# On self-affine tiles that are homeomorphic to a ball

Jörg M. Thuswaldner  and  Shu-Qin Zhang  Chair of Mathematics and Statistics,
University of Leoben, Franz-Josef-Strasse 18, A-8700 Leoben, Austria
<EMAIL_ADDRESS>School of Mathematics and Statistics,
Zhengzhou University, 100 Science Avenue, Zhengzhou, Henan 45001, People’s
Republic of China<EMAIL_ADDRESS>
###### Abstract.

Let $M$ be a $3\times 3$ integer matrix which is expanding in the sense that
each of its eigenvalues is greater than $1$ in modulus and let
$\mathcal{D}\subset\mathbb{Z}^{3}$ be a digit set containing $|\det M|$
elements. Then the unique nonempty compact set $T=T(M,\mathcal{D})$ defined by
the set equation $MT=T+\mathcal{D}$ is called an integral self-affine tile if
its interior is nonempty. If $\mathcal{D}$ is of the form
$\mathcal{D}=\\{0,v,\ldots,(|\det M|-1)v\\}$ we say that $T$ has a collinear
digit set. The present paper is devoted to the topology of integral self-
affine tiles with collinear digit sets. In particular, we prove that a large
class of these tiles is homeomorphic to a closed $3$-dimensional ball.
Moreover, we show that in this case $T$ carries a natural CW complex structure
that is defined in terms of the intersections of $T$ with its neighbors in the
lattice tiling $\\{T+z\,:\,z\in\mathbb{Z}^{3}\\}$ induced by $T$. This CW
complex structure is isomorphic to the CW complex defined by the truncated
octahedron.

###### Key words and phrases:

Self-affine sets, tiles and tilings, low dimensional topology, truncated
octahedron

###### 2020 Mathematics Subject Classification:

Primary: 28A80, 57M50. Secondary: 51M20, 52C22, 54F65.

Part of this paper was written during the conference “Numeration 2019” which
took place at the Erwin Schrödinger Institute in Vienna. The authors
acknowledge the hospitality and the convenient working conditions that were
provided there. Both authors are supported by the grant FWF W1230 funded by
the Austrian Science Fund.

##  1\. introduction

###  1.1. Context of the paper

The present paper is devoted to the study of the topology of $3$-dimensional
self-affine tiles.

Let $M\in\mathbb{Z}^{n\times n}$ be an integer matrix which is expanding in
the sense that each of its eigenvalues has modulus strictly greater than one.
Moreover, let $\mathcal{D}\subset\mathbb{Z}^{n}$ be a digit set with $|\det
M|$ elements.

Figure 1. Two examples of $3$-dimensional self-affine tiles.

Then it follows from the theory of iterated function systems (see e.g.
Hutchinson [17]) that there is a unique nonempty compact set
$T=T(M,\mathcal{D})$ such that

(1.1) $MT=T+\mathcal{D}.$

If $T$ has nonempty interior then it is called an integral self-affine tile,
or just a self-affine tile for short. If $\mathcal{D}$ is a complete set of
coset representatives of the residue class ring
$\mathbb{Z}^{n}/M\mathbb{Z}^{n}$, it is called a standard digit set. For
standard digit sets it is known that the nonempty compact set $T$ defined by
(1.1) always has nonempty interior (see Bandt [2]).

Self-affine tiles have been studied systematically since the 1990ies when
Bandt [2], Kenyon [19], Gröchenig and Haas [12], as well as Lagarias and Wang
[22, 23, 24] proved fundamental results on these objects. Since that time the
research on self-affine tiles developed in many different directions and they
play a role in various branches of mathematics like in the theory of dynamical
systems, in number theory, and in Fourier analysis and the construction of
wavelets. The present paper is concerned with the topology of self-affine
tiles. Since the seminal paper of Hata [15], the topology of self-affine sets
in general, and of self-affine tiles in particular, has been thoroughly
studied. Connectivity properties of self-affine tiles can be treated in a
satisfactory way in arbitrary dimension $n$ (see for instance Kirat and Lau
[20]). Further investigation of their topology often relies on the Jordan
curve theorem. For this reason, many papers on the topology of self-affine
tiles are restricted to the $2$-dimensional case. We refer for instance to
Bandt and Wang [4] or Leung and Lau [27] where homeomorphy to a disk was
investigated, or to Ngai and Tang [31, 32] for the study of self-affine tiles
with disconnected interior. Another interesting direction of research which
has relations to the Fuglede conjecture (cf. e.g. [11, 34]) is the
characterization of all digit sets $\mathcal{D}$ that give rise to a self-
affine tile $T(M,\mathcal{D})$ for a given expanding integer matrix $M$, see
for instance An and Lau [1], Lai et al. [26], and the survey by Lai and Lau
[25].

The present paper is devoted to the topology of $3$-dimensional self-affine
tiles. The systematic topological study of the $3$-dimensional case was
initiated some years ago when Bandt [3] considered the combinatorial topology
of some $3$-dimensional self-affine tiles. Later, Conner and Thuswaldner [6]
gave criteria for a self-affine tile to be a closed $3$-dimensional ball and
Deng et al. [9] dealt with self-affine tiles of a special form and showed that
they are $3$-dimensional balls. Kamae et al. [18] investigated a particular
class of $n$-dimensional self-affine tiles. Recently, Thuswaldner and Zhang
[35] studied a natural class of $3$-dimensional self-affine tiles and proved
that their boundary is homeomorphic to a $2$-sphere. It is this class of tiles
that we are interested in. Indeed, we want to explore if these tiles are
indeed homeomorphic to a $3$-dimensional ball, which means that we have to
exclude pathologies like the Alexander horned sphere which is known to occur
in the context of self-affine tiles (see [6, Section 8.2]).

###  1.2. Description of the main results

Our aim is to prove that a large class of well-known $3$-dimensional self-
affine tiles is homeomorphic to a closed $3$-dimensional ball. Moreover, we
will show that each tile in this class carries a natural CW complex structure
(see e.g. Hatcher [16, p. 5] for the definition of a CW complex).

Before we state our main results, we introduce some notation. Let $M$ be an
expanding $3\times 3$ integer matrix and let
$\mathcal{D}\subset\mathbb{Z}^{3}$ be a digit set such that the unique
nonempty compact set $T=T(M,\mathcal{D})$ defined by the set equation

(1.2) $T=\bigcup_{d\in\mathcal{D}}M^{-1}(T+d)$

has nonempty interior. Then $T$ is a self-affine tile. Define the set of
_neighbors_ of $T$ by

(1.3)
$\mathcal{S}=\\{\alpha\in\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}\;:\;T\cap(T+\alpha)\neq\emptyset\\}.$

Here

$\mathbb{Z}[M,\mathcal{D}]=\mathbb{Z}[\mathcal{D},M\mathcal{D},M^{2}\mathcal{D}]\subseteq\mathbb{Z}^{3}$

is the smallest $M$-invariant lattice containing $\mathcal{D}$. This
definition is motivated by the fact that the collection
$\\{T+z\,:\,z\in\mathbb{Z}[M,\mathcal{D}]\\}$ often tiles the space
$\mathbb{R}^{3}$ with overlaps of Lebesgue measure $0$ (cf. e.g. Lagarias and
Wang [24]). The translated tiles $T+\alpha$ with $\alpha\in\mathcal{S}$ are
then those tiles which “touch” (i.e., have nonempty intersection with) the
“central tile” $T$ in this tiling. It is clear that $\mathcal{S}$ is a finite
set since $T$ is compact by definition and $\mathbb{Z}[M,\mathcal{D}]$ is
discrete. For the sets in which $T$ intersects with one given other tile we
use the notation

(1.4)
$\boldsymbol{B}_{\alpha}=T\cap(T+\alpha)\qquad(\alpha\in\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}).$

More generally, for $\ell\geq 0$ we define the set of points in which $T$
intersects $\ell$ given other tiles by

(1.5)
$\boldsymbol{B}_{\boldsymbol{\alpha}}=\boldsymbol{B}_{\\{\alpha_{1},\ldots,\alpha_{\ell}\\}}=T\cap(T+\alpha_{1})\cap\cdots\cap(T+\alpha_{\ell})\qquad(\boldsymbol{\alpha}=\\{\alpha_{1},\ldots,\alpha_{\ell}\\}\subset\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}).$

Note in particular that $\boldsymbol{B}_{\emptyset}=T$. Compactness of $T$ and
discreteness of $\mathbb{Z}[M,\mathcal{D}]$ again ensures that there exist
only finitely many
$\boldsymbol{\alpha}\subset\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}$
satisfying $\boldsymbol{B}_{\boldsymbol{\alpha}}\not=\emptyset$.

We will be interested in the following class of self-affine tiles. Let $M$ be
an expanding $3\times 3$ integer matrix. We call
$\mathcal{D}\subset\mathbb{Z}^{3}$ a collinear digit set for $M$ if there is a
vector $v\in\mathbb{Z}^{3}\setminus\\{0\\}$ such that

(1.6) $\mathcal{D}=\\{0,v,2v,\ldots,(|\det M|-1)v\\}.$

If $\mathcal{D}$ has this form, a self-affine tile111Note that we assume here
that $T(M,\mathcal{D})$ is a self-affine tile. This does not follow from the
collinearity of $\mathcal{D}$. $T=T(M,\mathcal{D})$ is called a self-affine
tile with collinear digit set (such tiles have been studied intensively in
recent years, cf. e.g. [27, 35]).

For $k\geq 0$ denote the $k$-dimensional unit ball by
$\mathbb{D}^{k}=\\{x\in\mathbb{R}^{k}:\|x\|_{2}\leq 1\\}\subset\mathbb{R}^{k}$
($\|\cdot\|_{2}$ is the Euclidean norm). We note that $\mathbb{D}^{0}$ is a
single point. A closed $k$-cell or $k$-ball is a topological space that is
homeomorphic to $\mathbb{D}^{k}$.

Our first main result shows that a large class of self-affine tiles with
collinear digit sets are $3$-balls.

######  Theorem 1.1.

Let $T=T(M,\mathcal{D})$ be a $3$-dimensional self-affine tile with collinear
digit set and assume that the characteristic polynomial
$\chi(x)=x^{3}+Ax^{2}+Bx+C$ of $M$ satisfies $1=A\leq B<C$. If $T$ has $14$
neighbors then $T$ is a $3$-ball.

######  Remark 1.2.

Let $T=T(M,D)$ be a $3$-dimensional self-affine tile with collinear digit set.
If the coefficients $A,B,C$ of the characteristic polynomial
$\chi(x)=x^{3}+Ax^{2}+Bx+C$ of $M$ satisfy $1=A\leq B<C$ then the matrix $M$
is expanding (see [35, Lemma 2.2]). Moreover, according to [35, Theorem 1.1]
the collection $\\{T+\alpha\,:\,\alpha\in\mathbb{Z}[M,\mathcal{D}]\\}$ tiles
the space $\mathbb{R}^{3}$ with overlaps of Lebesgue measure $0$.

######  Remark 1.3.

According to [35, Theorem 1.4] a $3$-dimensional self-affine tile $T=T(M,D)$
with collinear digit set and characteristic polynomial
$\chi(x)=x^{3}+Ax^{2}+Bx+C$ of $M$ satisfying $1\leq A\leq B<C$ has $14$
neighbors if and only if one of the following conditions holds:

  * •

$1\leq A<B<C$ and $B\geq 2A-1,C\geq 2(B-A)+2$;

  * •

$1\leq A<B<C$ and $B<2A-1,C\geq A+B-2$.

We believe that similar criteria can also be established if negative
coefficients are allowed.

######  Remark 1.4.

We conjecture that, apart from sporadic cases (as, for instance, the ones
studied in [3]), $3$-dimensional self-affine tiles with collinear digit set
having more than $14$ neighbors are not homeomorphic to a $3$-ball. In the
$2$-dimensional case, only self-affine tiles with a small number of neighbors
have a nice topological structure (see [4]; we refer to [29, 31, 32] for
$2$-dimensional tiles with wild topology).

Our second main result shows that the sets
$\boldsymbol{B}_{\boldsymbol{\alpha}}$ defined in (1.5) provide a natural CW
complex structure on $T$.

######  Theorem 1.5.

Let $T=T(M,\mathcal{D})$ be a $3$-dimensional self-affine tile with collinear
digit set and assume that the characteristic polynomial
$\chi(x)=x^{3}+Ax^{2}+Bx+C$ of $M$ satisfies $1=A\leq B<C$. If $T$ has $14$
neighbors then $T$ carries the following natural CW complex structure.

  * •

The closed $0$-cells are the $24$ nonempty sets
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}$ with
$\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}\subset\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}$.

  * •

The closed $1$-cells are the $36$ nonempty sets
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}$ with
$\\{\alpha_{1},\alpha_{2}\\}\subset\mathbb{Z}[M,\mathcal{D}]\setminus\\{0\\}$.

  * •

The closed $2$-cells are the $14$ nonempty sets $\boldsymbol{B}_{\alpha_{1}}$
with $\alpha_{1}\in\mathcal{S}$.

  * •

The closed $3$-cell is $\boldsymbol{B}_{\emptyset}$.

For $i\in\\{1,2,3\\}$, the closed $i$-cell
$\boldsymbol{B}_{\boldsymbol{\alpha}}$, $\\#\boldsymbol{\alpha}=3-i$, is
attached to the $(i-1)$-skeleton $T^{i-1}$ by attaching its boundary
$\partial\boldsymbol{B}_{\boldsymbol{\alpha}}$ (as a manifold) to the
$(i-1)$-sphere

$\bigcup_{\alpha\not\in\boldsymbol{\alpha}}\boldsymbol{B}_{\boldsymbol{\alpha}\cup\\{\alpha\\}}.$

This CW complex is isomorphic to the natural CW complex structure of a
truncated octahedron.

######  Remark 1.6.

In the literature (see e.g. Hatcher [16, p. 5]), an (open) $k$-cell of a CW
complex is a topological space that is homeomorphic to the open unit ball in
$\mathbb{R}^{k}$ for $k\geq 0$ (a $0$-cell is a single point). The $k$-cells
of the CW complex defined in Theorem 1.5 are the nonempty sets
$\mathrm{Int}(\boldsymbol{B}_{\boldsymbol{\alpha}})$ with $|\alpha|=3-k$
($0\leq k\leq 3$). Here, for a $k$-manifold $\mathcal{M}$ with boundary,
$\mathrm{Int}(\mathcal{M})$ denotes the set of $x\in\mathcal{M}$ having a
neighborhood that is homeomorphic to a $k$-cell (contrary to the topological
interior $X^{\circ}$ of a set $X$ w.r.t. some ambient space). We use closed
cells for notational convenience.

Figure 2. The CW complex structure of a self-affine tile $T$.

In Figure 2 we visualize the CW complex structure of the self-affine tile $T$
on the right hand side of Figure 1. The whole tile
$T=\boldsymbol{B}_{\emptyset}$ is a closed $3$-cell. Each of the patches is
homeomorphic to a closed $2$-cell $\boldsymbol{B}_{\alpha}$ for some
$\alpha\in\mathcal{S}$. The union of these patches forms the $2$-sphere
$\partial T$. Two distinct closed $2$-cells meet in a closed $1$-cell
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}$, and three closed $2$-cells
meet in a single point of the form
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}$. If we consider open
cells, then clearly $T$ can be written as the disjoint union

$T=\coprod_{\boldsymbol{\alpha}\subset\mathbb{Z}^{3}}\mathrm{Int}(\boldsymbol{B}_{\boldsymbol{\alpha}}).$

In our proofs we need new ideas because the criterion for the homeomorphy of a
self-affine tile to a $3$-ball established in [6] is applicable only to single
tiles, while the theories developed in [9, 18] just cover tiles of a
particular shape. Our proofs use the theory of Bing [5] that leads to a
topological characterization of $k$-spheres for $k\leq 3$. However, since our
conditions differ from the ones of Bing, our proof differs from Bing’s proofs
and exploits the self-affinity of our tiles.

We have some hope that our theory can be applied to the case $A\geq 2$ as
well. However, this generalization would require more case studies and tedious
calculations. If negative coefficients $A,B,C$ are permitted, according
classes of expanding matrices can be studied. Moreover, Kwun [21] and O. G.
Harrold [13, 14] establish higher dimensional generalizations of the results
of Bing [5] that we are using here. These results can probably be used to
extend our theory to higher dimensions.

###  1.3. Outline of the paper

The paper is organized as follows. In Section 2 we provide preliminaries and
basic notions that will be of importance in the proofs of our main results.
This includes some graphs that are commonly used in the study of the topology
of self-affine tiles and a description of a tiling induced by the truncated
octahedron. This tiling is used as a model for the tiling induced by a self-
affine tile taken from the class we are interested in. Section 3 describes
intersections of subtiles of a self-affine tile. These intersections, which
will play an important role in the proofs of our main results, are captured by
a large graph, that will be studied in some detail. Finally, Section 4 gives
an account on the theory of partitionings due to Bing [5] and defines
particular sequences of partitionings that are suitable for our purposes.
Finally, these sequences of partitionings are used to establish Theorem 1.1.
Combining Theorem 1.1 with results from [35] finally leads to the proof of
Theorem 1.5.

##  2\. Intersections of self-affine tiles and CW complexes

In this section we set up some preliminaries. In Section 2.1 we provide some
basic properties of self-affine tiles that will be needed in the sequel. In
Section 2.2 we recall that each $3$-dimensional self-affine tile with
collinear digit set has a normal form, a so-called $ABC$-tile. This entails
that in the sequel we can restrict ourselves to the investigation of this
class of tiles without loss of generality. After that, in Section 2.3 we
recall the notion of neighbor graph that permits us to study intersections of
the form $T\cap(T+\alpha)$ for an $ABC$-tile $T$. Section 2.4 is devoted to
the Hata graph, a graph that surveys the intersections between the sets
$T+\alpha$, $\alpha\in\mathcal{S}$, and gives some results related to this
graph. Finally, in Section 2.5 we relate an $ABC$-tile $T$ with $14$ neighbors
and its lattice tiling to the so-called bitruncated cubic honeycomb, a lattice
tiling of $\mathbb{R}^{3}$ by truncated octahedra.

###  2.1. Basic properties of self-affine tiles

Let $M\in\mathbb{Z}^{3\times 3}$ and $\mathcal{D}\subset\mathbb{Z}^{3}$ be
given in a way that $T=T(M,\mathcal{D})$ is a self-affine tile. Let

(2.1)
$\mathcal{D}_{i}=\mathcal{D}+M\mathcal{D}+\dots+M^{i-1}\mathcal{D}\qquad(i\in\mathbb{N})$

and define the empty sum $\mathcal{D}_{0}$ to be equal to the vector
$0\in\mathbb{R}^{3}$. Iterating the set equation (1.2) for $i\in\mathbb{N}$
times yields

(2.2) $T=\bigcup_{d\in\mathcal{D}_{i}}M^{-i}(T+d).$

If $\mu$ denotes the Lebesgue measure in $\mathbb{R}^{3}$ we have

(2.3)
$\mu((T+d_{1})\cap(T+d_{2}))=0\qquad(d_{1},d_{2}\in\mathcal{D}_{i},\,d_{1}\neq
d_{2}),$

i.e., the sets in the union on the right hand side of (2.2) are mutually
essentially disjoint (cf. [23, (3.11)]). For this reason each set of the form
$M^{-i}(T+d)$ with $i\in\mathbb{N}$ and $d\in\mathcal{D}_{i}$ is called a
subtile of $T$. Accordingly, $M^{-k}(t+z)$ is called a subtile of
$M^{-k}(T+z)$ if $t$ is a subtile of $T$ ($k\in\mathbb{N}$ and
$z\in\mathcal{D}_{k}$).

Because $T$ is a self-affine tile it has nonempty interior. Thus the following
is true by [23, Theorem 1.1].

######  Lemma 2.1.

A self-affine tile $T$ is equal to the closure of its interior. Its boundary
$\partial T$ has zero Lebesgue measure.

Let $t_{1},t_{2}$ be two distinct subtiles of $T$. It is clear from the
measure disjointness of the union in (2.2) that either $t_{1}\subset t_{2}$,
or $t_{2}\subset t_{1}$, or $\mu(t_{1}\cap t_{2})=0$. Lemma 2.1 implies that

(2.4) $\mu(t_{1}\cap t_{2})=0\;\Longleftrightarrow\;t_{1}^{\circ}\cap
t_{2}^{\circ}=\emptyset\;\Longleftrightarrow\;t_{1}\cap t_{2}=\partial
t_{1}\cap\partial t_{2}.$

In the sequel we will often tacitly make use of these equivalences.

