2.1 3-Point configurations in $\mathbb {R}$ & triangles in the plane part I

First, we demonstrate the following more general version of Proposition 1.1, which recovers the original result when $\lambda =\frac {1}{2}$ .

Proposition 2.1 (Convex combinations in $\mathbb {R}$ )

Let $C\subset \mathbb {R}$ be a compact set with ${\tau (C)\geq 1}$ . Then, for each $\lambda \in (0, 1)$ , the set C contains a nondegenerate $3$ -term progression of the form

$$ \begin{align*}\{a, (1-\lambda)a + \lambda b , b\}.\end{align*} $$

In other words, any 3-point subset of the line is realized in C.

The proof of this result relies on demonstrating that $C\cap \left ( (1-\lambda ) C+ \lambda C \right )\neq \emptyset $ and is found in Section 4.

As a consequence of Proposition 2.1 combined with the fact that the interior of the difference set

$$ \begin{align*}C-C= \{x-y: x,y\in C\}\end{align*} $$

has non-empty interior, we have the following geometric consequence for triangles.

It follows from Theorem 2.2 that the Cartesian product $C\times C$ contains the vertices of a similar copy of any 3-point configuration whenever $C \subset \mathbb {R}$ is a compact set satisfying $\tau (C) \geq 1$ . For emphasis, we state the result for equilateral triangles (see Figure 1).

Figure 1: We see that $C\times C$ contains an equilateral triangle by combining two facts: (i) C contains an arithmetic progression $\mathcal {A} = \{x, x+t, x+2t\}$ , where $t>0$ can be taken arbitrarily small; (ii) the distance set $\Delta (C)$ contains an interval $[0,\ell ]$ for some $\ell>0$ .

The proofs for the results in this section appear in Section 4.

Remark 2.5 (Comparison between Newhouse thickness and Hausdorff dimension)

As mentioned above, Hausdorff dimension alone is not enough to guarantee the realization of similar triangles in subsets of $\mathbb {R}^2$ . Our result gives a class of compact Lebesgue null subsets of $\mathbb {R}^2$ and explicit criteria, mainly $\tau (C)\geq 1$ , that guarantees the realization of a similar copy of any 3-point configuration in $C\times C$ . Because Hausdorff dimension and Newhouse thickness obey the following relationship [Reference Palis and Takens33] for $\tau (C)>0$ ,

$$ \begin{align*} \dim_H(C) \geq \frac{\log(2)}{\log(2+\frac{1}{\tau(C)})}, \end{align*} $$

we can calculate a lower bound for the Hausdorff dimension of $C\times C$ using its thickness; i.e., since $\tau (C)\geq 1$ , we know $\dim _H(C\times C)\geq 2\dim _H(C) \geq 2\frac {\log {2}}{\log {3}}$ .

Remark 2.6 (Longer progressions)

For longer progressions, higher thickness is required. It is known that the middle- $\epsilon $ Cantor set $C_\epsilon $ does not contain arithmetic progressions of length $\lfloor \frac {1}{\epsilon }\rfloor +2$ or larger. In particular, Broderick, Fishman, and Simmons [Reference Broderick, Fishman and Simmons7] proved that if $L_{\text {AP}}(S)$ denotes the maximal length of an arithmetic progression in a set $S\subset \mathbb {R}$ , then for all $\epsilon>0$ sufficiently small and $n\in \mathbb {N}$ sufficiently large,

$$ \begin{align*} L_{\text{AP}}(C_\epsilon)\leq 1/\epsilon + 1. \end{align*} $$

It is also not hard to see that $L_{\text {AP}}(C_\epsilon ) =2 $ for all $\epsilon>\frac 13$ .

