# Explosion Points and Topology of Julia Sets of Zorich maps

ATHANASIOS TSANTARIS  School of Mathematical Sciences, University of
Nottingham, Nottingham NG7 2RD<EMAIL_ADDRESS>
###### Abstract.

Zorich maps are higher dimensional analogues of the complex exponential map.
For the exponential family $\lambda e^{z}$, $\lambda>0$, it is known that for
small values of $\lambda$ the Julia set is an uncountable collection of
disjoint curves. The same was shown to hold for Zorich maps by Bergweiler and
Nicks.

In this paper we introduce a topological model for the Julia sets of certain
Zorich maps, similar to the so called straight brush of Aarts and Oversteegen.
As a corollary we show that $\infty$ is an explosion point for the set of
endpoints of the Julia sets. Moreover we introduce an object called a hairy
surface which is a compactified version of the Julia set of Zorich maps and we
show that those objects are not uniquely embedded in $\mathbb{R}^{3}$, unlike
the corresponding two dimensional objects which are all ambiently
homeomorphic.

##  1\. Introduction

††2020 Mathematics Subject Classification. Primary 30C65, 37F10; Secondary
54G15, 54F65

One of the most well-studied families of holomorphic dynamical systems is the
exponential family $E_{\lambda}(z)=\lambda e^{z}$, $\lambda>0$. The primary
object of study is the Julia set $\mathcal{J}(E_{\lambda})$, the set of points
where the iterates $E_{\lambda}^{n}(z)$ do not form a normal family. It is
well known (see the survey [12]) that when $0<\lambda<1/e$ the Julia set of
$E_{\lambda}$ is a "Cantor Bouquet". This roughly means that the Julia set
consists of uncountably many disjoint curves each connecting a point to
infinity. There are a number of striking phenomena concerning the Julia sets
of maps in the exponential family.

Recently there has been a systematic attempt to generalize results from the
theory of iteration of holomorphic maps to the higher dimensional setting of
quasiregular mappings. In [6, 7] Bergweiler and Nicks defined a Julia set for
quasiregular maps in $\mathbb{R}^{n}$ that are analogous to rational or
transcendental maps in $\mathbb{C}$ and proved that this Julia set shares many
of the properties of the classical Julia set.

In this higher dimensional setting there is a map, first constructed by Zorich
in [29], called the Zorich map which can be thought of as the quasiregular
version of the exponential map on the plane. In this paper we will restrict
ourselves in $\mathbb{R}^{3}$ for simplicity. We will denote this map by
$\mathcal{Z}$ and consider the family
$\mathcal{Z}_{\lambda}=\lambda\mathcal{Z},$ $\lambda>0$. In [5] (see also [7,
Section 7]) Bergweiler showed that the Julia set of $\mathcal{Z}_{\lambda}$,
for small values of $\lambda$, has a similar structure with
$\mathcal{J}(E_{\lambda})$ for $\lambda<1/e$. In other words the Julia set
consists of uncountably many disjoint curves each connecting a point to
infinity. It is also worth mentioning here that the dynamics of Zorich maps
have been studied for large values of the parameter $\lambda$. In [27] it was
proven that for a slightly modified version of the Zorich maps and for large
values of $\lambda$ the Julia set is $\mathbb{R}^{3}$. This can be seen as the
analogous theorem to that of Misiurewicz for the exponential family in
$\mathbb{C}$, see [20].

Concerning now the case of small values of the parameter $\lambda$, Bergweiler
also proved that Zorich maps exhibit a dimension paradox, also known as
Karpinska’s paradox (see [15, 16]), which appears in the exponential family as
well. This paradox concerns the endpoints of the curves that make up the Julia
set. The striking fact is that while the Hausdorff dimension of the endpoints
is 3 the curves without the endpoints have Hausdorff dimension 1.

Another amazing fact, proven by Mayer in [19], concerning the endpoints of
those curves is that the point at $\infty$ is what is called a dispersion
point for the set of endpoints. In other words if we denote the set of
endpoints by $\mathcal{E}\left(\mathcal{J}(E_{\lambda})\right)$ then this set
is totally disconnected while
$\mathcal{E}\left(\mathcal{J}(E_{\lambda})\right)\cup\\{\infty\\}$ is
connected. That this amazing fact is even possible for a planar set is usually
exhibited through a set known as Cantor’s leaky tent or Knaster-Kuratowski fan
(see for example [26]). In fact something even stronger is true here, the work
of Aarts and Oversteegen in [1] shows that the point at $\infty$ is an
explosion point. This means that the set
$\mathcal{E}\left(\mathcal{J}(E_{\lambda})\right)$ is in fact totally
separated, meaning that for any two of its points, $x,y$ there is a clopen
subset $U\subset\mathcal{E}\left(\mathcal{J}(E_{\lambda})\right)$ such that
$x\in U$ but $y\not\in U$. Note that this property implies that
$\mathcal{E}\left(\mathcal{J}(E_{\lambda})\right)$ is totally disconnected and
thus any explosion point is also a dispersion point. We also warn the reader
that dispersion points have been sometimes called explosion points in the
literature.

Moreover, it is worth mentioning that explosion and dispersion points for
exponential maps have been studied for the subset of escaping points of all
endpoints (endpoints that escape to $\infty$ under iteration) in [2, 13].
Also, in [11] the presence of Cantor bouquets is proved for generalized
exponential maps on the complex plane that are more general than Zorich maps
(not necessarily quasiregular) and in fact those bouquets also have $\infty$
as an explosion point for their set of endpoints.

In the higher dimensional setting of Zorich maps now, Bergweiler in [5]
pointed out that $\infty$ might also be an explosion point for the set of
endpoints of the hairs of $\mathcal{J}(\mathcal{Z}_{\lambda})$. One of the
goals of this paper is to prove that this indeed holds.

Before we state our theorems let us define the Zorich map that we will work
with. Although the construction can be made in arbitrary dimensions we confine
ourselves in $\mathbb{R}^{3}$ for simplicity. First consider an $L$ bi-
Lipschitz, sense-preserving map $h$ that maps the square

$Q:=\Big{\\{}(x_{1},x_{2})\in\mathbb{R}^{2}:|x_{1}|\leq 1,|x_{2}|\leq
1\Big{\\}}$

to the upper hemisphere

$\\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1,x_{3}\geq
0\\}.$

Then define $\mathcal{Z}:Q\times\mathbb{R}\to\mathbb{R}^{3}$ as

$\mathcal{Z}(x_{1},x_{2},x_{3})=e^{x_{3}}h(x_{1},x_{2}).$

The map $\mathcal{Z}$ maps the square beam $Q\times\mathbb{R}$ to the upper
half-space. By repeatedly reflecting now, across the sides of the square beam
and the $x_{1}x_{2}$ plane, we get a map
$\mathcal{Z}:\mathbb{R}^{3}\to\mathbb{R}^{3}$. Note that this map is doubly
periodic meaning that
$\mathcal{Z}(x_{1}+4,x_{2},x_{3})=\mathcal{Z}(x_{1},x_{2}+4,x_{3})=\mathcal{Z}(x_{1},x_{2},x_{3})$
and also $\mathcal{Z}\circ R=\mathcal{Z}$, where $R$ is a half-turn around the
lines $x_{1}=2n+1,x_{2}=2m+1$, $n,m\in\mathbb{Z}$. Moreover, this map is not
locally injective everywhere. The lines $x_{1}=2n+1,x_{2}=2m+1$,
$n,m\in\mathbb{Z}$ belong to the branch set, namely the set

$\mathcal{B}_{\mathcal{Z}}:=\\{x\in\mathbb{R}^{3}:\mathcal{Z}\hskip
5.69054pt\text{is not locally homeomorphic at}\hskip 5.69054ptx\\}.$

Also it can be shown that this map has an essential singularity at infinity,
just like the exponential map on the plane, and it is quasiregular, see [14].
We call such quasiregular maps of transcendental type.

We say that a subset $H$ of $\mathbb{C}$ (or $\mathbb{R}^{n}$) is a hair if
there exists a homeomorphism $\gamma~{}:~{}[0,\infty)\to H$ such that
$\gamma(t)\to\infty$ as $t\to\infty$. We call $\gamma(0)$ the endpoint of the
hair $H$.

Consider now the family $\lambda\mathcal{Z}$, $\lambda>0$.

######  Theorem 1 (Bergweiler and Nicks, [5, 7]).

For small enough values of $\lambda>0$ the Julia set
$\mathcal{J}(\mathcal{Z}_{\lambda})$ of a map in the Zorich family consists of
uncountably many disjoint hairs. Each of the hairs tends to $\infty$ by having
their third coordinate go to $\infty$ while the other two coordinates remain
bounded. Moreover, for each point
$x\not\in\mathcal{J}(\mathcal{Z}_{\lambda})$, $\mathcal{Z}^{n}_{\lambda}(x)$
converges to a fixed point.

######  Remark.

In [5, 7] Theorem 1 is shown to be true for small enough values of $\lambda$
without stating an explicit estimate for suitable values of $\lambda$. In
section 2 we make this more precise by proving that Theorem 1 holds for
$0<\lambda<e^{-\left(\log L+L\right)}$.

We now state our result on explosion points.

######  Theorem 2.

Let $0<\lambda<e^{-\left(\log L+L\right)}$. Then $\infty$ is an explosion
point for the set of endpoints of hairs in the Julia set,
$\mathcal{E}(\mathcal{J}(\mathcal{Z}_{\lambda}))$.

In the process of proving the above theorem we will need to introduce a
topological model for $\mathcal{J}(\mathcal{Z}_{\lambda})$. That model is a
generalization, in three dimensions, of a model that Aarts and Oversteegen
introduced in [1] in order to study the topology of Julia sets. They called
their model a straight brush and proved that the Julia set of the exponential
map $E_{\lambda}$, $0<\lambda<1/e$ is homeomorphic to a straight brush. We
call our new three dimensional model a 3-d straight brush (see section 2 for
details) and we will prove the following.

######  Theorem 3.

Let $0<\lambda<e^{-\left(\log L+L\right)}$. Then there is a 3-d straight brush
$B$ and a homeomorphism of $B$ onto the Julia set
$\mathcal{J}(\mathcal{Z}_{\lambda})$ of the Zorich map. Moreover, this
homeomorphism extends to a homeomorphism between $B\cup\\{\infty\\}$ and
$\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$.

