Badly approximable triangles and Delone sets

Shigeki Akiyama Institute of Mathematics, University of Tsukuba, 1-1-1 Tennodai, Tsukuba, Ibaraki, 305-8571 Japan [email protected] , Emily R. Korfanty Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1, Canada [email protected] and Yanli Xu^∗ Department of Mathematics and Statistics, Central China Normal University, Wuhan, 430079, China [email protected]

Abstract.

We investigate the linear equation $x+y+z=1$ where $\pi x$ , $\pi y$ , $\pi z$ are three angles of a triangle and the numbers $x$ , $y$ , $z$ are badly approximable. We show that there are exactly two solutions which have the smallest partial quotients by lexicographical ordering. Lastly, we give a construction of associated Delone sets which are expected to have rotationally invariant diffraction.

Key words and phrases: Badly approximable numbers, Hall’s ray, Iterated Function System, Delone sets, Chabauty–Fell topology.

* Corresponding author.

1. Introduction

Let $\lVert x\rVert=\min_{a\in\mathbb{Z}}|a-x|$ denote the distance from a real number $x$ to the nearest integer. A real irrational number $x$ is called badly approximable if

\inf_{q\in\mathbb{Z}_{+}}q\lVert qx\rVert>0.

It is a well known fact that an irrational number $x\in(0,1)$ is badly approximable if and only if the partial quotients in the continued fraction expansion

x=[a_{1}(x),a_{2}(x),\dots]=\cfrac{1}{a_{1}(x)+\cfrac{1}{a_{2}(x)+\cfrac{1}{\ddots}}}\,,\quad a_{j}(x)\in\mathbb{Z}_{+}\,,\ j=1,2,\ldots\,,

are bounded, i.e. if $\sup_{k\geq 1}a_{k}(x)<\infty$ . By using the Gauss map

T(x)=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\,,

we have

T^{k-1}(x)=[a_{k}(x),a_{k+1}(x),a_{k+2}(x),\dots]\,,

and $a_{k}(x)=\lfloor 1/T^{k-1}(x)\rfloor$ for all $k\geq 1$ .

In this work, we are interested in planar triangles where every angle is the product of $\pi$ and a badly approximable number. More specifically, our target is the equation

x+y+z=1\,,

where $\pi x,\pi y,\pi z$ are the angles of the corresponding triangle and $x,y,z$ are badly approximable numbers. We are especially interested in solutions such that the partial quotients of $x$ , $y$ , $z$ are small by lexicographical ordering. In this case, we call $\pi x,\pi y,\pi z$ optimal badly approximable angles.

Badly approximable angles often appear in phyllotaxis. One such example is the golden angle $\pi\omega$ where

\omega=\frac{\sqrt{5}-1}{2}=[1,1,\dots]\,.

When considering the irrational rotation $t\mapsto t+\psi\pmod{1}$ , the choice $\psi=\omega$ achieves the minimum discrepancy of the sequence $n\psi\pmod{1}$ , i.e. it gives the most uniform distribution in $[0,1]$ . In general, if the maximum of the partial quotients of $\psi$ is small, then the discrepancy becomes small (c.f. [11, Chapter 2, Theorem 3.4]).

We first encountered this topic while exploring constructions of Delone sets with rotationally invariant diffraction patterns. Delone sets with this property can be derived from tilings with statistical circular symmetry involving angles that are incommensurable with $\pi$ . There are many tilings with statistical circular symmetry [8, 9, 15, 16]; however, in all of these examples, the angles are not known to be badly approximable. Motivated by this, we produce new Delone sets by triangles with optimal badly approximable angles. We essentially follow the threshold method for multiscale substitution schemes considered in [19], but we use contractions described by a graph directed iterated function system to give a concise presentation. The main idea is to subdivide the triangles until the areas reach a given threshold, and then renormalize them to obtain larger and larger patches. A different construction of Delone sets via badly approximable angles and Fermat Spirals is discussed in [1, 5, 13, 20].

We call the continued fraction

x=[a_{1},a_{2},\dots]\,,

periodic if there exist positive integers $k_{0}$ and $h$ such that, for arbitrary $k\geq k_{0}$ ,

a_{k+h}=a_{k}\,.

In analogy with the common notation for periodic decimal expansions, we shall write such a periodic continued fraction as:

x=[a_{1},a_{2},\dots,a_{k_{0}-1},\overline{a_{k_{0}},a_{k_{0}+1},\dots,a_{k_{0}+h-1}}]\,.

We use the notation $(a_{k_{0}},a_{k_{0}+1},\dots,a_{k_{0}+h-1})^{\ell}$ to denote the repetition of the numbers $a_{k_{0}},a_{k_{0}+1},\dots a_{k_{0}+h-1}$ in the continued fraction $\ell\geq 0$ many times. We write $(a_{j})^{\ell}$ for the repetition of a single number $a_{j}$ . For convenience, in the case where $x\in(0,1)\cap{\mathbb{Q}}$ we use the notation

x=[a_{1},a_{2},\dots,a_{n},\infty]=\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\ddots+\frac{1}{a_{n}}}}}\,.

Definition 1.1.

Define the cylinder set of $b_{1},\dots,b_{n}\in\mathbb{N}$ by

I(b_{1},\dots,b_{n})=\{x\in(0,1)\,:\,x=[x_{1},x_{2},\dots]\,,x_{i}=b_{i}\ for\ 1\leq i\leq n\}\,.

The set $I(b_{1},\dots,b_{n})$ is an interval with endpoints

\frac{P_{n}+P_{n{\color[rgb]{0,0,1}-}1}}{Q_{n}+Q_{n{\color[rgb]{0,0,1}-}1}}\quad and\quad\frac{P_{n}}{Q_{n}}\,,

for $n\geq 1$ , where

P_{n}=b_{n}P_{n-1}+P_{n-2}\,,\quad Q_{n}=b_{n}Q_{n-1}+Q_{n-2}\,,

with

\begin{pmatrix}P_{-1}&P_{0}\\ Q_{-1}&Q_{0}\end{pmatrix}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\,.

Let us define our linear problem for badly approximable numbers more precisely. An irrational number $x\in(0,1)$ is $B$ -bad if $a_{k}(x)\leq B$ holds for all $k\geq 1$ . Let ${\mathcal{B}}_{B}$ be the set of all $B$ -bad numbers in $(0,1)\backslash{\mathbb{Q}}$ . For $j\geq 0$ , we define the set

{\mathcal{B}}_{B,j}={\mathcal{B}}_{B+1}\cap T^{-j}({\mathcal{B}}_{B})\,,

i.e., ${\mathcal{B}}_{B,j}$ is the set of irrational numbers which satisfy

\begin{cases}a_{k}\leq B+1&k\leq j\\ a_{k}\leq B&k>j\,.\end{cases}

Clearly, we have

{\mathcal{B}}_{B}={\mathcal{B}}_{B,0}\subset{\mathcal{B}}_{B,1}\subset{\mathcal{B}}_{B,2}\subset\cdots\,.

Further, we define ${\mathcal{B}}^{*}_{B}=\bigcup_{j=0}^{\infty}{\mathcal{B}}_{B,j}$ to be the set of eventually $B$ -bad numbers in ${\mathcal{B}}_{B+1}$ . In this paper, we are interested in the additive structure of ${\mathcal{B}}_{B,j}$ and ${\mathcal{B}}^{*}_{B}$ . We begin with a simple lemma.

Lemma 1.2.

For $x=[a_{1},a_{2},a_{3},\dots]\in(0,1)$ , we have

1-x=\begin{cases}[1,a_{1}-1,a_{2},a_{3},\dots]&a_{1}\geq 2\\ [1+a_{2},a_{3},\dots]&a_{1}=1\,.\end{cases}

Proof.

Putting $x=1/(a_{1}+y)$ with $y\in(0,1)$ , we see that

1-x=\cfrac{1}{1+\frac{1}{a_{1}-1+y}}\,,

from which the result easily follows. ∎

Corollary 1.3.

An irrational number $x$ is in ${\mathcal{B}}_{2,1}$ if and only if $1-x$ is also in ${\mathcal{B}}_{2,1}$ .

Remark 1.4.

The property of ${\mathcal{B}}_{2,1}$ described in Corollary 1.3 does not hold in ${\mathcal{B}}_{2}$ or in ${\mathcal{B}}_{2,j}$ for any $j\geq 2$ .

Remark 1.5.

Lemma 1.2 shows that the equation $x+y=1\ (x,y\in{\mathcal{B}}_{2},\ x\leq y)$ is trivially solved and has the set of solutions

{\color[rgb]{0,0,1}\{(x,1-x)\ |\ x\in{\mathcal{B}}_{2}\cap[0,1/2)\}\,.}

In particular, the equation has uncountably many different solutions. However, our equation of interest $x+y+z=1$ has no solutions in ${\mathcal{B}}_{2}$ . Indeed, if $x,y,z\in{\mathcal{B}}_{2}$ , then we also have $x,y,z\in I(1)\cup I(2)=[\frac{1}{3},1)$ . However, if we also have $x+y+z=1$ , then the only possible solution is $x=y=z=\frac{1}{3}\in\mathbb{Q}$ , which contradicts irrationality of $x,y,z\in{\mathcal{B}}_{2}$ .

Our main results are as follows:

Theorem 1.6.

The equality $x+y+z=1\ (x,y,z\in{\mathcal{B}}_{2,1},\ x\leq y\leq z)$ has exactly two solutions

x=2-\sqrt{3}=[3,\overline{1,2}],\ y=z=\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,,

and

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}],\ z=\sqrt{2}-1=[\overline{2}]\,.

By using Lemma 1.2, we may rephrase Theorem 1.6 as follows:

Theorem 1.7.

The equality $x+y=z\ (x,y,z\in{\mathcal{B}}_{2,1}\,,x\leq y)$ has exactly four solutions

x=2-\sqrt{3}=[3,\overline{1,2}],\ y=\frac{\sqrt{3}-1}{2}=[\overline{2,1}],\ z=\frac{3-\sqrt{3}}{2}=[1,1,1,\overline{2,1}]\,,

x=y=\frac{\sqrt{3}-1}{2}=[\overline{2,1}],\ z=\sqrt{3}-1=[\overline{1,2}]\,,

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}],\ z=2-\sqrt{2}=[1,1,\overline{2}]\,,

and

x=\frac{2-\sqrt{2}}{2}=[3,\overline{2}],\ y=\sqrt{2}-1=[\overline{2}],\ z=\frac{\sqrt{2}}{2}=[1,\overline{2}]\,.