Further, using Newhouse’s gap lemma and symmetry, $L_{\text {AP}}(C_\epsilon ) \geq 4$ for all $0<\epsilon \le \frac 13$ (this lower bound is attributed to Shmerkin and pointed out in [Reference Broderick, Fishman and Simmons7, Footnote 4]). Indeed, it follows by the gap lemma that there exists a $t \in (C_\epsilon -\frac 12) \cap \frac 13(C_\epsilon - \frac 12)\neq \emptyset $ , and so we have $c_1, c_2 \in C_\epsilon $ so that

$$ \begin{align*}c_1 =\frac12 +t \,\,\, \text{ and } \,\,\, c_2 = \frac12+3t.\end{align*} $$

By symmetry of $C_\epsilon $ about $\frac 12$ , we further have $c_3, c_4 \in C_\epsilon $ so that

$$ \begin{align*}c_3 = \frac12-t \,\,\, \text{ and } \,\,\, c_4 = \frac12 -3t,\end{align*} $$

which yields the desired $4$ -AP.

Remark 2.7 (Sharpness of our result: the off-center Cantor set has thickness 1 and no $4$ -AP)

We now construct a set of thickness $1$ that contains no $4$ -term arithmetic progressions. We call our construction the off-center Cantor set and denote it $C_a$ .

Let $0<a<\frac {1}{3}$ . We construct $C_a$ iteratively, starting with $[0,1]$ , by removing from each interval I an open interval of length $a|I|$ , call this open interval G. Then, $I\setminus G$ is split into two closed intervals: the left interval L of length $a|I|$ and the right interval R of length $(1-2a)|I|$ . Hence, we remove $(a,2a)$ from $[0,1]$ to obtain $[0,a]\cup [2a,1]$ . From these intervals, we remove $(a^2,2a^2)$ and $(3a-2a^2,4a-4a^2)$ to obtain $[0,a^2]\cup [2a^2,a]\cup [2a,3a-2a^2]\cup [4a-4a^2,1]$ , and so on, as illustrated in Figure 2.

Figure 2: First two iterations of off-center Cantor set $C_a$ for $a=\frac {3}{10}$ .

Observe that by construction $|L|=|G|< |R|$ ; as this is true for every step in the construction of $C_a$ ,

$$ \begin{align*}\tau(C_a) =\frac{\min\{|L|,|R|\}}{|G|}=1.\end{align*} $$

Thus, $C_a$ is a self-similar set of thickness $1$ .

We will now show $C_a$ does not contain any $4$ -term arithmetic progressions for $\frac {2-\sqrt {2}}{2}<a<\frac {3-\sqrt {3}}{4}$ . Assume to the contrary that $C_a$ contains a $4$ -AP, call it $P=\{x,x+y,x+2y,x+3y\}$ . Then, there exists some smallest interval in the construction of $C_a$ that contains P but P is split over two levels in the next step of the construction. Because of the self-similarity of $C_a$ , we assume without loss of generality that P is contained in $[0,1]$ but splits over $[0,a]\cup [2a,1]$ .

For P to be contained in $[0,1]$ and split over $[0,a]\cup [2a,1]$ , we must have $0\leq x\leq a$ , $y\geq a$ (otherwise the arithmetic progression cannot jump the gap $(a,2a)$ ), and $x+3y\leq 1$ . Then, observe that $x+y\ngeq 2a$ because if it was then

$$ \begin{align*}x+3y\geq 2y+2a\geq 4a>1,\end{align*} $$

a contradiction. We also have $x+y\notin (a,2a)$ ; otherwise, P would not be contained in $C_a$ . Thus, it must be the case that $x+y\in [0,a]$ . Because $0\leq x\leq a$ , $y\geq a$ , and $0\leq x+y\leq a$ , we conclude that $x=0$ and $y=a$ , so $P=\{0,a,2a,3a\}$ .

Notice that $0,a,2a\in C_a$ , so it remains to show $3a\notin C_a$ . By construction of $C_a$ , the interval $[4a-4a^2,1]$ is split by the gap $(5a-8a^2+4a^3,6a-12a^2+8a^3)$ , and by the assumption that $\frac {2-\sqrt {2}}{2}<a<\frac {3-\sqrt {3}}{4}$ , we know $3a\in (5a-8a^2+4a^3,6a-12a^2+8a^3)$ . Consequently, $C_a$ cannot contain P. By self-similarity, $C_a$ cannot contain any $4$ -term arithmetic progressions.

In the next section, we introduce higher-dimensional variants of Proposition 2.1 (on 3-point configurations on the line) and Theorem 2.2 (on triangles in the plane) that do not depend on Cartesian product structure.