Another object that Aarts and Oversteegen introduced in the same paper was the
straight one-sided hairy arc which is compact object living in $[0,1]^{2}$. It
turns out that if we suitably embed a straight brush in the square $[0,1]^{2}$
and then compactify we get such an object. It also turns out that any two such
objects are ambiently homeomorphic in the complex plane so that there is
homeomorphism between them that also extends to the entire complex plane.
Moreover, if we compactify $\mathcal{J}(E_{\lambda})$, $0<\lambda<1/e$ in a
certain way then we get a space which is homeomorphic to a straight one-sided
hairy arc. Aarts and Oversteegen call the homeomorphic images of straight one-
sided hairy arcs just hairy arcs. Moreover, they show that if we embed a hairy
arc to the plane in a way that it has the extra property of one-sidedness then
it is ambiently homeomorphic, this time the ambient space being the Riemann
sphere, with a straight one-sided hairy arc. As a result the compactified
version of $\mathcal{J}(E_{\lambda})$ is a hairy arc which when suitably
embedded in the Riemann sphere is ambiently homeomorphic to a straight one-
sided hairy arc.

Such considerations make sense in higher dimensions too. Thus in section 5 we
define a straight one-sided hairy square and a hairy surface which are the
analogous objects to straight one-sided hairy arcs and hairy arcs respectively
(see section 5 for details).

Similarly with the case of exponential maps on the plane we can show

######  Theorem 4.

Let $0<\lambda<e^{-\left(\log L+L\right)}$. Then there is a compactification
$\widetilde{\mathcal{J}(\mathcal{Z}_{\lambda})}$ of
$\mathcal{J}(\mathcal{Z}_{\lambda})$ which is a hairy surface.

Two other natural questions to ask in this higher dimensional setting would be
whether or not any two straight one-sided hairy squares are ambiently
homeomorphic (the ambient space being $\mathbb{R}^{3}$) and whether all one-
sided hairy surfaces are ambiently homeomorphic to a straight-one sided hairy
square. Both of these are true in the two dimensional setting as we have
already mentioned.

Although we will not give an answer to the first question here, we can show
that in higher dimensions not all one-sided hairy surfaces are ambiently
homeomorphic to a straight one-sided hairy square as the next theorem shows.

######  Theorem 5.

There exists a straight one-sided hairy square $S$ and a homemorphism

$H:S\to H(S)\subset\mathbb{R}^{3}$

such that $H(S)$ is one sided and $H(S)$ and $S$ are not ambiently
homeomorphic, that is there is no homeomorphism of $\mathbb{R}^{3}$ that maps
$S$ to $H(S)$.

The structure of the rest of paper is as follows. In section 2 we give some
definitions, some preliminary results and the general idea for the proof of
Theorem 2. In section 3 we construct the 3-d straight brush that corresponds
to the Julia set and we prove Theorem 3. In section 4 we prove that
$\mathcal{J}(\mathcal{Z}_{\nu})\cup\\{\infty\\}$ is a Lelek fan (see section 2
for the definition). In Section 5 we define straight one sided hairy squares
and we prove Theorem 4. Finally, in section 6 we discuss the proof of Theorem
5.

######  Acknowledgements.

_I would like to thank my supervisor, Daniel Nicks, for all his help and
encouragement while writing this paper._

##  2\. Main Idea and preliminaries for the proof of Theorem 2

Let us briefly mention what quasiregular maps are. For the general theory of
quasiregular maps we refer to [24] and [28]. Also, see [4] for a survey on
quasiregular dynamics.

If $d\geq 2$ and $G\subset\mathbb{R}^{d}$ is a domain, then for $1\leq
p<\infty$ the Sobolev space $W^{1}_{p,loc}(G)$ consists of functions
$f=(f_{1},f_{2},\cdots,f_{d}):G\to\mathbb{R}^{d}$ for which the first order
weak partial derivatives $\partial_{i}f_{j}$ exist and are locally in $L^{p}$.
A continuous map $f\in W^{1}_{d,loc}(G)$ is called quasiregular if there
exists a constant $K_{O}\geq 1$ such that

(1) $\left|Df(x)\right|^{d}\leq K_{O}J_{f}(x)\hskip 5.69054pta.e.,$

where $Df(x)$ denotes the total derivative at point $x$,

$|Df(x)|=\sup_{|h|=1}|Df(x)(h)|$

denotes the operator norm of the derivative, and $J_{f}(x)$ denotes the
Jacobian determinant. Also let

$\ell(Df(x))=\inf_{|h|=1}|Df(x)(h)|.$

The condition that (1) is satisfied for some $K_{O}\geq 1$ implies that

$K_{I}\ell(Df(x))^{d}\geq J_{f}(x),\hskip 5.69054pta.e.,$

for some $K_{I}\geq 1$. The smallest constants $K_{O}$ and $K_{I}$ for which
those two conditions hold are called the outer dilatation and inner dilatation
respectively. We call the maximum of those two numbers the dilatation of $f$
and we denote it by $K(f)$.

In [6] Bergweiler developed a Fatou-Julia theory for quasiregular self-maps of
$\overline{\mathbb{R}^{d}}$ which can be thought of as analogues of rational
maps, while in [7] Bergweiler and Nicks did the same but for transcendental
type quasiregular maps. To define a Julia set for such maps we need the notion
of conformal capacity for which we refer to [24, Chapter II, Section 10].
Following those two papers we define the Julia set of
$f:\mathbb{R}^{d}\to\mathbb{R}^{d}$, denoted ${\mathcal{J}}(f)$, to be the set
of all those $x\in\mathbb{R}^{d}$ such that

$\operatorname{cap}\left(\mathbb{R}^{d}\setminus\bigcup_{k=1}^{\infty}f^{k}(U)\right)=0$

for every neighbourhood $U$ of $x$, where $\operatorname{cap}$ denotes the
conformal capacity. Following [23] we call the complement of
${\mathcal{J}}(f)$ the quasi-Fatou set, and we denote it by $QF(f)$.

It turns out that the Julia set we defined enjoys many of the properties that
the classical Julia set for holomorphic maps has. In particular, a property
that we will use in this paper is that the Julia set is a completely invariant
set, meaning that $x\in\mathcal{J}(f)$ if and only if $f(x)\in\mathcal{J}(f)$.
For more details and motivation behind the definition of the Julia set we
refer to [4, 6, 7].

As we already have noted, the Zorich map is a quasiregular map of
transcendental type and thus the above definitions make sense for this map.

The main idea of the proof of Theorem 2 is to first show that the Julia set
$\mathcal{J}(\mathcal{Z}_{\lambda})$ is a so called Lelek fan. Let us first
define Lelek fans before we explain why we need this.

######  Definition (Fans).

Let $X$ be a continuum (compact and connected metric space). Then $X$ is a fan
with top $x_{0}$ when the following conditions are satisfied.

  1. (i)

$X$ is hereditarily unicoherent, meaning that $K\cap L$ is connected for every
pair of subcontinua $K$, $L$.

  2. (ii)

$X$ is arcwise connected. This together with (i) implies that $X$ is uniquely
arcwise connected.

  3. (iii)

$x_{0}$ is the only point that is the common endpoint of at least three
different arcs that are otherwise disjoint.

Let $X$ be a fan and $x$, $y\in X$. With $[x,y]$ we will denote the unique arc
connecting $x$ and $y$. Also if a point $x\in X$ is an endpoint of every arc
in $X$ containing it then we call this point an endpoint. We will use the same
symbol as in the introduction to denote the set of endpoints $\mathcal{E}(X)$
of a fan.

######  Definition (Lelek fans).

A fan with top $x_{0}$ is called a Lelek fan if it has the following two
properties.

  1. (i)

Smoothness: For any sequence $y_{n}\in X$ converging to $y\in X$ the arcs
$[x_{0},y_{n}]$ converge to $[x_{0},y]$ in the Hausdorff metric.

  2. (ii)

Density of endpoints: The endpoints of $X$ are dense in $X$.

With this terminology we have the following.

######  Theorem 6.

Let $0<\lambda<e^{-\left(\log L+L\right)}$ be such that Theorem 1 holds. Then
$\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$ is a Lelek fan with top
at $\infty$.

The reason why this theorem is important for what we want to prove is that
Lelek in [18] gave an example of a Lelek fan with top $\\{x_{0}\\}$ and showed
that $x_{0}$ is an explosion point for the set of endpoints $\mathcal{E}(X)$.
Much later Charatonik and independently Bula and Oversteegen proved the
following.

######  Theorem 7 (Charatonik, Bula-Oversteegen [8, 10]).

Any two Lelek fans are homeomorphic.

The above theorem combined with Theorem 6 allows us to prove Theorem 2.

###### Proof of Theorem 2.

Since, by Theorem 6, $\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$ is a
Lelek fan it follows by Theorem 7 that it is homeomorphic to the one that
Lelek constructed. Hence, because endpoints get mapped to endpoints and the
top gets mapped to the top we get that
$\mathcal{E}(\mathcal{J}(\mathcal{Z}_{\lambda}))\cup\\{\infty\\}$ is connected
while $\mathcal{E}(\mathcal{J}(\mathcal{Z}_{\lambda}))$ is totally separated.
∎

So the only thing left to prove now is Theorem 6. For that we will need to
construct a topological model for the Julia set for which it will be easier to
show that it is a Lelek fan. For the exponential map a topological model for
the Julia set is given by the so called straight brush which was first
introduced by Aarts and Oversteegen in [1]. Following them, we define a three
dimensional version of a straight brush.

######  Definition (3-d Straight Brush).

A 3-d Straight Brush $B$ is a subset of

$\\{(y,a_{1},a_{2})\in\mathbb{R}^{3}:y\geq
0,(a_{1},a_{2})\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}\\}$

with the following properties:

  1. (i)

Hairiness: For every $(a_{1},a_{2})\in\mathbb{R}^{2}$, there is a
$t_{(a_{1},a_{2})}\in[0,\infty]$ such that $(t,a_{1},a_{2})\in B$ if and only
if $t\geq t_{(a_{1},a_{2})}$

  2. (ii)

Density: The set of $(a_{1},a_{2})$ with $t_{(a_{1},a_{2})}<\infty$ is dense
in $\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$. Also, for any such
$(a_{1},a_{2})$ there exist sequences $(a_{1},a_{n,2})$, $(a_{1},b_{n,2})$,
$(c_{n,1},a_{2})$, $(d_{n,1},a_{2})$, such that $a_{n,2}\uparrow a_{2}$,
$b_{n,2}\downarrow a_{2}$, $c_{n,1}\uparrow a_{1}$, $d_{n,1}\downarrow a_{1}$.
Moreover it is true that $t_{(a_{1},a_{n,2})}\to t_{(a_{1},a_{2})}$ and
similarly for the other sequences.

  3. (iii)

Compact Sections: $B$ is a closed subset of $\mathbb{R}^{3}$.

