
a^{(n-1)},h_{n},a_{s(n)!+1},\dots,a_{s})\rvert^{-\rho}$ $\displaystyle\geq$
$\displaystyle\frac{\tilde{C}(m+3)^{1-\rho}}{C_{m+3}^{\rho}}\lvert\tilde{a}_{1}-\tilde{a}_{2}\rvert.$

Therefore the map $\theta^{-1}:\theta(K^{u}(\Lambda^{0}))\rightarrow
K^{u}(\Lambda^{0})$ is locally a Hölder map with exponent $1-\rho$ and then

$\displaystyle
HD(K^{u}(\Lambda^{0}))=HD(\theta^{-1}(\theta(K^{u}(\Lambda^{0}))))$
$\displaystyle\leq$ $\displaystyle 1/(1-\rho)HD(\theta(K^{u}(\Lambda^{0})))$
$\displaystyle\leq$ $\displaystyle 1/(1-\rho)HD(k^{-1}(\eta)).$

Letting $\rho$ go to zero, we obtain

$HD(K^{u}(\Lambda^{0}))\leq HD(k^{-1}(\eta)).$

Now, as we indicated before, for $s\leq\max f|_{\Lambda}$ one has

$HD(k^{-1}(-\infty,s])=\frac{1}{2}HD(\Lambda_{s}),$

therefore

$\displaystyle
HD(k^{-1}(-\infty,\eta-\epsilon])=\frac{1}{2}HD(\Lambda_{\eta-\epsilon})$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}HD(\Lambda_{t_{0}})\leq\frac{1}{2}HD(\Lambda^{0})$
$\displaystyle=$ $\displaystyle HD(K^{u}(\Lambda^{0}))\leq HD(k^{-1}(\eta)).$

Letting $\epsilon$ tend to zero we have

$HD(k^{-1}(-\infty,\eta])\leq HD(k^{-1}(\eta))$

and as the other inequality is clearly true, the first part of the result is
proven for
$\eta\in(m+1+[0;\overline{1}]+[0;1,m+2,\overline{1,m+3}],m+4)\cap\overline{T}$
which is accumulated from the left by points of $T$.

For the second part of the theorem, we need the following lemma whose proof is
essentially the same as the proof of lemma 2.5 of [8].

######  Lemma 3.8.

Given $(K,\mathcal{P},\psi)$ a $C^{\alpha}$-regular Cantor set, if
$\mathcal{P}^{{}^{\prime}}\neq\mathcal{P}$ is a finite sub collection of
$\mathcal{P}$ that is also a Markov partition of $\psi$, then the Cantor set
determined by $\psi$ and $\mathcal{P}^{{}^{\prime}}$

$\tilde{K}=\bigcap\limits_{n\geq
0}\psi^{-n}\left(\bigcup\limits_{I\in\mathcal{P}^{{}^{\prime}}}I\right)$

satisfies $HD(\tilde{K})<HD(K).$

######  Corollary 3.9.

Let $\Lambda$ be a mixing horseshoe associated with a $C^{2}$-diffeomorphism
$\varphi:S\rightarrow S$ of some surface $S$. Then for any proper mixing
subhorsehoe $\tilde{\Lambda}\subset\Lambda$

$HD(\tilde{\Lambda})<HD(\Lambda).$

###### Proof.

Refine the original Markov partition $\mathcal{P}$ of $\Lambda$ in such a way
that some $\mathcal{P}^{{}^{\prime}}\subset\mathcal{P}$,
$\mathcal{P}^{{}^{\prime}}\neq\mathcal{P}$ is a Markov partition for
$\tilde{\Lambda}$. Use the lemma 3.8 with the maps $\psi_{s}$ and $\psi_{u}$
that define the stable and unstable Cantor sets, in order to obtain

$HD(\tilde{\Lambda})=HD(K^{s}(\tilde{\Lambda}))+HD(K^{u}(\tilde{\Lambda}))<HD(K^{s}(\Lambda))+HD(K^{u}(\Lambda))=HD(\Lambda).$

∎

Given any $t<\eta$, take $n\in\mathbb{N}$ big enough such that $t<t_{n}$. Now,
as $\max f|_{\Lambda^{n}}<t_{n+1}$ and $t_{n+1}<\max f|_{\Lambda^{n+1}}$ then
$\Lambda^{n}$ is a proper subhorseshoe of $\Lambda^{n+1}$, therefore

$\displaystyle HD(k^{-1}(-\infty,t])=\frac{1}{2}HD(\Lambda_{t})$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}HD(\Lambda^{n})<\frac{1}{2}HD(\Lambda^{n+1})$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}HD(\Lambda_{t_{n+2}})\leq\frac{1}{2}HD(\Lambda_{\eta})$
$\displaystyle=$ $\displaystyle HD(k^{-1}(-\infty,\eta]).$

Then the map is strictly monotone.