###  2.2. A normal form

For the tiles of our main results we now define a simple normal form. Let
$A,B,C\in\mathbb{N}$ with $1\leq A\leq B<C$ be given and set

(2.5) $M=\begin{pmatrix}0&0&-C\\\ 1&0&-B\\\ 0&1&-A\\\
\end{pmatrix}\quad\hbox{and}\quad\mathcal{D}=\left\\{\begin{pmatrix}0\\\ 0\\\
0\\\ \end{pmatrix},\begin{pmatrix}1\\\ 0\\\ 0\\\
\end{pmatrix},\dots,\begin{pmatrix}C-1\\\ 0\\\ 0\\\ \end{pmatrix}\right\\}.$

The matrix $M$ is expanding by [35, Lemma 2.2]. Moreover, $\mathcal{D}$ is a
complete set of coset representatives of $\mathbb{Z}^{3}/M\mathbb{Z}^{3}$ and
$\mathbb{Z}[M,\mathcal{D}]=\mathbb{Z}^{3}$. Define $T$ by $MT=T+\mathcal{D}$.
Then $T$ is a self-affine tile. We call such a tile $T$ an _$ABC$ -tile_. We
know from [35, Lemma 2.4] that an $ABC$-tile $T$ tiles $\mathbb{R}^{3}$ by
$\mathbb{Z}^{3}$-translates in the sense that
$T+\mathbb{Z}^{3}=\mathbb{R}^{3}$, where $(T+\alpha_{1})\cap(T+\alpha_{2})$
has Lebesgue measure $0$ for all $\alpha_{1},\alpha_{2}\in\mathbb{Z}^{3}$ with
$\alpha_{1}\not=\alpha_{2}$. We thus say that
$\\{T+z\,:\,z\in\mathbb{Z}^{3}\\}$ forms a tiling of $\mathbb{R}^{3}$.

It turns out that we can confine ourselves to the study of $ABC$-tiles.
Indeed, let $M^{\prime}$ be a $3\times 3$ integer matrix with characteristic
polynomial $\chi(x)=x^{3}+Ax^{2}+Bx+C$ satisfying $1\leq A\leq B<C$. By [35,
Lemma 2.2] we know that $M^{\prime}$ is an expanding matrix. Let
$v\in\mathbb{Z}^{3}\setminus\\{0\\}$ and let
$\mathcal{D}^{\prime}=\\{0,v,2v,\dots,(C-1)v\\}\subset\mathbb{Z}^{3}$ be a
collinear digit set such that
$T^{\prime}=T^{\prime}(M^{\prime},\mathcal{D}^{\prime})$ is a self-affine
tile. Let $T=T(M,\mathcal{D})$ be the $ABC$-tile with characteristic
polynomial $\chi$. From [35, Section 2.1] we know that there is a regular
matrix $E$ such that the linear mapping $E:\mathbb{R}^{3}\to\mathbb{R}^{3}$
maps $\mathbb{Z}^{3}$ bijectively onto
$\mathbb{Z}[M^{\prime},\mathcal{D}^{\prime}]$, that $T^{\prime}=ET$, and that
for each
$\\{\alpha_{1},\ldots,\alpha_{\ell}\\}\subset\mathbb{Z}[M^{\prime},\mathcal{D}^{\prime}]\setminus\\{0\\}$
we have

$T^{\prime}\cap(T^{\prime}+E\alpha_{1})\cap\dots\cap(T^{\prime}+E\alpha_{\ell})=E(T\cap(T+\alpha_{1})\cap\dots\cap(T+\alpha_{\ell})).$

It is therefore sufficient to prove Theorems 1.1 and 1.5 for $ABC$-tiles and
in all what follows we may restrict our attention to this class of self-affine
tiles.

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile and let $z\in\mathcal{D}_{i}$ for
some $i\geq 0$. Because $\mathcal{D}$ is a standard digit set there exist
unique elements $e_{0},\ldots,e_{i-1}\in\\{0,\ldots,C-1\\}$ such that

$z=\sum_{j=0}^{i-1}M^{j}\begin{pmatrix}e_{j}\\\ 0\\\ 0\end{pmatrix}.$

In this case we write

(2.6) $z=(e_{i-1},\ldots,e_{0})_{M}.$

This notation will prove particularly useful for $i=1$ to write digits in a
space-saving way.

###  2.3. The neighbor graph

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile. In the sequel we will need the so-
called _neighbor graph_ (see e.g. [33]), a graph that can be used to describe
the intersections $\boldsymbol{B}_{\alpha}=T\cap(T+\alpha)$ for
$\alpha\in\mathcal{S}$. We begin by recalling some definitions from graph
theory. For a directed labeled graph $G$ with set of nodes $V$, set of edges
$E$, and set of edge-labels $L$, we write an edge leading from $v\in V$ to
$v^{\prime}\in V$ labeled by $\ell\in L$ as $v\xrightarrow{\ell}v^{\prime}$.
In this case $v$ is called a _predecessor_ of $v^{\prime}$ and $v^{\prime}$ is
called a _successor_ of $v$. Following e.g. Diestel [10, Chapter 1] a (finite
or infinite) sequence
$v_{0}\xrightarrow{\ell_{1}}v_{1}\xrightarrow{\ell_{2}}v_{2}\xrightarrow{\ell_{3}}\cdots$
of consecutive edges in $G$ is called a _walk_. A walk whose nodes
$v_{0},v_{1},v_{2},\ldots$ are mutually distinct is called a _path_. If $G$ is
undirected and not labeled, then an edge of $G$ connecting the nodes $v$ and
$v^{\prime}$ is denoted by $v\,\mbox{---}\,v^{\prime}$. Walks and paths in $G$
are defined as in the directed case as sequences of consecutive edges with or
without possible repetitions, respectively. The length of a walk is its number
of edges. A walk of length $n$ in $G$ that starts and ends at the same node is
called an $n$-cycle if it contains a path of length $n-1$. $G$ is called
connected if for each pair $(v,v^{\prime})$ of distinct nodes of $G$ there is
a path of the form $v\,\mbox{---}\cdots\mbox{---}\,v^{\prime}$.

######  Definition 2.2 (Neighbor graph; cf. [33, Section 2]).

Let $M\in\mathbb{Z}^{3\times 3}$ and $\mathcal{D}\subset\mathbb{Z}^{3}$ be
given in a way that $T=T(M,\mathcal{D})$ is an $ABC$-tile with neighbor set
$\mathcal{S}$. Define the directed labeled neighbor graph $G(\mathcal{S})$ as
follows. The nodes of $G(\mathcal{S})$ are the neighbors $\mathcal{S}$, and
there is a labeled edge

(2.7) $\displaystyle\alpha\xrightarrow{d|d^{\prime}}\alpha^{\prime}\quad\text{
if and only if }M\alpha+d^{\prime}-d=\alpha^{\prime}\text{ with
}\alpha,\alpha^{\prime}\in\mathcal{S}\text{ and }d,d^{\prime}\in\mathcal{D}.$

In (2.7) the vector $d^{\prime}$ is determined by $\alpha,\alpha^{\prime},d$.
Thus we will often just write $\alpha\xrightarrow{d}\alpha^{\prime}$ instead
of $\alpha\xrightarrow{d|d^{\prime}}\alpha^{\prime}$. The notation $\alpha\in
G(\mathcal{S})$ means that $\alpha$ is a node of $G(\mathcal{S})$ and
$\alpha\xrightarrow{d}\alpha^{\prime}\in G(\mathcal{S})$ means that
$\alpha\xrightarrow{d}\alpha^{\prime}$ is an edge of $G(\mathcal{S})$. For
walks we will use an analogous notation.

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile. Because
$\\{T+z\;:\;z\in\mathbb{Z}^{3}\\}$ forms a tiling of $\mathbb{R}^{3}$, we have

(2.8) $\partial T=\bigcup_{\alpha\in\mathcal{S}}\boldsymbol{B}_{\alpha}.$

Here $\mathcal{S}$ and $\boldsymbol{B}_{\alpha}$, $\alpha\in\mathcal{S}$, are
given by (1.3) and (1.4), respectively (note that
$\mathbb{Z}[M,\mathcal{D}]=\mathbb{Z}^{3}$ in these definitions because $T$ is
an $ABC$-tile). One can show (see e.g. [33, Proposition 2.2]) that the
nonempty compact sets $\boldsymbol{B}_{\alpha}$, $\alpha\in\mathcal{S}$, are
uniquely determined by the set equations

(2.9)
$\boldsymbol{B}_{\alpha}=\bigcup_{\begin{subarray}{c}d\in\mathcal{D},\alpha^{\prime}\in\mathcal{S}\\\
\alpha\xrightarrow{d}\alpha^{\prime}\in
G(\mathcal{S})\end{subarray}}M^{-1}(\boldsymbol{B}_{\alpha^{\prime}}+d)\qquad(\alpha\in\mathcal{S}).$

Here the union on the right hand side of (2.9) is extended over all
$d,\alpha^{\prime}$ with $\alpha\xrightarrow{d}\alpha^{\prime}\in
G(\mathcal{S})$. The defining equation (2.9) is an instance of a graph-
directed iterated function system. These objects were first studied in [30].
By (2.8) and (2.9) the boundary $\partial T$ is determined by the graph
$G(\mathcal{S})$.

Figure 3.  The neighbor graph $G(\mathcal{S})$ for an $ABC$-tile $T$ with
$1\leq A\leq B<C$ having $14$ neighbors. Here we set
$P=(1,0,0)^{t},~{}Q=(A,1,0)^{t},~{}N=(B,A,1)^{t}$. To save space we write
$\alpha\xrightarrow{e}\alpha^{\prime}$ instead of
$\alpha\xrightarrow{(e)_{M}}\alpha^{\prime}$ in this figure (recall the
notation (2.6)). Multiple labels correspond to multiple edges. If an edge has
labels $d,\ldots,d^{\prime}$ with $d>d^{\prime}$ then the edge has to be
deleted.

The set $\mathcal{S}$ as well as the neighbor graph $G(\mathcal{S})$ of an
$ABC$-tile $T=T(M,\mathcal{D})$ can be calculated explicitly. In the present
paper we are interested in $ABC$-tiles having $14$ neighbors (observe the
characterization in Remark 1.3). In [35, Section 2.4] the following results
have been proved. Suppose that $T$ has $14$ neighbors. Then the neighbor set
$\mathcal{S}$ and the neighbor graph $G(\mathcal{S})$ are given as follows.
Set

$\mathcal{S}_{1}=\\{P,Q,N,Q-P,N-P,N-Q,N-Q+P\\},$

where $P=(1,0,0)^{t}$, $Q=(A,1,0)^{t}$, and $N=(B,A,1)^{t}$. Then the
$ABC$-tile $T$ has the neighbors
$\mathcal{S}=\mathcal{S}_{1}\cup(-\mathcal{S}_{1})$. Moreover, in this case
the neighbor graph $G(\mathcal{S})$ is given by the graph in Figure 3.

######  Remark 2.3.

This neighbor graph is strongly related to the de Bruijn graph $N_{4}$ of
binary words of length $4$ (see [8, Section 3]). Indeed, if we delete the
nodes corresponding to the words $0000$ and $1111$ in $N_{4}$ we get the graph
in Figure 3 (apart from the edge labels).

###  2.4. The Hata graph and Peano continua

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile. The Hata graph $H(\mathcal{S})$ of
the neighbors of $T$ is defined as follows. The nodes of $H(\mathcal{S})$ are
the elements of $\mathcal{S}$ and there is an undirected edge between two
distinct elements $\alpha_{1},\alpha_{2}\in\mathcal{S}$ if and only if
$(T+\alpha_{1})\cap(T+\alpha_{2})\not=\emptyset$. For an $ABC$-tile with $14$
neighbors the Hata graph $H(\mathcal{S})$ is depicted in Figure 4. It can be
determined by using [35, Lemma 2.16] (see also [35, Figure 9]). The following
lemma is a reformulation of some basic results from [35, Section 2].

Figure 4. The Hata graph $H(\mathcal{S})$ (left) which is isomorphic to the
graph of vertices and edges of the so-called tetrakis hexahedron. The tetrakis
hexahedron (right) is a Catalan polyhedron which is the dual of the truncated
octahedron (see e.g. [7, p. 284]).

######  Lemma 2.4.

Let $T$ be an $ABC$-tile with $14$ neighbors. Let
$\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{Z}^{3}\setminus\\{0\\}$ be
mutually distinct. We have

  * (1)

$\boldsymbol{B}_{\alpha_{1}}\not=\emptyset$ if and only if $\alpha_{1}$ is a
node of $H(\mathcal{S})$.

  * (2)

$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}\not=\emptyset$ if and only if
$\alpha_{1}\relbar\\!\relbar\\!\relbar\alpha_{2}$ is an edge in
$H(\mathcal{S})$.

  * (3)

$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}\not=\emptyset$ if and
only if there is a 3-cycle with nodes $\alpha_{1},\alpha_{2},\alpha_{3}$ in
$H(\mathcal{S})$.

  * (4)

If $\boldsymbol{\alpha}\subset\mathbb{Z}^{3}\setminus\\{0\\}$ has more than
three elements then $\boldsymbol{B}_{\boldsymbol{\alpha}}=\emptyset$.

###### Proof.

Item (1) follows because the nodes of $H(\mathcal{S})$ are the neighbors of
$T$. Item (2) follows from the definition of the edges of $H(\mathcal{S})$.
Items (3) and (4) follow from [35, Lemma 2.16]. For (3) one just has to check
that the nodes of the graph $G_{3}(\mathcal{S})$ defined in [35, Figure 6] are
in one-to-one correspondence with the $3$-cycles of $H(\mathcal{S})$. ∎

The Hata graph $H(\mathcal{S})$ and some other Hata graphs are used in the
proof of the following lemma.

######  Lemma 2.5.

Let $T$ be an $ABC$-tile with $14$ neighbors. Then $T$ and $\partial T$ are
Peano continua.

###### Proof.

Since $P\in\mathcal{S}$, we have $M^{-1}\boldsymbol{B}_{P}=M^{-1}T\cap
M^{-1}(T+P)\not=\emptyset$. Thus $T$ is a Peano continuum by (1.2) and [15,
Theorem 4.6].

Next we prove that $\boldsymbol{B}_{\alpha}$ is a Peano continuum for each
$\alpha\in\mathcal{S}$. For $\alpha\in\mathcal{S}$ let

$Z_{\alpha}=\\{M^{-1}(\boldsymbol{B}_{\alpha^{\prime}}+d)\;:\;d\in\mathcal{D},\alpha^{\prime}\in\mathcal{S}\hbox{
such that }\alpha\xrightarrow{d}\alpha^{\prime}\in G(\mathcal{S})\hbox{
exists}\\}$

be the collection of the sets in the union on the right hand side of (2.9).
The Hata graph of $Z_{\alpha}$ is the undirected graph $H_{\alpha}$ whose
nodes are the elements of $Z_{\alpha}$ and that has an edge between two
distinct elements of $b_{1},b_{2}\in Z_{\alpha}$ if and only if $b_{1}\cap
b_{2}\not=\emptyset$. According to [35, Lemma 3.3] (see also [28, Theorem
4.1]), to establish the claim we have to prove that $H_{\alpha}$ is connected
for each $\alpha\in\mathcal{S}$. To this matter we have to construct the
graphs $H_{\alpha}$. This is done by checking whether intersections of the
form $b_{1}\cap b_{2}$ with distinct $b_{1},b_{2}\in Z_{\alpha}$ are empty or
not. Since $b_{1}=M^{-1}((T+d_{1})\cap(T+d_{1}+\alpha_{1}))$ and
$b_{2}=M^{-1}((T+d_{2})\cap(T+d_{2}+\alpha_{2}))$ with some
$d_{1},d_{2}\in\mathcal{D}$ and some $\alpha_{1},\alpha_{2}\in\mathcal{S}$,

(2.10) $b_{1}\cap
b_{2}=M^{-1}((T+d_{1})\cap(T+d_{1}+\alpha_{1})\cap(T+d_{2})\cap(T+d_{2}+\alpha_{2})).$

Set
$\boldsymbol{\alpha}=\\{\alpha_{1},d_{2}-d_{1},\alpha_{2}+d_{2}-d_{1}\\}\setminus\\{0\\}$.
Then $b_{1}\cap b_{2}$ is an affine image of
$\boldsymbol{B}_{\boldsymbol{\alpha}}$, where
$|\boldsymbol{\alpha}|\in\\{2,3\\}$ depending on whether the four translates
$\\{d_{1},d_{1}+\alpha_{1},d_{2},d_{2}+\alpha_{2}\\}$ in (2.10) are mutually
distinct or not. But if $\boldsymbol{B}_{\boldsymbol{\alpha}}$ is empty or not
can be read off the Hata graph $H(\mathcal{S})$ in view of Lemma 2.4. For
$\alpha=P$ we see from Figure 3 that the nodes of $H_{P}$ are

$Z_{P}=\\{M^{-1}(\boldsymbol{B}_{Q}+(e)_{M})\;:\;0\leq e\leq
C-A-1\\}\cup\\{M^{-1}(\boldsymbol{B}_{Q-P}+(e)_{M})\;:\;0\leq e\leq C-A\\}.$

Let $b_{1},b_{2}\in Z_{P}$ be distinct. Inspecting $H(\mathcal{S})$ (or
directly from [35, Corollary 3.23]) we see that the Hata graph $H_{P}$ is the
line given in Figure 5, and, hence, $H_{P}$ is connected.

Figure 5. The Hata graph $H_{P}$ (we omit the multiplication by $M^{-1}$ and
write $e$ instead of $(e)_{M}$ to save space).

Analogously we see that $H_{\alpha}$ is a line or a single node and, hence,
connected for each $\alpha\in\mathcal{S}\setminus\\{P\\}$ as well. Thus [35,
Lemma 3.3] yields that $\boldsymbol{B}_{\alpha}$ is a Peano continuum222It is
easy to see from (2.9) that $\boldsymbol{B}_{\alpha}$ is not a single point
($\alpha\in\mathcal{S}$). for each $\alpha\in\mathcal{S}$.

Since $T$ is connected, $\partial T$ is connected as well by [28, Theorem
1.2]. Therefore, by (2.8), $\partial T$ is a connected union of finitely many
Peano continua and, hence, a Peano continuum. ∎

The fact that $\partial T$ is a Peano continuum is not used in the present
paper. However, it is tacitly used in [35, Section 3.4] without giving a
formal proof (although in [35, Corollary 3.23 and Lemma 3.3] all ingredients
for the proof are provided). Thus we decided to prove it here before we state
the following version of the main result of [35], which is formulated by using
$H(\mathcal{S})$.

######  Proposition 2.6.

Let $T$ be an $ABC$-tile with $14$ neighbors and let
$\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{Z}^{3}\setminus\\{0\\}$ be
mutually distinct. Then the following assertions hold.

  * (1)

$\boldsymbol{B}_{\alpha_{1}}$ is a $2$-ball if $\alpha_{1}\in\mathcal{S}$, and
$\boldsymbol{B}_{\alpha_{1}}=\emptyset$ otherwise.

  * (2)

$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}$ is a $1$-ball if there is an
edge $\alpha_{1}\relbar\\!\relbar\\!\relbar\alpha_{2}$ in $H(\mathcal{S})$,
and $\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}=\emptyset$ otherwise.
Moreover, for each $\alpha_{1}\in\mathcal{S}$ we have

$\bigcup_{\alpha_{2}:\;\alpha_{1}\relbar\\!\relbar\\!\relbar\alpha_{2}\in
H(\mathcal{S})}\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2}\\}}\simeq\mathbb{S}^{1}.$

  * (3)

$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}$ is a $0$-ball if
there is a 3-cycle
$\alpha_{1}\relbar\\!\relbar\\!\relbar\alpha_{2}\relbar\\!\relbar\\!\relbar\alpha_{3}\relbar\\!\relbar\\!\relbar\alpha_{1}$
in $H(\mathcal{S})$, and
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}=\emptyset$ otherwise.