In order to prove Theorem 6 we will need to prove Theorem 3 first.

Here let us also introduce some notation that we will need later. For
$(r_{1},r_{2})\in\mathbb{Z}^{2}$ we set

$P(r_{1},r_{2}):=\\{(x_{1},x_{2})\in\mathbb{R}^{2}:|x_{1}-2r_{1}|<1,|x_{2}-2r_{2}|<1\\}.$

For any $c\in\mathbb{R}$ we also define the half space

$H_{>c}:=\\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x_{3}>c\\}.$

Note now that $\mathcal{Z}_{\lambda}$ maps $P(r_{1},r_{2})\times\mathbb{R}$
bijectively onto $H_{>0}$ or $H_{<0}$ depending on whether $r_{1}+r_{2}$ is
even or odd. Also we mention here that from Bergweiler’s and Nicks’ work in
[5, 7] it follows that for the values of $\lambda$ they consider

(2) $\overline{P(r_{1},r_{2})}\hskip 2.84526pt\text{belongs to the quasi-Fatou
set when}\hskip 2.84526ptr_{1}+r_{2}=\text{odd}\hskip 2.84526pt.$

We will also need the notion of the itinerary of a point
$x\in\mathcal{J}(\mathcal{Z}_{\lambda})$. To each such point we can associate
a sequence

$\Delta(x)=n_{0}n_{1}n_{2}\dots,$

where $n_{k}=(n_{k,1},n_{k,2})\in\mathbb{Z}\times\mathbb{Z}$ and
$n_{k,1}+n_{k,2}=\text{even}$, in such a way that

$\mathcal{Z}^{k}_{\lambda}(x)\in P(n_{k})\times\mathbb{R},$

for all $k\in\mathbb{N}$. That sequence we will call the itinerary of $x$.

We also set $p:\mathbb{R}^{3}\to\mathbb{R}^{2}$,
$p_{3}\mathbb{R}^{3}\to\mathbb{R}$ to be the projection maps defined by

$p(x_{1},x_{2},x_{3})=(x_{1},x_{2})\hskip 5.69054pt\text{and}\hskip
5.69054ptp_{3}(x_{1},x_{2},x_{3})=x_{3}.$

Let us note here again that in [5] it is shown that Theorem 1 holds for all
sufficiently small values of the parameter $\lambda$ without an explicit
estimate for those values. In what follows we make this more precise.

######  Lemma 1.

(3) $\frac{\lambda
e^{x_{3}}}{L}\leq\ell\left(D\mathcal{Z}_{\lambda}(x_{1},x_{2},x_{3})\right)\leq\left|D\mathcal{Z}_{\lambda}(x_{1},x_{2},x_{3})\right|\leq\lambda
Le^{x_{3}}\hskip 5.69054pt\text{a.e.}$

###### Proof.

We assume that $(x_{1},x_{2})\in Q$ since the other case can be handled
similarly. We note that

(4)
$D\mathcal{Z}_{\lambda}(x)=e^{x_{3}}D\mathcal{Z}_{\lambda}(x_{1},x_{2},0).$

Then $D\mathcal{Z}_{\lambda}(x_{1},x_{2},0)$ is the linear map induced by the
matrix

$\lambda\begin{pmatrix}\frac{\partial h_{1}}{\partial
x_{1}}(p(x))&\frac{\partial h_{1}}{\partial x_{2}}(p(x))&h_{1}(p(x))\\\ \\\
\frac{\partial h_{2}}{\partial x_{1}}(p(x))&\frac{\partial h_{2}}{\partial
x_{2}}(p(x))&h_{2}(p(x))\\\ \\\ \frac{\partial h_{3}}{\partial
x_{1}}(p(x))&\frac{\partial h_{3}}{\partial
x_{2}}(p(x))&h_{3}(p(x))\end{pmatrix},$

where $h=(h_{1},h_{2},h_{3})$ is the bi-Lipschitz map we used in the
construction of the Zorich map.

We now set

$A=\begin{pmatrix}\frac{\partial h_{1}}{\partial x_{1}}\left(p(x)\right)\\\
\\\ \frac{\partial h_{2}}{\partial x_{1}}\left(p(x)\right)\\\ \\\
\frac{\partial h_{3}}{\partial x_{1}}\left(p(x)\right)\end{pmatrix},\hskip
5.69054ptB=\begin{pmatrix}\frac{\partial h_{1}}{\partial
x_{2}}\left(p(x)\right)\\\ \\\ \frac{\partial h_{2}}{\partial
x_{2}}\left(p(x)\right)\\\ \\\ \frac{\partial h_{3}}{\partial
x_{2}}\left(p(x)\right)\end{pmatrix},\hskip
5.69054ptC=\begin{pmatrix}h_{1}\left(p(x)\right)\\\ \\\
h_{2}\left(p(x)\right)\\\ \\\ h_{3}\left(p(x)\right)\end{pmatrix}$

and $v=(v_{1},v_{2},v_{3})\in\mathbb{R}^{3}$. Then by using (4)

$\displaystyle\ell\left(D\mathcal{Z}_{\lambda}(x)\right)^{2}=\inf_{|v|=1}|D\mathcal{Z}_{\lambda}(x)(v)|^{2}=e^{2x_{3}}\lambda^{2}\inf_{|v|=1}|v_{1}A+v_{2}B+v_{3}C|^{2}.$

Notice now that $C$ is orthogonal to $A$ and $B$ and thus the above equation
becomes

$\displaystyle\ell\left(D\mathcal{Z}_{\lambda}(x)\right)^{2}=$ $\displaystyle
e^{2x_{3}}\lambda^{2}\inf_{|v|=1}\left(|v_{1}A+v_{2}B|^{2}+|v_{3}C|^{2}\right)$
$\displaystyle=$ $\displaystyle
e^{2x_{3}}\lambda^{2}\inf_{|v|=1}\left(\left|Dh(p(x))(v_{1},v_{2})\right|^{2}+|v_{3}|^{2}|C|^{2}\right)$
$\displaystyle\geq$ $\displaystyle
e^{2x_{3}}\lambda^{2}\inf_{|v|=1}\left(\frac{1}{L^{2}}|(v_{1},v_{2})|^{2}+|v_{3}|^{2}\right)$
$\displaystyle\geq$ $\displaystyle e^{2x_{3}}\frac{\lambda^{2}}{L^{2}}\hskip
5.69054pt\text{a.e.}$

Similarly we can show that
$\left|D\mathcal{Z}_{\lambda}(x_{1},x_{2},x_{3})\right|^{2}\leq\lambda^{2}L^{2}e^{2x_{3}}$
so that (3) readily follows. ∎

######  Lemma 2.

For all $0<\lambda<e^{-\left(\log L+L\right)}$ Theorem 1 is true.

###### Proof.

First we note that in [5, 7] a different parametrization for the Zorich family
is used, namely $\mathscr{Z}_{\kappa}(x)=\mathcal{Z}(x)+(0,0,\kappa$),
$\kappa\in\mathbb{R}$. Moreover, Theorem 1 is shown to hold for all

$\kappa\leq M_{2}-e^{M_{1}},$

where $M_{1}:=M_{1}(\alpha)>0$ is such that $x_{3}\geq M_{1}\hskip
5.69054pt\text{implies}\hskip
5.69054pt\ell\left(D\mathcal{Z}(x_{1},x_{2},x_{3})\right)\geq\frac{1}{\alpha}$,
$M_{2}:=M_{2}(\alpha)$ is such that $x_{3}\leq M_{2}$ implies that
$\left|D\mathcal{Z}(x_{1},x_{2},x_{3})\right|\leq\alpha$ and $\alpha$ is any
number in $(0,1)$ (see [5, (1.4),(1.5)]). It is easy to see now using Lemma 1,
for $\lambda=1$, that we can take

$M_{1}=\log\frac{L}{\alpha}\hskip 5.69054pt\text{and}\hskip
5.69054ptM_{2}=\log\frac{\alpha}{L}.$

Of course the two parametrizations are conjugate with $\kappa=\log\lambda$.
Converting the facts of the above paragraph in our setting we get that Theorem
1 holds for

$0<\lambda\leq e^{M_{2}-e^{M_{1}}}=e^{\log\frac{\alpha}{L}-\frac{L}{\alpha}}.$

Taking $\alpha\to 1$ we get that Theorem 1 is true for all
$0<\lambda<e^{-\left(\log L+L\right)}$. ∎

Later we will also need the fact that the Zorich maps are expansive in a
suitable upper half space. This is analogous to the fact that the exponential
map is expansive in a right half plane. The next lemma makes this precise.

######  Lemma 3.

For any $0<\lambda<e^{-\left(\log L+L\right)}$ there is an $0<\alpha<1$ such
that if $M=\log\frac{L}{\lambda\alpha}$ and $\Lambda:H_{\geq M}\to
P(r_{1},r_{2})$, for some $(r_{1},r_{2})\in\mathbb{Z}^{2}$, is an inverse
branch of the Zorich map $\mathcal{Z}_{\lambda}$ defined in $H_{\geq M}$, then
$\mathcal{Z}_{\lambda}(H_{\leq M})\subset H_{<M}$ and for all $x,y\in H_{\geq
M}$

(5) $|\Lambda(x)-\Lambda(y)|\leq\alpha|x-y|.$

###### Proof.

First note that

$D\Lambda(x)=D\mathcal{Z}_{\lambda}(\Lambda(x))^{-1}.$

The Zorich map is absolutely continuous on any line segment since it is
locally Lipschitz. Using the fundamental theorem of calculus for the Lebesgue
integral now, it is not too hard to prove that a version of the finite
increment theorem, see [30, 10.4.1, Theorem 1], is true for such functions. In
other words, if $\gamma$ is the segment that connects $x$ and $y$ then

(6)
$|\Lambda(x)-\Lambda(y)|\leq\operatorname*{ess\,sup}_{z\in\gamma}|D\Lambda(z)||x-y|.$

It is true that

(7)
$\operatorname*{ess\,sup}_{z\in\gamma}|D\Lambda(z)|\leq\frac{1}{\operatorname*{ess\,inf}_{z\in\gamma}\ell\left(D\mathcal{Z}_{\lambda}(\Lambda(z))\right)}.$

By Lemma 1 we have that

(8)
$\ell\left(D\mathcal{Z}_{\lambda}(\Lambda(x))\right)\geq\frac{e^{p_{3}(\Lambda(x))}\lambda}{L}\hskip
5.69054pt\text{a.e.}$

Hence for any $0<\alpha<1$ and for all $x$ such that

(9) $p_{3}(\Lambda(x))\geq\log\frac{L}{\lambda\alpha}$

we have $\ell(D\mathcal{Z}_{\lambda}(\Lambda(x)))\geq 1/\alpha$.