In 1954, Hall [10] proved that ${\mathcal{B}}_{4}+{\mathcal{B}}_{4}$ contains an interval. Subsequently, Freiman [7] and Schecker [17] showed that ${\mathcal{B}}_{3}+{\mathcal{B}}_{3}$ contains an interval. This implies that the equalities $x+y+z=1\ (x,y,z\in{\mathcal{B}}_{3},\ x\leq y\leq z)$ and $x+y=z\ (x,y,z\in{\mathcal{B}}_{3},\ x\leq y)$ both have uncountably many solutions.

These results were proven using a criterion for Cantor sets which, when satisfied, implies that the arithmetic sum of two Cantor sets contains an interval. By this method, it may be challenging to make the solutions explicit, i.e. we know that one can choose any $z\in{\mathcal{B}}_{3}$ in the interval, but it is not clear which numbers $x,y\in{\mathcal{B}}_{3}$ satisfy the equation. In contrast, our method gives explicit solutions in ${\mathcal{B}}^{*}_{2}$ and ${\mathcal{B}}_{3}$ .

Theorem 1.8.

The equality $x+y+z=1\ (x,y,z\in{\mathcal{B}}^{*}_{2},\ x\leq y\leq z)$ has infinitely many solutions. Furthermore, uncountably many solutions of the equality $x+y+z=1\ (x,y,z\in{\mathcal{B}}_{3},\ x\leq y\leq z)$ can be constructed explicitly.

The paper is organized as follows. In Section 2, we prove Theorem 1.6 and 1.7. In Section 3, we prove Theorem 1.8. In Section 4, we return to the original motivation and discuss the construction of Delone sets using the optimal badly approximable angles found in Theorem 1.6. Finally, in Section 5, we give several open problems.

2. Proof of Theorems 1.6 and 1.7

To prove Theorem 1.6 and 1.7, we will need the following lemmas:

Lemma 2.1 (Forbidden patterns).

Let $z\in(0,1)\backslash{\mathbb{Q}}$ be such that $z\in[r_{1},r_{2}]$ for some $r_{1},r_{2}\in(0,1)\cap{\mathbb{Q}}$ . If the simple continued fractions of $r_{1}$ and $r_{2}$ are of the form

r_{1}=[(2)^{2k-1},(2,1)^{\ell},s,a_{1},\dots,a_{n_{1}},\infty]\,,\quad r_{2}=[(2)^{2k-1},b_{1},\dots,b_{n_{2}},\infty]\,,

(2.1)

r_{1}=[(2)^{2k},a_{1},\dots,a_{n_{1}},\infty]\,,\quad r_{2}=[(2)^{2k},(2,1)^{\ell},s,b_{1},\dots,b_{n_{2}},\infty]\,,

(2.2)

where $k\geq 1$ , $\ell\geq 0$ , and $s\geq 3$ , then $z\notin{\mathcal{B}}_{2}$ .

Proof.

Case 1. Assume that $r_{1}$ and $r_{2}$ are of the form (2.1) where $k\geq 1$ , $\ell\geq 0$ , and $s\geq 3$ .

Let $z=[c_{1},c_{2},\dots]$ . Suppose by contradiction that $z\in{\mathcal{B}}_{2}$ . First, for simplicity, observe that (2.1) implies

[(2)^{2k-1},(2,1)^{\ell},s,\infty]\leq z\leq[(2)^{2k-1},\infty]\,.

(2.3)

Since (2.1) requires $c_{1},\dots,c_{2k-1}\geq 2$ and $z\in{\mathcal{B}}_{2}$ requires $c_{1},\dots,c_{2k-1}\leq 2$ , we must have

c_{1}=\dots=c_{2k-1}=2\,,

(2.4)

i.e. $z\in I((2)^{2k-1})$ .

Next, consider $u=[c_{2k},c_{2k+1},\dots]$ . By (2.3) and (2.4), we have

u\leq[(2,1)^{\ell},s,\infty]\,.

(2.5)

From this, we can deduce that $u\in I((2,1)^{\ell})$ . To see this, observe that (2.5) gives us $c_{2k}\geq 2$ , so we must have $c_{2k}=2$ because $u\in{\mathcal{B}}_{2}$ . We also get that $c_{2k+1}\leq 1$ , so we must have $c_{2k+1}=1$ . Proceeding inductively, we find that $c_{2q}=2$ and $c_{2q+1}=1$ for $k\leq q\leq k+\ell-1$ .

Now consider $v=[c_{2(k+\ell)},c_{2(k+\ell)+1},\dots]$ . From (2.5) we get

v\leq[s,\infty]=\frac{1}{s}\,,

so $c_{2(k+\ell)}=\lfloor 1/v\rfloor\geq s\geq 3$ , contradicting $v\in{\mathcal{B}}_{2}$ . Therefore, $z$ cannot have the form (2.1).

Case 2. The proof when $r_{1}$ and $r_{2}$ are of the form (2.2) is similar to Case 1. ∎

We call any interval of the form (2.1) or (2.2) a forbidden pattern. Moreover, if $z\in[r_{1},r_{2}]$ for some forbidden pattern $[r_{1},r_{2}]$ , we say that $z$ is contained in a forbidden pattern.

Lemma 2.2.

The equality $x+y=z\ (x,y,z\in{\mathcal{B}}_{2})$ has exactly one solution

x=y=\frac{\sqrt{3}-1}{2}=[\overline{2,1}],\ z=\sqrt{3}-1=[\overline{1,2}]\,.

(2.6)

Proof.

Suppose $x=[a_{1},a_{2},\cdots]$ where $a_{i}\in\{1,2\}$ with $i\geq 1$ and $y=[b_{1},b_{2},\cdots]$ where $b_{j}\in\{1,2\}$ with $j\geq 1$ . It is very easy to get that $x<y$ if and only if $(-1)^{n}a_{n}<(-1)^{n}b_{n}$ for the first $n$ such that $a_{n}\neq b_{n}$ . As a consequence, we have

\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,{\color[rgb]{0,0,1}=\min({\mathcal{B}}_{2})}\,,

and

\sqrt{3}-1=[\overline{1,2}]\,{\color[rgb]{0,0,1}=\max({\mathcal{B}}_{2})}\,.

With this preparation, we show that (2.6) is the unique solution of $x+y=z\ (x,y,z\in{\mathcal{B}}_{2})$ by considering the sum of $x$ and $y$ .

If $x+y<\sqrt{3}-1$ , we get that one of them is less than $\frac{\sqrt{3}-1}{2}$ , which contradicts $\frac{\sqrt{3}-1}{2}$ being the minimum of ${\mathcal{B}}_{2}$ . Thus, $x+y\geq\sqrt{3}-1$ . Similarly, if $x+y>\sqrt{3}-1$ , then $z=x+y>\sqrt{3}-1$ , which contradicts $\sqrt{3}-1$ being the maximum of ${\mathcal{B}}_{2}$ .

Therefore, $z=x+y=\sqrt{3}-1$ and $x=y=\frac{\sqrt{3}-1}{2}$ . ∎

We can now provide a concise proof of Theorem 1.6. Indeed, we will see that the main part of the proof can be summarized by the careful analysis of cases shown in Table 1 and Table 2, which we present below. However, the reader may also wish to refer to (3.1) and Remark 3.2 for an intuitive explanation of why this case analysis is stable as the parameter $n$ increases.

Proof of Theorem 1.6.

Assume that $x\leq y\leq z$ . We divide the proof into 4 cases:

Case 1. $x,y,z\in{\mathcal{B}}_{2}$ .

This case is impossible by Remark 1.5.

Case 2. $x\in{\mathcal{B}}_{2,1}\setminus{\mathcal{B}}_{2}$ , and $y,z\in{\mathcal{B}}_{2}$ .

Setting $x=\frac{1}{3+X}$ with $X\in{\mathcal{B}}_{2}$ , we have

y+z=1-\frac{1}{3+X}=\frac{1}{1+\frac{1}{2+X}}\in{\mathcal{B}}_{2}\,.

By Lemma 2.2, we obtain

y=z=\frac{\sqrt{3}-1}{2}\,,\quad 1-\frac{1}{3+X}=\sqrt{3}-1\,,

i.e.

\displaystyle x=\frac{1}{3+X}=2-\sqrt{3}=[3,\overline{1,2}]\,,\quad y=z=\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,,

(2.7)

is the solution of $x+y+z=1$ .

Case 3. $x,y\in{\mathcal{B}}_{2,1}\setminus{\mathcal{B}}_{2}$ , and $z\in{\mathcal{B}}_{2}$ .

In this case, the continued fraction expansions of $x$ and $y$ both satisfy $a_{1}=3$ , $a_{j}\leq 2$ for all $j\geq 2$ . Under these assumptions, we aim to show that

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}]\,,\quad z=\sqrt{2}-1=[\overline{2}]\,,

is the only possible solution. First, we introduce variables $X_{n},Y_{n},Z_{n}$ to represent arbitrary numbers satisfying a certain property that depends on a given nonnegative integer $n$ . More specifically, we will prove the following statement:

For each $n\geq 0$ , if $X_{n}\notin I(3,(2)^{n+1})$ or $Y_{n}\notin I(3,(2)^{n+1})$ , then $Z_{n}=1-X_{n}-Y_{n}$ is contained in a forbidden pattern.

We divide the proof into 2 cases according to the parity of $n$ .

Case 3.1. $n$ is even.

The computations when $n$ is even are summarized in Table 1. There are 17 cases that must be considered for the cylinder sets of $X_{n}$ and $Y_{n}$ .