  * (4)

If $\boldsymbol{\alpha}\subset\mathbb{Z}^{3}\setminus\\{0\\}$ has more than
three elements then $\boldsymbol{B}_{\boldsymbol{\alpha}}=\emptyset$.

###### Proof.

Assertion (1) is the content of [35, Theorem 1.1 (2)]. Assertion (2) follows
from [35, Proposition 3.10 (2)] and Lemma 2.4 (2). To see assertion (3)
observe that in [35, Section 3.1] it is shown that
$\boldsymbol{B}_{\\{\alpha_{1},\alpha_{2},\alpha_{3}\\}}$ is either a
singleton or empty. Thus (3) follows from Lemma 2.4 (3). Assertion (4) is just
Lemma 2.4 (4). ∎

###  2.5. On the topology of certain subsets of $\partial T$

Let $M\in\mathbb{Z}^{3\times 3}$ and $\mathcal{D}\subset\mathbb{Z}^{3}$ be
given in a way that $T=T(M,\mathcal{D})$ is an $ABC$-tile. Suppose that $T$
has $14$ neighbors. In what follows we will need precise information on the
topology of the subsets

(2.11) $U(R)=\bigcup_{\alpha\in
R}\boldsymbol{B}_{\alpha}\qquad(R\subseteq\mathcal{S})$

of the boundary $\partial T$.

Let $O$ be a truncated octahedron whose sides are labeled by the elements of
$\mathcal{S}$ in the way shown on the left hand side of Figure 6, with the
convention that the side opposite to the side labeled with
$\alpha\in\mathcal{S}$ is labeled with $-\alpha$. We denote the face of $O$
labeled with $\alpha\in\mathcal{S}$ by $O_{\alpha}$. Moreover, for
$\boldsymbol{\alpha}\subseteq\mathcal{S}$ we define the intersections

(2.12)
$O_{\boldsymbol{\alpha}}=\bigcap_{\alpha\in\boldsymbol{\alpha}}O_{\alpha}$

with the convention that $O_{\emptyset}=O$.

Figure 6. A truncated octahedron and a patch of the bitruncated cubic
honeycomb.

It is well-known that $O$ induces a tiling of the $3$-dimensional Euclidean
space: the so-called bitruncated cubic honeycomb (see the left hand side of
Figure 6 for a patch of this tiling). This tiling has the same “intersection
structure” as $\\{T+z\,:\,z\in\mathbb{Z}^{3}\\}$. In particular, comparing the
labeled octahedron $O$ from Figure 6 with Proposition 2.6 we see that the
following result holds.

######  Lemma 2.7.

Let $T$ be an $ABC$-tile with $14$ neighbors. For each nonempty
$\boldsymbol{\alpha}\subseteq\mathcal{S}$ we have

$\boldsymbol{B}_{\boldsymbol{\alpha}}\simeq O_{\boldsymbol{\alpha}}.$

Moreover, we get the following topological characterization of the sets
$U(R)$.

######  Lemma 2.8.

Let $T$ be an $ABC$-tile with $14$ neighbors. Let $R\subseteq\mathcal{S}$ be
given. Then

(2.13) $U(R)\simeq\bigcup_{\alpha\in R}O_{\alpha}.$

Here $U(R)$ is as in (2.11).

###### Proof.

Denote the right hand side of (2.13) by $U^{\prime}(R)$. It is easy to see
that $U^{\prime}(R)$ is a CW complex333Again we use closed cells instead of
open ones for convenience. (cf. [16, p. 5]). Indeed, for $i\in\\{0,1,2\\}$ the
closed $i$-cells are given by the nonempty sets $O_{\boldsymbol{\alpha}}$ with
$\boldsymbol{\alpha}\subseteq\mathcal{S}$, $\boldsymbol{\alpha}\cap
R\neq\emptyset$, and $\\#\boldsymbol{\alpha}=3-i$. Thus the $0$-skeleton
$U^{\prime}(R)^{0}$ is the set of vertices of $U^{\prime}(R)$. Each closed
$1$-cell $O_{\\{\alpha_{1},\alpha_{2}\\}}$ is attached to the two closed
$0$-cells $O_{\boldsymbol{\alpha}}$ satisfying
$\boldsymbol{\alpha}\supset\\{\alpha_{1},\alpha_{2}\\}$ and
$\\#\boldsymbol{\alpha}=3$. This yields the $1$-skeleton $U^{\prime}(R)^{1}$.
To get $U^{\prime}(R)$ we attach each closed $2$-cell $O_{\alpha_{1}}$,
$\alpha_{1}\in R$, to the circle
$\bigcup_{\alpha_{2}\in\mathcal{S}:\alpha_{2}\not=\alpha_{1}}O_{\\{\alpha_{1},\alpha_{2}\\}}$.

From Proposition 2.6 we see that the set $U(R)$ is a CW complex whose closed
$i$-cells are given by the nonempty sets
$\boldsymbol{B}_{\boldsymbol{\alpha}}$ with
$\boldsymbol{\alpha}\subseteq\mathcal{S}$, $\boldsymbol{\alpha}\cap
R\neq\emptyset$, and $\\#\boldsymbol{\alpha}=3-i$ for $i\in\\{0,1,2\\}$ with
analogous attaching rules as above.

Thus, by Lemma 2.7, $U(R)$ and $U^{\prime}(R)$ have isomorphic CW complex
structures, hence, they are isomorphic as topological spaces. ∎

This lemma reduces the problem of determining the topology of $U(R)$ to a
simple combinatorial problem. In Figure 7 we give two examples. The one on the
left hand side shows that $U(R)$ is a $2$-ball if $R=\\{P,N-Q,N-Q+P\\}$, from
the second one we immediately see that $U(R)$ is the union of $2$ disjoint
$2$-balls if $R=\\{N,N-P,N-Q,N-Q+P,Q-N-P\\}$.

Figure 7. The set $\bigcup_{\alpha\in R}O_{\alpha}$ for two choices of
$R\subseteq\mathcal{S}$.

##  3\. Types of intersections

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile with $14$ neighbors. In Section 3.1
we study basic properties of intersections of the form $t_{1}\cap t_{2}$ where
$t_{1}$ and $t_{2}$ are essentially disjoint subtiles of $T$ ($t_{1}$ may also
be equal to $\overline{\mathbb{R}^{3}\setminus T}$). We will show that we can
attach to $t_{1}\cap t_{2}$ a set $R\subseteq\mathcal{S}$ such that $t_{1}\cap
t_{2}\simeq U(R)$. According to Lemma 2.8, the topology of $U(R)$ is easy to
determine. Knowing the topology of such intersections will be important in
order to apply the results of Bing [5] that will be needed in the proof of
Theorem 1.1. Section 3.2 shows a way to choose the set $R\subseteq\mathcal{S}$
for each intersection $t_{1}\cap t_{2}$ in a unique way (up to sign changes).
This set is, by definition, the type of the intersection. In Section 3.3 we
define a graph that will help us to survey the possible types of intersection
(i.e., the possible subsets $R$) that will occur in this context.

###  3.1. Basic properties of intersections

The definition of the type of an intersection requires some preparation. Let
$T=T(M,\mathcal{D})$ be an $ABC$-tile. Let

$t_{\infty}=\overline{\mathbb{R}^{3}\setminus
T}=T+(\mathbb{Z}^{3}\setminus\\{0\\})$

be the closure of the complement of $T$. We define the collection (recall that
$\mathcal{D}_{i}$ is defined in (2.1))

$\mathcal{C}=\\{M^{-i}(T+d)\;:\;i\in\mathbb{N},\,d\in\mathcal{D}_{i}\\}\cup\\{t_{\infty}\\}$

that contains $t_{\infty}$ as well as each of the subtiles of $T$. If
$t\in\mathcal{C}$ we define

(3.1) $\mathrm{level}(t)=\begin{cases}i,&\hbox{if $t$ is of the form
$M^{-i}(T+d)$ for $i\in\mathbb{N}$ and $d\in\mathcal{D}_{i}$},\\\
-\infty,&\hbox{if $t=t_{\infty}$}.\end{cases}$

We provide the following simple result. Recall that $U(R)$ is defined in
(2.11).

######  Lemma 3.1.

Let $T$ be an $ABC$-tile with $14$ neighbors. Let $t_{1},t_{2}\in\mathcal{C}$
be essentially disjoint. Then there is $R\subseteq\mathcal{S}$ (possibly
empty) such that $t_{1}\cap t_{2}=M^{-\ell}(U(R)+d)$ for some
$\ell\in\mathbb{N}$ and some $d\in\mathbb{Z}^{3}$.

###### Proof.

Assume w.l.o.g. that $\mathrm{level}(t_{1})\leq\mathrm{level}(t_{2})$. Set
$\ell_{i}=\mathrm{level}(t_{i})$. Then $\ell_{2}\in\mathbb{N}$ and
$t_{2}=M^{-\ell_{2}}(T+d)$ for some $d\in\mathcal{D}_{\ell_{2}}$ and, by
possibly subdividing $t_{1}$, we see that $t_{1}$ is a union of sets of the
form $M^{-\ell_{2}}(T+z_{k})$ with $z_{k}\in\mathbb{Z}^{3}\setminus\\{d\\}$
(this union is infinite if and only if $t_{1}=t_{\infty}$). Thus

$t_{1}\cap t_{2}=\bigcup_{k}M^{-\ell_{2}}(T+z_{k})\cap
M^{-\ell_{2}}(T+d)=M^{-\ell_{2}}\bigcup_{k}(\boldsymbol{B}_{z_{k}-d}+d).$

Because $\boldsymbol{B}_{\alpha}\neq\emptyset$ holds if and only if
$\alpha\in\mathcal{S}$ there is a set $R\subseteq\mathcal{S}$ such that

$t_{1}\cap t_{2}=M^{-\ell_{2}}\bigcup_{\alpha\in
R}(\boldsymbol{B}_{\alpha}+d)=M^{-\ell_{2}}(U(R)+d).\qed$

By this lemma the topology of the intersection of two essentially disjoint
elements of $\mathcal{C}$ can be described in terms of a subset
$R\subseteq\mathcal{S}$. Using the notation (2.6), from (2.7) we gain

(3.2) $\alpha\xrightarrow{d}\alpha^{\prime}\in G(\mathcal{S})\quad\hbox{if and
only if}\quad-\alpha\xrightarrow{(C-1)_{M}-d}-\alpha^{\prime}\in
G(\mathcal{S}).$

Thus (2.9) yields $\boldsymbol{B}_{-\alpha}=x_{C}-\boldsymbol{B}_{\alpha}$
with $x_{C}=\sum_{i\geq 1}M^{-i}(C-1)_{M}$ for each $\alpha\in\mathcal{S}$
and, hence, $U(-R)=x_{C}-U(R)$. This implies that $U(-R)\simeq U(R)$, and we
therefore want to identify $R$ with $-R$ in this description. To this matter
we define the equivalence relation $\approx$ on the power set
$2^{\mathcal{S}}$ of $\mathcal{S}$ by $R\approx R^{\prime}$ if and only if
$R^{\prime}=\pm R$. The equivalence classes of this relation are denoted by
$\overline{R}$ for $R\subseteq\mathcal{S}$. Since this notation is only used
for (finite) subsets $R$ of $\mathcal{S}$, there is no risk of confusion with
the closure $\overline{X}$ of a set $X$, for which the same notation is used.

######  Remark 3.2.

Let $t_{1},t_{2}\in\mathcal{C}$ be essentially disjoint. By Lemma 3.1,
$t_{1}\cap t_{2}=M^{-\ell}(U(R)+d)$ for some $\ell\in\mathbb{N}$,
$d\in\mathbb{Z}^{3}$, and $R\subset\mathcal{S}$. We could define
$\overline{R}$ as the type of the intersection of $t_{1}\cap t_{2}$. However,
a priori $\overline{R}$ is not uniquely defined by this equality and we would
have to prove unicity. To circumvent this, in Section 3.2 we give another
(equivalent) definition of type that is obviously unique and better suited to
our purposes. Roughly speaking, we pick the “right” class $\overline{R}$ by
using the neighbor graph. The additional effort we need in order to state this
definition will pay off later.

Before we can define the type of an intersection, we need one more lemma.

######  Lemma 3.3.

Let $T$ be an $ABC$-tile. Let $\alpha\in\mathbb{Z}^{3}\setminus\\{0\\}$,
$i\geq 0$, and $d=d_{i}+Md_{i-1}+\dots+M^{i-1}d_{1}\in\mathcal{D}_{i}$. Then

(3.3) $\begin{split}(T+\alpha)\cap
M^{-i}(T+d)&=M^{-i}\bigcup_{\alpha_{i}:\;\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\in
G(\mathcal{S})}(\boldsymbol{B}_{\alpha_{i}}+d),\end{split}$

where the union is extended over all $\alpha_{i}\in\mathcal{S}$ for which
there exist $\alpha_{1},\ldots,\alpha_{i-1}\in\mathcal{S}$ such that there is
a walk
$\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\in
G(\mathcal{S})$.

Note that the union in (3.3) may well be empty. This is certainly the case if
$\alpha\not\in\mathcal{S}$.

###### Proof.

For $i=0$ we have $d=0$ and (3.3) is trivial. For $i\geq 1$ we prove (3.3) by
induction on $i$. For the induction start let $i=1$ and observe that for each
fixed $d\in\mathcal{D}$ we get, by the set equation (1.2) and the definition
of the edges in $G(\mathcal{S})$ provided in (2.7),

(3.4) $\begin{split}(T+\alpha)\cap
M^{-1}(T+d)&=M^{-1}\big{(}(MT+M\alpha)\cap(T+d)\big{)}\\\
&=M^{-1}\bigcup_{d^{\prime}\in\mathcal{D}}\big{(}(T+d^{\prime}+M\alpha)\cap(T+d)\big{)}\\\
&=M^{-1}\bigcup_{d^{\prime}\in\mathcal{D}}\big{(}\big{(}(T+M\alpha+d^{\prime}-d)\cap
T\big{)}+d\big{)}\\\
&=M^{-1}\bigcup_{\alpha^{\prime}:\;\alpha\xrightarrow{d}\alpha^{\prime}\in
G(\mathcal{S})}(\boldsymbol{B}_{\alpha^{\prime}}+d).\end{split}$

For the induction step assume that (3.3) holds for $i-1$ instead of $i$, let
$d^{\prime}=d_{i-1}+Md_{i-2}+\dots+M^{i-2}d_{1}\in\mathcal{D}_{i-1}$ and
$d=d_{i}+Md^{\prime}$. The set equation (1.2) implies that $M^{-i}(T+d)\subset
M^{-i+1}(T+d^{\prime})$. Thus by the induction hypothesis

$\begin{split}(T+\alpha)\,\cap\,&M^{-i}(T+d)=(T+\alpha)\cap
M^{-i+1}(T+d^{\prime})\cap M^{-i}(T+d)\\\
&=M^{-i+1}\bigcup_{\alpha_{i-1}:\;\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i-1}}\alpha_{i-1}\in
G(\mathcal{S})}\big{(}(\boldsymbol{B}_{\alpha_{i-1}}+d^{\prime})\cap
M^{-1}(T+d)\big{)}\\\
&=M^{-i+1}\bigcup_{\alpha_{i-1}:\;\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i-1}}\alpha_{i-1}\in
G(\mathcal{S})}\big{(}\big{(}(T+\alpha_{i-1})\cap
M^{-1}(T+d_{i})\big{)}+d^{\prime}\big{)}.\end{split}$

Applying (3.4) to the last intersection yields (3.3) and the induction is
finished. ∎

###  3.2. The type of an intersection

We are now ready to define the type of an intersection. Let $T$ be an
$ABC$-tile with $14$ neighbors, let
$t\in\mathcal{C}\setminus\\{t_{\infty}\\}$, and set $i=\mathrm{level}(t)$.
Then there is $d=d_{i}+Md_{i-1}+\dots+M^{i-1}d_{1}\in\mathcal{D}_{i}$ such
that $t=M^{-i}(T+d)\subseteq T$. Thus Lemma 3.3 implies that

(3.5) $t_{\infty}\cap t=\bigcup_{\alpha\in\mathcal{S}}((T+\alpha)\cap
M^{-i}(T+d))=M^{-i}\bigcup_{\alpha\in\mathcal{S}}\bigcup_{\alpha_{i}\;:\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\in
G(\mathcal{S})}(\boldsymbol{B}_{\alpha_{i}}+d).$

We say that the intersection $t_{\infty}\cap t$ is of type
$\overline{R(t_{\infty},t)}$ with

(3.6) $R(t_{\infty},t)=\\{\alpha_{i}\;:\,\hbox{there is $\alpha\in\mathcal{S}$
with
}\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\in
G(\mathcal{S})\\}.$

Note that (3.5) implies that $t_{\infty}\cap t\simeq U(R(t_{\infty},t))\simeq
U(-R(t_{\infty},t))$. Thus the type $\overline{R(t_{\infty},t)}$ determines
the topology of the intersection $t_{\infty}\cap t$.

Let $t_{1},t_{2}\in\mathcal{C}\setminus\\{t_{\infty}\\}$ be essentially
disjoint and ordered such that
$i=\mathrm{level}(t_{1})\leq\mathrm{level}(t_{2})=j$. We can uniquely choose
$z\in\mathbb{Z}^{3}$, $\alpha\in\mathbb{Z}^{3}\setminus\\{0\\}$,
$d=d_{j-i}+Md_{j-i-1}+\dots+M^{j-i-1}d_{1}\in\mathcal{D}_{j-i}$ in a way that

$M^{i}(t_{1}\cap t_{2})+z=(T+\alpha)\cap M^{i-j}(T+d).$

Thus Lemma 3.3 implies that

(3.7) $t_{1}\cap
t_{2}=M^{-j}\Big{(}\bigcup_{\alpha_{j-i}:\,\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{j-i}}\alpha_{j-i}\in
G(\mathcal{S})}(\boldsymbol{B}_{\alpha_{j-i}}+d)\Big{)}-M^{-i}z.$

We say that the intersection $t_{1}\cap t_{2}$ is of type444If $i=j$ we could
switch the roles of $t_{1}$ and $t_{2}$. But since it is easy to see that in
this case $R(t_{1},t_{2})=-R(t_{2},t_{1})$, the type
$\overline{R(t_{1},t_{2})}$ is well defined also in this case.
$\overline{R(t_{1},t_{2})}$ with

(3.8)
$R(t_{1},t_{2})=\\{\alpha_{j-i}\;:\;\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{j-i}}\alpha_{j-i}\in
G(\mathcal{S})\\}.$

Note that (3.7) implies that $t_{1}\cap t_{2}\simeq U(R(t_{1},t_{2}))\simeq
U(-R(t_{1},t_{2}))$. Thus the type $\overline{R(t_{1},t_{2})}$ determines the
topology of the intersection $t_{1}\cap t_{2}$. Summing up we have the
following lemma.

######  Lemma 3.4.

Let $t_{1},t_{2}\in\mathcal{C}$ be essentially disjoint. If $t_{1}\cap t_{2}$
is of type $\overline{R}$ for some $R\subseteq\mathcal{S}$ then $t_{1}\cap
t_{2}\simeq U(R)$.

Let $t_{1},t_{2}\in\mathcal{C}$ be essentially disjoint. If $t_{1}\cap t_{2}$
has a certain type, we want to know how this influences the type of $t_{1}\cap
t_{2}^{\prime}$ for $t_{2}^{\prime}\in\mathcal{C}$ with $t_{2}^{\prime}\subset
t_{2}$. This will be studied in the next section.