We now claim that for $M=\log\frac{L}{\lambda\alpha}$ we have that for all
$x\in H_{\geq M}$ equation (9) holds. Indeed if $x=(x_{1},x_{2},x_{3})\in
H_{\leq M}$ we have that

(10) $p_{3}\left(\mathcal{Z}_{\lambda}(x)\right)=p_{3}\left(\lambda
e^{x_{3}}h(x_{1},x_{2})\right)\leq\lambda\frac{L}{\lambda\alpha}=\frac{L}{\alpha}.$

Notice now that by continuity we can choose $0<\alpha<1$ so that

$\lambda<e^{-\left(\log
L+L\right)}=\frac{e^{-L}}{L}\leq\frac{e^{-\frac{L}{\alpha}}L}{\alpha},$

which after rearranging and using (10) implies that

$p_{3}\left(\mathcal{Z}_{\lambda}(x)\right)<\log\frac{L}{\lambda\alpha}.$

Hence $\mathcal{Z}_{\lambda}(H_{\leq M})\subset H_{<M}$ as we wanted.
Moreover, this implies that whenever $x\in H_{\geq M}$ we have that
$\Lambda(x)\in H_{\geq M}$ so that (9) is true.

Putting everything together we have that for $x\in H_{\geq M}$

$|D\Lambda(x)|\leq\frac{1}{\ell(D\mathcal{Z}_{\lambda}(\Lambda(x)))}\leq\alpha\hskip
5.69054pt\text{a.e.}$

Combined with (6) and (7) this gives us what we wanted. ∎

The next lemma introduces a number $p_{\lambda}$ which is going to play an
important role in the subsequent sections.

######  Lemma 4.

Let $0<\lambda<e^{-\left(\log L+L\right)}$ and $\alpha$, $M$ be as in Lemma 3.
Then there is a number $p_{\lambda}>1$ such that (5) holds for all $x,y\in
H_{\geq p_{\lambda}}$ and such that all planes $x_{3}=c$, where
$c>p_{\lambda}+1$, intersect the Julia set
$\mathcal{J}(\mathcal{Z_{\lambda}})$.

###### Proof.

From Lemma 3 we know that $\mathcal{Z}_{\lambda}(H_{\leq M})\subset H_{<M}$ so
that $H_{\leq M}$ is in the quasi-Fatou set. Moreover, Theorem 1 implies that
there is a minimum number $J$ such that for all $c\geq J$ the plane
$\\{(x_{1},x_{2},x_{3}):x_{3}=c\\}$ intersects the Julia set
$\mathcal{J}(\mathcal{Z}_{\lambda})$ and thus $J\geq M$. Also all the points
in $H_{<J}$ are in the quasi-Fatou set and converge to a fixed point.

We set

$p_{\lambda}=\max\\{J-1,M\\}.$

Also notice that

$p_{\lambda}\geq M=\log\frac{L}{\lambda\alpha}\geq\log\frac{L}{e^{-(\log
L+L)}\alpha}\geq 2\log L+L-\log\alpha>1,$

where in the last inequality we used the fact that $L>1$ and $0<\alpha<1$. ∎

##  3\. Proving Theorem 3

Our proof will closely follow that of the corresponding result for the
exponential map which is due to Aarts and Oversteegen (see [1, Theorem 1.4]).

###  3.1. Construction of the 3-d straight brush

First we will need to define a correspondence between the set
$\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$ and
$\prod_{i=0}^{\infty}\mathbb{Z}\times\mathbb{Z}$. This can be done in many
ways but let us mention here a method used by Devaney based on Farey trees. In
[12, Section 5.3] Devaney finds for every irrational number $\zeta$ a sequence
of integers $n_{0}n_{1}n_{2}\dots$ by doing the following. We break the real
line into intervals $I_{k}=(k,k+1)$ for each $k\in\mathbb{Z}$. Then we further
subdivide each $I_{k}$ in intervals $I_{kl}$, $l\in\mathbb{Z}$ in a certain
way and so on. Specifically, assuming that $I_{n_{0}n_{1}\dots n_{k}}$ has
been defined we define $I_{n_{0}\dots n_{k}j}$ as follows. Let

$I_{n_{0}n_{1}\dots n_{k}}=\left(\frac{a}{b},\frac{c}{d}\right)$

and $\frac{p_{0}}{q_{0}}=\frac{a+c}{b+d}$. We then define
$\frac{p_{n}}{q_{n}}=\frac{p_{n-1}+c}{q_{n-1}+d}$ and
$\frac{p_{-n}}{q_{-n}}=\frac{p_{-n+1}+a}{q_{-n+1}+b}$, for all
$n\in\mathbb{N}$. Finally, define

$I_{n_{0}\dots
n_{k}j}=\left(\frac{p_{j}}{q_{j}},\frac{p_{j+1}}{q_{j+1}}\right).$

We refer to [12] for more details but the whole construction implies that the
intervals we get satisfy the following.

  1. (1)

$I_{n_{0}n_{1}\dots n_{k+1}}\subset I_{n_{0}n_{1}\dots n_{k}}.$

  2. (2)

The endpoints of each interval $I_{n_{0}n_{1}\dots n_{k}}$ are rational.

  3. (3)

$\\{\zeta\\}=\bigcap_{k=1}^{\infty}I_{n_{0}n_{1}\dots n_{k}}$ and every
irrational can be obtained this way.

In our case for any pair of irrational numbers $(a_{1},a_{2})$ we can apply
this method twice and get a pair of sequences
$\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}}\in\prod_{i=0}^{\infty}\mathbb{Z}\times\mathbb{Z}$.
We also equip $\prod_{i=0}^{\infty}\mathbb{Z}\times\mathbb{Z}$ with the
product topology of the spaces $\mathbb{Z}\times\mathbb{Z}$ equipped with the
discrete topology. The correspondence between
$\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$ and
$\prod_{i=0}^{\infty}\mathbb{Z}\times\mathbb{Z}$ can be shown to be a
homeomorphism.

Let now $x\in[0,\infty)$ and $(s_{1},s_{2})\in\mathbb{Z}^{2}$ with
$s_{1}+s_{2}=\text{even}$. We define the cubes

$S(x,s_{1},s_{2}):=\\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:x\leq x_{3}\leq
x+1,(x_{1},x_{2})\in\overline{P(s_{1},s_{2})}\\}.$

Note that because the Zorich map is doubly periodic with periods $(4,0,0)$,
$(0,4,0)$ and because $\mathcal{Z}\circ R=\mathcal{Z}$, where $R$ is a half-
turn around the lines $x_{1}=2n+1,x_{2}=2m+1$, $n,m\in\mathbb{Z}$ the Julia
set will be periodic and invariant under $R$ as well. Also note that by (2)
the Julia set lies in the square beams $P(s_{1},s_{2})\times\mathbb{R}$ with
$s_{1}+s_{2}=$ even and thus when $s_{1}+s_{2}$ is even

(11)
$S(x,s_{1},s_{2})\cap\mathcal{J}(\mathcal{Z}_{\lambda})\not=\emptyset,\hskip
2.84526pt\text{for all}\hskip 5.69054ptx\geq p_{\lambda},$

where $p_{\lambda}$ was defined in Lemma 4.

Let us now construct the 3-d Straight Brush $B$ of Theorem 3 which will be
homeomorphic to our Julia set. We will suitably modify the construction done
in [1]. Let $x\geq p_{\lambda}$ and
$(a_{1},a_{2})\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$ with the
corresponding sequence $\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}\cup\\{0\\}}$.
Set $x_{0}=x$ and $R_{0}(x,a_{1},a_{2})=S(x,n_{0,1},n_{0,2})$. By induction on
$k$ now we define $R_{k}=R_{k}(x,a_{1},a_{2})$ and $x_{k}$. We consider now
two cases:

  1. (i)

$R_{k}\not=\emptyset$ and there is a $\xi$ with

(12)
$S(\xi,n_{k+1,1},n_{k+1,2})\subset\mathcal{Z}_{\lambda}\left(R_{k}\right).$

Let $\xi^{\prime}=\min\\{\xi:\xi\hskip 2.84526pt\text{satisfies}\hskip
2.84526pt(\ref{eq1})\\}$. If $\xi^{\prime}\geq p_{\lambda}$, we set
$x_{k+1}=\xi^{\prime}$ and

$R_{k+1}=S(\xi^{\prime},n_{k+1,1},n_{k+1,2}).$

If $\xi^{\prime}<p_{\lambda}$, we set $x_{k+1}=x_{k}$ and $R_{k+1}=\emptyset$.

  2. (ii)

If the case (i) does not apply, we set $x_{k+1}=x_{k}$ and
$R_{k+1}=\emptyset$.

For each $k\in\mathbb{N}$ now we set

$B_{k}=B_{k}(x):=\\{y\in R_{0}:\mathcal{Z}^{j}_{\lambda}(y)\in R_{j},\hskip
5.69054pt\text{for}\hskip 5.69054pt1\leq j\leq k\\}.$

The 3-d straight brush is then defined as

$B:=\\{(x,a_{1},a_{2}):R_{k}\not=\emptyset,\hskip 2.84526pt\text{for
all}\hskip 2.84526ptk\in\mathbb{N}\\}.$

The proof that this construction defines a 3-d straight brush and that this is
homeomorphic to the Julia set will be now split in several lemmas. First we
need to show that the set $B$ we defined is indeed a 3-d straight brush.

######  Lemma 5.

$B$ is a 3-d straight brush.

###### Proof.

First we prove hairiness.

Hairiness.
Let $(x,a_{1},a_{2})\in B$ with $\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}}$ the
corresponding sequence of $(a_{1},a_{2})$. Suppose that $x<y$ then
$(y,a_{1},a_{2})\in B$. Indeed, by induction $x_{k}<y_{k}$ and
$R_{k}(y,a_{1},a_{2})\not=\emptyset$ for all $k\in\mathbb{N}$. Hence
$(y,a_{1},a_{2})\in B$. Suppose now that $(z,a_{1},a_{2})\not\in B$. Consider
the smallest $k$ with $R_{k}(z,a_{1},a_{2})\not=\emptyset$ but
$R_{k+1}(z,a_{1},a_{2})=\emptyset$. Then either there is no $\xi$ with

$S(\xi,n_{k+1,1},n_{k+1,2})\subset\mathcal{Z}_{\lambda}(R_{k}(z,a_{1},a_{2}))$

or there is such a $\xi$ but for the minimal $\xi_{m}$ we have that
$\xi_{m}<p_{\lambda}$. In both cases the same holds for all $y$ slightly
larger than $z$.