Case	Cylinder set for $X_{n}$	Cylinder set for $Y_{n}$	Left endpoint of the forbidden pattern	Right endpoint of the forbidden pattern
1	$I(3,(2)^{n},1)$	$I(3,(2)^{n},1)$	$[(2)^{n+1},3,\infty]$	$[(2)^{n+1},\infty]$
2.1	$I(3,(2)^{n},1,2)$	$I(3,(2)^{n},2,2)$	$[(2)^{n+1},3,\infty]$	$[(2)^{n+1},3,1,\infty]$
2.2	$I(3,(2)^{n},1,2)$	$I(3,(2)^{n},2,1)$	$[(2)^{n+1},(2,1),12,\infty]$	$[(2)^{n+1},3,1,\infty]$
2.3.1	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,2,1)$	$[(2)^{n+1},3,\infty]$	$[(2)^{n+1},3,2,\infty]$
2.3.2	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,2,2)$	$[(2)^{n+1},3,\infty]$	$[(2)^{n+1},3,3,\infty]$
2.3.3	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,2,1)$	$[(2)^{n+1},(2,1),14,\infty]$	$[(2)^{n+1},3,10,\infty]$
2.3.4	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,2,2)$	$[(2)^{n+1},(2,1),10,\infty]$	$[(2)^{n+1},3,20,\infty]$
2.4.1	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,1,1)$	$[(2)^{n+1},(2,1),6,\infty]$	$[(2)^{n+1},3,4,\infty]$
2.4.2	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,1,2)$	$[(2)^{n+1},(2,1),4,\infty]$	$[(2)^{n+1},3,8,\infty]$
2.4.3	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,1,1)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.1.1	$I(3,(2)^{n},1,1,2,1,1)$	$I(3,(2)^{n},2,1,2,1,1)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.1.2	$I(3,(2)^{n},1,1,2,1,1)$	$I(3,(2)^{n},2,1,2,1,2)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.1.3	$I(3,(2)^{n},1,1,2,1,2)$	$I(3,(2)^{n},2,1,2,1,2)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.1.4	$I(3,(2)^{n},1,1,2,1,2)$	$I(3,(2)^{n},2,1,2,1,1)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.2	$I(3,(2)^{n},1,1,2,1)$	$I(3,(2)^{n},2,1,2,2)$	$[(2)^{n+1},(2,1)^{2},20,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.3	$I(3,(2)^{n},1,1,2,2)$	$I(3,(2)^{n},2,1,2,1)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$
2.4.4.4	$I(3,(2)^{n},1,1,2,2)$	$I(3,(2)^{n},2,1,2,2)$	$[(2)^{n+1},(2,1),3,\infty]$	$[(2)^{n+1},2,1,\infty]$

Table 1. Case analysis showing the forbidden patterns containing

Z_{n}=1-X_{n}-Y_{n}

when

n

is even.

The computations are lengthy and we only detail Case 2.1 of Table 1 in this paper.

To prove the result for Case 2.1 in Table 1, assume that

X_{n}\in I(3,(2)^{n},1,2)\,,\quad Y_{n}\in I(3,(2)^{n},2,2)\,.

Let

A_{n}=1-\tfrac{\sqrt{2}}{2}+\tfrac{12-19\sqrt{2}}{-19+6\sqrt{2}+17(1+\sqrt{2})^{2n+4}}\,,\quad B_{n}=1-\tfrac{\sqrt{2}}{2}+\tfrac{10+\sqrt{2}}{1+5\sqrt{2}-7(1+\sqrt{2})^{2n+5}}\,,

and

C_{n}=1-\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{-(1+\sqrt{2})^{2n+8}+1}\,,\quad D_{n}=1-\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{(1+\sqrt{2})^{2n+8}+1}\,.

By calculating, we get

I(3,(2)^{n},1,2)=(A_{n},B_{n}]=1-\tfrac{\sqrt{2}}{2}+\left(\tfrac{12-19\sqrt{2}}{-19+6\sqrt{2}+17(1+\sqrt{2})^{2n+4}},\tfrac{10+\sqrt{2}}{1+5\sqrt{2}-7(1+\sqrt{2})^{2n+5}}\right]\,,

\ I(3,(2)^{n},2,2)=(C_{n},D_{n}]=1-\tfrac{\sqrt{2}}{2}+\left(\tfrac{\sqrt{2}}{-(1+\sqrt{2})^{2n+8}+1},\tfrac{\sqrt{2}}{(1+\sqrt{2})^{2n+8}+1}\right]\,.

We claim that

\displaystyle Z_{n}=1-X_{n}-Y_{n}\in\left[[(2)^{n+1},3,\infty]\,,[(2)^{n+1},3,1,\infty]\right]\,.

Since $n+1$ is odd, by Lemma 2.1, this implies that the bounds on $Z_{n}$ are contained in a forbidden pattern. Indeed,

Z_{n}=1-X_{n}-Y_{n}\in[1-B_{n}-D_{n},1-A_{n}-C_{n})\,,

by direct computation, we get

1-B_{n}-D_{n}=\sqrt{2}-1-\tfrac{10+\sqrt{2}}{1+5\sqrt{2}-7(1+\sqrt{2})^{2n+5}}-\tfrac{\sqrt{2}}{(1+\sqrt{2})^{2n+8}+1}\,,

1-A_{n}-C_{n}=\sqrt{2}-1-\tfrac{12-19\sqrt{2}}{-19+6\sqrt{2}+17(1+\sqrt{2})^{2n+4}}-\tfrac{\sqrt{2}}{-(1+\sqrt{2})^{2n+8}+1}\,.

By assumption $n$ is even, so

	$\displaystyle\ \ \ \ (1-B_{n}-D_{n})-[(2)^{n+1},3,\infty]$
	$\displaystyle=\sqrt{2}-1-\tfrac{10+\sqrt{2}}{1+5\sqrt{2}-7(1+\sqrt{2})^{2n+5}}-\tfrac{\sqrt{2}}{(1+\sqrt{2})^{2n+8}+1}-\tfrac{(-3+2\sqrt{2})^{n+2}+1}{(1-\sqrt{2})(-3+2\sqrt{2})^{n+2}+1+\sqrt{2}}$
	$\displaystyle=\tfrac{(1+\sqrt{2})^{n}(24+16\sqrt{2})-(1-\sqrt{2})^{n}(-24+16\sqrt{2})}{\left(-(1-\sqrt{2})^{n+3}-(1+\sqrt{2})^{n+3}\right)\left(10+(2\sqrt{2}-1)(1-\sqrt{2})^{2n+6}-(2\sqrt{2}+1)(1+\sqrt{2})^{2n+6}\right)}\,,$

is positive. We can see this by checking that the numerator and the denominator are both positive. Indeed, since $\left|1+\sqrt{2}\right|>\left|1-\sqrt{2}\right|$ and $\left|24+16\sqrt{2}\right|>\left|-24+16\sqrt{2}\right|$ , the numerator is positive. Similarly, the first term in the denominator $-(1-\sqrt{2})^{n+3}-(1+\sqrt{2})^{n+3}$ is negative since $\left|1-\sqrt{2}\right|<\left|1+\sqrt{2}\right|$ . Now, since $\left|2\sqrt{2}-1\right|<\left|2\sqrt{2}+1\right|$ and $\left|1-\sqrt{2}\right|<\left|1+\sqrt{2}\right|$ , the second term in the denominator

10+(2\sqrt{2}-1)(1-\sqrt{2})^{2n+6}-(2\sqrt{2}+1)(1+\sqrt{2})^{2n+6}\,,

is negative, so the denominator is positive. Therefore, the quotient is positive, as desired.

Next, we verify that

	$\displaystyle\ \ \ \ \ [(2)^{n+1},3,1,\infty]-(1-A_{n}-C_{n})$
	$\displaystyle=\tfrac{\frac{1}{7}(9+4\sqrt{2})(-3+2\sqrt{2})^{n+2}+1}{\frac{1}{7}(1-5\sqrt{2})(-3+2\sqrt{2})^{n+2}+1+\sqrt{2}}-\left(\sqrt{2}-1-\tfrac{12-19\sqrt{2}}{-19+6\sqrt{2}+17(1+\sqrt{2})^{2n+4}}-\tfrac{\sqrt{2}}{-(1+\sqrt{2})^{2n+8}+1}\right)$
	$\displaystyle=\tfrac{(1+\sqrt{2})^{n}(72+60\sqrt{2})-(1-\sqrt{2})^{n}(-72+60\sqrt{2})}{\left((1-\sqrt{2})^{n+3}(4+\sqrt{2})+(1+\sqrt{2})^{n+3}(4-\sqrt{2})\right)\left(-28+(1-\sqrt{2})^{2n+6}(6-\sqrt{2})+(1+\sqrt{2})^{2n+6}(6+\sqrt{2})\right)}\,,$

is positive. As before, we can see this by checking that the numerator and the denominator are both positive. This time, we have $\left|1+\sqrt{2}\right|>\left|1-\sqrt{2}\right|$ and $\left|72+60\sqrt{2}\right|>\left|72-60\sqrt{2}\right|$ , so the numerator is positive. Similarly, it is easy to check that the first term and the second term in the denominator are positive, so the denominator is positive. Therefore, the quotient is positive, as desired. Thus, we obtain that

Z_{n}=1-X_{n}-Y_{n}\in\left[[(2)^{n+1},3,\infty],[(2)^{n+1},3,1,\infty]\right]\,,

which implies that the bounds on $Z_{n}$ are contained in a forbidden pattern. The other cases in Table 1 can be proven in the same way.

Case 3.2. $n$ is odd.

The computations of forbidden patterns containing $Z_{n}=1-X_{n}-Y_{n}$ when $n$ is odd are summarized in Table 2, where $X_{n}\notin I(3,(2)^{n+1})$ or $Y_{n}\notin I(3,(2)^{n+1})$ . Again, there are 17 cases that must be considered. The proofs are similar to those for the even case.

Case	Cylinder set for $X_{n}$	Cylinder set for $Y_{n}$	Left endpoint of the forbidden pattern	Right endpoint of the forbidden pattern
1	$I(3,(2)^{n},1)$	$I(3,(2)^{n},1)$	$[(2)^{n+1},\infty]$	$[(2)^{n+1},3,\infty]$
2.1	$I(3,(2)^{n},1,2)$	$I(3,(2)^{n},2,2)$	$[(2)^{n+1},3,1,\infty]$	$[(2)^{n+1},3,\infty]$
2.2	$I(3,(2)^{n},1,2)$	$I(3,(2)^{n},2,1)$	$[(2)^{n+1},3,1,\infty]$	$[(2)^{n+1},(2,1),12,\infty]$
2.3.1	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,2,1)$	$[(2)^{n+1},3,2,\infty]$	$[(2)^{n+1},3,\infty]$
2.3.2	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,2,2)$	$[(2)^{n+1},3,3,\infty]$	$[(2)^{n+1},3,\infty]$
2.3.3	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,2,1)$	$[(2)^{n+1},3,10,\infty]$	$[(2)^{n+1},(2,1),14,\infty]$
2.3.4	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,2,2)$	$[(2)^{n+1},3,20,\infty]$	$[(2)^{n+1},(2,1),10,\infty]$
2.4.1	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,1,1)$	$[(2)^{n+1},3,4,\infty]$	$[(2)^{n+1},(2,1),6,\infty]$
2.4.2	$I(3,(2)^{n},1,1,1)$	$I(3,(2)^{n},2,1,2)$	$[(2)^{n+1},3,8,\infty]$	$[(2)^{n+1},(2,1),4,\infty]$
2.4.3	$I(3,(2)^{n},1,1,2)$	$I(3,(2)^{n},2,1,1)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.1.1	$I(3,(2)^{n},1,1,2,1,1)$	$I(3,(2)^{n},2,1,2,1,1)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.1.2	$I(3,(2)^{n},1,1,2,1,1)$	$I(3,(2)^{n},2,1,2,1,2)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.1.3	$I(3,(2)^{n},1,1,2,1,2)$	$I(3,(2)^{n},2,1,2,1,2)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.1.4	$I(3,(2)^{n},1,1,2,1,2)$	$I(3,(2)^{n},2,1,2,1,1)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.2	$I(3,(2)^{n},1,1,2,1)$	$I(3,(2)^{n},2,1,2,2)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1)^{2},20,\infty]$
2.4.4.3	$I(3,(2)^{n},1,1,2,2)$	$I(3,(2)^{n},2,1,2,1)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$
2.4.4.4	$I(3,(2)^{n},1,1,2,2)$	$I(3,(2)^{n},2,1,2,2)$	$[(2)^{n+1},2,1,\infty]$	$[(2)^{n+1},(2,1),3,\infty]$

Table 2. Case analysis showing the forbidden patterns containing

Z_{n}=1-X_{n}-Y_{n}

when

n

is odd.