###  3.3. A graph that governs the types of intersections

Let $T$ be an $ABC$-tile with $14$ neighbors. We want to know which classes
$\overline{R}$ are needed to describe all possible intersections of
essentially disjoint elements of $\mathcal{C}$. To this end we introduce the
following notation. For a subset $R\subseteq\mathcal{S}$ and a digit
$d\in\mathcal{D}$, we define

(3.9)
$n_{d}(R):=\\{\alpha^{\prime}\;:\;\alpha\xrightarrow{d}\alpha^{\prime}\in
G(\mathcal{S})\text{ for }\alpha\in R\\}.$

Then $n_{d}(R)$ contains the successors of elements of $R$ in the neighbor
graph that can be reached by an edge with label $d$. Of course, $n_{d}(R)$ is
a subset of $\mathcal{S}$. By the symmetry property (3.2) we have

(3.10)
$n_{d}(R)=-n_{(C-1)_{M}-d}(-R)\qquad(R\subseteq\mathcal{S},\,d\in\mathcal{D}).$

Let $N_{0}=\\{\overline{\mathcal{S}}\\}$ be the set containing the residue
class of the full set of neighbors and recursively define a nested sequence
$(N_{k})_{k\geq 0}$ of subsets of the power set $2^{\mathcal{S}}$ by

(3.11) $N_{k}=\\{\overline{n_{d}(R)}\;:\;\overline{R}\in
N_{k-1},d\in\mathcal{D}\\}\cup N_{k-1}\qquad(k\geq 1).$

By (3.10), $N_{k}$ is well-defined because nothing changes if we replace $R$
by $-R$ in the argument of $n_{d}$ on the right hand side of (3.11). Because
$2^{\mathcal{S}}$ is finite there exists a minimal $k_{0}\in\mathbb{N}$ such
that $N_{k_{0}+1}=N_{k_{0}}$ and, hence, $N_{k}=N_{k_{0}}$ for each $k\geq
k_{0}$. This leads to the following definition.

######  Definition 3.5 (Intersection graph).

Let $T$ be an $ABC$-tile with $14$ neighbors. The _intersection graph_
$\mathcal{I}$ is the graph whose nodes are the elements of555We leave away the
empty set for practical reasons. It would cause many additional edges in the
intersection graph. $N_{k_{0}}\setminus\\{\overline{\emptyset}\\}$. And whose
edges are defined by

(3.12) $\overline{R}\rightarrow\overline{R^{\prime}}\in\mathcal{I}\quad\hbox{
if and only if }\quad R^{\prime}=\pm n_{d}(R)\hbox{ for some }d\in\mathcal{D}$

(which is again well-defined because of (3.10)).

We note that the in-out graph defined in [6, Section 7] is used for a similar
purpose as our intersection graph $\mathcal{I}$. However, $\mathcal{I}$ has a
simpler structure than the in-out graph.

######  Lemma 3.6.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Then we
have the following two cases for $\mathcal{I}$.

  * (1)

For $A=1$, $B=2$, and $C\geq 4$ the graph $\mathcal{I}$ is given by Figure 8.
In particular, we have $\\#\mathcal{I}=55$.

  * (2)

For $A=1$, $B\geq 3$, and $C\geq 2B$ the graph $\mathcal{I}$ is given by
Figure 9. In particular, we have $\\#\mathcal{I}=57$.

By Remark 1.3 the constellations $A,B,C$ covered in (1) and (2) exhaust all
$ABC$-tiles with $14$ neighbors having $A=1$.

######  Remark 3.7.

The graphs $\mathcal{I}$ are rather large. Thus we cannot draw them directly.
Figure 8 contains a tree. The quotient graph we obtain by identifying nodes
with the same node-label in this tree equals $\mathcal{I}$ for
$A=1,\;B=2,\;C\geq 4$. Similarly if we quotient the tree in Figure 9 by
identifying nodes with the same node-label we obtain $\mathcal{I}$ for
$A=1,\;B\geq 3,\;C\geq 2B$.

Figure 8. The graph $\mathcal{I}$ for $A=1,B=2,C\geq 4$ is obtained as a
quotient graph of this tree; see Remark 3.7. Figure 9. The graph $\mathcal{I}$
for $A=1,B\geq 3,C\geq 2B$ is obtained as a quotient graph of this tree; see
Remark 3.7.

###### Proof.

This proof is just a lengthy but easy calculation. Since
$N_{1}=N_{1}^{\prime}\cup N_{0}$ where
$N_{1}^{\prime}=\\{\overline{n_{d}(\mathcal{S})}\,:\,d\in\mathcal{D}\\}$, we
have to determine $n_{d}(\mathcal{S})$ for each $d\in\mathcal{D}$. Recall the
notation (2.6). From the neighbor graph we see that, for $d=0$, there exists
an edge of the form $\alpha\xrightarrow{0|d^{\prime}}\alpha^{\prime}$ for each
$\alpha^{\prime}\in\mathcal{S}\setminus\\{P\\}$. Thus
$\overline{\mathcal{S}\setminus\\{P\\}}\in N_{1}^{\prime}$. For $d=(e)_{M}$
with $1\leq e\leq C-2$, an edge of the form
$\alpha\xrightarrow{(e)_{M}|d^{\prime}}\alpha^{\prime}$ exists for each
$\alpha^{\prime}\in\mathcal{S}\setminus\\{{P,-P}\\}$, hence,
$\overline{\mathcal{S}\setminus\\{P,-P\\}}\in N_{1}^{\prime}$. Finally, for
$d=(C-1)_{M}$ an edge of the form
$\alpha\xrightarrow{(C-1)_{M}|d^{\prime}}\alpha^{\prime}$ exists for each
$\alpha^{\prime}\in\mathcal{S}\setminus\\{-P\\}$. Thus also
$\overline{\mathcal{S}\setminus\\{-P\\}}$ is an element of $N_{1}^{\prime}$.
However, since
$\overline{\mathcal{S}\setminus\\{-P\\}}=\overline{\mathcal{S}\setminus\\{P\\}}$
we already got this element before. The sets $\overline{R}$ with
$R\subseteq\mathcal{S}$ contained in $N_{1}^{\prime}$ are listed in the second
column of Table LABEL:tab_N1. Table LABEL:tab_N1, as well as all the other
tables666We provide all these tables in order to illustrate the proof and
because we need them for later reference. in this proof, has the following
columns: The first column contains the name of the node $\overline{R}$ in the
graphs in Figures 8 and 9 corresponding to the subset $R\subseteq\mathcal{S}$
in the second column. The third column indicates the condition under which
this subset occurs. Finally, the fourth column describes the topology of
$U(R)$. Recall that according to Lemma 2.8 the topology of $U(R)$ can be
obtained by easy combinatorial arguments which can (as we did) be easily
checked by a computer program. Summing up we have shown that
$N_{1}=N_{1}^{\prime}\cup
N_{0}=\\{\overline{\mathcal{S}},\overline{\mathcal{S}\setminus\\{{P}\\}},\overline{\mathcal{S}\setminus\\{{P,-P}\\}}\\}$.

Table 1. The set $N_{1}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{1}$ | $\mathcal{S}\setminus\\{{P}\\}$ | — |  $2$-ball
$b^{1}$ | $\mathcal{S}\setminus\\{P,-P\\}$ | — |  $\mathbb{S}^{1}\times[0,1]$ (a “ribbon”)

Now we can calculate $N_{k}$ for $k\geq 1$ in an analogous way as follows.

Table 2. The set $N_{2}^{\prime}$ consists of the subsets in the second column in this table.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{2}$ | $\mathcal{S}\setminus\\{P,Q,Q-P\\}$ | — |  $2$-ball
$b^{2}$ | $\mathcal{S}\setminus\\{P,Q,Q-P,-P\\}$ | — |  $2$-ball
$c^{2}$ | $\mathcal{S}\setminus\\{P,P-Q\\}$ | — |  $2$-ball
$d^{2}$ | $\mathcal{S}\setminus\\{P,Q,Q-P,P-Q\\}$ | — |  $2$-ball
$e^{2}$ | $\mathcal{S}\setminus\\{P,Q,Q-P,-P,-Q,P-Q\\}$ | — |  $2$ disjoint $2$-balls

Starting from $N_{1}$ we use (3.11) and the neighbor graph $G(\mathcal{S})$ to
calculate $N_{2}$. This yields that $N_{2}=N_{2}^{\prime}\cup N_{1}$ where the
set $N_{2}^{\prime}$ corresponds to the subsets indicated in Table
LABEL:tab:N2.

Table 3. The set $N_{3}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{3}$ | $\mathcal{S}_{1}$ | — |  $2$-ball
$b^{3}$ | $\mathcal{S}_{1}\setminus\\{P\\}$ | — |  $2$-ball
$c^{3}$ | $(\mathcal{S}_{1}\setminus\\{P\\})\cup\\{-N,Q-N-P\\}$ | — |  $2$-ball
$d^{3}$ | $\mathcal{S}_{1}\setminus\\{Q-P\\}$ | — |  $2$-ball
$e^{3}$ | $\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\}$ | — |  $2$-ball
$f^{3}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{-N,Q-N-P\\}$ | — |  $2$ disjoint $2$-balls
$g^{3}$ | $\\{Q,N,Q-P,N-P\\}\cup(-\mathcal{S}_{1}\setminus\\{P-Q\\})$ | — |  $2$-ball
$h^{3}$ | $\\{Q,N,Q-P,N-P\\}\cup(-\mathcal{S}_{1}\setminus\\{-P,-Q,P-Q\\})$ | — |  $2$-ball
$i^{3}$ | $\mathcal{S}_{1}\setminus\\{P,N-Q\\}\cup(-\mathcal{S}_{1}\setminus\\{-P,-Q,P-Q\\})$ | — |  $2$-ball
$j^{3}$ | $\\{N,N-Q+P\\}$ | — |  $2$-ball
$k^{3}$ | $\\{Q,N,Q-P,N-P,-N,Q-N-P\\}$ | — |  $2$-ball
$l^{3}$ | $\mathcal{S}_{1}\setminus\\{P,N-Q\\}$ | — |  $2$-ball
$m^{3}$ | $\\{Q,N,Q-P,N-P\\}$ | $C>2B$ |  $2$-ball

We now go on in the same way. If $N_{3}^{\prime}$ consists of the sets in the
second column of Table LABEL:tab:N3 then, using (3.11), a somewhat lengthy but
easy calculation shows that $N_{3}=N_{3}^{\prime}\cup N_{2}$. Here we have to
be careful about the node $m^{3}$. This node only occurs in $N_{3}^{\prime}$
if $C>2B$. If $C=2B$, it occurs in $N_{5}^{\prime}$. Thus we have
$\\#N_{3}=21$ for $C>2B$ and $\\#N_{3}=20$ for $C=2B$.

Table 4. The set $N_{4}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{4}$ | $\\{N-P,N-Q\\}$ | — |  $2$-ball
$b^{4}$ | $\\{N-Q+P\\}$ | — |  $2$-ball
$c^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{Q-N,Q-N-P\\}$ | (C2) |  $2$ disjoint $2$-balls
$d^{4}$ | $\\{P,N-Q,N-Q+P\\}$ | — |  $2$-ball
$e^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{-P,-Q,P-Q\\}$ | — |  $2$-ball
$f^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{-Q,P-Q,P-N\\}$ | — |  $2$-ball
$g^{4}$ | $\\{Q-P\\}$ | — |  $2$-ball
$h^{4}$ | $\\{Q,Q-P,N-P\\}$ | — |  $2$-ball
$i^{4}$ | $\\{N-P,N-Q,N-Q+P\\}$ | — |  $2$-ball
$j^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{-P,-Q\\}$ | — |  $2$-ball
$k^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{Q-N-P\\}$ | — |  $2$ disjoint $2$-balls
$l^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup(-\mathcal{S}_{1}\setminus\\{-N,P-Q,P-N\\})$ | — |  $2$-ball
$m^{4}$ | $\\{Q,Q-P,N-P,N-Q\\}$ | — |  $2$-ball
$n^{4}$ | $\\{Q,Q-P\\}$ | (C2) |  $2$-ball
$o^{4}$ | $\\{N-Q,N-Q+P\\}$ | (C2) |  $2$-ball
$p^{4}$ | $(\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\})\cup\\{-Q,P-Q\\}$ | (C2) |  $2$-ball

From the next step onwards we need to distinguish between the cases

$\begin{split}&\mathrm{(C1)}\qquad\qquad\qquad A=1,\;B=2,\;C\geq 4;\\\
&\mathrm{(C2)}\qquad\qquad\qquad A=1,\;B\geq 3,\;C\geq 2B.\end{split}$

With $N_{4}^{\prime}$ as in Table LABEL:tab:N4 we gain
$N_{4}=N_{4}^{\prime}\cup N_{3}$. This entails that $\\#N_{4}=33$ for
$A=1,B=2,C>4$ ($\\#N_{4}=32\text{ for }C=4$) and $\\#N_{4}=37$ for $A=1,B\geq
3,C>2B$ ($\\#N_{4}=36\text{ for }C=2B$).

Table 5. The set $N_{5}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{5}$ | $\\{N\\}$ | — |  $2$-ball
$b^{5}$ | $\\{Q,N,N-P\\}$ | — |  $2$-ball
$c^{5}$ | $\\{N-Q,N-Q+P\\}$ | (C1) |  $2$-ball
$d^{5}$ | $\\{Q,Q-P\\}$ | (C1) |  $2$-ball
$e^{5}$ | $\\{N,N-Q+P,-Q,P-Q\\}$ | — |  $2$-ball
$f^{5}$ | $\\{N,N-P,N-Q,N-Q+P,-Q,P-Q\\}$ | (C1) |  $2$-ball
$g^{5}$ | $\\{\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\}\cup\\{-Q\\}$ | — |  $2$-ball
$h^{5}$ | $\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\}\cup\\{-Q,-N,P-N\\}$ | — |  $2$-ball
$i^{5}$ | $\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\}\cup\\{-N\\}$ | — |  $2$ disjoint $2$-balls
$j^{5}$ | $\mathcal{S}_{1}\setminus\\{P,Q-P\\}$ | — |  $2$-ball
$k^{5}$ | $\\{N-Q\\}$ | — |  $2$-ball
$l^{5}$ | $\\{N,N-P\\}$ | (C2) |  $2$-ball
$m^{5}$ | $\mathcal{S}_{1}\setminus\\{P,Q,Q-P\\}\cup\\{-N,P-N\\}$ | (C2) |  $2$ disjoint $2$-balls
$m^{3}$ | $\\{Q,N,Q-P,N-P\\}$ | $C=2B$ |  $2$-ball

Now, $N_{5}=N_{5}^{\prime}\cup N_{4}$, where $N_{5}^{\prime}$ is given by
Table LABEL:tab:N5. Then $\\#N_{5}=44$ for (C1) and $\\#N_{5}=47$ for $(C2)$.
As we indicated at the step that lead to $N_{3}$, at this stage the node
$m^{3}$ is contained in $N_{5}$ in all the cases. Thus from now onwards we do
not have to distinguish the cases $2B<C$ and $2B=C$.

Table 6. The set $N_{6}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{6}$ | $\\{N,N-P,-P,-Q\\}$ | — |  $2$-ball
$b^{6}$ | $\\{N,N-P\\}$ | (C1) |  $2$-ball
$c^{6}$ | $\\{N-P\\}$ | — |  $2$-ball
$d^{6}$ | $\\{P\\}$ | — |  $2$-ball
$e^{6}$ | $-\mathcal{S}_{1}\setminus\\{P-Q\\}\cup\\{N,N-P\\}$ | — |  $2$-ball
$f^{6}$ | $\\{N,N-P,-N,Q-N-P\\}$ | — |  $2$ disjoint $2$-balls
$g^{6}$ | $\\{N,N-P,N-Q+P\\}$ | — |  $2$-ball
$h^{6}$ | $\\{P,Q\\}$ | (C2) |  $2$-ball

We get $N_{6}=N_{6}^{\prime}\cup N_{5}$ with $N_{6}^{\prime}$ as in Table
LABEL:tab:N6. Thus $\\#N_{6}=51$ for $(C1)$ and $\\#N_{6}=54$ for $(C2)$.

Table 7. The set $N_{7}^{\prime}$.  Node |  Subset $R$ | Condition |  Topology of $U(R)$
---|---|---|---
$a^{7}$ | $\\{P,Q,Q-P\\}$ | — |  $2$-ball
$b^{7}$ | $\\{Q\\}$ | — |  $2$-ball
$c^{7}$ | $\\{P,Q\\}$ | (C1) |  $2$-ball
$d^{7}$ | $\\{P,Q,N-Q,N-Q+P\\}$ | — |  $2$-ball

The next step of the iteration yields $N_{7}=N_{7}^{\prime}\cup N_{6}$ with
$N_{7}$ as in Table LABEL:tab:N7. Thus $\\#N_{7}=55$ for $A=1,B=2,C\geq 4$ and
$\\#N_{7}=57$ for $A=1,B\geq 3,C\geq 2B$. We repeat the procedure once more
and observe that $N_{8}=N_{7}$ for both conditions, so we have reached the end
with $\\#\mathcal{I}=55$ for $A=1,B=2,C\geq 4$ and $\\#\mathcal{I}=57$ for
$A=1,B\geq 3,C\geq 2B$.

It just remains to insert the edges of $\mathcal{I}$ according to (3.12) in
order to end up with the graphs depicted in Figure 8 and Figure 9 (and
observing Remark 3.7). ∎

The set $U^{\prime}(R)=\bigcup_{\alpha\in R}O_{\alpha}$ with $R$ as in $d^{4}$
and $k^{4}$ is depicted on the left and right hand side of Figure 7,
respectively.

######  Remark 3.8.

For each $\alpha\in\mathcal{S}$, $\overline{\\{\alpha\\}}$ is a node of
$\mathcal{I}$. In particular, $d^{6}$$=\overline{\\{P\\}}$,
$b^{7}$$=\overline{\\{Q\\}}$, $a^{5}$$=\overline{\\{N\\}}$,
$g^{4}$$=\overline{\\{Q-P\\}}$, $c^{6}$$=\overline{\\{N-P\\}}$,
$k^{5}$$=\overline{\\{N-Q\\}}$, and $b^{4}$$=\overline{\\{N-Q+P\\}}$. This
will be of importance later.

######  Remark 3.9.

The number in the superscript of the labels of the nodes of $\mathcal{I}$
indicates in which level of Figure 8 and Figure 9 a node occurs for the first
time. For some nodes this happens at different levels in Figure 8 and Figure
9. In these cases we gave different names to this node in the two graphs. So
we have $n^{4}$= $d^{5}$, $o^{4}$= $c^{5}$, $l^{5}$= $b^{6}$, $h^{6}$=
$c^{7}$. This fact is of no relevance in the sequel. Only the classes
$\overline{R}$ corresponding to $c^{4}$ and $m^{5}$ are in $\mathcal{I}$ under
condition $(C2)$ but not under $(C1)$. We just did it that way because it
makes it easier to locate the first occurrence of a given node in the figures.

######  Lemma 3.10.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let
$t_{1},t_{2}\in\mathcal{C}$ be essentially disjoint with
$\mathrm{level}(t_{1})\leq\mathrm{level}(t_{2})$. Let
$t_{2}^{\prime}\in\mathcal{C}$ with

(3.13)
$\mathrm{level}(t_{2}^{\prime})=\mathrm{level}(t_{2})+1\quad\hbox{and}\quad
t_{2}^{\prime}\subset t_{2}.$

Assume that the type of $t_{1}\cap t_{2}$ is $\overline{R}\in\mathcal{I}$.
Then the type $\overline{R^{\prime}}$ of $t_{1}\cap t_{2}^{\prime}$ is either
$\overline{\emptyset}$ or
$\overline{R}\to\overline{R^{\prime}}\in\mathcal{I}$.

###### Proof.

Assume first that $t_{1}=t_{\infty}$. Set $\mathrm{level}(t_{2})=i$. Then
there is $d=d_{i}+\cdots+M^{i-1}d_{1}\in\mathcal{D}_{i}$ and
$d_{i+1}\in\mathcal{D}$ such that $t_{2}=M^{-i}(T+d)$ and
$t_{2}^{\prime}=M^{-i-1}(T+d_{i+1}+Md)$. Now (3.5) and (3.6) yield

$\begin{split}R&=\\{\alpha_{i}\;:\,\hbox{there is $\alpha\in\mathcal{S}$ with
}\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\in
G(\mathcal{S})\\},\\\ R^{\prime}&=\\{\alpha_{i+1}\;:\,\hbox{there is
$\alpha\in\mathcal{S}$ with
}\alpha\xrightarrow{d_{1}}\alpha_{1}\xrightarrow{d_{2}}\cdots\xrightarrow{d_{i}}\alpha_{i}\xrightarrow{d_{i+1}}\alpha_{i+1}\in
G(\mathcal{S})\\}.\end{split}$

Thus (3.9) yields $R^{\prime}=n_{d_{i+1}}(R)$, and by (3.11) and (3.12) either
$\overline{R^{\prime}}=\overline{\emptyset}$ or $\overline{R^{\prime}}$
satisfies $\overline{R}\to\overline{R^{\prime}}\in\mathcal{I}$.