Combining now what we have proven we see that the set $\\{t:(t,a_{1},a_{2})\in
B\\}$ is of the form $[t_{(a_{1},a_{2})},\infty)$.

Secondly, we prove that $B$ is closed.
Compact Sections.
Let $(x,a_{1},a_{2})\in B^{\complement}$. We will show that $B^{\complement}$
is an open set by constructing an open box $S$ such that $(x,a_{1},a_{2})\in
S\subset\mathbb{R}^{3}\setminus B$. The case where
$(a_{1},a_{2})\not\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$ is the
hardest one and the one we will prove here.

Without loss of generality suppose that $a_{1}\in\mathbb{Q}$. Consider the
minimal $k\in\mathbb{N}$ such that

$(a_{1},a_{2})\not\in\bigcup_{i,j\in\mathbb{N}}I_{n_{0}\dots n_{k-1}i}\times
J_{m_{0}\dots m_{k-1}j},$

where $I_{n_{0}\dots n_{k}}$ and $J_{m_{0}\dots m_{k}}$ are the intervals we
get using the method we described at the start of this section. There is a
$n_{k}$ such that for all $m\in\mathbb{Z}$, if we denote by $\zeta$ and
$\zeta^{\prime}$ the irrationals that correspond to the sequences
$(n_{0},\dots,n_{k},m,\dots)$ and $(n_{0},\dots,n_{k}+1,m,\dots)$ respectively
then

(13) $\zeta<a_{1}<\zeta^{\prime}.$

We can now choose $N$ large enough so that no point
$(x,\gamma_{1},\gamma_{2})$ belongs to $B$, where $\gamma_{1}$ satisfies

(14) $(n_{0},\dots,n_{k},N,\dots)<\gamma_{1}<(n_{0},\dots,n_{k}+1,-N,\dots),$

$\gamma_{2}\in\mathbb{R}$ and we have abused the notation in the obvious way.
The reason we can do this is because for such $(\gamma_{1},\gamma_{2})$ we
have that $R_{k+1}(x,\gamma_{1},\gamma_{2})=\emptyset$. Notice that by (13) we
have that $a_{1}$ satisfies (14). It follows that there is a $\delta>0$ such
that for all $y$ with $|y-x|<\delta$ we have that
$R_{k+1}(y,\gamma_{1},\gamma_{2})=\emptyset$, for all $\gamma_{1}$ satisfying
equations (14) and $\gamma_{2}\in\mathbb{R}$.

Finally we prove density.
Density.
We want to show that

$A:=\\{(a_{1},a_{2}):t_{a_{1},a_{2}}<\infty\\}$

is dense in $\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}$. It is enough to
show that the corresponding sequences in $\prod_{i=0}^{\infty}\mathbb{Z}^{2}$
of such points are dense in the space of all sequences. We know that the
periodic sequences, meaning sequences with
$(n_{k,1},n_{k,2})=(n_{k+N,1},n_{k+N,2})$, for some $N\in\mathbb{N}$ and all
$k\in\mathbb{N}$, are dense in the space of all sequences. Hence, it is enough
to prove that periodic sequences correspond to points in $A$. Indeed, for any
periodic sequence it not hard to see that when $x$ is large enough the
sequence $x_{k}$ is increasing and $R_{k}(x,a_{1},a_{2})\not=\emptyset$ for
$k\leq N+1$ and as a result, since the sequence $(n_{k,1},n_{k,2})$ is
periodic, for all $k\in\mathbb{N}$. This implies that
$t_{(a_{1},a_{2})}<\infty$ for that $(a_{1},a_{2})$.

Suppose now that $(x,a_{1},a_{2})\in B$ and $(a_{1},a_{2})$ has a
corresponding sequence $\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}}$. Choose
$y_{i}$ with $x<y_{i}<x+\frac{1}{i}$, for all $i\in\mathbb{N}$. Then for all
$i$ the inner radius of the shell
$\mathcal{Z}_{\lambda}(R_{k-1}(y_{i},a_{1},a_{2}))$ will be much larger than
the inner radius of $\mathcal{Z}_{\lambda}(R_{k-1}(x,a_{1},a_{2}))$, for $k$
large enough. It follows that $(y_{i},a_{1},a_{i,2})\in B$ where
$(a_{1},a_{i,2})$ is chosen so that it has a corresponding sequence

$(n_{0,1},n_{0,2}),\dots(n_{i,1},n_{i,2}-1),(n_{i+1,1},n_{i+1,2}),(n_{i+2,1},n_{i+2,2}),\dots$

and thus $a_{i,2}$ is an increasing sequence. Hence, $y_{i}\to x$ while
$a_{i,2}\uparrow a_{2}$ as $i\to\infty$. Note that closedness of $B$ implies
now that $t_{(a_{1},a_{n,2})}\to t_{(a_{1},a_{2})}$.

Similarly we will have that $(y_{i},a_{1},b_{i,2})$, $(y_{i},c_{i,1},a_{2})$,
$(y_{i},d_{i,1},a_{2})$$\in B$, where $(a_{1},b_{i,2})$, $(c_{i,1},a_{2})$,
$(d_{i,1},a_{2})$ have corresponding sequences

$(n_{0,1},n_{0,2}),\dots(n_{i,1},n_{i,2}+1),(n_{i+1,1},n_{i+1,2}),(n_{i+2,1},n_{i+2,2}),\dots,$
$(n_{0,1},n_{0,2}),\dots(n_{i,1}-1,n_{i,2}),(n_{i+1,1},n_{i+1,2}),(n_{i+2,1},n_{i+2,2}),\dots,$
$(n_{0,1},n_{0,2}),\dots(n_{i,1}+1,n_{i,2}),(n_{i+1,1},n_{i+1,2}),(n_{i+2,1},n_{i+2,2}),\dots$

respectively and $b_{i,2}$ is decreasing, $c_{i,1}$ is increasing and
$d_{i,1}$ is decreasing.

∎

Next comes the construction of a suitable homemorphism $\varphi$ from $B$ to
the Julia set $\mathcal{J}(\mathcal{Z}_{\lambda})$.

###  3.2. Proving that $B$ is homeomorphic to
$\mathcal{J}(\mathcal{Z}_{\lambda})$

For each $(x,a_{1},a_{2})\in B$ we define

$\varphi(x,a_{1},a_{2}):=\bigcap_{k=0}^{\infty}B_{k}.$

Note that $B_{k}\subset\dots\subset B_{0}$, for all $k\in\mathbb{N}$ and also
that $\operatorname{diam}(B_{k})\to 0$ since all boxes $R_{k}(x,a_{1},a_{2})$
are inside the half space $\\{(x_{1},x_{2},x_{3}):x_{3}\geq p_{\lambda}\\}$ on
which, by Lemma 3 and by Lemma 4, the Zorich map is expanding. Thus

$\operatorname{diam}(B_{k})=\sup_{z,y\in
R_{k}(x,a_{1},a_{2})}\left|\Lambda^{k}(z)-\Lambda^{k}(y)\right|\leq\alpha^{k}\operatorname{diam}(R_{k}(x,a_{1},a_{2})).$

######  Lemma 6.

$\varphi$ is an injective continuous map from $B$ into the Julia set
$\mathcal{J}(\mathcal{Z}_{\lambda})$.

###### Proof.

Let us first show that $\varphi$ is injective. Let
$(x,a_{1},a_{2}),(y,b_{1},b_{2})$ be two different points of $B$ and let
$\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}}$ and
$\\{(m_{k,1},m_{k,2})\\}_{k\in\mathbb{N}}$ their corresponding sequences. Now
either $x\not=y$ or $x=y$ and $(a_{1},a_{2})\not=(b_{1},b_{2})$. In the first
case we may assume $x<y$ and thus $x_{k}<y_{k}$ for all $k\in\mathbb{N}$. We
can easily see by the mapping properties of the Zorich map that in fact
$y_{k}-x_{k}\to\infty$ for $k\to\infty$. Thus for large enough $k$ we can find
cubes $R_{k}(x,a_{1},a_{2})\cap R_{k}(y,b_{1},b_{2})=\emptyset$, which implies
that $\varphi(x,a_{1},a_{2})\not=\varphi(y,b_{1},b_{2})$. In the second case,
we have that $(n_{k,1},n_{k,2})\not=(m_{k,1},m_{k,2})$ for some
$k\in\mathbb{N}$ and thus $R_{k}(x,a_{1},a_{2})\cap
R_{k}(y,b_{1},b_{2})=\emptyset$.

Now for the continuity of $\varphi$ let $(x,a_{1},a_{2})$ and
$(y,b_{1},b_{2})$ be two points in $B$ that are "close" meaning that their
corresponding sequences satisfy $n_{k,i}=m_{k,i}$, $i=1,2$, for $k=0,\dots,N$
where $N\in\mathbb{N}$ and $|x-y|<\delta$ for some small $\delta>0$. If
$\delta$ is small enough this implies that

$R_{k}(x,a_{1},a_{2})\cap R_{k}(y,b_{1},b_{2})\not=\emptyset,\hskip
2.84526pt\text{for all}\hskip 2.84526ptk\leq N.$

Hence, $B_{k}(x)\cap B_{k}(y)\not=\emptyset$ for all $k\leq N$. This implies
that

$|\varphi(x,a_{1},a_{2})-\varphi(y,b_{1},b_{2})|\leq\operatorname{diam}B_{N}(x)+\operatorname{diam}B_{N}(y)$

and because those diameters tend to zero we get that when $N$ is large enough
$\varphi(x,a_{1},a_{2})$ and $\varphi(y,b_{1},b_{2})$ are close.

Finally, we want to show that
$\varphi(x,a_{1},a_{2})\in\mathcal{J}(\mathcal{Z}_{\lambda})$. If
$\\{(n_{k,1},n_{k,2})\\}_{k\in\mathbb{N}}$ is the corresponding sequence of
$(a_{1},a_{2})$ then, by construction and (11), we will have that

$R_{k}(x,a_{1},a_{2})\cap\mathcal{J}(\mathcal{Z}_{\lambda})\not=\emptyset,\hskip
2.84526pt\text{for all}\hskip 2.84526ptk\in\mathbb{N}.$

This implies that
$d\left(\varphi(x,a_{1},a_{2}),\mathcal{J}(\mathcal{Z}_{\lambda})\right)=0$
and because $\mathcal{J}(\mathcal{Z}_{\lambda})$ is closed we have that
$\varphi(x,a_{1},a_{2})\in\mathcal{J}(\mathcal{Z}_{\lambda})$. ∎

Next we need to find the inverse of $\varphi$ and show that it is continuous.
Let us first define this function and then show that indeed it is the inverse
of $\varphi$.