Immediately, from Table 1 and Table 2, we see that

\displaystyle X_{n}\in I(3,(2)^{n},1)\,,\quad Y_{n}\in I(3,(2)^{n},1)\,,

is impossible by Case 1. Together, Cases 2.3.1 to 2.3.4 show that

\displaystyle X_{n}\in I(3,(2)^{n},1,1)\,,\quad Y_{n}\in I(3,(2)^{n},2,2)\,,

(2.8)

is impossible. Next, Cases 2.4.4.1.1 to 2.4.4.1.4 show that it is impossible to have $X_{n}\in I(3,(2)^{n},1,1,2,1)$ and $Y_{n}\in I(3,(2)^{n},2,1,2,1)$ . By this and Cases 2.4.4.2 to 2.4.4.4, we get that it is impossible to have $X_{n}\in I(3,(2)^{n},1,1,2)$ and $Y_{n}\in I(3,(2)^{n},2,1,2)$ . This, together with Cases 2.4.1 to 2.4.3, shows that

\displaystyle X_{n}\in I(3,(2)^{n},1,1)\,,\quad Y_{n}\in I(3,(2)^{n},2,1)\,,

(2.9)

is impossible. Next, from Case 2.1, Case 2.2, (2.8), and (2.9), we obtain that

\displaystyle X_{n}\in I(3,(2)^{n},1)\,,\quad Y_{n}\in I(3,(2)^{n},2)\,,

is impossible. This analysis of cases shows that any solution of $X_{n}+Y_{n}+Z_{n}=1$ must satisfy

X_{n}\,,Y_{n}\in I(3,(2)^{n+1})\,.

Therefore,

\displaystyle x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}]\,,\quad z=\sqrt{2}-1=[\overline{2}]\,,

(2.10)

is the only possible solution of $x+y+z=1$ for this case.

Case 4. $x,y,z\in{\mathcal{B}}_{2,1}\setminus{\mathcal{B}}_{2}$ .

We claim that this case is impossible. In fact, $I(3)=[\frac{1}{4},\frac{1}{3})$ , so $x+y+z<1$ , which is a contradiction. Therefore, by (2.7) and (2.10), the theorem is proven. ∎

Proof of Theorem 1.7.

Observe that $x,y,z\in{\mathcal{B}}_{2,1}$ satisfy the equality $x+y+z=1$ if and only if the three equalities

x+y=1-z\,,\quad x+z=1-y\,,\quad y+z=1-x\,,

(2.11)

are also satisfied. Moreover, by Corollary 1.3, the numbers $1-z$ , $1-y$ , and $1-x$ must also be in ${\mathcal{B}}_{2,1}$ . Next, recall from Theorem 1.6 that

x=2-\sqrt{3}=[3,\overline{1,2}]\,,\quad y=z=\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,,

(2.12)

and

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}]\,,\quad z=\sqrt{2}-1=[\overline{2}]\,,

(2.13)

are the only solutions of the equality $x+y+z=1\,,\ (x,y,z\in{\mathcal{B}}_{2,1}\,,\ x\leq y\leq z)$ .

In (2.12), the solution of $x+y+z=1$ happens to have $y=z$ , so (2.12) provides two solutions of our target equality due to (2.11). Specifically, we get that

x=2-\sqrt{3}=[3,\overline{1,2}]\,,\quad y=\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,,\quad z=\frac{3-\sqrt{3}}{2}=[1,1,1,\overline{2,1}]\,,

and

x=y=\frac{\sqrt{3}-1}{2}=[\overline{2,1}]\,,\quad z=\sqrt{3}-1=[\overline{1,2}]\,,

are both solutions of the equality $x+y=z\,,\ (x,y,z\in{\mathcal{B}}_{2,1}\,,x\leq y)$ .

In (2.13), the solution of $x+y+z=1$ happens to have $x=y$ , so (2.13) provides two more solutions of our target equality due to (2.11). Specifically, we get that

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}]\,,\quad z=2-\sqrt{2}=[1,1,\overline{2}]\,,

and

x=\frac{2-\sqrt{2}}{2}=[3,\overline{2}]\,,\quad y=\sqrt{2}-1=[\overline{2}]\,,\quad z=\frac{\sqrt{2}}{2}=[1,\overline{2}]\,,

are also solutions of the equality $x+y=z\,,\ (x,y,z\in{\mathcal{B}}_{2,1}\,,x\leq y)$ .

Finally, we know that these four solutions are the only possibilities because any additional solution to the equality $x+y=z\,,\ (x,y,z\in{\mathcal{B}}_{2,1}\,,x\leq y)$ would produce an additional solution to $x+y+z=1\,,\ (x\leq y\leq z)$ , which is impossible by Theorem 1.6. ∎

Remark 2.3.

Observe that Table 2 can be obtained from Table 1 by exchanging the left and the right endpoints of the forbidden patterns in the second-last and last columns.

3. Proof of Theorem 1.8

Proof.

By Theorem 1.6, we know that there are exactly two solutions of the equality $x+y+z=1$ under the restrictions $x,y,z\in{\mathcal{B}}_{2,1}$ and $x\leq y\leq z$ . Starting from the second solution

x=y=\frac{2-\sqrt{2}}{2}=[3,\overline{2}],\ z=\sqrt{2}-1=[\overline{2}]\,,

we can construct further explicit solutions of the equation $x+y+z=1$ under the weaker restrictions $x,y,z\in{\mathcal{B}}_{2}^{*}$ and $x,y,z\in{\mathcal{B}}_{3}$ . We do this by making certain insertions into the continued fraction expansions of $x$ , $y$ , and $z$ .

Consider the result of inserting a 2 after the number 3 in the continued fraction expansions of $x$ and $y$ , and inserting a 2 foremost in the continued fraction expansion of $z$ . This produces new numbers $X$ , $Y$ , $Z$ that satisfy

X=\cfrac{1}{3+\cfrac{1}{\cfrac{1}{x}-1}}\,,\quad Y=\cfrac{1}{3+\cfrac{1}{\cfrac{1}{y}-1}}\,,\quad Z=\cfrac{1}{2+z}\,.

From these expressions, we obtain the equality

1-X-Y-Z=\frac{(x-y)^{2}}{(3-2x)(3-2y)(3-x-y)}\,,

(3.1)

which implies that $X+Y+Z=1$ because $x=y$ .

Similarly, consider a different type of insertion where we insert $3,1,3$ after the number $3$ in the continued fraction expansions of $x$ and $y$ , and $1,1,2,1,1$ after the number $2$ in the continued fraction expansion of $z$ . In this case, we get new numbers $X$ , $Y$ , and $Z$ that satisfy

X=\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+x}}}\,,\quad Y=\cfrac{1}{3+\cfrac{1}{3+\cfrac{1}{1+y}}}\,,\quad Z=\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{\cfrac{1}{z}-1}}}}}}\,.

This gives us the equality

1-X-Y-Z=\frac{-5(x-y)^{2}}{(10x+13)(10y+13)(5x+5y+13)}\,,

(3.2)

which again implies that $X+Y+Z=1$ because $x=y$ .

Therefore, there are at least two different types of insertions that result in further solutions of the equality. Moreover, $S=\{2,11211\}$ is a code, i.e., any word generated by $S$ can be uniquely decomposed into a word over $S$ (see [12, Chapter 1] for a definition). From this observation, we immediately get that there are infinitely many solutions of the equality for $x,y,z\in{\mathcal{B}}_{2}^{*}$ , since ${\mathcal{B}}_{2}^{*}=\bigcup_{j=1}^{\infty}{\mathcal{B}}_{2,j}$ . Furthermore, we obtain uncountably many explicit solutions of the equality for $x,y,z\in{\mathcal{B}}_{3}$ by this method. ∎

Remark 3.1.

We can also prove Theorem 1.8 by starting from the first solution of Theorem 1.6 and using the same method with a small modification to the insertions.

Remark 3.2.

By the formula (3.1), we can reprove Theorem 1.6 by induction. The right side of (3.1) gives the error term after the number $2$ is inserted into the continued fraction expansions of $X_{n}$ , $Y_{n}$ , and $Z_{n}$ in Tables 1 and 2. In particular, this error term is small enough that the number of cases to consider is always 17.

4. Construction of Delone sets by badly approximable triangles

In this section, we construct Delone sets from tilings of the plane by triangles involving angles $\alpha$ , $\beta$ , $\gamma$ , where $\alpha+\beta+\gamma=\pi$ . In the general case when $\alpha$ , $\beta$ , $\gamma$ are arbitrary, we have a subdivision rule for a scalene triangle with angles $\alpha$ , $\beta$ , $\gamma$ and an isosceles triangle with angles $\gamma$ , $\gamma$ , $\pi-2\gamma$ , as illustrated in Figure 1.

Refer to caption — Figure 1. Subdivision rule for triangles with angles $\alpha$ , $\beta$ , $\gamma$ . The triangle on the left is scalene, and the triangle on the right is isosceles. In this illustration, we have $\alpha=\frac{\pi}{3}$ , $\beta=\frac{\pi}{4}$ , $\gamma=\frac{5\pi}{12}$ , but any solution of $\alpha+\beta+\gamma=\pi$ can be used.

Our objective is to obtain Delone sets associated with the unique solutions to the equality $x+y+z=1\ (x,y,z\in{\mathcal{B}}_{2,1},\ x\leq y\leq z)$ found in Theorem 1.6:

x=2-\sqrt{3}\,,\quad y=z=\frac{\sqrt{3}-1}{2}\,,

and

x=y=\frac{2-\sqrt{2}}{2}\,,\quad z=\sqrt{2}-1\,.

For the first solution, by choosing

\alpha=(2-\sqrt{3})\pi\,,\quad\beta=\gamma=\frac{(\sqrt{3}-1)\pi}{2}\,,

(4.1)

the subdivision rule in Figure 1 reduces to a subdivision rule involving only the isosceles triangle with angles $\alpha$ and $\beta=\gamma$ as shown in Figure 2.