If $t_{1}\not=t_{\infty}$ the result follows analogously by using (3.7) and
(3.8) instead of (3.5) and (3.6). ∎

######  Proposition 3.11.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. If
$t_{1},t_{2}\in\mathcal{C}$ are essentially disjoint then the type
$\overline{R}$ of $t_{1}\cap t_{2}$ is either $\overline{\emptyset}$ or
$\overline{R}\in\mathcal{I}$. In particular, the following assertions hold.

  * (1)

Let $t_{1}=t_{\infty}$ and $t_{2}\in\mathcal{C}\setminus\\{t_{\infty}\\}$ with
$\mathrm{level}(t_{2})=i$. Then either $t_{\infty}\cap t_{2}=\emptyset$ or
there is a walk
$\overline{\mathcal{S}}\rightarrow\overline{R_{1}}\rightarrow\overline{R_{2}}\rightarrow\cdots\rightarrow\overline{R_{i}}$
of length $i$ in $\mathcal{I}$ such that $t_{\infty}\cap t_{2}$ is of type
$\overline{R_{i}}$ and, hence, $t_{\infty}\cap t_{2}\simeq U(R_{i})$.

  * (2)

Let $t_{1},t_{2}\in\mathcal{C}\setminus\\{t_{\infty}\\}$ be essentially
disjoint with $\mathrm{level}(t_{1})\leq\mathrm{level}(t_{2})$ and let
$i=\mathrm{level}(t_{2})-\mathrm{level}(t_{1})$. Then either $t_{1}\cap
t_{2}=\emptyset$ or there is $\alpha\in\mathcal{S}$ and a walk
$\overline{\\{\alpha\\}}\rightarrow\overline{R_{1}}\rightarrow\overline{R_{2}}\rightarrow\cdots\rightarrow\overline{R_{i}}$
of length $i$ in $\mathcal{I}$ such that $t_{1}\cap t_{2}$ is of type
$\overline{R_{i}}$ hence, $t_{1}\cap t_{2}\simeq U(R_{i})$.

###### Proof.

We prove $(1)$. The proof is done by induction on $i=\mathrm{level}(t_{2})$.
If $i=0$ then $t_{2}=T$ and, hence, the type of $t_{\infty}\cap t_{2}$ is
$\overline{R}=\overline{\mathcal{S}}\in\mathcal{I}$.

For the induction hypothesis assume that the result holds for all
$t_{2}\in\mathcal{C}$ with $0\leq\mathrm{level}(t_{2})\leq i-1$.

For the induction step let $t_{2}^{\prime}\in\mathcal{C}$ with
$\mathrm{level}(t_{2}^{\prime})=i$ be given and assume that $t_{2}$ satisfies
(3.13). Then by the induction hypothesis the type $\overline{R_{i-1}}$ of
$t_{\infty}\cap t_{2}$ is either $\overline{\emptyset}$ or there is a walk
$\overline{\mathcal{S}}\rightarrow\overline{R_{1}}\rightarrow\overline{R_{2}}\rightarrow\cdots\rightarrow\overline{R_{i-1}}$
of length $i-1$ in $\mathcal{I}$. If $\overline{R_{i-1}}=\overline{\emptyset}$
then (3.13) implies that $t_{\infty}\cap t_{2}^{\prime}=\emptyset$, hence, its
type is $\overline{\emptyset}$ as well, and we are done. If
$\overline{R_{i-1}}\in\mathcal{I}$ then by Lemma 3.10 the type
$\overline{R_{i}}$ of $t_{\infty}\cap t_{2}^{\prime}$ is either
$\overline{\emptyset}$ or satisfies
$\overline{R_{i-1}}\to\overline{R_{i}}\in\mathcal{I}$. In the latter case
there is a walk
$\overline{\mathcal{S}}\rightarrow\overline{R_{1}}\rightarrow\overline{R_{2}}\rightarrow\cdots\rightarrow\overline{R_{i}}$
of length $i$ in $\mathcal{I}$. This finishes the induction step.

The case $t_{1}\not=t_{\infty}$ follows analogously by induction on
$\mathrm{level}(t_{2})-\mathrm{level}(t_{1})$. Just note that, if
$\mathrm{level}(t_{1})=\mathrm{level}(t_{2})$, then $t_{1}\cap t_{2}$ is
either empty or has type $\overline{\\{\alpha\\}}\in\mathcal{I}$ for some
$\alpha\in\mathcal{S}$ (observe Remark 3.8).

The fact that $t_{1}\cap t_{2}\simeq U(R)$ if it has type $U(R)$ is already
contained in Lemma 3.4. ∎

##  4\. Proofs of the main results

This section is devoted to the proofs of our main results. In Section 4.1 we
recall the definition of partitionings in the sense of Bing [5] and give some
results on partitionings that will be needed in the sequel. In Section 4.2 we
define sequences of partitionings that are suitable for our purposes. In
Section 4.3 we make sure that in these sequences each atom is subdivided in a
way that certain connectivity properties are maintained. Finally, Sections 4.4
and 4.5 contain the proofs of Theorem 1.1 and Theorem 1.5, respectively.

###  4.1. Partionings

In this section we give the definitions and results of Bing’s theory of
partitionings [5] that will be relevant for the proof of Theorem 1.1. We start
with some terminology.

######  Definition 4.1 (Partitioning).

Let $X$ be a metric space. A partitioning of $X$ is a collection of mutually
disjoint open sets (so-called atoms) whose union is dense in $X$. A
partitioning is called _regular_ if each of its atoms is the interior of its
closure. Let $G$ and $G^{\prime}$ be two partitionings of $X$. $G^{\prime}$ is
a refinement of $G$ if for each $g^{\prime}\in G^{\prime}$ there exists $g\in
G$ with $g^{\prime}\subseteq g$. A sequence $(G_{i})_{i\geq 1}$ of
partitionings is called a decreasing sequence of partitionings if $G_{i+1}$ is
a refinement of $G_{i}$ and the maximum of the diameters of the atoms of
$G_{i}$ tends to $0$ as $i$ tends to infinity.

######  Definition 4.2 (Equivalent sequences of partitionings).

Let $X_{1}$ and $X_{2}$ be two metric spaces. Let $(G_{ij})_{j\geq 1}$ be a
sequence of partitionings of $X_{i}$ for each $i\in\\{1,2\\}$. We say that
$(G_{1j})$ and $(G_{2j})$ are equivalent partitionings, if for each $j\geq 1$
there exists a 1 to 1 correspondence between the atoms of $G_{1j}$ and
$G_{2j}$ such that

  * (1)

two atoms of $G_{1j}$ have a boundary point in common if and only if the
corresponding atoms of $G_{2j}$ have a boundary point in common.

  * (2)

corresponding atoms of $G_{1,j+1}$ and $G_{2,j+1}$ are subsets of
corresponding atoms of $G_{1j}$ and $G_{2j}$.

If $(G_{1j})$ and $(G_{2j})$ are equivalent we write $(G_{1j})\sim(G_{2j})$.

We say that two finite sequences $(G_{ij})_{j=1}^{n}$ of partitionings of
$X_{i}$ ($i\in\\{1,2\\}$) are equivalent if for each $j\in\\{1,\ldots,n\\}$
there exists a 1 to 1 correspondence between the atoms of $G_{1j}$ and
$G_{2j}$ such that (1) holds for $1\leq j\leq n$ and (2) holds for $1\leq
j<n$.

######  Remark 4.3.

It is easy to check that the relation “$\sim$” is an equivalence relation.

The following lemma, which can be easily proved, is just a reformulation of
[5, Theorem 6].

######  Lemma 4.4.

Two Peano continua $X_{1}$ and $X_{2}$ are homeomorphic if and only if for
each $i\in\\{1,2\\}$ there exists a decreasing sequence of partitionings
$(G_{ij})_{j\geq 1}$ for $X_{i}$ such that $(G_{1j})_{j\geq
1}\sim(G_{2j})_{j\geq 1}$.

This lemma will be used in the proof of Theorem 1.1. Indeed, we will construct
a decreasing sequence of partitionings for the self-affine tile $T$ (which is
a Peano continuum by Lemma 2.5) that is equivalent to a decreasing sequence of
partitionings of $\mathbb{D}^{3}$. In the course of our proof we will use the
following two results from [5]. The first one is about the extension of
homeomorphisms. Recall that a $2$-sphere $C$ in $\mathbb{R}$ is tame if there
is a homeomorphism from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ that maps $C$ to
the unit sphere $\mathbb{S}^{2}$ in $\mathbb{R}^{3}$.

######  Proposition 4.5 (see [5, Theorem 3]).

Let $S$ be a Peano continuum and $S_{2}\subset S$ a $2$-sphere. Let
$C\subset\mathbb{R}^{3}$ be a tame $2$-sphere and $F:S_{2}\to C$ a
homeomorphism. Assume that $G$ is a regular partitioning of $S$ satisfying the
following conditions.

  * (1)

If $g\in G$ then $\partial g\cong\mathbb{S}^{2}$.

  * (2)

If $g_{1},g_{2}\in G$ are distinct then $\partial g_{1}\cap\partial g_{2}$ is
either empty or a finite union of mutually disjoint $2$-balls.

  * (3)

If $g_{1},g_{2},g_{3}\in G$ are mutually distinct then $\partial
g_{1}\cap\partial g_{2}\cap\partial g_{3}$ is either empty or a finite union
of arcs.

  * (4)

There exist $g_{1},\ldots,g_{n}\in G$ such that
$S_{2}=\partial(\overline{g_{1}\cup\dots\cup g_{n}})$ and such that the
intersection $\partial g_{j}\cap(S_{2}\cup\partial g_{1}\cup\dots\cup\partial
g_{j-1})$ is connected for each $j\in\\{1,\ldots,n\\}$.

Then there is a partitioning $\\{h_{0},h_{1},\ldots,h_{n}\\}$ of
${\mathbb{R}}^{3}$ and a homeomorphism
$F^{\prime}:\partial_{S}(g_{1}\cup\dots\cup
g_{n})\to\partial_{\mathbb{R}^{3}}(h_{1}\cup\dots\cup h_{n})$ such that
$h_{0}$ is the exterior of $C$ and $\partial h_{i}$ is a tame $2$-sphere,
$F=F^{\prime}$ on $S_{2}$, and $F^{\prime}(\partial g_{i})=\partial h_{i}$
($1\leq i\leq n$).

The next result will be used in the proof of Theorem 1.1 in the context of
decreasing sequences of partitionings.

######  Proposition 4.6 (see [5, Theorem 5]).

Let $C\subset\mathbb{R}^{3}$ be a tame $2$-sphere and $(G_{i})_{i\geq 1}$ a
sequence of partitionings of ${\mathbb{R}}^{3}$ satisfying the following
conditions for each $i\geq 1$.

  * (1)

If $g\in G_{i}$ then $\partial g\cong\mathbb{S}^{2}$.

  * (2)

For each $g\in G_{i}$ with $\overline{g}\cap C\not=\emptyset$ the set
$\partial g\cap C$ is connected and does not separate $\partial g$.

  * (3)

$G_{i+1}$ is a refinement of $G_{i}$.

  * (4)

One atom $g_{0}\in G_{i}$ is the exterior of $C$.

  * (5)

For each $\varepsilon>0$ and each $i\in\mathbb{N}$ there is
$n=n(i,\varepsilon)\geq 1$ such that $\overline{g^{\prime}}\cap\bigcup_{g\in
G_{i}}\partial g$ has diameter less than $\varepsilon$ for each $g^{\prime}\in
G_{n}\setminus\\{g_{0}\\}$.

Then for each $\delta>0$ there is $m\geq 1$ and a homeomorphism
$F:{\mathbb{R}}^{3}\to{\mathbb{R}}^{3}$ such that $F$ leaves each point of
$g_{0}$ invariant and ${\rm diam}(F(g))<\delta$ for each $g\in
G_{m}\setminus\\{g_{0}\\}$.

###  4.2. Sequences of partitionings

Let $T=T(M,\mathcal{D})$ be an $ABC$-tile. In Section 3 it was convenient to
work with closed sets (the subtiles of $T$). When it comes to partitionings,
open sets are required. Therefore, in the sequel we will mainly work with the
interiors of subtiles. Moreover, we often use the one point compactification
$\mathbb{S}^{3}=\mathbb{R}^{3}\cup\\{\infty\\}$ as ambient space because
$\mathbb{S}^{3}$ is a Peano continuum. The following lemma provides a first
sequence of partitionings defined in terms of interiors of subtiles. We will
frequently use the notation $g_{\infty}=\mathbb{S}^{3}\setminus T$ in the
sequel. Note that $g_{\infty}=t_{\infty}^{\circ}\cup\\{\infty\\}$ and, hence,
$\partial_{\mathbb{S}^{3}}g_{\infty}=\partial_{\mathbb{R}^{3}}t_{\infty}$.

######  Lemma 4.7.

Let $T$ be an $ABC$-tile. Let $g_{\infty}=\mathbb{S}^{3}\setminus T$. Then for
each $i\geq 0$ the collection

(4.1)
$\mathcal{P}_{i}=\left\\{M^{-i}(T+z)^{\circ}\;:\;z\in\mathcal{D}_{i}\right\\}\cup\\{g_{\infty}\\}$

is a regular partitioning of $\mathbb{S}^{3}$. Moreover
$\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ is a regular partitioning of $T$.

###### Proof.

Let $i\geq 0$. By (2.2) we have $T=\bigcup_{d\in\mathcal{D}_{i}}M^{-i}(T+d)$.
Since each subtile $M^{-i}(T+d)$, $d\in\mathcal{D}_{i}$, is the closure of its
interior by Lemma 2.1, we see form (2.3) that
$\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ is a regular partitioning of $T$.
Thus $\mathcal{P}_{i}$ is a regular partitioning of $\mathbb{S}^{3}$. ∎

We use the definition of level from (3.1) also for the elements of
$\mathcal{P}_{i}$, $i\geq 0$. Indeed, we set

$\mathrm{level}(g)=\mathrm{level}(\overline{g}\setminus\\{\infty\\})$

for $g\in\bigcup_{i\geq 0}\mathcal{P}_{i}$. As usual, for a subset
$Y\subset\bigcup_{i\geq 0}\mathcal{P}_{i}$ we will write
$\mathrm{level}(Y)=\\{\mathrm{level}(g)\,:\,g\in Y\\}$.

In view of (2.4) an intersection $\partial g_{1}\cap\partial g_{2}$ for
disjoint atoms $g_{1},g_{2}\in\bigcup_{i\geq 0}\mathcal{P}_{i}$ is equal to
the intersection $\overline{g_{1}}\cap\overline{g_{2}}$ of the corresponding
elements $\overline{g_{1}},\overline{g_{2}}\in\mathcal{C}$.

We continue with topological properties of intersections of boundaries of the
atoms of $\mathcal{P}_{i}$.

######  Lemma 4.8.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let $i\geq
2$. For any $g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$, the intersection
$\partial g_{\infty}\cap\partial g$ is either empty, or a union of at most $2$
disjoint $2$-balls.

###### Proof.

Suppose that $\partial g_{\infty}\cap\partial g\neq\emptyset$. Then by
Proposition 3.11 (1) the intersection $\partial g_{\infty}\cap\partial g$ is
homeomorphic to $U(R)$, where $R\subseteq\mathcal{S}$ is a representative of a
node $\overline{R}$ of $\mathcal{I}$. We can now read off Figures 8 and 9 that
in this case $U(R)$ is either a union of at most $2$ disjoint $2$-balls, or
homeomorphic to $\mathbb{S}^{1}\times[0,1]$ (a “ribbon”), or homeomorphic to
$\mathbb{S}^{2}$. We need to exclude the last two cases. Suppose that
$\partial g_{\infty}\cap\partial g$ is homeomorphic to
$\mathbb{S}^{1}\times[0,1]$ or $\mathbb{S}^{2}$. Then, because
$g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ we have $\mathrm{level}(g)=i$
and, according to Proposition 3.11 (1) there is a walk
$\overline{\mathcal{S}}\rightarrow\overline{R_{1}}\rightarrow\cdots\rightarrow\overline{R_{i}}$
in $\mathcal{I}$ with $U(R_{i})$ being homeomorphic to
$\mathbb{S}^{1}\times[0,1]$ or $\mathbb{S}^{2}$. We know that
$\overline{\mathcal{S}}$ and $b^{1}$ are the only nodes of $\mathcal{I}$
homeomorphic to $\mathbb{S}^{1}\times[0,1]$ or $\mathbb{S}^{2}$. However, as
we see from Figure 8 and Figure 9, there is no walk of length $i\geq 2$ in
$\mathcal{I}$ ending at $b^{1}$ or $\mathcal{S}$. Thus $\partial
g_{\infty}\cap\partial g$ can neither be homeomorphic to
$\mathbb{S}^{1}\times[0,1]$ nor to $\mathbb{S}^{2}$. ∎

In view of Lemma 4.8 we can subdivide the atoms of
$\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$, $i\geq 2$, according to the way
they intersect $\partial g_{\infty}=\partial T$. In particular, for $i\geq 2$
set

(4.2) $\displaystyle\mathcal{P}_{i1}$
$\displaystyle=\\{g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}\;:\;\partial
g\cap\partial T=\emptyset\\},$ $\displaystyle\mathcal{P}_{i2}$
$\displaystyle=\\{g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}\;:\;\partial
g\cap\partial T\text{ is a single $2$-ball}\\},$
$\displaystyle\mathcal{P}_{i3}$
$\displaystyle=\\{g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}\;:\;\partial
g\cap\partial T\text{ is the union of $2$ disjoint $2$-balls}\\}.$

Then we have
$\mathcal{P}_{i}=\mathcal{P}_{i1}\cup\mathcal{P}_{i2}\cup\mathcal{P}_{i3}\cup\\{g_{\infty}\\}$.
We need partitionings whose atoms have intersections with $\partial T$ that
are either empty or a $2$-ball. To achieve this we further subdivide the atoms
of $\mathcal{P}_{i3}$ and put, again for $i\geq 2$,

$\displaystyle\mathcal{Q}_{i1}=$ $\displaystyle\mathcal{P}_{i1},$
$\displaystyle\mathcal{Q}_{i2}=$ $\displaystyle\mathcal{P}_{i2},$
$\displaystyle\mathcal{Q}_{i3}=$
$\displaystyle\\{g\in\mathcal{P}_{i+1}\;:\;g\subset g^{\prime}\text{ for
}g^{\prime}\in\mathcal{P}_{i3}\\}.$

Let $(\mathcal{Q}^{\prime}_{i})_{i\geq 1}$ be given by

(4.3) $\begin{split}\mathcal{Q}^{\prime}_{1}&=\\{T^{\circ}\\},\\\
\mathcal{Q}^{\prime}_{i}&=\mathcal{Q}_{i1}\cup\mathcal{Q}_{i2}\cup\mathcal{Q}_{i3}\qquad(i\geq
2),\end{split}$

and set $\mathcal{Q}_{i}=\mathcal{Q}^{\prime}_{i}\cup\\{g_{\infty}\\}$ for
$i\geq 1$. From this definition we immediately get

(4.4)
$\mathrm{level}(\mathcal{Q}^{\prime}_{1})=0\quad\hbox{and}\quad\mathrm{level}(\mathcal{Q}^{\prime}_{i})=\\{i,i+1\\}\hbox{
for }i\geq 2.$

######  Lemma 4.9.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. The
sequence $(\mathcal{Q}_{i}^{\prime})_{i\geq 1}$ given by (4.3) is a decreasing
sequence of regular partitionings of $T$.

###### Proof.

For $i=1$ the collection $\mathcal{Q}_{1}^{\prime}$ is clearly a regular
partitioning of $T$. Let now $i\geq 2$. Let
$g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ be given. Then
$g=M^{-i}(T+d)^{\circ}$ for some $d\in\mathcal{D}_{i}$. We claim that
$X_{g}=\\{h\in\mathcal{Q}_{i}^{\prime}\,:\,h\cap g\not=\emptyset\\}$ is a
regular partitioning of the Peano continuum $\overline{g}$. If
$g\in\mathcal{P}_{i1}\cup\mathcal{P}_{i2}$ then $X_{g}=\\{g\\}$ and the claim
is trivial. If $g\in\mathcal{P}_{i3}$ then
$X_{g}=\\{M^{-i-1}(T+d_{i}+Md)^{\circ}\,:\,d_{i}\in\mathcal{D}\\}$ and the
claim follows from Lemma 2.1 and (2.3) because
$\overline{g}=\bigcup_{d_{i}\in\mathcal{D}}M^{-i-1}(T+d_{i}+Md)$ by the set
equation (1.2). This proves the claim in all cases. Since

$\mathcal{Q}_{i}^{\prime}=\bigcup_{g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}}X_{g},$

$X_{g}$ is a regular partitioning of $\overline{g}$ for each
$g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$, and
$\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ is a regular partitioning of $T$ by
Lemma 4.7, we conclude that $\mathcal{Q}_{i}^{\prime}$ is a regular
partitioning of $T$ as well.