Let $w=(w_{1},w_{2},w_{3})\in\mathcal{J}(\mathcal{Z}_{\lambda})$. Remember now
that with each $w$ in the Julia set we can associate its itinerary
$\Delta(w)=n_{0}n_{1}n_{2}\dots$ and $\mathcal{Z}^{k}_{\lambda}(w)\in
P(n_{k})\times\mathbb{R}$. For any $k\in\mathbb{N}$ define the boxes
$T_{j}(k)$, $j\in\mathbb{N}$ as follows. First we set
$T_{k}(k)=S(u,n_{k,1},n_{k,2})$, where $u$ is minimal with respect to the
properties $u\geq p_{\lambda}$ and $\mathcal{Z}^{k}_{\lambda}(w)\in T_{k}(k)$.
We now define $T_{j}(k)$ for $j=0,\dots,k-1$ inductively as follows. Suppose
that $T_{j}(k)$ has been defined, let $T_{j-1}(k)=S(v,n_{j-1,1},n_{j-1,2})$,
where $v$ is maximal with respect to the property

(15) $T_{j}(k)\subset\mathcal{Z}_{\lambda}(T_{j-1}(k)).$

Note that a $v$ for which (15) holds exists since $u\geq p_{\lambda}>1$.
Moreover, (15) thanks to the continuity of $\mathcal{Z}_{\lambda}$, implies
that the box $T_{j}(k)$ hits the inner radius of the half-shell
$\mathcal{Z}_{\lambda}(T_{j-1}(k))$ in exactly one point. Also it implies that
the lowest (in terms of $x_{3}$ coordinate) side of the box $T_{j-1}(k)$ has a
third coordinate at least $p_{\lambda}$ for all $j$. This can be shown by
first noting that it is true for $T_{k-1}(k)$ and then we can use induction on
$j$ to prove it for all $j$. To see why it is true for $T_{k-1}(k)$ note that
if that was not the case and $p_{\lambda}=J-1$ (see Lemma 4) then $T_{k-1}(k)$
would not contain any points in the Julia set and since $T_{k}(k)$ does (by
Lemma 4) we get a contradiction by (15) and the invariance of the Julia set.
If on the other hand $p_{\lambda}=M$ then the lowest side of the box
$T_{k-1}(k)$ would be below $M$ and since $\mathcal{Z}_{\lambda}(H_{\leq
M})\subset H_{<M}$ we get a contradiction by the maximality condition on (15).
Thus in any case the lowest (in terms of $x_{3}$ coordinate) side of the box
$T_{k-1}(k)$ has a third coordinate at least $p_{\lambda}$.

We now define $z_{k}$ by the condition $T_{0}(k)=S(z_{k},n_{0,1},n_{0,2})$ and
note that $w\in T_{0}(k)$. This implies that

(16) $w_{3}-1\leq z_{0}\leq z_{k}\leq w_{3},$

for all $k\in\mathbb{N}$. We can also prove that

(17) $z_{k}\leq z_{k+1}.$

This follows by the fact that $T_{k}(k+1)$ is higher than $T_{k}(k)$ (in terms
of $x_{3}$ coordinate) and thus by the inductive construction (17) holds.

Finally, we set $z_{\infty}:=\lim_{k\to\infty}z_{k}$ and define $\psi$ by

$\psi(w)=\left(z_{\infty},a_{1},a_{2}\right),$

where $(a_{1},a_{2})$ is the pair of irrationals associated with the pair of
sequences $\Delta(w)$. The next lemma concludes the proof of Theorem 3.

######  Lemma 7.

The map $\psi$ is the inverse of $\varphi$ and $\varphi$ is a homeomorphism.
Moreover, $\varphi$ extends to a homeomorphism between $B\cup\\{\infty\\}$ and
$\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$.

###### Proof.

We will first show that

(18) $\varphi\circ\psi=\text{id}_{\mathcal{J}(\mathcal{Z}_{\lambda})}.$

Let $w\in\mathcal{J}(\mathcal{Z}_{\lambda})$ and we continue using notation as
above. We define $T_{0}(\infty)=S(z_{\infty},n_{0,1},n_{0,2})$ and note that
$T_{0}(\infty)=\lim_{k\to\infty}T_{0}(k)$ in the Hausdorff metric. Define also
$T_{j}(\infty):=\lim_{k\to\infty}T_{j}(k)$, where the limit is again in the
Hausdorff metric. Notice that by construction the box $T_{j}(\infty)$ hits the
inner radius of the half-shell $\mathcal{Z}_{\lambda}(T_{j-1}(\infty))$ in
exactly one point and that point has a third coordinate at least
$p_{\lambda}$. Applying now the construction of $\varphi$ to the point
$\psi(w)=\left(z_{\infty},a_{1},a_{2}\right)$, where $(a_{1},a_{2})$ is the
pair of irrationals associated with the pair of sequences $\Delta(w)$ and
using the notation of the construction of $B$ we get that $x_{0}=z_{\infty}$
and

$R_{k}(x,a_{1},a_{2})=T_{k}(\infty),$

for all $k\in\mathbb{N}$. Hence, $\psi(w)\in B$ and

$\varphi(\psi(w))=\varphi(z_{\infty},a_{1},a_{2})=\bigcap_{k\in\mathbb{N}}\mathcal{Z}^{-k}_{\lambda}(T_{k}(\infty)),$

where ${Z}^{-k}_{\lambda}$ is the composition of inverse branches of
$\mathcal{Z}_{\lambda}$ to appropriate square beams (see construction of
$\varphi$). Now note that by equation (16) taking limits we have that $w\in
T_{0}(\infty)$. We can prove similar equations with (16), (17) for all boxes
$T_{j}(k)$. Hence, we will also have that $\mathcal{Z}^{k}_{\lambda}(w)\in
T_{k}(\infty)$. This implies that

$w\in\bigcap_{k\in\mathbb{N}}\mathcal{Z}^{-k}_{\lambda}(T_{k}(\infty))$

and since that intersection contains only one point we get that
$\varphi(\psi(w))=w$.

Equation (18) implies now that $\varphi$ is onto the Julia set and thus a
bijection. Since we have already proven that $\varphi$ is continuous we can
now extend $\varphi$ to a continuous map $\hat{\varphi}$, from
$B\cup\\{\infty\\}$ to the one point compactification
$\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$ of
$\mathcal{J}(\mathcal{Z}_{\lambda})$, with the spherical metric by setting
$\hat{\varphi}(\infty)=\infty$, since $\varphi(b)\to\infty$ as $b\to\infty$,
$b\in B$. We can now use a well known lemma (see for example [25, Theorem
4.17]) and immediately conclude that in fact $\hat{\varphi}$ is a
homeomorphism (with the spherical metric) and $\hat{\psi}$, defined as the
extension of $\psi$ on $\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}$ by
setting $\hat{\psi}(\infty)=\infty$, is its inverse. This implies that $\psi$
is continuous with the euclidean metric. ∎

##  4\. Proving Theorem 6

First we will show that the model of the Julia set, namely the 3-d straight
brush, is a Lelek fan.

######  Lemma 8.

The one point compactification $B\cup\\{\infty\\}$ of a 3-d straight brush $B$
is a Lelek fan with top at $\infty$.

###### Proof.

Any subcontinuum of $B\cup\\{\infty\\}$ is a collection of segments of
straight lines together with $\infty$ or a segment of one such line. Obviously
the intersection of any two such subcontinua is again of that form and thus
connected. Properties (ii) and (iii) of the definition of a fan are obviously
satisfied for $B\cup\\{\infty\\}.$ Smoothness is also quite easy to prove
since arcs $[y_{n},\infty]$ in $B\cup\\{\infty\\}$ are straight lines and thus
converge to a straight line $[y,\infty]$ in the Hausdorff metric, when
$y_{n}\to y$.

Finally, we prove that the set of endpoints $\mathcal{E}(B)$ is dense in $B$.
Choose a point $(x,b_{1},b_{2})\in B$. We will show that there are hairs with

(19) $(t_{b_{1,n},b_{2,n}},b_{1,n},b_{2,n})\to(x,b_{1},b_{2}),$

as $n\to\infty$. In other words their endpoints converge to $(x,b_{1},b_{2})$
and this obviously gives us the density of endpoints.

Consider the length function
$\mathcal{L}:\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}\to[0,\infty)$
which measures the spherical length of the hair at $(a_{1},a_{2})$. It is easy
to see that thanks to the closedness of $B$, $\mathcal{L}$ is upper semi-
continuous. Let

$M:=\sup\\{\mathcal{L}(a_{1},a_{2}):(a_{1},a_{2})\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}\\}.$

Take now any $c\in(0,M)$ and consider the set

$V:=\\{(a_{1},a_{2})\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}:\mathcal{L}(a_{1},a_{2})>c\\}.$

Fix now a point $(b_{1},b_{2})$ in $V$ and consider the set

$W:=\\{(b_{1},a_{2})\in\overline{V}:|a_{2}|\leq
t,\mathcal{L}(b_{1},a_{2})>c\\},$

where $t>0$ a constant.

From the upper semi-continuity of $\mathcal{L}$ we conclude that the set

$\\{(a_{1},a_{2})\in\left(\mathbb{R}\setminus\mathbb{Q}\right)^{2}:\mathcal{L}(a_{1},a_{2})\geq
c\\}$

is closed and thus contains $\overline{W}$.

The set $\overline{W}$ is compact and lies on a line. By property (ii) of the
definition of a 3-d straight brush now we will have that it is also perfect.
Moreover, it is a totally disconnected set, since it is a subset of a totally
disconnected set. Hence, by the characterization of the Cantor set (see for
example [9, Theorem 6.17] or the more general [17, Theorem 7.4]) we get that
$\overline{W}$ is a Cantor set (i.e. homeomorphic to the standard ternary
Cantor set $\mathcal{C}$). In fact, the sets $\overline{W}$ and $\mathcal{C}$
are ambiently homeomorphic (meaning here that the homeomorphism extends to the
line on which $\overline{W}$ lies). This implies that the obvious order in the
sets is either preserved or reversed under the homeomorphism. This in turn
implies, that a point $(b_{1},a_{2})\in\overline{W}$ for which
$\mathcal{L}(b_{1},a_{2})>c$ cannot get mapped to an endpoint of the Cantor
set since, thanks to property (ii) of a 3-d straight brush, it can be
approximated from both sides by other points of $\overline{W}$ (unless of
course $a_{2}=\pm t$ but those are just two points). Hence, only points with
$\mathcal{L}(b_{1},a_{2})=c$ get mapped to endpoints and since endpoints are
dense in the Cantor set so are points with $\mathcal{L}(b_{1},a_{2})=c$ dense
in $\overline{W}$.