Similarly, for the second solution, by choosing

\alpha=(\sqrt{2}-1)\pi\,,\quad\beta=\gamma=\frac{(2-\sqrt{2})\pi}{2}\,,

(4.2)

we again get a subdivision rule involving only an isosceles triangle, exactly as in Figure 2 except with different values for $\alpha$ and $\beta$ .

4.1. Delone sets and the Chabauty–Fell topology

We restrict our attention to ${\mathbb{R}}^{2}$ . Throughout, we denote the Euclidean norm on ${\mathbb{R}}^{2}$ by $\|\cdot\|$ , and we write $B(x,r)$ to denote the open ball of radius $r$ centered at $x$ , i.e. $B(x,r)=\{y\in{\mathbb{R}}^{2}\,:\,\|y-x\|<r\}$ . Similarly, we write $\overline{B(x,r)}$ for the closed ball, i.e. $\overline{B(x,r)}=\{y\in{\mathbb{R}}^{2}\,:\,\|y-x\|\leq r\}$ . We begin by recalling the definition of a Delone set.

Definition 4.1.

Let $\Lambda$ be a closed subset of $X\subseteq{\mathbb{R}}^{2}$ .

(i)

If there exists an $r>0$ such that for any $x\in X$ , $\text{card}(B(x,r)\cap\Lambda)\leq 1$ , then $\Lambda$ is $r$ -uniformly discrete in $X$ .
(ii)

If there exists an $R>0$ such that for any $x\in X$ , $\overline{B(x,R)}\cap\Lambda\neq\emptyset$ , then $\Lambda$ is $R$ -relatively dense in $X$ .
(iii)

We say that $\Lambda$ is a $(r,R)$ -Delone set in $X$ if $\Lambda$ is both $r$ -uniformly discrete in $X$ and $R$ -relatively dense in $X$ .

Remark 4.2.

To clarify, in the above definition, the sets $B(x,r)$ and $\overline{B(x,R)}$ denote open and closed balls in ${\mathbb{R}}^{2}$ , and not open and closed balls in the relative topology for $X$ .

Definition 4.3.

Let $X\subseteq{\mathbb{R}}^{2}$ be closed. We define the following spaces of point sets:

(i)

Denote by $2^{X}$ the set of all closed subsets of $X$ ;
(ii)

Denote by ${\mathcal{W}}_{r,R}(X)$ the set of closed subsets $\Lambda\subseteq{\mathbb{R}}^{2}$ such that $\Lambda$ is an $(r,R)$ -Delone set in $X$ .

Remark 4.4.

If $X$ and $Y$ are closed subsets of ${\mathbb{R}}^{2}$ and $Y\subseteq X$ , then any set which is $r$ -uniformly discrete and $R$ -relatively dense in $X$ must also be $r$ -uniformly discrete and $R$ -relatively dense in $Y$ . In particular, we have ${\mathcal{W}}_{r,R}(X)\subseteq{\mathcal{W}}_{r,R}(Y)$ .

In the next section, we will construct a sequence of finite point sets and show the existence of a subsequence converging to a Delone set in the Chabauty–Fell topology [4, 6]. This topology is also commonly referred to as the Chabauty topology or the Fell topology, which are defined more generally in any topological space. In the case of locally compact groups, it is also called the local rubber topology [3]. Since we work in ${\mathbb{R}}^{2}$ , the Chabauty–Fell topology is metrizable and induced by the following metric (see [18, Appendix A]):

Definition 4.5 (Chabauty–Fell Topology).

For each $\Lambda_{1},\Lambda_{2}\in 2^{X}$ , define

d(\Lambda_{1},\Lambda_{2})=\inf\ \{\{1\}\cup\{\varepsilon>0\,:\,\Lambda_{1}\cap B(0,\textstyle{\frac{1}{\varepsilon}})\subseteq\Lambda_{2}+B(0,\varepsilon)\ \text{and}\ \Lambda_{2}\cap B(0,\textstyle{\frac{1}{\varepsilon}})\subseteq\Lambda_{1}+B(0,\varepsilon)\}\}\,.

The map $d:2^{X}\times 2^{X}\rightarrow[0,\infty)$ is a metric on $2^{X}$ inducing the Chabauty–Fell topology.

Remark 4.6.

It has long been known that the space $2^{X}$ is compact in the Fell topology [6]. Furthermore, it is easy to show that the space ${\mathcal{W}}_{r,R}(X)$ is also compact in the Fell topology. As we will use this fact, we give a proof for $X={\mathbb{R}}^{2}$ in Appendix A.

We will deduce the existence of a subsequence converging to a Delone set using the compactness of $\mathcal{W}_{r,R}({\mathbb{R}}^{2})$ ; however, our sequence will not be contained in $\mathcal{W}_{r,R}({\mathbb{R}}^{2})$ because no finite set is relatively dense in ${\mathbb{R}}^{2}$ . Thus, we will use the following easy observation about the Chabauty–Fell topology:

Proposition 4.7.

Let $\{\Lambda_{n}\}_{n\geq 1}$ be a sequence in $2^{X}$ and $\Lambda\in 2^{X}$ . Let $\{R_{n}\}_{n\geq 1}$ be a sequence of positive real numbers with $\lim_{n\rightarrow\infty}R_{n}=\infty$ . Then $\Lambda_{n}$ converges to $\Lambda$ if and only if $\Lambda_{n}\cap B(0,R_{n})$ converges to $\Lambda$ in the Chabauty–Fell topology.

Proof.

First, assume that $\Lambda_{n}$ converges to $\Lambda$ in the Chabauty–Fell topology. Then, for any $\varepsilon>0$ , there exists a constant $N_{1}$ such that for any $n\geq N_{1}$ , we have $d(\Lambda_{n},\Lambda)<\varepsilon$ . That is,

\Lambda_{n}\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda+B(0,\varepsilon)\,,

(4.3)

and

\Lambda\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda_{n}+B(0,\varepsilon)\,.

(4.4)

For the above $\varepsilon$ , there must exist a constant $N_{2}$ such that for any $n\geq N_{2}$ , $R_{n}>\tfrac{1}{\varepsilon}+\varepsilon$ . Let $M=\max\{N_{1},N_{2}\}$ . Then by (4.3), we have

\Lambda_{n}\cap B(0,R_{n})\cap B(0,\tfrac{1}{\varepsilon})=\Lambda_{n}\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda+B(0,\varepsilon)\quad\forall n\geq M\,.

(4.5)

Furthermore, we get

\Lambda\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda_{n}\cap B(0,R_{n})+B(0,\varepsilon)\,.

(4.6)

Next, let $x\in\Lambda\cap B(0,\tfrac{1}{\varepsilon})$ . By (4.4), we have $x\in\Lambda_{n}+B(0,\varepsilon)$ . Thus, there exists some $y\in\Lambda_{n}$ and some $z\in B(0,\varepsilon)$ such that $x=y+z$ . Since $R_{n}>\tfrac{1}{\varepsilon}+\varepsilon$ and $x\in B(0,\tfrac{1}{\varepsilon})$ we have

\|y\|\leq\|x\|+\|x-y\|=\|x\|+\|z\|<\frac{1}{\varepsilon}+\varepsilon<R_{n}\,,

so $y\in\Lambda_{n}\cap B(0,R_{n})$ . In particular, we have $x=y+z\in\Lambda_{n}\cap B(0,R_{n})+B(0,\varepsilon$ ).

By (4.5) and (4.6), we have shown that $d(\Lambda_{n}\cap B(0,R_{n}),\Lambda)<\varepsilon$ for all $n\geq M$ , so $\Lambda_{n}\cap B(0,R_{n})$ converges to $\Lambda$ in the Chabauty–Fell topology.

Conversely, suppose that $\Lambda_{n}\cap B(0,R_{n})$ converges to $\Lambda$ in the Chabauty–Fell topology. For any $\varepsilon>0$ , there exists a constant $N_{1}$ such that for any $n\geq N_{1}$ , we have $d(\Lambda_{n}\cap B(0,R_{n}),\Lambda)<\varepsilon$ . For the above $\varepsilon$ , there must exist a constant $N_{2}$ such that $R_{n}>\tfrac{1}{\varepsilon}$ for any $n\geq N_{2}$ . Let $M=\max\{N_{1},N_{2}\}$ . Then, for all $n\geq M$ , we get

\Lambda_{n}\cap B(0,\tfrac{1}{\varepsilon})=\Lambda_{n}\cap B(0,R_{n})\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda+B(0,\varepsilon)\,,

and

\Lambda\cap B(0,\tfrac{1}{\varepsilon})\subseteq\Lambda_{n}\cap B(0,R_{n})+B(0,\varepsilon)\subseteq\Lambda_{n}+B(0,\varepsilon)\,,

so $d(\Lambda_{n},\Lambda)<\varepsilon$ . Therefore, $\Lambda_{n}$ converges to $\Lambda$ in the Chabauty–Fell topology. ∎

In other words, the convergence of a sequence in the Chabauty–Fell topology is equivalent to the convergence of its restriction to larger and larger patches. Provided that we can extend our finite point sets to elements of $\mathcal{W}_{r,R}({\mathbb{R}}^{2})$ , we can use this fact and the compactness of $\mathcal{W}_{r,R}({\mathbb{R}}^{2})$ to prove the existence of a Delone limit set.

4.2. Threshold method

To any subdivision rule on a finite set of tiles in ${\mathbb{R}}^{2}$ , there is an associated graph-directed iterated function system (GIFS) consisting of similitudes. In this section, starting from a GIFS of this type, we construct a corresponding Delone set. In particular, starting from some initial tile, we apply the GIFS until the area of every tile is below some threshold $\varepsilon>0$ . Once there are no tiles of area greater than $\varepsilon$ , we inflate the finite patch by $\frac{1}{\sqrt{\varepsilon}}$ . We call this an $\varepsilon$ -rule, which is defined more precisely below.

4.2.1. GIFS

Let $(V,\Gamma)$ be a directed graph with vertex set $V$ and directed-edge set $\Gamma$ where both $V$ and $\Gamma$ are finite. Denote the set of edges from $j$ to $i$ by $\Gamma_{j,i}$ and assume that for any $j\in V$ , there is at least one edge starting from vertex $j$ . Furthermore, assume that for each edge $e\in\Gamma$ , there is a corresponding contractive similitude $g_{e}:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ . We call $(g_{e})_{e\in\Gamma}$ a graph-directed IFS (GIFS) (see [14]). The invariant sets of this GIFS, also called graph-directed sets, are the unique non-empty compact sets $(T_{j})_{j\in V}$ satisfying

T_{j}=\bigcup\limits_{i\in V}\bigcup\limits_{e\in\Gamma_{j,i}}g_{e}(T_{i}),\quad j\in V\,.