Because $M^{-1}$ is a uniform contraction,
$\max\\{\mathrm{diam}\,g\,:\,g\in\mathcal{P}_{i}\setminus\\{g_{\infty}\\}\\}=\mathrm{diam}\,M^{-i}T\to
0$ for $i\to\infty$. The fact that $(\mathcal{Q}_{i}^{\prime})_{i\geq 1}$ is
decreasing now follows because by (4.4), $\mathcal{Q}_{2}^{\prime}$ is a
refinement of $\mathcal{Q}_{1}^{\prime}$ and, for each $i\geq 2$,
$\mathcal{Q}^{\prime}_{i+1}$ is a refinement of
$\mathcal{P}_{i+1}\setminus\\{g_{\infty}\\}$ and
$\mathcal{P}_{i+1}\setminus\\{g_{\infty}\\}$ is a refinement of
$\mathcal{Q}^{\prime}_{i}$. ∎

Let $g\in\bigcup_{i\geq 0}\mathcal{P}_{i}\setminus\\{g_{\infty}\\}$ be given.
Then there is $k\geq 0$ and $d\in\mathcal{D}_{k}$ such that
$g=M^{-k}(T^{\circ}+d)$. In this case we associate with $g$ the mapping
$[g]:\mathbb{R}^{3}\to\mathbb{R}^{3}$, $x\mapsto M^{-k}(x+d)$. If $H$ is a
collection of sets, then we set $[g](H)=\\{[g](h)\,:\,h\in H\\}$. Clearly, if
$H$ is a partitioning of a Peano continuum $X$, then $[g](H)$ is a
partitioning of $[g](X)$. We need the following generalization of
$(\mathcal{Q}^{\prime}_{i})$ and $(\mathcal{Q}_{i})$. Let
$\mathbf{n}=(n_{j})_{j\geq 1}$ be a sequence with
$n_{j}\in\mathbb{N}\cup\\{\infty\\}$ satisfying $n_{1}\geq 3$ and
$n_{j+1}-n_{j}\geq 3$ (we allow that $\mathbf{n}$ can become ultimately
$\infty$, i.e., for each $n\in\mathbb{N}$ we define $n<\infty$ and
$\infty+n\leq\infty$). We define the sequence of partitionings
$(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))_{i\geq 1}$ by

(4.5)
$\begin{array}[]{rcll}\displaystyle\mathcal{Q}^{\prime}_{i}(\mathbf{n})&=&\displaystyle\mathcal{Q}^{\prime}_{i}&(1\leq
i<n_{1}),\\\\[5.69054pt]
\displaystyle\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})&=&\displaystyle\mathcal{Q}^{\prime}_{n_{j}-1}(\mathbf{n})&(j\geq
1),\\\\[5.69054pt]
\displaystyle\mathcal{Q}^{\prime}_{i}(\mathbf{n})&=&\displaystyle\bigcup_{g\in\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})}[g](\mathcal{Q}^{\prime}_{i-\mathrm{level}(g)})&(n_{j}<i<n_{j+1},\;j\geq
1).\end{array}$

Moreover, set
$\mathcal{Q}_{i}(\mathbf{n})=\mathcal{Q}^{\prime}_{i}(\mathbf{n})\cup\\{g_{\infty}\\}$
for $i\geq 1$. Note that $(\mathcal{Q}_{i})_{i\geq
1}=(\mathcal{Q}_{i}(\mathbf{n}))_{i\geq 1}$ if $\mathbf{n}=(n_{j})_{j\geq 1}$
satisfies $n_{j}=\infty$ for each $j\geq 1$.

######  Remark 4.10.

The definition of $(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))$ is a bit technical.
Its main feature is a repetitivity property. After $n_{j}$ steps each atom of
$\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ is subdivided in the same way as
$T$ itself (i.e. by using partitionings equivalent to
$\mathcal{Q}_{i}^{\prime}$) for $n_{j+1}-n_{j}-1$ steps. Sloppily speaking, in
$\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ each atom is subdivided by the
“nice” subdivision equivalent to $\mathcal{Q}_{i}^{\prime}$ for some time.
This repetitivity, which is not present in $(\mathcal{Q}_{i}^{\prime})$, will
be of importance later.

The next result contains basic properties of the sequence of partitionings
$(\mathcal{Q}_{i}(\mathbf{n}))_{i\geq 1}$.

######  Lemma 4.11.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let
$\mathbf{n}=(n_{j})_{j\geq 1}$ be a sequence with
$n_{j}\in\mathbb{N}\cup\\{\infty\\}$ satisfying $n_{1}\geq 3$ and
$n_{j+1}-n_{j}\geq 3$ and let $(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))_{i\geq
1}$ be as in (4.5). Then

  * (i)

$g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ implies
$\mathrm{level}(g)\in\\{i-1,i,i+1\\}\setminus\\{1\\}$ ($i\geq 1$).

  * (ii)

$\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ is a regular partitioning of $T$
($i\geq 1$).

  * (iii)

$(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))_{i\geq 1}$ is a decreasing sequence of
partitionings of $T$.

###### Proof.

To prove (i) we first prove the following more detailed result (set $n_{0}=0$
for convenience).

  * (a)

If $n_{j-1}+1<i<n_{j}$ then $g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ implies
$\mathrm{level}(g)\in\\{i,i+1\\}$ ($j\geq 1$).

  * (b)

If $i=n_{j}$ then $g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ implies
$\mathrm{level}(g)\in\\{i-1,i\\}$ ($j\geq 1$).

  * (c)

If $i=n_{j}+1$ then $g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ implies
$\mathrm{level}(g)\in\\{i-1,i,i+1\\}$ ($j\geq 0$).

This is proved by induction on $j$. For $1\leq i\leq n_{1}$ we have
$\mathcal{Q}^{\prime}_{i}(\mathbf{n})=\mathcal{Q}^{\prime}_{\min\\{i,n_{1}-1\\}}$
and the result follows from (4.4). Suppose that the result holds for $i\leq
n_{j}$. If $n_{j}<i<n_{j+1}$ then

(4.6)
$\mathcal{Q}^{\prime}_{i}(\mathbf{n})=\bigcup_{g\in\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})}[g](\mathcal{Q}^{\prime}_{i-\mathrm{level}(g)})$

Let $g^{\prime}\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$. Assume first that
$i=n_{j}+1$. Then, because
$\mathrm{level}(\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n}))=\\{n_{j}-1,n_{j}\\}$
this implies that either $g^{\prime}\in[g](\mathcal{Q}^{\prime}_{2})$ for some
$g$ with $\mathrm{level}(g)=n_{j}-1$, hence, by (4.4),
$\mathrm{level}(g^{\prime})\in\\{n_{j}+1,n_{j}+2\\}$, or
$g^{\prime}\in[g](\mathcal{Q}^{\prime}_{1})$ for some $g$ with
$\mathrm{level}(g)=n_{j}$, hence, by (4.4), $\mathrm{level}(g^{\prime})=n_{j}$
which is (c). If $n_{j}+1<i<n_{j+1}$ then, because
$\mathrm{level}(\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n}))=\\{n_{j}-1,n_{j}\\}$
this implies that either $g^{\prime}\in[g](\mathcal{Q}^{\prime}_{i-n_{j}+1})$
for some $g\in\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ with
$\mathrm{level}(g)=n_{j}-1$, hence, by (4.4),
$\mathrm{level}(g^{\prime})\in\\{i,i+1\\}$, or
$g^{\prime}\in[g](\mathcal{Q}^{\prime}_{i-n_{j}})$ for some
$g\in\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ with $\mathrm{level}(g)=n_{j}$,
hence, by (4.4), $\mathrm{level}(g^{\prime})\in\\{i,i+1\\}$ which is (a). If
$i=n_{j+1}$, (b) follows immediately from (a). This finishes the induction
proof of (a), (b), and (c).

Finally let $g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ for some $i\geq 1$. Now
(i) follows from (a), (b), and (c) because for $i=1$ we have
$\mathrm{level}(g)=0$, for $i=2$ we have $\mathrm{level}(g)\in\\{2,3\\}$
(since $n_{1}\geq 3$), and for $i\geq 3$ we have $\mathrm{level}(g)\geq 2$.
Thus $\mathrm{level}(g)=1$ cannot occur for any
$g\in\mathcal{Q}^{\prime}_{i}(\mathbf{n})$, $i\geq 1$.

To prove (ii) we use again induction on $j$. For $1\leq i\leq n_{1}$ the
collection $\mathcal{Q}_{1}^{\prime}(\mathbf{n})$ is a regular partitioning of
$T$ by Lemma 4.9. Let now $j\geq 2$. Since
$\mathcal{Q}^{\prime}_{n_{j+1}}(\mathbf{n})=\mathcal{Q}^{\prime}_{n_{j+1}-1}(\mathbf{n})$
we may assume that $n_{j}<i<n_{j+1}$. In this case
$[g](\mathcal{Q}^{\prime}_{i-\mathrm{level}(g)})$ is a regular partitioning of
$\overline{g}$ for each $g\in\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ by
Lemma 4.9, and $\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$ is a regular
partitioning of $T$ by the induction hypothesis. Thus by (4.6) also
$\mathcal{Q}^{\prime}_{i}(\mathbf{n})$ is a regular partitioning of $T$, and
the induction is finished. This proves (ii).

For (iii) we first show that $\mathcal{Q}^{\prime}_{i+1}(\mathbf{n})$ is a
refinement of $\mathcal{Q}^{\prime}_{i}(\mathbf{n})$. For
$n_{j}+1<i<n_{j+1}-1$ this follows from (a). For $i=n_{j}-1$ it follows
because
$\mathcal{Q}^{\prime}_{n_{j}-1}(\mathbf{n})=\mathcal{Q}^{\prime}_{n_{j}}(\mathbf{n})$,
for $i=n_{j}$ it follows form (b) and (c), and for $i=n_{j}+1$ it follows from
(c) and (a). The fact that $(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))_{i\geq 1}$
is decreasing follows from (i) because $M^{-1}$ a uniform contraction. ∎

The following result contains some topological properties of
$(\mathcal{Q}_{i}(\mathbf{n}))_{i\geq 1}$ that are related to some of the
conditions of Propositions 4.5 and 4.6.

######  Proposition 4.12.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let
$\mathbf{n}=(n_{j})_{j\geq 1}$ be a sequence with
$n_{j}\in\mathbb{N}\cup\\{\infty\\}$ satisfying $n_{1}\geq 3$ and
$n_{j+1}-n_{j}\geq 3$. Then the following conditions hold for $i\geq 2$.

  * (1)

For each $g\in\mathcal{Q}_{i}(\mathbf{n})$ we have $\partial
g\cong\mathbb{S}^{2}$.

  * (2)

If $g_{1},g_{2}\in\mathcal{Q}_{i}(\mathbf{n})$ are distinct then $\partial
g_{1}\cap\partial g_{2}$ is either empty or a union of at most $2$ disjoint
$2$-balls.

  * (3)

If $g_{1},g_{2}\in\mathcal{Q}_{i}$ are distinct then $\partial
g_{1}\cap\partial g_{2}$ is either empty or a single $2$-ball.777We need this
item only for $g_{1}=g_{\infty}$ but give the more general case for the sake
of completeness.

  * (4)

If $g_{1},g_{2},g_{3}\in\mathcal{Q}_{i}(\mathbf{n})$ are distinct then
$\partial g_{1}\cap\partial g_{2}\cap\partial g_{3}$ is either empty or a
finite union of arcs.

###### Proof.

Throughout the proof we assume that $i\geq 2$.

Each $g\in\mathcal{Q}_{i}(\mathbf{n})$ either satisfies
$g=\mathbb{S}^{3}\setminus T$ or $g=M^{-j}(T+z)^{\circ}$ for some
$j\in\\{i-1,i,i+1\\}\setminus\\{1\\}$ and some $z\in\mathbb{Z}^{3}$ by Lemma
4.11 (i). In any case $\partial g$ is homeomorphic to $\partial T$ and, hence,
item (1) follows from [35, Theorem 1.1 (1)].

If $g_{1},g_{2}\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})$ then, after possibly
exchanging $g_{1}$ and $g_{2}$, Lemma 4.11 (i) implies that there are
$k,l\in\\{i-1,i,i+1\\}\setminus\\{1\\}$ such that $k\leq l$,
$\mathrm{level}(g_{1})=k$ and $\mathrm{level}(g_{2})=l$. Assume that $\partial
g_{1}\cap\partial g_{2}\not=\emptyset$. Thus Proposition 3.11 (2) implies that
the intersection $\partial g_{1}\cap\partial g_{2}$ is homeomorphic to $U(R)$
for a node $\overline{R}$ of $\mathcal{I}$ which can be reached from one of
the nodes $\overline{\\{\alpha\\}}$, $\alpha\in\mathcal{S}$, by a walk of
length zero, one, or two. Since we see from Figures 8 and 9 (recall Remark
3.8) that all these nodes correspond to a $2$-ball, item (2) follows for this
case. It remains to show item (2) for the case $g_{1}=g_{\infty}$. Because
$i\geq 2$ we have $\mathrm{level}(g_{2})\geq 2$ by Lemma 4.11 (i) and (2)
follows from Lemma 4.8

To prove (3) let first $g_{1},g_{2}\in\mathcal{Q}_{i}^{\prime}$ then, after
possibly exchanging $g_{1}$ and $g_{2}$, by the definition of
$\mathcal{Q}_{i}^{\prime}$ there are $k,l\in\\{i,i+1\\}$ such that $k\leq l$,
$\mathrm{level}(g_{1})=k$ and $\mathrm{level}(g_{2})=l$. Assume that $\partial
g_{1}\cap\partial g_{2}\not=\emptyset$. Then Proposition 3.11 (2) implies that
the intersection $\partial g_{1}\cap\partial g_{2}$ is homeomorphic to $U(R)$
for a node $\overline{R}$ of $\mathcal{I}$ which can be reached from one of
the nodes $\overline{\\{\alpha\\}}$, $\alpha\in\mathcal{S}$, by a walk of
length zero or one. Since we see from Figures 8 and 9 (recall again Remark
3.8) that all these nodes correspond to a $2$-ball, item (2) follows for this
case. Let now $g_{1}=g_{\infty}$ and assume that $\partial
g_{\infty}\cap\partial g_{2}\not=\emptyset$. If
$g_{2}\in\mathcal{Q}_{i1}\cup\mathcal{Q}_{i2}$ then by the definition of
$\mathcal{Q}_{i1}$ and $\mathcal{Q}_{i2}$, $\partial g_{\infty}\cap\partial
g_{2}$ is clearly a $2$-ball. If $g_{2}\in\mathcal{Q}_{i3}$ then by the
definition of $\mathcal{Q}_{i3}$, there is $g_{2}^{\prime}\in\mathcal{P}_{i3}$
with $\mathrm{level}(g_{2})=\mathrm{level}(g_{2}^{\prime})+1$ and
$g_{2}\subset g_{2}^{\prime}$ such that $\partial g_{\infty}\cap\partial
g_{2}^{\prime}$ is a union of $2$ disjoint $2$-balls. By Proposition 3.11 (1)
the intersection $\partial g_{\infty}\cap\partial g_{2}^{\prime}$ is of type
$\overline{R^{\prime}}$ where $\overline{R^{\prime}}\in\mathcal{I}$. Lemma
3.10 now implies that there is an edge
$\overline{R^{\prime}}\rightarrow\overline{R}$ in $\mathcal{I}$ such that
$U(R^{\prime})$ is a union of at least $2$ disjoint $2$-balls and $\partial
g_{\infty}\cap\partial g_{2}\simeq U(R)$. An inspection of the graph
$\mathcal{I}$ in Figures 8 and 9 shows that each successor of a node
corresponding to $2$ disjoint $2$-balls corresponds to a single $2$-ball. Thus
$\partial g_{\infty}\cap\partial g_{2}$ is a $2$-ball and item (3) is proved.

To prove item (4) we note that by Lemma 4.11 (i) each of the atoms
$g_{j}\in\mathcal{Q}_{i}(\mathbf{n})$ ($1\leq j\leq 3$) is a union of sets of
the form $M^{-i-1}(T+z)^{\circ}$ with $z\in\mathbb{Z}^{3}$. This union is
finite unless $g_{j}=g_{\infty}$. Thus $\partial g_{1}\cap\partial
g_{2}\cap\partial g_{3}$ is a finite (possibly empty) union of intersections
of the form $M^{-i-1}((T+z_{1})\cap(T+z_{2})\cap(T+z_{3}))$ with
$z_{1},z_{2},z_{3}\in\mathbb{Z}^{3}$. By Proposition 2.6 (2) each of these
intersections is either empty or homeomorphic to an arc. This proves item (4).
∎

###  4.3. An order on the subsets of an atom

Let $T$ be an $ABC$-tile with $A=1$ having $14$ neighbors. Let
$\mathbf{n}=(n_{j})_{j\geq 1}$ be a sequence with
$n_{j}\in\mathbb{N}\cup\\{\infty\\}$ satisfying $n_{1}\geq 3$ and
$n_{j+1}-n_{j}\geq 3$ and let $(\mathcal{Q}_{i}(\mathbf{n}))_{i\geq 1}$ be the
associated sequence of regular partitionings defined in (4.5) (see Lemma
4.11). In this section we define an order on the sets
$\\{g^{\prime}\in\mathcal{Q}_{i+1}(\mathbf{n})\,:\,g^{\prime}\subseteq g\\}$
of atoms in $\mathcal{Q}_{i+1}(\mathbf{n})$ that are contained in some fixed
$g\in\mathcal{Q}_{i}(\mathbf{n})$ and prove some connectivity properties of
related intersections ($i\geq 1$).

Let $k\in\mathbb{N}$. If $z=(e_{k-1},\ldots,e_{0})_{M}$ and
$z^{\prime}=(e^{\prime}_{k-1},\ldots,e^{\prime}_{0})_{M}$ are elements of
$\mathcal{D}_{k}$ we say that $z\prec z^{\prime}$ if and only if
$(e_{k-1},\ldots,e_{0})<_{\mathrm{lex}}(e^{\prime}_{k-1},\ldots,e^{\prime}_{0})$
in lexicographic order (so, for instance $(2,1,4)_{M}\prec(3,0,0)_{M}$ and
$(0,2,3)_{M}\prec(0,2,4)_{M}$). This defines an order on $\mathcal{D}_{k}$. By
definition, this order has the following property. Let
$k,k^{\prime}\in\mathbb{N}$ with $k\leq k^{\prime}$ be given. Let
$M^{-k}(T+d_{1})$, $M^{-k}(T+d_{2})$ with $d_{1},d_{2}\in\mathcal{D}_{k}$,
$d_{1}\neq d_{2}$. If $M^{-k^{\prime}}(T+d^{\prime}_{\ell})$ with
$d^{\prime}_{\ell}\in\mathcal{D}_{k^{\prime}}$ is a subtile of
$M^{-k}(T+d_{\ell})$ for $\ell\in\\{1,2\\}$, then

(4.7) $d_{1}\prec d_{2}\quad(\hbox{in
}\mathcal{D}_{k})\qquad\Longleftrightarrow\qquad d^{\prime}_{1}\prec
d^{\prime}_{2}\quad(\hbox{in }\mathcal{D}_{k^{\prime}}).$

We continue with two lemmas that will be needed in the proof of the
connectivity result stated in Proposition 4.15.

######  Lemma 4.13.

Let $T$ be an $ABC$-tile and assume that $A=1$. Let
$z=(e_{2},e_{1},e_{0})_{M}\in\mathcal{D}_{3}$ be given. Then the following
assertions hold (where “$\prec$” denotes the order on $\mathcal{D}_{3}$).