Since this argument works for any $(b_{1},b_{2})\in V$ we have that points
with $\mathcal{L}(a_{1},a_{2})=c$ are dense in $\overline{V}$.

Take now any point $(x,b_{1},b_{2})\in B$, which is not an endpoint, and let
$c$ be the spherical length of the hair segment $\\{(t,b_{1},b_{2}):t\geq
x\\}$. This implies that $\mathcal{L}(b_{1},b_{2})>c.$ As we have shown then
there is a sequence of hairs $(t_{b_{1,n},b_{2,n}},b_{1,n},b_{2,n})$ of length
$c$ for which $(b_{1,n},b_{2,n})\to(b_{1},b_{2})$ and thus the endpoints of
the hairs $(t_{b_{1,n},b_{2,n}},b_{1,n},b_{2,n})$ converge to
$(x,b_{1},b_{2})$ as we wanted. ∎

###### Proof of Theorem 6.

From Theorem 3 we know that there a homeomorphism

$\hat{\varphi}:B\cup\\{\infty\\}\to\mathcal{J}(\mathcal{Z}_{\lambda})\cup\\{\infty\\}.$

By Lemma 8 we know that $B\cup\\{\infty\\}$ is a Lelek fan. It is quite easy
to see that all the properties of a Lelek fan are preserved under a
homeomorphism and thus $\mathcal{J}(\mathcal{Z}_{\lambda})$ will also be a
Lelek fan with top at $\infty$. ∎

##  5\. Hairy squares and hairy surfaces

In this section we generalize the notion of a hairy arc that Aarts and
Oversteegen first introduced in [1]. Our exposition closely follows theirs but
as we shall see some things are different in three dimensions.

First we will introduce the notion of a straight one-sided hairy square
(abbreviated soshs) which will be a generalization of straight one-sided hairy
arcs (abbreviated sosha) on the plane (see [1] for that definition). Let
$I=[0,1]$.

######  Definition (Straight one-sided hairy square).

A straight one-sided hairy square $X$ is a compact subset of $I^{3}$
satisfying the following properties. There is a function $\ell:I^{2}\to I$,
called the length function, such that

  1. (i)

For all $(x,y,z)\in I^{3}$ we have $(x,y,z)\in X$ if and only if $0\leq
z\leq\ell(x,y)$.

  2. (ii)

The sets

$\\{(x,y)\in I^{2}:\ell(x,y)>0\\},\hskip 2.84526pt\\{x\in
I:\ell(x,y)=0,\forall y\in I\\}\hskip 2.84526pt\text{and}\hskip
2.84526pt\\{y\in I:\ell(x,y)=0,\forall x\in I\\}$

are dense in $I^{2}$, $I$ and $I$ respectively and
$\ell(0,t)=\ell(1,t)=\ell(t,0)=\ell(t,1)=0$ for all $t\in[0,1]$.

  3. (iii)

For each $(x,y)\in I^{2}$ with $\ell(x,y)>0$ there exist sequences $a_{n}$,
$b_{n}$, $c_{n}$, $d_{n}$ such that $a_{n}\uparrow y$, $b_{n}\downarrow y$,
$c_{n}\uparrow x$, $d_{n}\downarrow x$. Moreover it is true that
$\ell\left(x,a_{n}\right)\to\ell(x,y)$ and similarly for the other sequences.

For each $(x,y)\in I^{2}$, the set $\\{(x,y,z):0\leq z\leq\ell(x,y)\\}$ will
be called the hair at $(x,y)$ while the set $I^{2}\times\\{0\\}$ will be
called the base.

The usefulness of the above object lies in the fact that when we suitably
embed a 3-d straight brush to $I^{3}$ with the usual topology and then
compactify that embedding we get a soshs.

Indeed, let $\mathcal{H}:\mathbb{R}^{3}\to I^{3}$ be defined as

(20) $\mathcal{H}(x,y,z)=\left(\frac{\arctan y}{\pi}+\frac{1}{2},\frac{\arctan
z}{\pi}+\frac{1}{2},\frac{\mathcal{L}\left([x,\infty)\times\\{(y,z)\\}\right)}{\pi}\right)$

where $\mathcal{L}$ is once again the spherical length of the half line
$[x,\infty)\times\\{(y,z)\\}$. We note here that we are viewing
$\overline{\mathbb{R}^{3}}$ as the sphere of centre $(0,0,1/2)$ and radius
$1/2$ so the spherical length of any straight line is less or equal than
$\pi$.

It is easy to see now that $\mathcal{H}$ is an embedding of $\mathbb{R}^{3}$
to $I^{3}$ and that if $B$ is a 3-d straight brush then the compactification
of $\mathcal{H}(B)$ in $I^{3}$, with usual topology is a soshs.

After they defined straight one-sided hairy arcs, Aarts and Oversteegen went
on to define the notions of a hairy arc and a one-sided hairy arc which are a
homeomorphic image of a sosha and a homeomorphic image of a sosha on the plane
with all of the hairs attached on the same side respectively. The importance
of those objects becomes apparent when we notice that compactified versions of
Julia sets of many entire transcendental functions are one-sided hairy arcs.
Initially, Aarts and Oversteegen showed this for some functions in the
exponential family and some functions in the sine and cosine families. That
was extended to much larger classes of transcendental entire maps by Baranski,
Jarque and Rempe in [3].

Generalizing in $\mathbb{R}^{3}$ we define the notions of a hairy surface and
a one-sided hairy surface.

######  Definition (Hairy Surface).

A hairy surface is any homeomorphic image of a soshs. The base of the hairy
surface is the image of $I^{2}\times\\{0\\}$ under that homeomorphism and the
hairs are the images of the hairs of the soshs. A one-sided hairy surface is
an embedding $\varphi$ of a hairy surface $X$, with base $D$, in
$\mathbb{R}^{3}$ such that all hairs are attached to the same side of the base
$\varphi(D)$.

The notion of same sidedness is intuitively clear. Rigorously we can define it
as follows. There is a surface $\mathcal{S}$ which contains $\varphi(D)$ and
all of the hairs are contained in the same bounded complementary component of
$\mathcal{S}$.

Next we prove that when we suitably compactify the Julia sets of some Zorich
maps we do get hairy surfaces.

###### Proof of Theorem 4.

We know from section 3 that there is a homeomorphism
$\psi:\mathcal{J}(\mathcal{Z}_{\lambda})\to B$, where $B$ is a 3-d straight
brush. Consider now the embedding $f:\mathcal{J}(\mathcal{Z}_{\lambda})\to
I^{3}$, with $f=\mathcal{H}\circ\psi$, where $\mathcal{H}:B\to I^{3}$ was
defined in (20). The compactification we are looking for is the one induced by
the embedding $f$ (see for example [22, Chapter 5.3]). Also, by construction,
$f$ extends to a homeomorphism $\widetilde{f}$ between the compactification of
$\mathcal{J}(\mathcal{Z}_{\lambda})$,
$\widetilde{\mathcal{J}(\mathcal{Z}_{\lambda})}$ and
$\overline{\mathcal{H}(B)}$ the closure of $\mathcal{H}(B)$ in the usual
topology of $I^{3}$. Hence, $\widetilde{\mathcal{J}(\mathcal{Z}_{\lambda})}$
is a hairy surface. ∎

##  6\. Wild one sided hairy surfaces

When studying embeddings of a subset $X$ of $\mathbb{R}^{d}$ in
$\mathbb{R}^{d}$ an important notion is that of tameness of $X$. In other
words, whether or not there is a homeomorphism
$H:\mathbb{R}^{d}\to\mathbb{R}^{d}$ sending $X$ to $h(X)$, where
$h:X\to\mathbb{R}^{d}$ is a homeomorphism onto its image. We also say that $X$
and $h(X)$ are then ambiently homeomorphic.

Aarts and Oversteegen in [1] showed that all one-sided hairy arcs are tame. In
other words any embedding $\varphi:X\to\mathbb{R}^{2}$ of a sosha $X$ for
which the hairs of $\varphi(X)$ are all on the same side extends to a
homeomorphism of the whole plane. This is reminiscent of the tameness of the
Cantor set and the arc in the plane. Once we move however to higher dimensions
the situation is different. It is well known that in $\mathbb{R}^{3}$, for
example, there are wild arcs, wild Cantor sets and wild spheres meaning sets
that are homeomorphic images of $[0,1]$, the standard ternary Cantor set and
the unit sphere respectively and yet they are not ambiently homeomorphic to
those sets. Classical examples of such sets are the Wild arc, Antoine’s
necklace and Alexander’s horned sphere (see for example [21] and references
therein).

The situation is similar for soshs’ and one sided hairy surfaces as Theorem 5
shows.

In order to prove our theorem we will first construct a soshs $S$ and then a
homeomorphism which takes that soshs to a wild one sided hairy surface.

###  6.1. Constructing a soshs

For convenience we will construct our soshs as a subset of
$I^{2}\times[0,3/2]$ instead of $I^{3}$.

Our soshs will be constructed as an intersection of subsets $T_{n}$ of
$I^{2}\times[0,3/2]$ each of which comprises of a square base $I^{2}$ and a
set of cuboids $R(n,i,j)$. The cuboids will have a square base on $I^{2}$ and
will have their sides parallel with the axis $x$, $y$ and $z$. The
construction proceeds inductively. Set $T_{0}=T_{1}=I^{2}\times[0,3/2]$ and
partition $I^{2}$ in 9 equal squares $Q(2,i,j)$, $i,j=1,2,3$ which will be the
base of our cuboids $R(2,i,j)$. On each of those squares we erect a cuboid of
height as in Figure 1. The bottom left corner square is $Q(2,1,1)$ and the top
right one is $Q(2,3,3)$.

1/2 |  1/2 |  1/2
---|---|---
1/2 |  3/2 |  1/2
1/2 |  1/2 |  1/2
Figure 1. The squares $Q(2,i,j)$ and the heights of the corresponding cuboids

We now name $T_{2}=\bigcup_{i,j\in\\{1,2,3\\}}R(2,i,j)$. Suppose now that we
have constructed $T_{n}$ and it is a finite union of cuboids $R(n,i,j)$ with
bases $Q(n,i,j)$. Take each square base $Q(n,i,j)$ and partition it in
$(2n+1)^{2}$ equal squares, which we will now call $Q(n,i,j,k,l)$. After we
define $T_{n+1}$ we relabel those squares as $Q(n+1,i,j).$ Notice that we have
used the same indices $i,j$ although they take different values for different
values of $n$. Also, let us denote by $h(n,i,j)$ the height of the cuboid
$R(n,i,j)$.