(4.7)

Let $\mathcal{F}=\{T_{1},T_{2},\dots,T_{m}\}$ denote the invariant sets of a GIFS $(g_{e})_{e\in\Gamma}$ with vertex set $V=\{1,2,\dots,m\}$ . We make the following additional assumptions on the GIFS:

1.

Each tile in ${\mathcal{F}}$ has a nonempty interior.
2.

Each tile in ${\mathcal{F}}$ satisfies the open set condition, i.e. each union in (4.7) has no interior-overlap.
3.

The directed graph $(V,\Gamma)$ is strongly connected, i.e. for all $i,j\in V$ , there is similar copy of $T_{i}$ in the subdivision of $T_{j}$ .

4.2.2. $\bm{\varepsilon}$ -rule.

For convenience, let $\Sigma_{N}$ denote the set of edge sequences of length $N$ on $(V,\Gamma)$ . In other words, $\Sigma_{N}$ is the collection of sequences $e_{1},e_{2},\dots,e_{N}\in\Gamma$ such that the composition $g_{e_{1}}\circ g_{e_{2}}\circ\dots\circ g_{e_{N}}$ is permitted by the GIFS. We iterate the GIFS starting from an initial tile $T_{j}\in{\mathcal{F}}$ with the following $\varepsilon$ -rule:

(i)

Fix $0<\varepsilon<1$ . Given a finite sequence $(e_{1},e_{2},\dots,e_{m(\varepsilon)})\in\Sigma_{m(\varepsilon)}$ , we continue applying the GIFS to the tile

$T=g_{e_{1}}\circ g_{e_{2}}\circ\dots\circ g_{e_{m(\varepsilon)}}(T_{j})\,,$

while $\text{Area}(T)>\varepsilon$ , and stop applying the GIFS to $T$ when $\text{Area}(T)\leq\varepsilon$ .
(ii)

Once the process in (i) terminates, we inflate the resulting collection of tiles by $\frac{1}{\sqrt{\varepsilon}}$ . We denote the resulting finite patch by $\mathcal{P}_{\varepsilon}(T_{j})$ .

Remark 4.8.

Let $|g_{e}^{\prime}|$ denote the contraction factor of the similitude $g_{e}$ in the GIFS. After applying the $\varepsilon$ -rule, we are guaranteed to have $\text{Area}(T)\in[a,1]$ for every tile $T\in{\mathcal{P}}_{\varepsilon}(T_{j})$ , where

a=\min\{|g_{e}^{\prime}|^{2}\,:\,e\in\Gamma\}\,.

4.2.3. Associated point sets.

Our next step is to define a point set $\Lambda_{j}(\varepsilon)$ associated with the patch ${\mathcal{P}}_{\varepsilon}(T_{j})$ by placing one point inside each tile. Since ${\mathcal{F}}$ is a finite set, we can choose fixed constants $r_{0}\,,R_{0}>0$ which do not depend on $T_{j}$ and distinguished points $\lambda_{j}^{(0)}\in T_{j}$ for each $j\in V$ such that

B(\lambda_{j}^{(0)},r_{0})\subset T_{j}\subset B(\lambda_{j}^{(0)},R_{0})\,.

(4.8)

Let $0<\varepsilon<1$ and $T_{j}\in{\mathcal{F}}$ . Starting from $\lambda_{j}^{(0)}$ , we construct the elements of $\Lambda_{j}(\varepsilon)$ recursively as follows. For each application of the GIFS

T_{j}=\bigcup\limits_{i\in V}\bigcup\limits_{e\in\Gamma_{j,i}}g_{e}(T_{i})\,,\quad j\in V\,,

in the $\varepsilon$ -rule, we produce new points

\lambda_{j}^{(n)}=\bigcup\limits_{i\in V}\bigcup\limits_{e\in\Gamma_{j,i}}g_{e}(\lambda_{i}^{(n-1)})\,,\quad j\in V\,,

that replace the previous points $\lambda_{i}^{(n-1)}$ . This process terminates when there are no further applications of the GIFS. Finally, we inflate the resulting collection of points by $\frac{1}{\sqrt{\varepsilon}}$ to obtain the associated point sets $\Lambda_{j}(\varepsilon)$ .

In other words, since any tile $T$ in ${\mathcal{P}}_{\varepsilon}(T_{j})$ is similar to a unique $T_{i}\in{\mathcal{F}}$ , we place exactly one point in $T$ via similarity at the same location as $\lambda_{i}^{(0)}\in T_{i}$ . Though one can choose any points $\{\lambda_{j}^{(0)}\}_{j\in V}$ such that (4.8) holds, for simplicity we choose the centroid of each tile. In this case, $\Lambda_{j}(\varepsilon)$ is simply the set of centroids of the tiles in ${\mathcal{P}}_{\varepsilon}(T_{j})$ . Furthermore, we introduce the additional assumption that

\displaystyle\min\{\text{Area}(T_{j})\,:\,T_{j}\in{\mathcal{F}}\}=1,\quad\max\{\text{Area}(T_{j})\,:\,T_{j}\in{\mathcal{F}}\}=S\geq 1\,,

(4.9)

as this will simplify subsequent proofs. The value $S$ is related to the GIFS $(g_{e})_{e\in\Gamma}$ ; however, one can easily modify a given GIFS to satisfy (4.9) by simultaneously enlarging or shrinking all tiles $T_{j}$ by a fixed similitude.

4.2.4. Existence of a Delone limit set

Next, we consider a decreasing sequence $\{\varepsilon_{n}\}_{n\geq 1}$ in $(0,1)$ and show that the corresponding sequence of point sets $\{\Lambda_{j}(\varepsilon_{n})\}_{n\geq 1}$ has a subsequence converging to a Delone set in the Chabauty–Fell topology.

Lemma 4.9.

For any decreasing sequence $\{\varepsilon_{n}\}_{n\geq 1}$ in $(0,1)$ with $\lim_{n\rightarrow\infty}\varepsilon_{n}=0$ , there exists a sequence $\{K_{n}\}_{n\geq 1}$ of compact sets such that

(i)

$K_{n}\subseteq K_{n+1}$ for all $n\geq 1$ ,
(ii)

$\cup_{n\geq 1}K_{n}={\mathbb{R}}^{2}$ , and
(iii)

$\Lambda_{j}(\varepsilon_{n})\in{\mathcal{W}}_{r,R}(K_{n})$ for all $n\geq 1$ , where $r=\sqrt{\tfrac{a}{S}}r_{0}$ and $R=R_{0}$ .

In other words, $\Lambda_{j}(\varepsilon_{n})$ is an $(r,R)$ -Delone set in $K_{n}$ for each $n\geq 1$ .

Proof.

Recall from (4.8) that we have

B(\lambda_{j}^{(0)},r_{0})\subset T_{j}\subset B(\lambda_{j}^{(0)},R_{0})\,,\quad\forall\,T_{j}\in{\mathcal{F}}\,.

Given $0<\varepsilon_{n}<1$ , consider an arbitrary tile $T\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})$ . There must exist a unique finite sequence $(e_{1},e_{2},\ldots,e_{m(\varepsilon_{n})})\in\Sigma_{m(\varepsilon_{n})}$ and one tile $T_{i_{0}}\in{\mathcal{F}}$ such that

T=\tfrac{1}{\sqrt{\varepsilon_{n}}}g_{e_{1}}\circ g_{e_{2}}\circ\dots\circ g_{e_{m(\varepsilon_{n})}}(T_{i_{0}})\,.

Define $x_{T}$ to be the point of $\Lambda_{j}(\varepsilon_{n})$ that lies in $T$ , i.e.

x_{T}=\tfrac{1}{\sqrt{\varepsilon_{n}}}g_{e_{1}}\circ g_{e_{2}}\circ\dots\circ g_{e_{m(\varepsilon_{n})}}(\lambda_{i_{0}}^{(0)})\,.

Let $c=\text{Area}(T)$ and $A=\text{Area}(T_{i_{0}})$ . Hence, the linear scaling factor from $T_{i_{0}}$ to $T$ is $\sqrt{\frac{c}{A}}$ . Thus, scaling $B(\lambda_{i_{0}}^{(0)},r_{0})$ by the linear factor $\sqrt{\frac{c}{A}}$ will result in a ball of radius $\sqrt{\frac{c}{A}}r_{0}$ that fits inside $T$ when centered at $x_{T}$ . Hence, we have $B(x_{T},\sqrt{\frac{c}{A}}r_{0})\subseteq T$ . Similarly, we have $T\subseteq B(x_{T},\sqrt{\frac{c}{A}}R_{0})$ . Moreover, by Remark 4.8, we have $a\leq c\leq 1$ . By assumption 4.9, we get $A\in[1,S]$ . From this, we obtain

B(x_{T},\textstyle{\sqrt{\tfrac{a}{S}}}r_{0})\subseteq B(x_{T},\textstyle{\sqrt{\tfrac{c}{A}}}r_{0})\subseteq T\subseteq B(x_{T},\textstyle{\sqrt{\tfrac{c}{A}}}R_{0})\subseteq B(x_{T},R_{0})\quad\forall\,T\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})\,.

(4.10)

Assume without loss of generality that $\lambda_{j}^{(0)}=0$ . Then, for each $\varepsilon_{n}$ , we have that

\Lambda_{j}(\varepsilon_{n})\subseteq K_{n}=\bigcup_{T\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})}T\,.

For this choice of $K_{n}$ , it is easy to see that conditions (i) and (ii) are satisfied. To prove condition (iii), we must show that for every $x\in K_{n}$ , $\text{card}(B(x,r)\cap\Lambda_{j}(\varepsilon_{n})\leq 1$ and $\overline{B(x,R)}\cap\Lambda_{j}(\varepsilon_{n})\neq\emptyset$ , when $r=\sqrt{\tfrac{a}{S}}r_{0}$ and $R=R_{0}$ .

To this end, let $x\in K_{n}$ be arbitrary. By our choice of $K_{n}$ , there exists $T\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})$ such that $x\in T$ . First, notice that (4.10) gives us $x\in B(x_{T},R)$ , which implies that $x_{T}\in B(x,R)$ , so $\overline{B(x,R)}\cap\Lambda_{j}(\varepsilon_{n})\neq\emptyset$ holds. It remains to prove that $\text{card}(B(x,r)\cap\Lambda_{j}(\varepsilon_{n}))\leq 1$ . We consider two cases:

Case 1. Assume that $x$ lies in $B(x_{T},r)$ . Then $x_{T}\in B(x,r)$ . We claim that $x_{T^{\prime}}\notin B(x,r)$ for any $T^{\prime}\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})\backslash T$ . Indeed, we have

2r\leq\|x_{T}-x_{T^{\prime}}\|\leq\|x_{T}-x\|+\|x-x_{T^{\prime}}\|\leq r+\|x-x_{T^{\prime}}\|\,.