  * •

$z+P\in\mathcal{D}_{3}$ with $z\prec z+P$ if and only if $e_{0}<C-1$.

  * •

$z+Q\in\mathcal{D}_{3}$ with $z\prec z+Q$ if and only if $e_{0}<C-1$ and
$e_{1}<C-1$.

  * •

$z+N\in\mathcal{D}_{3}$ with $z\prec z+N$ if and only if $e_{0}<C-B$,
$e_{1}<C-1$, and $e_{2}<C-1$.

  * •

$z+Q-P\in\mathcal{D}_{3}$ with $z\prec z+Q-P$ if and only if $e_{1}<C-1$.

  * •

$z+N-P\in\mathcal{D}_{3}$ with $z\prec z+N-P$ if and only if $e_{0}<C-B+1$,
$e_{1}<C-1$, and $e_{2}<C-1$.

  * •

$z+N-Q\in\mathcal{D}_{3}$ with $z\prec z+N-Q$ if and only if $e_{0}<C-B+1$ and
$e_{2}<C-1$.

  * •

$z+N-Q+P\in\mathcal{D}_{3}$ with $z\prec z+N-Q+P$ if and only if $e_{0}<C-B$
and $e_{2}<C-1$.

  * •

Let $\alpha\in-\mathcal{S}_{1}$. Then $z+\alpha\in\mathcal{D}_{3}$ and $z\prec
z+\alpha$ cannot hold simultaneously.

The proof is done easily by direct calculation; note that
$(e_{2},e_{1},e_{0})_{M}=\begin{pmatrix}e_{0}\\\ e_{1}\\\ e_{2}\end{pmatrix}$.

######  Lemma 4.14.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let
$j\in\\{1,2,3\\}$ and $z\in\mathcal{D}_{j}$ be given. Then

$U_{z,j}=(T+z)\cap\bigg{(}\partial(M^{j}T)\cup\bigcup_{\begin{subarray}{c}y\prec
z\\\ y\in\mathcal{D}_{j}\end{subarray}}(T+y)\bigg{)}$

is a connected set (here “$\prec$” denotes the order on $\mathcal{D}_{j}$).

###### Proof.

The intersection in the statement of the lemma can be written as

$U_{z,j}=(T+z)\cap\bigg{(}\bigcup_{y\not\in\mathcal{D}_{j}}(T+y)\cup\bigcup_{\begin{subarray}{c}y\prec
z\\\
y\in\mathcal{D}_{j}\end{subarray}}(T+y)\bigg{)}=\bigcup_{\alpha\in\mathcal{S}\setminus\mathcal{S}^{\prime}}(T+z)\cap(T+z+\alpha),$

where

$\mathcal{S}^{\prime}=\\{\alpha\in\mathcal{S}\;:\;z+\alpha\in\mathcal{D}_{j}\hbox{
and }z\prec z+\alpha\\}.$

We prove the case $j=3$. To this end let
$z=(e_{2},e_{1},e_{0})_{M}\in\mathcal{D}_{3}$. We have to distinguish 12 cases
according to the inequalities occurring in Lemma 4.13.

  * (i)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}\in\\{0,\ldots,C-2\\}$,
$e_{0}\in\\{0,\ldots,C-B-1\\}$. According to Lemma 4.13 in this case we have
$\mathcal{S}^{\prime}=\mathcal{S}_{1}$ and, hence,
$U_{z,3}=\bigcup_{\alpha\in-\mathcal{S}_{1}}(T+z)\cap(T+z+\alpha)$ is
homeomorphic to the (connected) $2$-ball $a^{3}$.

  * (ii)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}\in\\{0,\ldots,C-2\\}$, $e_{0}=C-B$. Here
Lemma 4.13 yields $\mathcal{S}^{\prime}=\\{P,Q,Q-P,N-P,N-Q\\}$, hence,
$\mathcal{S}\setminus\mathcal{S}^{\prime}=-S_{1}\cup\\{N,N-Q+P\\}$ and
$U_{z,3}$ easily seen to be a $2$-ball by using Lemma 2.8.

  * (iii)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}\in\\{0,\ldots,C-2\\}$,
$e_{0}=\\{C-B+1,\ldots,C-2\\}$. Lemma 4.13 yields
$\mathcal{S}^{\prime}=\\{P,Q,Q-P\\}$ and $U_{z,3}$ is homeomorphic to the
$2$-ball $a^{2}$.

  * (iv)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}\in\\{0,\ldots,C-2\\}$, $e_{0}=C-1$. Here
$\mathcal{S}^{\prime}=\\{Q-P\\}$ and $U_{z,3}$ is homeomorphic to the $2$-ball
$\partial T\setminus$ $g^{4}$.

  * (v)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}=C-1$, $e_{0}\in\\{0,\ldots,C-B-1\\}$.
Here $\mathcal{S}^{\prime}=\\{P,N-Q,N-Q+P\\}$ and $U_{z,3}$ is homeomorphic to
the $2$-ball $\partial T\setminus$ $d^{4}$.

  * (vi)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}=C-1$, $e_{0}=C-B$. Here
$\mathcal{S}^{\prime}=\\{P,N-Q\\}$ and $U_{z,3}$ is homeomorphic to
$\mathbb{S}^{1}\times[0,1]$ by Lemma 2.8 and, hence, it is connected.

  * (vii)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}=C-1$, $e_{0}=\\{C-B+1,\ldots,C-2\\}$.
Here $\mathcal{S}^{\prime}=\\{P\\}$ and $U_{z,3}$ is homeomorphic to the
$2$-ball $a^{1}$.

  * (viii)

$e_{2}\in\\{0,\ldots,C-2\\}$, $e_{1}=C-1$, $e_{0}=C-1$. Here
$\mathcal{S}^{\prime}=\emptyset$ and $U_{z,3}$ is homeomorphic to the
$2$-sphere by Lemma 2.8 and, hence, it is connected.

  * (ix)

$e_{2}=C-1$, $e_{1}\in\\{0,\ldots,C-2\\}$, $e_{0}\in\\{0,\ldots,C-2\\}$. Here
$\mathcal{S}^{\prime}=\\{P,Q,Q-P\\}$ and, hence, $U_{z,3}$ is homeomorphic to
the $2$-ball $a^{2}$.

  * (x)

$e_{2}=C-1$, $e_{1}\in\\{0,\ldots,C-2\\}$, $e_{0}=C-1$. Here
$\mathcal{S}^{\prime}=\\{Q-P\\}$ and $U_{z,3}$ is homeomorphic to the $2$-ball
$\partial T\setminus$ $g^{4}$.

  * (xi)

$e_{2}=C-1$, $e_{1}=C-1$, $e_{0}\in\\{0,\ldots,C-2\\}$. Here
$\mathcal{S}^{\prime}=\\{P\\}$ and $U_{z,3}$ is homeomorphic to the $2$-ball
$a^{1}$.

  * (xii)

$e_{2}=C-1$, $e_{1}=C-1$, $e_{0}=C-1$. Here $\mathcal{S}^{\prime}=\emptyset$
and $U_{z,3}$ is homeomorphic to the $2$-sphere.

The proof for the cases $j\in\\{1,2\\}$ is similar but easier than the case
$j=3$ and we omit it. ∎

We are now ready to prove the following proposition. Note that the property
proved in this result is related to the condition stated in Proposition 4.5
(4).

######  Proposition 4.15.

Let $T$ be an $ABC$-tile with $14$ neighbors and assume that $A=1$. Let
$\mathbf{n}=(n_{j})_{j\geq 1}$ be a sequence with
$n_{j}\in\mathbb{N}\cup\\{\infty\\}$ satisfying $n_{1}\geq 3$ and
$n_{j+1}-n_{j}\geq 3$. Let $i\geq 1$ and let $g\in\mathcal{Q}_{i}(\mathbf{n})$
be given. The set
$\\{g_{1},\ldots,g_{n}\\}\subseteq\mathcal{Q}_{i+1}(\mathbf{n})$ of all atoms
of $\mathcal{Q}_{i+1}(\mathbf{n})$ that are subsets of $g$ can be ordered in a
way that $\partial g_{j}\cap(\partial g\cup\partial g_{1}\cup\dots\cup\partial
g_{j-1})$ is connected for each $j\in\\{1,\ldots,n\\}$.

###### Proof.

If $n=1$ (which is true in particular for $g=g_{\infty}$) the result is
trivial. If $n>1$ then $g=M^{-k}(T^{\circ}+z)$ for some $k\geq 0$ and some
$z\in\mathcal{D}_{k}$. For convenience, we set $g^{\prime}=M^{k}g-z=T^{\circ}$
and $g_{j}^{\prime}=M^{k}g_{j}-z$ for $j\in\\{1,\ldots,n\\}$. By Lemma 4.11
(i) we know that $g^{\prime}_{j}=M^{-k_{j}}(T^{\circ}+y_{j})$ with
$k_{j}\in\\{1,2,3\\}$ and $y_{j}\in\mathcal{D}_{k_{j}}$. We assume that
$\\{g^{\prime}_{1},\ldots,g^{\prime}_{n}\\}$ is ordered in a way that the
following is true: For each $j$ subdivide $\overline{g_{j}^{\prime}}$ in
subtiles of the form $M^{-3}(T+d)$ with $d\in\mathcal{D}_{3}$ by the set
equation (2.2). Let $j_{1},j_{2}\in\\{1,\ldots,n\\}$ be distinct and let
$M^{-3}(T+d_{\ell})$ be a subtile of $\overline{g_{j_{\ell}}^{\prime}}$
($\ell\in\\{1,2\\}$). Then $d_{1}\prec d_{2}$ w.r.t. the order in
$\mathcal{D}_{3}$ if and only if $j_{1}<j_{2}$.

Note that

(4.8) $\partial g_{j}\cap(\partial g\cup\partial g_{1}\cup\dots\cup\partial
g_{j-1})\simeq\partial g^{\prime}_{j}\cap(\partial g^{\prime}\cup\partial
g^{\prime}_{1}\cup\dots\cup\partial
g^{\prime}_{j-1})=\overline{g^{\prime}_{j}}\cap(\partial{g^{\prime}}\cup\overline{g^{\prime}_{1}}\cup\dots\cup\overline{g^{\prime}_{j-1}})$

holds for each $j\in\\{1,\ldots,n\\}$ (the equality holds because the sets
$\overline{g^{\prime}_{1}},\ldots,\overline{g^{\prime}_{n}}$ cover
$\overline{g^{\prime}}$ overlapping only at their boundaries). Moreover, we
have

$\displaystyle\overline{g^{\prime}_{j}}\cap(\partial
g^{\prime}\cup\overline{g^{\prime}_{1}}\cup\dots\cup\overline{g^{\prime}_{j-1}})$
$\displaystyle=M^{-k_{j}}(T+y_{j})\cap\bigg{(}\partial
T\cup\bigcup_{\ell=1}^{j-1}M^{-k_{\ell}}(T+y_{\ell})\bigg{)}$ (4.9)
$\displaystyle=M^{-k_{j}}(T+y_{j})\cap\bigg{(}\partial
T\cup\bigcup_{\begin{subarray}{c}y\prec y_{j}\\\
y\in\mathcal{D}_{k_{j}}\end{subarray}}M^{-k_{j}}(T+y)\bigg{)}$
$\displaystyle=M^{-k_{j}}\bigg{(}(T+y_{j})\cap\bigg{(}\partial(M^{k_{j}}T)\cup\bigcup_{\begin{subarray}{c}y\prec
y_{j}\\\ y\in\mathcal{D}_{k_{j}}\end{subarray}}(T+y)\bigg{)}\bigg{)}.$

In the second equality we used the set equation (2.2) to subdivide (if
$k_{\ell}<k_{j}$) or group (if $k_{\ell}>k_{j}$) the sets
$M^{-k_{\ell}}(T+y_{\ell})$ into sets that are all of the form
$M^{-k_{j}}(T+y)$ for some $y\in\mathcal{D}_{k_{j}}$. By the ordering of
$\\{g^{\prime}_{1},\ldots,g^{\prime}_{n}\\}$ and by property (4.7) of
“$\prec$” this yields the union over all sets of the form $M^{-k_{j}}(T+y)$
with $y\prec y_{j}$, $y\in\mathcal{D}_{k_{j}}$.

Because $k_{j}\in\\{1,2,3\\}$, the last set in (4.9) is connected by Lemma
4.14 and the result follows from (4.8) and (4.9). ∎

###  4.4. Proof of Theorem 1.1

We have to show that $T$ is homeomorphic to $\mathbb{D}^{3}$ under the
conditions of Theorem 1.1. In view of Section 2.2 we may assume that $T$ is an
$ABC$-tile with $14$ neighbors and that $A=1$. Throughout the proof we will
use the fact that $T$ is a Peano continuum by Lemma 2.5.

Our strategy is to construct a decreasing sequence of partitionings of
$\mathbb{D}^{3}$ that is equivalent to
$(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))_{i\geq 1}$ for a suitable sequence
$\mathbf{n}=(n_{j})_{j\geq 1}$ with $n_{j}\in\mathbb{N}$ (finite) satisfying
$n_{1}\geq 3$ and $n_{j+1}-n_{j}\geq 3$. Then the result will follow from
Lemma 4.4. We will use the theory of partitionings due to Bing [5]. Bing gives
a topological characterization of $3$-spheres in terms of decreasing sequences
of regular partitionings. In Theorem 1.1 we deal with $3$-balls instead of
$3$-spheres. However, the main difference between Bing’s setting and our’s is
that, contrary to his assumptions (see [5, Theorem 1 (1.2)] and the discussion
in [5, p. 25]), we do not have that for $g_{1},g_{2}\in\bigcup_{i\geq
1}\mathcal{Q}_{i}^{\prime}(\mathbf{n})$ the intersection $\partial
g_{1}\cap\partial g_{2}$ is either empty or homeomorphic to $\mathbb{D}^{2}$.
We have to settle for the weaker results in Proposition 4.12 (2) and (3). To
make up for this we will exploit the self-affinity of $T$. This difference is
the reason why we cannot use Bing’s original proof here.

The following lemma contains the crucial tool for the proof of Theorem 1.1.

######  Lemma 4.16.

There is a sequence $\mathbf{n}=(n_{j})_{j\geq 1}$ with $n_{j}\in\mathbb{N}$,
$n_{j}\geq 3$ and $n_{j+1}-n_{j}\geq 3$ such that there are sequences
$(H_{i})_{i\geq 1}$ and $(K_{n_{j}})_{j\geq 1}$ of partitionings of
$\mathbb{R}^{3}$ with the following properties.

  * (i)

For each $h\in H_{i}$ the boundary $\partial h$ is a tame $2$-sphere in
$\mathbb{R}^{3}$ ($i\geq 1$).

  * (ii)

$H_{i+1}$ is a refinement of $H_{i}$ for each $i\geq 1$.

  * (iii)

$h_{0}=\mathbb{R}^{3}\setminus\mathbb{D}^{3}$ is an atom of $H_{i}$ for each
$i\geq 1$.

  * (iv)

$(H_{i}\setminus\\{h_{0}\\})_{i\geq 1}$ is equivalent to
$(\mathcal{Q}^{\prime}_{i}(\mathbf{n}))_{i\geq 1}$ in the sense of Definition
4.2.

  * (v)

There is a sequence $(F_{i})_{i\geq 1}$ where
$F_{i}:\bigcup_{g\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})}\partial
g\to\bigcup_{h\in H_{i}\setminus\\{h_{0}\\}}\partial h$ is a homeomorphism
with the following properties: If $i>1$ then the restriction of $F_{i}$ to
$\bigcup_{g\in\mathcal{Q}_{i-1}^{\prime}(\mathbf{n})}\partial g$ is equal to
$F_{i-1}$. If $i\geq 1$ then for each
$g\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})$ we have $F_{i}(\partial g)=\partial
h$, where $h\in H_{i}\setminus\\{h_{0}\\}$ is the atom corresponding to $g$
under the equivalence in (iv).

  * (vi)

For each $k\in K_{n_{j}}$ the boundary $\partial k$ is a tame $2$-sphere in
$\mathbb{R}^{3}$ ($j\geq 1$).

  * (vii)

$K_{n_{j+1}}$ is a refinement of $K_{n_{j}}$ for each $j\geq 1$.

  * (viii)

$h_{0}$ is an atom of $K_{n_{j}}$ for each $j\geq 1$.

  * (ix)

$(K_{n_{j}})_{j\geq 1}$ is equivalent to $(H_{n_{j}})_{j\geq 1}$ in the sense
of Definition 4.2.

  * (x)

$(K_{n_{j}}\setminus\\{h_{0}\\})_{j\geq 1}$ is a decreasing sequence of
partitionings of $\mathbb{D}^{3}$.

###### Proof.

The proof splits in two parts. The first part is an induction proof in which
we construct the sequences $(H_{i})_{i\geq 1}$ and $(K_{n_{j}})_{j\geq 1}$,
where the second sequence might a priori be finite. In the second part of the
proof we show that $(K_{n_{j}})_{j\geq 1}$ is in fact an infinite sequence.

We say that $\mathcal{A}(m)$ holds if there exist
$j_{0}=j_{0}(m)\in\mathbb{N}$, $n_{1},\ldots,n_{j_{0}}\leq m$, $n_{j}=\infty$
for $j>j_{0}$, and finite sequences $(H_{i})_{i=1}^{m}$,
$(K_{n_{j}})_{j=1}^{j_{0}}$ of partitionings of $\mathbb{R}^{3}$ such that the
following properties hold. (We set $n_{0}=0$ and
$K_{0}=\\{h_{0},\mathbb{R}^{3}\setminus\\{h_{0}\\}\\}$ for convenience.)

  * (i-$m$)

For each $h\in H_{i}$ the boundary $\partial h$ is a tame $2$-sphere in
$\mathbb{R}^{3}$ $(1\leq i\leq m)$.

  * (ii-$m$)

$H_{i+1}$ is a refinement of $H_{i}$ ($1\leq i<m$).

  * (iii-$m$)

$h_{0}=\mathbb{R}^{3}\setminus\mathbb{D}^{3}$ is an atom of $H_{i}$ ($1\leq
i\leq m$).

  * (iv-$m$)

$(H_{i}\setminus\\{h_{0}\\})_{i=1}^{m}$ is equivalent to
$(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))_{i=1}^{m}$ in the sense of Definition
4.2.

  * (v-$m$)

There is a sequence $(F_{i})_{i=1}^{m}$ where
$F_{i}:\bigcup_{g\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})}\partial
g\to\bigcup_{h\in H_{i}\setminus\\{h_{0}\\}}\partial h$ is a homeomrophism
with the following properties: If $1<i\leq m$ then the restriction of $F_{i}$
to $\bigcup_{g\in\mathcal{Q}_{i-1}^{\prime}(\mathbf{n})}\partial g$ is equal
to $F_{i-1}$. If $1\leq i\leq m$ then for each
$g\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})$ we have $F_{i}(\partial g)=\partial
h$, where $h\in H_{i}\setminus\\{h_{0}\\}$ is the atom corresponding to $g$
under the equivalence in (iv-$m$).

  * (vi-$m$)

For each $k\in K_{n_{j}}$ the boundary $\partial k$ is a tame $2$-sphere in
$\mathbb{R}^{3}$ ($1\leq j\leq j_{0}$).

  * (vii-$m$)

$K_{n_{j+1}}$ is a refinement of $K_{n_{j}}$ $(1\leq j<j_{0})$.

  * (viii-$m$)

$h_{0}$ is an atom of $K_{n_{j}}$ $(1\leq j\leq j_{0})$.

  * (ix-$m$)

$(K_{n_{j}})_{j=1}^{j_{0}}$ is equivalent to $(H_{n_{j}})_{j=1}^{j_{0}}$ in
the sense of Definition 4.2.

  * (x-$m$)

There exist homeomorphisms
$f_{1},\ldots,f_{j_{0}}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that each
boundary point of each atom of $K_{n_{j-1}}$ is invariant under $f_{j}$ and
$f_{j}\circ\cdots\circ f_{1}$ keeps $\mathbb{R}^{3}\setminus\mathbb{D}^{3}$
invariant and carries each other atom of $H_{n_{j}}$ into a set of diameter
less than $\frac{1}{2^{j}}$. Moreover,

$K_{n_{j}}=\\{f_{j}\circ\cdots\circ f_{2}\circ f_{1}(h)\;:\;h\in H_{n_{j}}\\}$

($1\leq j\leq j_{0}$). Thus $K_{n_{j}}\setminus\\{h_{0}\\}$ is a partitioning
of $\mathbb{D}^{3}$ with $\max\\{{\rm diam}(k)\,:\,k\in
K_{n_{j}}\\}<\frac{1}{2^{j}}$.