As $m\geq 1$ was arbitrary, we have the result for
$\eta\in(2+[0;\overline{1}]+[0;1,3,\overline{1,4}],\infty)\cap\overline{T}=(3,4209...,\infty)\cap\overline{T}$
which is accumulated from the left by points of $T$.

For $\eta\in(t^{*}_{1},3,4209...]\cap\overline{T}$ accumulated from the left
by points of $T$, consider the horseshoe $\Lambda=\Lambda(2)$ (note that $\max
f|_{\Lambda(2)}=\sqrt{12}>3,4209...$). As before, given
$t\in(t^{*}_{1},\eta)\cap T$, $0<\epsilon<\eta-t$ and $\delta>0$ we consider
the set

$\tilde{P}(t,\epsilon)=\bigcup\limits_{\begin{subarray}{c}x\in\mathcal{X}:\
\tilde{\Lambda}_{x}\ is\\\
subhorseshoe\end{subarray}}\tilde{\Lambda}_{x}=\bigcup\limits_{i\in\mathcal{I}}\tilde{\Lambda}_{i}\cup\bigcup\limits_{i\in\mathcal{J}}\tilde{\Lambda}_{j}$

where for $i\in\mathcal{I}$, $\tilde{\Lambda}_{i}$ connects with $\xi$ before
$t+\epsilon/2+\delta$ and for $j\in\mathcal{J}$, $\tilde{\Lambda}_{j}$ doesn’t
connect with $\xi$ before $t+\epsilon/2+\delta$. One more time, given
$j\in\mathcal{J}$, we will suppose that there is no $x\in
W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)$ with $m_{\varphi,f}(x)\leq
t+\epsilon/2+\delta/2$.

Following the above procedure, given $k\in\mathbb{N}$ large enough we
construct the covers $\mathcal{C}_{k}$ of $K^{u}(\tilde{\Lambda}_{j})$ in such
a way that given $I(a_{0};a_{1},\dots,a_{k})\in\mathcal{C}_{k}$, if
$(a_{0},a_{1},\dots,a_{k})$ has continuations with forced first letter
$a_{k+1}$ we replace the interval $I(a_{0};a_{1},\dots,a_{k})$ with the
interval $I(a_{0};a_{1},\dots,a_{k},a_{k+1})$.

On the other hand, if $(a_{0},a_{1},\dots,a_{k})$ has two continuations with
different initial letter, said $\gamma_{k+1}=(1,a_{k+2},\dots)$ and
$\beta_{k+1}=(2,a^{*}_{k+2},\dots)$. Take
$\alpha=(\alpha_{n})_{n\in\mathbb{Z}}\in\Pi(\tilde{\Lambda}_{j})$ and
$\tilde{\alpha}=(\tilde{\alpha}_{n})_{n\in\mathbb{Z}}\in\Pi(\tilde{\Lambda}_{j})$,
such that
$\alpha=(\dots,\alpha_{-2},\alpha_{-1};a_{0},a_{1},\dots,a_{k},\gamma_{k+1})$
and
$\tilde{\alpha}=(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};a_{0},a_{1},\dots,a_{k},\beta_{k+1})$.
We affirm that $a_{k+2}=1$ because in other case by lemma 3.5, as
$[0;\beta_{k+1}]<[0;\overline{1}]<[0;\gamma_{k+1}]$, we would have for all
$j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\overline{1}))<t+\epsilon/2+\delta/2,$

and for $j\geq k+1$

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\overline{1}))<3<t+\epsilon/2.$

Then for
$x=\Pi^{-1}((\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},i,\overline{1}))$
one would get the contradiction

$x\in W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)\ \mbox{and}\
m_{\varphi,f}(x)\leq t+\epsilon/2+\delta/2.$

In a similar way, we must have $a^{*}_{k+2}=2$ because if $a^{*}_{k+2}=1$,
then $[0;\beta_{k+1}]<[0;2,2,\overline{1}]<[0;\gamma_{k+1}]$ and by lemma 3.5
we would have for all $j\leq k$

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},2,2,\overline{1}))<t+\epsilon/2+\delta/2,$

for $j=k+1$,

$\displaystyle\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},2,2,\overline{1}))$
$\displaystyle=$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+2+$
$\displaystyle[0;2,\overline{1}]$ $\displaystyle<$
$\displaystyle[0;\tilde{\alpha}_{k},\dots,\tilde{\alpha}_{0},\tilde{\alpha}_{-1},\dots]+a^{*}_{k+1}$
$\displaystyle+[0;a^{*}_{k+2},a^{*}_{k+3},\dots]$ $\displaystyle=$
$\displaystyle\lambda_{0}(\sigma^{k+1}(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1}))$
$\displaystyle\leq$ $\displaystyle
m(\dots,\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},\beta_{k+1})$
$\displaystyle\leq$ $\displaystyle t+\epsilon/2$

and for $j>k+1$,

$\lambda_{0}(\sigma^{j}(\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},2,2,\overline{1}))<2+[0;\overline{1}]+[0;2,\overline{2,1}]=3,0406...<t+\epsilon/2$

then taking
$x=\Pi^{-1}((\dots,\tilde{\alpha}_{-2},\tilde{\alpha}_{-1};\tilde{\alpha}_{0},\dots,\tilde{\alpha}_{k},2,2,\overline{1}))$
one would have

$x\in W^{u}(\tilde{\Lambda}_{j})\cap W^{s}(\xi)\ \mbox{and}\
m_{\varphi,f}(x)\leq t+\epsilon/2+\delta/2$

that is again a contradiction.