(4.11)

the first inequality holds since $T^{\prime}$ and $T$ are non-overlapping, and by (4.11) we have $\|x-x_{T^{\prime}}\|\geq r$ .

Case 2. Assume that $x$ lies outside of $B(x_{T},r)$ , i.e. $x_{T}\notin B(x,r)$ . We need to check that there can be at most one tile $T^{\prime}\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})\backslash T$ such that $x_{T^{\prime}}\in B(x,r)$ . On the contrary, suppose that there exist two distinct tiles $T^{\prime},T^{\prime\prime}\in{\mathcal{P}}_{\varepsilon_{n}}(T_{j})\backslash T$ with $x_{T^{\prime}},x_{T^{\prime\prime}}\in B(x,r)$ . Since $T^{\prime}$ and $T^{\prime\prime}$ have no overlap, we must have $\|x_{T^{\prime}}-x_{T^{\prime\prime}}\|\geq 2r$ . On the other hand, since $x_{T^{\prime}},x_{T^{\prime\prime}}\in B(x,r)$ , the triangle inequality gives us

\|x_{T^{\prime}}-x_{T^{\prime\prime}}\|\leq\|x_{T^{\prime}}-x\|+\|x-x_{T^{\prime\prime}}\|<r+r=2r\,,

a contradiction.

Thus, we have shown that $\text{card}(B(x,r)\cap\Lambda_{j}(\varepsilon_{n}))\leq 1$ and $\overline{B(x,R)}\cap\Lambda_{j}(\varepsilon_{n})\neq\emptyset$ . As $x$ was arbitrary, this completes the proof. ∎

Theorem 4.10.

There exists a subsequence of $\Lambda_{j}(\varepsilon_{n})$ that converges to a $(r,R)$ -Delone set in ${\mathbb{R}}^{2}$ in the Chabauty–Fell topology.

Proof.

By Lemma 4.9, we have $\Lambda_{j}(\varepsilon_{n})\subseteq\mathcal{W}_{r,R}(K_{n})$ , so there must exist a point set $\Lambda_{n}\in\mathcal{W}_{r,R}({\mathbb{R}}^{2})$ such that $\Lambda_{n}\cap K_{n}=\Lambda_{j}(\varepsilon_{n})$ . Next, since ${\mathcal{W}}_{r,R}({\mathbb{R}}^{2})$ is compact, there exists a subsequence $\{\Lambda_{n_{i}}\}_{i\geq 1}$ of $\{\Lambda_{n}\}_{n\geq 1}$ converging to some $\Lambda\in{\mathcal{W}}_{r,R}({\mathbb{R}}^{2})$ in the Chabauty–Fell topology. Moreover, we have $K_{n}\subseteq K_{n+1}$ for all $n$ and $\bigcup_{n\geq 1}K_{n}={\mathbb{R}}^{2}$ . By compactness of $K_{n_{i}}$ in ${\mathbb{R}}^{2}$ (in the Euclidean topology), we can pick a sequence $R_{n_{i}}$ of positive real numbers such that $K_{n_{i}}\subseteq B(0,R_{n_{i}})$ for all $i$ . Therefore, by Proposition 4.7, we also have that $\Lambda_{n_{i}}\cap B(0,R_{n_{i}})$ converges to $\Lambda$ in the Chabauty–Fell topology. This proves the theorem. ∎

4.3. GIFS for arbitrary angles

Next, we consider the subdivision rule given in Figure 1 for arbitrary angles $\alpha,\beta,\gamma$ . Below, we describe the GIFS associated with this subdivision rule. Then, as illustrated in Section 4.4, we can use this GIFS and the methods in Section 4.2 to produce a sequence of finite patches leading to a Delone set.

Let $T_{1}$ and $T_{2}$ be the scalene and isosceles triangle of unit area, respectively. We compute the GIFS for ${\mathcal{F}}=\{T_{1},T_{2}\}$ . We use $e^{i\theta}$ to denote rotation about the origin by the angle $\theta$ . Suppose that the bottom left corners of the tiles are at the origin. The contractive similitudes are as follows:

		$\displaystyle f_{1}(x,y)=\tfrac{1}{1+t^{2}}(x,y)\,,$		(4.12)
		$\displaystyle f_{2}(x,y)=\tfrac{t}{s(1+t^{2})}(-x,y)\cdot e^{i(\pi-\beta)}+(a,0)+at(\cos\gamma,\sin\gamma)\,,$
		$\displaystyle f_{3}(x,y)=\tfrac{t}{1+t^{2}}(x,y)\cdot e^{i\gamma}+(a,0)\,,$
		$\displaystyle f_{4}(x,y)=\tfrac{\sqrt{C}t}{1+t^{2}}(-x,y)\cdot e^{i\alpha}+(bu,0)\,,$
		$\displaystyle f_{5}(z)=\tfrac{\sqrt{C}t}{1+t^{2}}(-x,y)\cdot e^{i(\pi-\alpha)}+(bu,0)+bt(\cos\alpha,\sin\alpha)\,,$
		$\displaystyle g_{1}(x,y)=\tfrac{u}{\sqrt{C}(st+u)}(x,y)\cdot e^{i(\pi-\beta)}+(a,0)+at(\cos(\gamma),\sin(\gamma))\,,$
		$\displaystyle g_{2}(x,y)=\tfrac{u}{st+u}(x,y)\,,$
		$\displaystyle g_{3}(x,y)=\tfrac{st}{st+u}(x,y)+bst^{2}(\cos\gamma,\sin\gamma)\,,$

where

s=\tfrac{\sin\beta}{\sin\gamma}\,,\quad t=\tfrac{\sin\alpha}{\sin\beta}\,,\quad u=2st^{2}\cos\gamma=\tfrac{2\sin^{2}\gamma}{\sin\beta\tan\gamma}\,,\quad C=\tfrac{2(1+t^{2})^{2}\sin\alpha\cot\gamma}{t(st+u)(1+2t\cos\gamma)}\,,

and

a=\tfrac{1}{1+t^{2}}\sqrt{\tfrac{2}{s\sin(\alpha)}}\,,\quad b=2\sqrt{\tfrac{\cot(\gamma)}{st(st+u)(2t\cos(\gamma)+1)}}\,.

Then we get

T_{1}=g_{1}(T_{2})\cup f_{1}(T_{1})\cup f_{2}(T_{1})\cup f_{3}(T_{1})\,,

and

T_{2}=g_{2}(T_{2})\cup g_{3}(T_{2})\cup f_{4}(T_{1})\cup f_{5}(T_{1})\,.

4.4. Tilings with optimal badly approximable angles

In this section, we provide some illustrations of our construction in Section 4.3 in the specific case when $\alpha$ and $\beta=\gamma$ are as in $\eqref{eq:optimal1-angles}$ and (4.2). First, in Figure 3, we illustrate the $\varepsilon$ -rule process for $\varepsilon=0.2$ as an example. Then, in Figure 4 and Figure 5, we show the final patches for a few different values of $\varepsilon$ .

As described in [19, Remark 4.11], one can produce a stationary tiling by applying an additional isometry after each $\varepsilon$ -rule. For our subdivision rule shown in Figure 1, we can do this by introducing a rotation by the badly approximable angle $-\gamma$ .

To make this more precise, choose the sequence $\varepsilon_{n}=\varepsilon_{0}^{n}$ where $\varepsilon_{0}=\frac{t^{2}}{(1+t^{2})^{2}}$ and $t=\frac{\sin(\alpha)}{\sin(\beta)}$ . Then $\varepsilon_{0}$ is the square of the contraction factor of the similitude $f_{3}$ in (4.12). This function maps the large scalene triangle $T_{1}$ shown on the left in Figure 1 to the smaller scalene triangle in its center. For this value of $\varepsilon_{0}$ , the finite patch ${\mathcal{P}}_{\varepsilon_{0}}(T_{1})$ contains a copy of $e^{i\gamma}T_{1}$ . With this choice of $\varepsilon_{n}$ , the sequence $e^{-in\gamma}({\mathcal{P}}_{\varepsilon_{n}}(T_{1}))$ will produce a stationary tiling. Figure 6 shows the resulting nested sequence of finite patches when $\alpha$ and $\beta=\gamma$ are as in $\eqref{eq:optimal1-angles}$ and (4.2).

Each finite patch in these nested sequences contains the previous finite patch, along with several new copies of the triangle rotated by badly approximable angles $\alpha$ and $\beta=\gamma$ . Due to these irrational rotations, we expect the diffraction from the resulting stationary tilings to be rotationally invariant. Moreover, the discrepancy of each irrational rotation is small and optimal for our two choices of angles.

We conclude this section with a remark on convergence.

Remark 4.11.

The tilings produced by our method have infinite local complexity; because of this, we utilized the Chabauty–Fell topology to show the existence of a limit Delone set via a non-constructive process of subsequence selection. However, by the method described above, we can always produce a nested sequence of finite patches converging to a stationary tiling. In this case, we also have convergence in the local topology, i.e. the finite patches overlap exactly out to larger and larger radii up to small translations (see [2, Chapter 5] for a formal definition). This stronger notion of convergence is primarily used in the study of tilings with finite local complexity, but is still applicable stationary constructions.

5. Open Problems

In Theorem 1.6, we prove that there are exactly two solutions of the equality $x+y+z=1$ in ${\mathcal{B}}_{2,1}$ and that these solutions are in ${\mathbb{Q}}(\sqrt{2})$ and ${\mathbb{Q}}(\sqrt{3})$ . In ${\mathcal{B}}_{2,2}$ , we obtain at least three additional solutions of the equation $x+y+z=1(x\leq y\leq z)$ :

	$\displaystyle x=y=[3,3,\overline{1,2}]\,,\quad z=[2,1,\overline{1,2}]\,;$
	$\displaystyle x=y=[3,1,\overline{1,2}]\,,\quad z=[2,3,\overline{1,2}]\,;$
	$\displaystyle x=[3,1,\overline{1,2}]\,,\quad y=[3,3,\overline{1,2}]\,,\quad z=[2,2,2,\overline{2,1}]\,.$

However, we do not know whether or not the number of solutions is finite in ${\mathcal{B}}_{2,j}({\color[rgb]{1,0,1}j\geq 2})$ and whether or not they are in a real quadratic field. In Theorem 1.8, we prove that there are infinitely many solutions of the equality $x+y+z=1$ in ${\mathcal{B}}_{2}^{*}$ , but we do not know how large the set of solutions is, or its Hausdorff dimension. We know that the number of solutions changes from finite to infinite as we go from ${\mathcal{B}}_{2,1}$ to ${\mathcal{B}}_{2}^{*}$ , but what happens in between?