To prove the lemma we first show by induction that $\mathcal{A}(m)$ is true
for all $m\geq 1$. In the course of this induction proof we construct
sequences $(H_{i})_{i\geq 1}$ and $(K_{n_{j}})_{j\geq 1}$ that satisfy (i-$m$)
– (x-$m$) for each $m\in\mathbb{N}$. This induction argument implies (i) –
(v). To gain (vi) – (x) we have to show that our construction leads to
$j_{0}(m)\nearrow\infty$ for $m\to\infty$.

For the induction start we prove $\mathcal{A}(1)$. Set $j_{0}(1)=0$. Thus
$n_{j}=\infty$ for all $j\geq 1$ and $(K_{n_{j}})_{j=1}^{0}$ is the empty
sequence. Set
$H_{1}=\\{(\mathbb{D}^{3})^{\circ},\mathbb{R}^{3}\setminus\mathbb{D}^{3}\\}$.
Then $\bigcup_{g\in\mathcal{Q}_{1}^{\prime}(\mathbf{n})}\partial
g=\bigcup_{g\in\mathcal{Q}_{1}^{\prime}}\partial g=\partial T$ and
$\bigcup_{h\in H_{1}\setminus\\{h_{0}\\}}\partial h=\partial\mathbb{D}^{3}$.
Since $\partial T$ is a $2$-sphere by [35, Theorem 1.1], there exists a
homeomorphism $F_{1}:\partial T\to\partial\mathbb{D}^{3}$. Thus $H_{1}$
satisfies (i-$1$), (ii-$1$) (which is empty for $m=1$), (iii-$1$), (iv-$1$)
(note that $\mathcal{Q}^{\prime}_{1}(\mathbf{n})=\\{T^{\circ}\\}$; thus
$T^{\circ}$ corresponds to $(\mathbb{D}^{3})^{\circ}$), and (v-$1$) (whose
first assertion is empty for $m=1$). Since $j_{0}(1)=0$, assertions (vi-$1$) –
(x-$1$) are empty. This concludes the induction start.

To perform the induction step, let $m\geq 1$ and assume that $\mathcal{A}(m)$
is true. We have to distinguish two cases.

Case 1: For $j_{0}=j_{0}(m)$ we have $m\geq n_{j_{0}}+2$ and there exists a
homeomorphism $f_{j_{0}+1}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ such that each
boundary point of each atom of $K_{n_{j_{0}}}$ is invariant under
$f_{j_{0}+1}$ and $f_{j_{0}+1}\circ\cdots\circ f_{1}$ keeps
$\mathbb{R}^{3}\setminus\mathbb{D}^{3}$ invariant and carries each other atom
of $H_{m}$ into a set of diameter less than $\frac{1}{2^{j_{0}+1}}$.

Case 2: $m<n_{j_{0}(m)}+2$ or a homeomorphism as in Case 1 does not exist
(this is the complement of Case 1).

If Case 1 is in force then set $j_{0}(m+1)=j_{0}(m)+1$ and
$n_{j_{0}(m+1)}=m+1$. This has no effect on the partitionings
$\mathcal{Q}_{1}^{\prime}(\mathbf{n}),\ldots,\mathcal{Q}_{m}^{\prime}(\mathbf{n})$.
By the definition of $(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))_{i\geq 1}$ in
(4.5) we have

$\mathcal{Q}_{m}^{\prime}(\mathbf{n})=\mathcal{Q}_{n_{j_{0}(m+1)-1}}^{\prime}(\mathbf{n})=\mathcal{Q}_{n_{j_{0}(m+1)}}^{\prime}(\mathbf{n})=\mathcal{Q}_{m+1}^{\prime}(\mathbf{n}).$

Thus, setting $H_{m+1}=H_{m}$ trivially yields (i-$(m+1)$) – (v-$(m+1)$) from
(i-$m$) – (v-$m$). Now let $f_{j_{0}(m)+1}=f_{j_{0}(m+1)}$ be the
homeomorphism having the properties specified in Case 1 and set
$K_{m+1}=\\{f_{j_{0}(m+1)}\circ\cdots\circ f_{2}\circ f_{1}(h)\,:\,h\in
H_{m+1}\\}$. The condition for Case 1 (here we use that $H_{m+1}=H_{m}$)
together with (x-$m$) implies that (x-$(m+1)$) is true. Because
$f_{j_{0}(m+1)}\circ\cdots\circ f_{2}\circ f_{1}$ is a homeomorphism that
keeps $\mathbb{R}^{3}\setminus\mathbb{D}^{3}$ invariant, (vi-$(m+1)$),
(viii-$(m+1)$), and (ix-$(m+1)$) follow. Finally, (vii-$(m+1)$) is true
because $f_{j_{0}(m+1)}$ leaves each boundary point of $K_{n_{j_{0}(m)}}$
invariant by (x-$(m+1)$). This finishes the induction step for Case 1.

If Case 2 holds, set $j_{0}(m+1)=j_{0}(m)$. Let
$a\in\mathcal{Q}_{m}^{\prime}(\mathbf{n})$ and let
$\\{g_{a,1},g_{a,2},\dots,g_{a,n(a)}\\}=\\{g\in\mathcal{Q}^{\prime}_{m+1}(\mathbf{n})\,:\,g\subseteq
a\\}$ be ordered in a way that they satisfy the conclusion of Proposition
4.15. Let $h(a)\in H_{m}$ be the element corresponding to $a$ via (iv-$m$). We
want to apply Proposition 4.5 with $S=\mathbb{S}^{3}$,
$S_{2}=\partial_{\mathbb{S}^{3}}a$, $C=\partial_{\mathbb{R}^{3}}h(a)$,
$G=\mathcal{Q}_{m+1}(\mathbf{n})$, and $F=F_{m}|_{\partial a}$. Therefore, we
have to check the conditions of this proposition. By Lemma 4.11 (ii),
$\mathcal{Q}_{m+1}(\mathbf{n})$ is a regular partitioning of $\mathbb{S}^{3}$
and Proposition 4.12 implies that $\mathcal{Q}_{m+1}(\mathbf{n})$ satisfies
conditions (1), (2), and (3) of Proposition 4.5 (note that $m+1\geq 2$). By
the order we chose for the elements $g_{a,1},\ldots,g_{a,n(a)}$ (using
Proposition 4.15), the set $\\{g_{a,1},\dots,g_{a,n(a)}\\}$ satisfies
condition (4) of Proposition 4.5 (observe that
$a=\big{(}\overline{\bigcup_{\ell=1}^{n(a)}g_{a,\ell}}\big{)}^{\circ}$). Thus
we can apply Proposition 4.5. This yields a partitioning
$H_{m+1,a}=\\{h_{a,0},h_{a,1},\dots,h_{a,n(a)}\\}$ of $\mathbb{R}^{3}$, where
$h_{a,0}$ is the exterior of $\partial h(a)$ and the boundaries of $h_{a,j}$
are tame $2$-spheres, and a homeomorphism

$F_{m+1,a}:\partial_{\mathbb{S}^{3}}(g_{a,1}\cup\dots\cup
g_{a,n(a)})\to\partial_{\mathbb{R}^{3}}(h_{a,1}\cup\dots\cup h_{a,n(a)})$

satisfying

(4.10) $F_{m+1,a}|_{\partial a}=F_{m}|_{\partial a}$

and

(4.11) $F_{m+1,a}(\partial{g_{a,j}})=\partial h_{a,j}\quad\hbox{for each
}j\in\\{1,2,\dots,n(a)\\}.$

Set
$H_{m+1}=\\{h_{0}\\}\cup\bigcup_{a\in\mathcal{Q}_{m}^{\prime}(\mathbf{n})}H_{m+1,a}^{\prime}$,
where $H_{m+1,a}^{\prime}=H_{m+1,a}\setminus\\{h_{a,0}\\}$. By construction,
$H_{m+1}$ is a partitioning of $\mathbb{R}^{3}$ whose atoms have tame
spherical boundary, which is a refinement of $H_{m}$, and which contains the
atom $h_{0}$. Thus, by the induction hypothesis $\mathcal{A}(m)$,
$(H_{i})_{i=1}^{m+1}$ satisfies (i-$(m+1)$), (ii-$(m+1)$), and (iii-$(m+1)$).
Observe that

$\displaystyle
F_{m+1}:\bigcup_{g\in\mathcal{Q}_{m+1}^{\prime}(\mathbf{n})}\partial g$
$\displaystyle\to\bigcup_{h\in H_{m+1}\setminus\\{h_{0}\\}}\partial h,\qquad
x\mapsto F_{m+1,a}(x)\quad\hbox{for}\quad
x\in\partial_{\mathbb{S}^{3}}(g_{a,1}\cup\dots\cup g_{a,n(a)})$

is a homeomorphism which is well-defined on the boundary of each atom of the
partitioning $\mathcal{Q}_{m+1}^{\prime}(\mathbf{n})$ because the
homeomorphisms $F_{m+1,a}$, $a\in\mathcal{Q}^{\prime}_{m}(\mathbf{n})$, agree
on the intersections of their domains. Thus (iv-$(m+1)$) holds with the
correspondence $g_{a,\ell}\leftrightarrow h_{a,\ell}$
($a\in\mathcal{Q}_{m}(\mathbf{n})$, $1\leq\ell\leq n(a)$; see in particular
(4.11)). To see that (v-$(m+1)$) is true, note that the restriction of
$F_{m+1}$ to the domain of $F_{m}$ equals $F_{m}$ by (4.10) and boundaries of
corresponding atoms are mapped bijectively to each other by (4.11). Thus
(i-$(m+1)$) – (v-$(m+1)$) hold also in Case 2.

In Case 2 we have $j_{0}(m+1)=j_{0}(m)$. Thus items (vi-$(m+1)$) – (x-$(m+1)$)
are the same as (vi-$m$) – (x-$m$) and there is nothing to prove. Thus the
induction step is finished also in Case 2. This completes the induction proof.

This induction proof already implies assertions (i) – (v) of the lemma. To get
(vi) – (x) it remains to show that our process defines an infinite sequence
$(n_{j})_{j\geq 1}$ of integers $n_{j}$, i.e., that $j_{0}(m)\nearrow\infty$
for $m\to\infty$. The monotonicity of $j_{0}(m)$ is clear from the
construction. Since $j_{0}(m+1)=j_{0}(m)+1$ whenever we are in Case 1, it
remains to prove the following claim.

Claim: Case 1 occurs for infinitely many $m$ in the above induction process.

To prove this assume on the contrary that Case 1 occurs only finitely many
times. Then either there is a largest $m$ that has $m=n_{j_{0}}$ for some
$j_{0}\geq 1$, or Case 1 never occurs; then we set $m=1$ and $j_{0}=0$. Let
$g\in\mathcal{Q}^{\prime}_{m}(\mathbf{n})$ and let $h$ be the element of
$H_{m}\setminus\\{h_{0}\\}$ corresponding to $g$. Let

$(K_{i}(h))_{i>m}=(\\{f_{j_{0}}\circ\cdots\circ
f_{1}(h^{\prime})\;:\;h^{\prime}\in H_{i}\hbox{ with }h^{\prime}\subseteq
h\\}\cup\\{\mathbb{R}^{3}\setminus\overline{f_{j_{0}}\circ\cdots\circ
f_{1}(h)}\\})_{i>m}.$

By the definition of $(\mathcal{Q}_{i}(\mathbf{n}))_{i\geq 1}$ and by (iv) we
have that

(4.12) $\begin{split}(K_{i}(h))_{i>m}&\sim(\\{h^{\prime}\in
H_{i}\;:\;h^{\prime}\subseteq
h\\}\cup\\{\mathbb{R}^{3}\setminus\overline{h}\\})_{i>m}\\\
&\sim(\\{g^{\prime}\in\mathcal{Q}_{i}^{\prime}(\mathbf{n})\;:\;g^{\prime}\subseteq
g\\}\cup\\{\mathbb{R}^{3}\setminus\overline{g}\\})_{i>m}\\\
&=([g](\mathcal{Q}^{\prime}_{i-\mathrm{level}(g)})\cup\\{\mathbb{R}^{3}\setminus\overline{g}\\})_{i>m}\\\
&\sim\begin{cases}(\mathcal{Q}_{i})_{i\geq 1}&\hbox{if
}\mathrm{level}(g)=m,\\\ (\mathcal{Q}_{i})_{i\geq 2}&\hbox{if
}\mathrm{level}(g)=m-1,\end{cases}\end{split}$

where the equivalences have the additional property that $\partial
a_{1}\cap\partial a_{2}\simeq\partial a_{1}^{\prime}\cap\partial
a_{2}^{\prime}$ if $a_{\ell}$ and $a_{\ell}^{\prime}$ are corresponding
elements ($1\leq\ell\leq 2$). Indeed, this homeomorpies hold by (v) and
because $f_{j_{0}}\circ\cdots\circ f_{1}$ and $[g]$ are homeomorphisms from
$\mathbb{R}^{3}$ to $\mathbb{R}^{3}$. Note that $(K_{i}(h))_{i>m}$ satisfies
the conditions of Proposition 4.6 with $C=\partial f_{j_{0}}\circ\cdots\circ
f_{1}(h)$: Proposition 4.6 (1) holds by (i), Proposition 4.6 (2)888It is this
condition (2) of Proposition 4.6 that required us to work with
$(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))$ rather than
$(\mathcal{Q}_{i}^{\prime})$. Indeed, the fact that we use
$(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))$ guarantees that (4.12) holds. holds
by (4.12) and Proposition 4.12 (3) (for $i\geq 2$; for
$\mathcal{Q}_{1}^{\prime}$ it is easy to see), Proposition 4.6 (3) is true by
(ii), Proposition 4.6 (4) is obviously true, and Proposition 4.6 (5) holds by
(v). Indeed, note that $(\mathcal{Q}_{i}^{\prime}(\mathbf{n}))$ is decreasing
and $F_{k}$ preserves $F_{k+n}$ on the boundaries of the elements of
$\mathcal{Q}_{k}^{\prime}(\mathbf{n})$ for each $n\in\mathbb{N}$. Applying
Proposition 4.6 to $(K_{i}(h))_{i>m}$ we see that there is an integer
$m^{\prime}(g)\geq m+2$ for which there is a homeomorphism
$f_{g,j_{0}+1}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ that leaves
$\mathbb{R}^{3}\setminus f_{j_{0}}\circ\cdots\circ f_{1}(h)$ pointwise
invariant and $\mathrm{diam}(f_{g,j_{0}+1}(k^{\prime}))<\frac{1}{2^{j_{0}+1}}$
holds for each $m^{\prime\prime}\geq m^{\prime}(g)$ and each $k^{\prime}\in
K_{m^{\prime\prime}}(h)\setminus\\{\mathbb{R}^{3}\setminus\overline{f_{j_{0}}\circ\cdots\circ
f_{1}(h)}\\}$. Doing this for each $g\in\mathcal{Q}^{\prime}_{m}(\mathbf{n})$
and choosing
$m^{\prime}=\max\\{m^{\prime}(g)\,:\,g\in\mathcal{Q}^{\prime}_{m}(\mathbf{n})\\}+3$
we can define the homeomorphism $f_{j_{0}+1}:\mathbb{R}^{3}\to\mathbb{R}^{3}$
by

$f_{j_{0}+1}(x)=f_{g,j_{0}+1}(x)\quad\hbox{for $x\in f_{j_{0}}\circ\cdots\circ
f_{1}(h)$, where $h\in H_{m}\setminus\\{h_{0}\\}$ corresponds to
$g\in\mathcal{Q}^{\prime}_{m}(\mathbf{n})$}$

(extending it continuously to $\mathbb{R}^{3}$ by the identity outside
$\mathbb{D}^{3}$). By construction, each boundary point of each atom of
$K_{n_{j_{0}}}$ is invariant under $f_{j_{0}+1}$ and
$f_{j_{0}+1}\circ\cdots\circ f_{1}$ keeps
$\mathbb{R}^{3}\setminus\mathbb{D}^{3}$ invariant and carries each other atom
of $H_{m^{\prime}}$ into a set of diameter less than $\frac{1}{2^{j_{0}+1}}$.
Because $m^{\prime}\geq m+3$ we are in Case 1 for $m^{\prime}>m$, a
contradiction to the maximality of $m$. This proves the claim and, hence, the
lemma. ∎

We can now easily finish the proof of Theorem 1.1. By Lemma 4.11 and Lemma
4.16 (x), there is a strictly increasing sequence $(n_{j})$ of positive
integers such that $(\mathcal{Q}_{n_{j}}^{\prime}(\mathbf{n}))_{j\geq 1}$ and
$(K_{n_{j}}\setminus\\{h_{0}\\})_{j\geq 1}$ are decreasing sequences of
partitionings of $T$ and $\mathbb{D}^{3}$, respectively. From Lemma 4.16 (iv)
and (ix) we obtain that $(\mathcal{Q}_{n_{j}}^{\prime}(\mathbf{n}))_{j\geq
1}\sim(H_{n_{j}}\setminus\\{h_{0}\\})_{j\geq
1}\sim(K_{n_{j}}\setminus\\{h_{0}\\})_{j\geq 1}$. Thus Lemma 4.4 (see also
Remark 4.3) implies that $T$ is homeomorphic to $\mathbb{D}^{3}$. This
concludes the proof of Theorem 1.1.

###  4.5. Proof of Theorem 1.5

In view of Section 2.2 we may assume that $T$ is an $ABC$-tile with $14$
neighbors and that $A=1$. As in the proof of Lemma 2.8 we see that the
truncated octahedron $O$ is a CW complex in the following natural sense. Let
$O_{\boldsymbol{\alpha}}$ (${\boldsymbol{\alpha}}\subseteq\mathcal{S}$) be as
in 2.12. For $i\in\\{0,1,2,3\\}$ the closed $i$-cells are given by the
nonempty sets $O_{\boldsymbol{\alpha}}$ with
$\boldsymbol{\alpha}\subseteq\mathcal{S}$ and $\\#\boldsymbol{\alpha}=3-i$.
Thus the $0$-skeleton $O^{0}$ is the set of vertices of $O$. Each closed
$1$-cell $O_{\\{\alpha_{1},\alpha_{2}\\}}$ is attached to the two closed
$0$-cells $O_{\boldsymbol{\alpha}}$ satisfying
$\boldsymbol{\alpha}\supset\\{\alpha_{1},\alpha_{2}\\}$ and
$\\#\boldsymbol{\alpha}=3$ (these $2$ closed $0$-cells form a $0$-sphere,
i.e., two points). This yields the $1$-skeleton $O^{1}$ (i.e., the edges of
$O$). To get the $2$-skeleton $O^{2}$ (whose support is $\partial O$) we
attach each closed $2$-cell $O_{\alpha_{1}}$, $\alpha_{1}\in\mathcal{S}$, to
the $1$-sphere
$\bigcup_{\alpha_{2}\in\mathcal{S}:\alpha_{2}\not=\alpha_{1}}O_{\\{\alpha_{1},\alpha_{2}\\}}$.
Finally, we attach the closed $3$-cell $O=O_{\emptyset}$ to the sphere
$O^{2}$.

From Proposition 2.6 and Theorem 1.1 we see that the set $T$ is a CW complex
whose closed $i$-cells are given by the nonempty sets
$\boldsymbol{B}_{\boldsymbol{\alpha}}$ with
$\boldsymbol{\alpha}\subseteq\mathcal{S}$ and $\\#\boldsymbol{\alpha}=3-i$ for
$i\in\\{0,1,2,3\\}$ with analogous attaching rules as above.

Thus, by Lemma 2.7 and Theorem 1.1, $T$ has the CW complex structure indicated
in the statement of Theorem 1.5. This CW complex structure is isomorphic to
the natural CW complex structure of $O$. The number of closed $i$-cells
asserted in Theorem 1.5 can immediately be counted on $O$: A truncated
octahedron has $14$ faces, $36$ edges, and $24$ vertices. This finishes the
proof of Theorem 1.5.

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