In particular, in this case, we can refine the cover $\mathcal{C}_{k}$ by
replacing the interval $I(a_{0};a_{1},\dots,a_{k})$ with the intervals
$I(a_{0};a_{1},\dots,a_{k},1,1)$ and $I(a_{0};a_{1},\dots,a_{k},2,2)$.

By lemma 3.5 we have the inequality

$\displaystyle\left(\frac{\lvert I(a_{1},\dots,a_{k},1,1)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}+\left(\frac{\lvert
I(a_{1},\dots,a_{k},2,2)\rvert}{\lvert
I(a_{1},\dots,a_{k})\rvert}\right)^{0.49}=$
$\displaystyle\left(\frac{1+r}{(2+r)(3+2r)}\right)^{0.49}+\left(\frac{1+r}{(5+2r)(7+3r)}\right)^{0.49}<$
$\displaystyle\left(\frac{2}{2\times 3}\right)^{0.49}+\left(\frac{2}{5\times
7}\right)^{0.49}<0.9$

that let us conclude that $HD(\tilde{\Lambda}_{j})<0.99$ again. The rest of
the proof follows the same lines as the previous one.

Finally, if $\eta\in\overline{T}$ is accumulated from the right by points of
$T$, as before, we can consider (depending on the region to which $\eta$
belongs) some horseshoe $\Lambda=\Lambda(\eta)$. Take any strictly decreasing
sequence $\\{t_{n}\\}_{n\geq 1}$ of points of $T$ such that
$\lim\limits_{n\rightarrow\infty}t_{n}=\eta$ and $t_{1}<\max f|_{\Lambda}$,
also $\epsilon>0$ small enough such that $HD(\Lambda_{\eta-\epsilon})>0.99$
and take any $t_{0}\in(\eta-\epsilon,\eta)$. The techniques we developed allow
us construct then, a sequence $\\{\Lambda^{n}\\}_{n\geq 0}$ of subhorseshoes
of $\Lambda$ with the following properties

  1. (1)

$\max f|_{\Lambda^{0}}<\eta$

  2. (2)

$\max f|_{\Lambda^{1}}<\max f|_{\Lambda}$

  3. (3)

$t_{n+1}<\max f|_{\Lambda^{n+1}}<t_{n}$, $\forall n\geq 1$

  4. (4)

$HD(\Lambda_{t_{n}})\leq HD(\Lambda^{n})$, $\forall n\geq 0$

  5. (5)

$\Lambda^{0}\subset\Lambda^{n+1}\subset\Lambda^{n}$, $\forall n\geq 1.$

Therefore, we can define a map

$\displaystyle\theta:K^{u}(\Lambda^{0})$ $\displaystyle\rightarrow$
$\displaystyle k^{-1}(\eta)$ $\displaystyle a$ $\displaystyle\rightarrow$
$\displaystyle\theta(a)$

given by

$\theta(a)=[a_{0};a_{1},\dots,a_{s(1)!},h_{1},a^{(1)},h_{2},a^{(2)},\dots,h_{n},a^{(n)},h_{n+1},\dots]$

where

$a=[a_{0};a_{1},a_{2},\dots]=[a_{0};a_{1},\dots,a_{s(1)!},a^{(1)},a^{(2)},\dots,a^{(n)},\dots]$

and the sequences $\\{s(n)\\}_{n\geq 1}$ and $\\{h_{n}\\}_{n\geq 1}$ are
defined as before and are such that
$(a^{(n)},h_{n+1},a^{(n+1)},h_{n+2},\dots)$ appears in the kneading sequence
of some point of $\Lambda^{n+1}$.

It is easy to see using the construction of $\theta$ that for every $a\in
K^{u}(\Lambda^{0})$, $k(\theta(a))=\eta$ and arguing as before, that $\theta$
is a local homeomorphism with local Hölder inverse and exponent arbitrarily
close to one. That let us show that
$HD(k^{-1}(-\infty,\eta])=HD(k^{-1}(\eta)).$

For the second part, corollary 3.9 let us conclude, one more time that
$HD(\Lambda^{n+1})<HD(\Lambda^{n})$, for $n\geq 1$ and then that
$HD(k^{-1}(-\infty,\eta])<HD(k^{-1}(-\infty,t])$, $\forall t>\eta.$ As we
wanted to see.

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