The proof of Theorem 1.8 relied on the two key identities (3.2) and (3.1). In fact, many similar identities can be found. For example,

	$\displaystyle 1-[2,1,3,\tfrac{1}{x}-1]-[2,1,3,\tfrac{1}{y}-1]-[3,1,1,\tfrac{1}{1-x-y}]$	$\displaystyle=\frac{4(x-y)^{2}}{(8x-11)(8y-11)(11-4x-4y)}\,,$
	$\displaystyle 1-[3,1,1,\tfrac{1}{x}]-[3,1,1,\tfrac{1}{y}]-[2,3,1,\tfrac{1}{1-x-y}-1]$	$\displaystyle=-\frac{2(x-y)^{2}}{(4x+7)(4y+7)(2x+2y+7)}\,,$
	$\displaystyle 1-[3,3,1,\tfrac{1}{x}-2]-[3,3,1,\tfrac{1}{y}-2]-[2,1,1,1,1-x-y]$	$\displaystyle=\frac{8(x-y)^{2}}{(16x-13)(16y-13)(13-8x-8y)}\,.$

From such identities whose right side is divisible by $x-y$ , we can construct further explicit solutions of $x+y+z=1$ in ${\mathcal{B}}_{2}^{*}$ or ${\mathcal{B}}_{3}$ . It may be an interesting problem to characterize the set of such identities. Such transformations on $x$ form a semi-group of integral Möbius transformations, but can we describe its generators? Are they finite? Moreover, our method produces many isosceles triangles, but we know very little about scalene triangles. Below, we point out a sporadic infinite family of solutions to $x+y+z=1$ of this type:

x=[3,(2)^{\ell},1,\overline{1,2}]\,,\quad y=[3,(2)^{\ell},3,\overline{1,2}]\,,\quad z=[(2)^{4+2\ell},\overline{1,2}]\qquad\ell=0,1,\dots\,,

which is shown by induction using the two lucky equalities:

	$\displaystyle 1-[3,\tfrac{1}{x}-1]-[3,\tfrac{1}{y}-1]-[2,2,\tfrac{1}{1-x-y}]$	$\displaystyle=\frac{2(x+y-3)(2xy-2x-2y+1)}{(2x-3)(2y-3)(7-2x-2y)}\,,$
	$\displaystyle 2[3,\tfrac{1}{x}-1][3,\tfrac{1}{y}-1]-2[3,\tfrac{1}{x}-1]-2[3,\tfrac{1}{y}-1]+1$	$\displaystyle=-\frac{2xy-2x-2y+1}{(2x-3)(2y-3)}\,.$

Lastly, regarding the diffraction from the Delone sets obtained in Section 4, an interesting open problem is to determine how the continued fraction expansions of the badly approximable numbers $x$ , $y$ , $z$ are related to the autocorrelation measures of the associated Delone sets; see [2, Chapter 9] for an overview of the theory of diffraction from point sets.

Appendix A Proof of compactness of $W_{r,R}({\mathbb{R}}^{2})$

For completeness, here we give a proof that $W_{r,R}({\mathbb{R}}^{2})$ is compact in the Chabauty–Fell topology. Note that to obtain the desired compactness of $W_{r,R}({\mathbb{R}}^{2})$ , it is necessary to use open balls for uniform discreteness and closed balls for relative denseness in the definition of an $(r,R)$ -Delone set.

Lemma A.1.

The set

{\mathcal{W}}_{r,R}({\mathbb{R}}^{2})=\{Y\in 2^{{\mathbb{R}}^{2}}\,:\,Y\ \text{is an}\ (r,R)\text{-Delone set in}\ {\mathbb{R}}^{2}\}\,,

is compact in the Chabauty–Fell topology.

Proof.

It follows from the early work of Fell [6] that the space $2^{X}$ is compact in the Chabauty–Fell topology for every topological space $X$ . Let $X={\mathbb{R}}^{2}$ . Since ${\mathcal{W}}_{r,R}(X)\subseteq 2^{X}$ , it suffices to show that ${\mathcal{W}}_{r,R}(X)$ is closed. To this end, let $\{\Lambda_{n}\}_{n\geq 1}$ be a sequence in ${\mathcal{W}}_{r,R}(X)$ converging to some $\Lambda\in 2^{X}$ . Our goal is to prove that $\Lambda$ is both $r$ -uniformly discrete and $R$ -relatively dense.

Proof of $r$ -uniform discreteness. Let $x\in X$ be arbitrary. Suppose by contradiction that there are two points $y,z\in\Lambda\cap B(x,r)$ with $y\neq z$ . Consider any

0<\varepsilon<\min(\textstyle{\frac{1}{r+\|x\|}},r-\|x-y\|,r-\|x-z\|,\textstyle{\frac{1}{2}}\|y-z\|)\,.

By convergence in the Chabauty–Fell topology, there exists an $N\geq 1$ such that

\Lambda\cap B(0,\textstyle{\frac{1}{\varepsilon}})\subseteq\Lambda_{n}+B(0,\varepsilon)\quad\forall n\geq N\,.

Now, since $\varepsilon<\frac{1}{r+\|x\|}$ , we have that $B(x,r)\subseteq B(0,\textstyle{\frac{1}{\varepsilon}})$ , so

y,z\in\Lambda\cap B(x,r)\subseteq\Lambda\cap B(0,\textstyle{\frac{1}{\varepsilon}})\subseteq\Lambda_{n}+B(0,\varepsilon)\quad\forall n\geq N\,.

In particular, there exist points $y_{N}$ and $z_{N}$ in $\Lambda_{N}$ such that $\|y-y_{N}\|<\varepsilon$ and $\|z-z_{N}\|<\varepsilon$ . By our choice of $\varepsilon$ , we have $\|y-z\|>2\varepsilon$ . From this, we see that

\|y_{N}-z_{N}\|\geq\|y-z\|-\|y-y_{N}\|-\|z-z_{N}\|>2\varepsilon-\varepsilon-\varepsilon=0\,,

so $y_{N}\neq z_{N}$ . Furthermore, we have $r-\|x-y\|>\varepsilon$ and $r-\|x-z\|>\varepsilon$ , so $y_{N}$ and $z_{N}$ are both in $B(x,r)$ . Indeed, we have

\|x-y_{N}\|\leq\|x-y\|+\|y-y_{N}\|<r-\varepsilon+\varepsilon=r\,,

and a similar inequality shows that $\|x-z_{N}\|<r$ . This contradicts $r$ -uniform discreteness of $\Lambda_{N}$ . Therefore, $\Lambda\cap B(x,r)$ has at most one element for every $x\in{\mathbb{R}}^{2}$ , i.e. $\Lambda$ is $r$ -uniformly discrete.

Proof of $R$ -relative denseness. Let $x\in X$ be arbitrary. We aim to show that $\Lambda\cap\overline{B(x,R)}\neq\emptyset$ . By assumption, we have that $\Lambda_{n}$ is $R$ -relatively dense for every $n$ , so there exists a sequence of points $y_{n}$ in $X$ such that $y_{n}\in\Lambda_{n}\cap\overline{B(x,R)}$ for all $n$ . Since $\overline{B(x,R)}$ is compact in $X$ (in the Euclidean topology), there exists a subsequence $y_{n_{i}}$ of $y_{n}$ and some $y\in\overline{B(x,R)}$ such that $\|y_{n_{i}}-y\|\rightarrow 0$ . Next, observe that for all $n\in{\mathbb{N}}$ with $n>R+\|x\|+1$ , we have $\overline{B(x,R)}\subseteq B(0,n)$ . From this and convergence in the Chabauty–Fell topology, there exists an $N>R+\|x\|+1$ such that

y_{n}\in\Lambda_{n}\cap\overline{B(x,R)}\subseteq\Lambda_{n}\cap B(0,n)\subseteq\Lambda+B(0,\textstyle{\frac{1}{n}})\quad\forall n\geq N\,.

Thus, there is a sequence $z_{n}$ in $\Lambda$ such that $\|y_{n}-z_{n}\|<\frac{1}{n}$ for all $n\geq N$ . In particular, we have $z_{n_{i}}\rightarrow y$ . Now, since $\Lambda$ is a closed set in the Euclidean topology, we must have $y\in\Lambda$ . Moreover, $y$ is also in $\overline{B(x,R)}$ , so we have shown that there is some $y\in\Lambda\cap\overline{B(x,R)}\neq\emptyset$ , as desired. This completes the proof. ∎

Acknowledgements

S.A. is supported by JSPS grants 20K03528, 24K06662 and RIMS in Kyoto University. E.R.K. is supported by the Natural Sciences and Engineering Research Council of Canada (NSERC) via grant 2024-0485 and the Canadian Graduate Scholarship - Doctoral. Y-L.X. is supported by the China Scholarship Council, China (No. 202306770085). We would also like to thank Nicolae Strungaru for his insights on the construction of Delone sets, and Noel Murasko for his assistance in producing the graphics.

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Badly approximable triangles and Delone sets

Abstract.

1. Introduction

Definition 1.1.

Lemma 1.2.

Proof.

Corollary 1.3.

Remark 1.4.

Remark 1.5.

Theorem 1.6.

Theorem 1.7.

Theorem 1.8.

2. Proof of Theorems 1.6 and 1.7

Lemma 2.1 (Forbidden patterns).

Proof.

Lemma 2.2.

Proof.

Proof of Theorem 1.6.

Proof of Theorem 1.7.

Remark 2.3.

3. Proof of Theorem 1.8

Proof.

Remark 3.1.

Remark 3.2.

4. Construction of Delone sets by badly approximable triangles

4.1. Delone sets and the Chabauty–Fell topology

Definition 4.1.

Remark 4.2.

Definition 4.3.

Remark 4.4.

Definition 4.5 (Chabauty–Fell Topology).

Remark 4.6.

Proposition 4.7.

Proof.

4.2. Threshold method

4.2.1. GIFS

4.2.2. 𝜺\bm{\varepsilon}-rule.

Remark 4.8.

4.2.3. Associated point sets.

4.2.4. Existence of a Delone limit set

Lemma 4.9.

Proof.

Theorem 4.10.

Proof.

4.3. GIFS for arbitrary angles

4.4. Tilings with optimal badly approximable angles

Remark 4.11.

5. Open Problems

Appendix A Proof of compactness of Wr,R​(ℝ2)W_{r,R}({\mathbb{R}}^{2})

Lemma A.1.

Proof.

Acknowledgements

References

4.2.2. $\bm{\varepsilon}$ -rule.

Appendix A Proof of compactness of $W_{r,R}({\mathbb{R}}^{2})$