We now erect cuboids $R(n,i,j,k,l)$ with bases $Q(n,i,j,k,l)$ with heights

$h(n,i,j,k,l)=\begin{cases}\frac{1}{n+1}h(n,i,j),&\hskip
5.69054pt\text{if}\hskip 5.69054ptl=1\hskip 5.69054pt\text{or}\hskip
5.69054ptl=2n+1\hskip 5.69054pt\text{or}\hskip 5.69054ptk=1\hskip
5.69054pt\text{or}\hskip 5.69054ptk=2n+1\\\ \\\ h(n,i,j),&\hskip
5.69054pt\text{if}\hskip 5.69054ptQ(n,i,j,k,l)\hskip 5.69054pt\text{is the
central square of}\hskip 5.69054ptQ(n,i,j)\\\ \\\
\frac{n}{n+1}h(n,i,j),&\hskip 5.69054pt\text{otherwise}.\par\end{cases}$

Then we define $T_{n+1}=\bigcup_{i,j,k,l}R(n,i,j,k,l).$

1/2 |  1/2 |  1/2 |  1/2 |  1/2
---|---|---|---|---
1/2 |  1 |  1 |  1 |  1/2
1/2 |  1 |  3/2 |  1 |  1/2
1/2 |  1 |  1 |  1 |  1/2
1/2 |  1/2 |  1/2 |  1/2 |  1/2
Figure 2. The squares $Q(2,2,2,k,l)$ and the heights $h(2,2,2,k,l)$ of the
corresponding cuboids

Our soshs will then be defined as $S=\bigcap_{n=0}^{\infty}T_{n}$. Since every
$T_{n}$ is a continuum and $T_{n+1}\subset T_{n}$, $S$ will also be a
continuum and it is easy to see that it will satisfy all of the properties of
a soshs. We will only prove here that property (iii) of the definition of a
soshs is satisfied.

Indeed, for any hair $\delta:[0,1]\to S$ of $S$ at the base point $(x,y)$ it
will be true that there is a sequence of cuboids $R(n,i_{n},j_{n})$ with
$R(k+1,i_{k+1},j_{k+1})\subset R(k,i_{k},j_{k})$, for all $k\in\mathbb{N}$ and

$\bigcap_{n}R(n,i_{n},j_{n})=\delta\left([0,1]\right).$

It is true now that there is a subsequence $n_{k}$ such that
$R(n_{k},i_{n_{k}},j_{n_{k}})$ is the central square of
$R(n_{k}-1,i_{n_{k}-1},j_{n_{k}-1})$ since otherwise there would not be a hair
at $(x,y)$.

We now find a sequence of hairs $\delta_{k}$, $k=1,2,\dots$ with base points
$(x_{k},y)$ such that $x_{k}\uparrow x$ and $\ell(x_{k},y)\to\ell(x,y)$. We
choose the hair $\delta_{k}$ as the one defined by the sequence of cuboids

$\displaystyle R(1,i_{1},j_{1}),$
$\displaystyle\dots,R(n_{k}-1,i_{n_{k}-1},j_{n_{k}-1}),R(n_{k},i_{n_{k}}-1,j_{n_{k}}),$
$\displaystyle
R(n_{k}+1,m_{1,k},j_{k+1}),R(n_{k}+2,m_{2,k},j_{n_{k}+2}),\dots,$

where $m_{i,k}$, $i\in\mathbb{N}$ is a sequence of integers such that the
sequence of squares

$Q(n_{k}+1,m_{1,k},j_{k+1}),Q(n_{k}+2,m_{2,k},j_{n_{k}+2}),\dots$

is the sequence of squares

$Q(n_{k}+1,i_{n_{k}+1},j_{k+1}),Q(n_{k}+2,i_{n_{k}+2},j_{n_{k}+2}),\dots$

translated to the left by the length of a side of a square at the $n_{k}$
level.

It is now easy to see that $\delta_{k}$ has its base point at $(x_{k},y)$ with
$x_{k}\uparrow x$. Also the hairs $\delta_{k}$ have length

$\ell(x_{k},y)=\frac{n_{k}}{n_{k}+1}\ell(x,y)$

so that $\ell(x_{k},y)\to\ell(x,y)$ as $k\to\infty$.

Similarly we can construct the other sequences of points from property (iii)
of the definition of a soshs.

###  6.2. Proof of Theorem 5

Before we proceed with the proof of Theorem 5 let us first introduce some
standard terminology taken from [21].

Let $V=\\{u_{0},u_{1},\dots,u_{n}\\}$ be a set of $n+1$ points in
$\mathbb{R}^{d}$ which are affinely independent, meaning $u_{1}-u_{0}$,
$u_{2}-u_{0}$, $\dots,u_{n}-u_{0}$ are linearly independent. Then the
$n$-simplex is defined as the convex hull of $V$. We will denote the
$n$-simplex defined by those points by $\sigma^{n}$. The convex hull $\tau$ of
a non empty subset $W$ of $V$ will be called a face of $\sigma^{n}$. A
(Euclidean) complex is a collection $\mathcal{K}$ of simplexes in
$\mathbb{R}^{d}$ such that

  1. (1)

$\mathcal{K}$ contains all faces of all elements of $\mathcal{K}$.

  2. (2)

If $\sigma$, $\tau\in\mathcal{K}$ are simplexes and
$\sigma\cap\tau\not=\emptyset$ then $\sigma\cap\tau$ is a face of both
$\sigma$ and $\tau$.

  3. (3)

Every $\sigma$ in $\mathcal{K}$ lies in an open set $U$ which intersects only
a finite number of members of $\mathcal{K}$.

If $\mathcal{K}$ is a complex then with $|\mathcal{K}|$ we denote the union of
the elements of $\mathcal{K}$. Such a set is called a polyhedron. Note that a
polyhedron can be seen as a manifold with boundary with the subspace topology
induced from the standard topology of $\mathbb{R}^{d}$. An $n$-cell is a space
homeomorphic to an $n$-simplex. A polyhedral $n$-cell is a polyhedron
homeomorphic to an $n$-simplex.

Let $M$ be a manifold with boundary. Then by $\operatorname{Bd}M$ and
$\operatorname{Int}M$ we denote its boundary and its interior respectively.

######  Lemma 9.

[21, Theorem 1, Chapter 19] Let $A$ be a polyhedral $1$-cell in
$\mathbb{R}^{3}$ with endpoints $P$ and $Q$. Then there is a polyhedral 3-cell
$C$ such that (1) $\operatorname{Int}A\subset\operatorname{Int}C$, (2)
$P,Q\in\operatorname{Bd}C$, and (3) there is a homeomorphism
$\phi:C\to\sigma^{2}\times[0,1]$, such that $A\mapsto R\times[0,1]$, for some
$R\in\operatorname{Int}\sigma^{2}$.

If $A$ and $C$ satisfy the conditions of the above lemma, then we say that $A$
is unknotted in $C$.

###### Proof of Theorem 5.

Consider the soshs $S$ we constructed in the previous subsection. We will
construct a homeomorphism by defining it on the sets $R(n,i,j)$ for all $n$.
Consider now cuboids $B_{k}$, $k\in\mathbb{N}$ with sides parallel to the $x$,
$y$ and $z$ axis which are constructed as follows:

  * •

$B_{1}$ has a square base $I^{2}\times\\{0\\}$ and height $h_{1}=1/2$.

  * •

$B_{2}$ has as base $Q(2,2,2)\times\\{1/2\\}$ and height $h_{2}=\frac{1}{4}$.
$\mathmakebox[\widthof{{}={}}][c]{\vdots}\\\ $

  * •

$B_{n}$ has as base $Q(n,i_{n},j_{n})\times\\{\sum_{k=1}^{n-1}h_{k}\\}$, where
$Q(n,i_{n},j_{n})$ is the central square of
$Q(n-1,i_{n-1},j_{n-1})$ and height $h_{n}=\frac{1}{2^{n}}$.
$\mathmakebox[\widthof{{}={}}][c]{\vdots}\\\ $

Figure 3. The polyhedral $1$-cells $A_{n}$ (left) and the polyhedral $3$-cells
$C_{n}$ (right) used in the Proof of Theorem 5.

Consider now the polyhedral $1$-cells $A_{n}$ inside $B_{n}$ that are overhand
knotted (see Figure 3) with their endpoints at
$(\frac{1}{2},\frac{1}{2},h_{n-1})$ and $(\frac{1}{2},\frac{1}{2},h_{n})$.
From Lemma 9 now, for each of those $1$-cells, we can find a polyhedral
$3$-cell $C_{n}$, inside $B_{n}$, in which $A_{n}$ is unknotted and
$\operatorname{Bd}C_{n}$ contains the bottom and the top square bases of
$B_{n}$.

Lemma 9 also gives us that there are homeomorphisms
$\phi_{n}:C_{n}\to\sigma^{2}\times[0,1]$. Since $\sigma^{2}\times[0,1]$ is
homeomorphic to $B_{n}$ this implies that there are homeomorphisms
$\varphi_{n}:~{}B_{n}\to C_{n}$ which can also be arranged so that they fix
the square bases of $B_{n}$.

Define now the map $H:S\to\mathbb{R}^{3}$ as

$H(x):=\begin{cases}\varphi_{n}(x),&\hskip 5.69054pt\text{for}\hskip
2.84526ptx\in B_{n},\\\ x,&\hskip 5.69054pt\text{otherwise}.\end{cases}$

It is easy to see, by our construction, that this map is a homeomorphism of
$S$ onto $H(S)$.

Suppose now that there exists a homeomorphism
$h:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that $h(S)=H(S)$. Any homeomorphism
that sends $S$ to $H(S)$ will map the base of $S$, $I^{2}\times\\{0\\}$ to the
base of $H(S)$ and the hairs of $S$ to the hairs of $H(S)$. Let $\gamma(t)$,
$t\in[0,1]$ be a parametrization of the longest hair of $S$ attached at
$(1/2,1/2,0)$. By [21, Theorem 4, Chapter 19] and our construction we have
that $H(\gamma(t))$ will be a wild arc and since it is a hair of $H(S)$ it
will be the image of a hair of $S$ under $h$. However this is a contradiction
since a wild arc cannot be the image of an arc under a homeomorphism of
$\mathbb{R}^{3}$ like $h$. ∎

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