2.2 3-Point configurations in $\mathbb {R}^d$ & triangles in the plane part II

In this section, we introduce results in dimensions $d\geq 2$ . Theorem 2.8 of this section yields conditions to guarantee the occurrence of arithmetic progressions and other linear 3-point configurations in $\mathbb {R}^d$ . Beyond linear combinations, Theorem 2.12 guarantees the occurrence of a similar copy of any 3-point configuration in higher dimensions.

Here, we use a higher-dimensional notion of thickness introduced by Yavicoli [Reference Yavicoli40]. We directly state the results of this section, and we delay formal introduction of Yavicoli thickness and the corresponding gap lemma to Section 3. We require the notion of a system of balls and r-uniformity, which will also be defined in Section 3.

Our first result says that a compact set C generated by a system of balls $\{S_I\}_I$ in $\mathbb {R}^d$ with Yavicoli thickness (Definition 3.2 below) satisfying

$$ \begin{align*} \tau\left( C,\{S_I\}\right) \geq \frac{2}{1-2r} \end{align*} $$

for some $0<r<\frac {1}{2}$ contains a $3$ -point arithmetic progression; e.g., any $\frac {1}{4}$ -uniformly compact set of thickness greater than $4$ contains an arithmetic progression of length $3$ .

The proof of this result is inspired by the proof of Proposition 1.1. For a compact set C, we take two disjoint subsets A and B and apply the Gap Lemma to show that $C\cap \left (\lambda A +(1-\lambda )B\right ) \neq \emptyset $ . The assumption that $0<r< \frac 12$ is used to apply the gap lemma in Theorem 3.2. Our proofs quickly diverge, though, as we lose the well-ordering of $\mathbb {R}$ in higher dimensions and the higher-dimensional Gap Lemma has a number of additional assumptions to verify over the one-dimensional Gap Lemma.

In particular, our method views compact sets in $\mathbb {R}^d$ as a sequence of nested balls, each generation of which is finite, and our method requires the existence of two first-generation children, call them $S_{1_A}$ and $S_{1_B}$ , out of the total $k_\emptyset $ first-generation children that are both disjoint from all first-generation children; these first-generation children $S_{1_A}$ and $S_{1_B}$ are used to construct the sets A, B mentioned above. This requirement is explained in Section 3.1.

Theorem 2.8 (Convex combinations in $\mathbb {R}^d$ )

Let C be a compact set in $(\mathbb {R}^d,\operatorname {\mathrm {dist}})$ generated by the system of balls $\{S_I\}_I$ such that C is r-uniformly dense where $ 0<r<\frac {1}{2}$ . Let $\lambda \in (0,\frac 12]$ , and suppose that

$$ \begin{align*}\tau\left( C,\{S_I\}\right) \geq \frac{2(1-\lambda)}{\lambda(1-2r)}.\end{align*} $$

Suppose that there exist distinct first-generation children disjoint from all other children: $S_{1_A}$ and $S_{1_B}$ with $1\leq 1_A<1_B\leq k_\emptyset $ such that $S_{1_A} \cap S_i = \emptyset $ and $S_{1_B}\cap S_i =\emptyset $ for all $i\neq 1_A,1_B$ where $1\leq i\leq k_\emptyset $ . Then, C contains a $3$ -point convex combination of the form

$$ \begin{align*}\{a,\lambda a+(1-\lambda)b,b\}.\end{align*} $$

In particular, under the hypotheses above with $\lambda = \frac {1}{2}$ , we have the following.

Corollary 2.9 ( $3$ -term arithmetic progressions in $\mathbb {R}^d$ )

Let C be a compact set in $(\mathbb {R}^d,\operatorname {\mathrm {dist}})$ generated by the system of balls $\{S_I\}_I$ such that C is r-uniformly dense, where $ 0<r<\frac {1}{2}$ . Suppose that

$$ \begin{align*}\tau\left( C,\{S_I\}\right) \geq \frac{2}{1-2r},\end{align*} $$

and suppose that there exist distinct first-generation children of $\{S_I\}$ disjoint from all other children. Then, C contains an arithmetic progression $\{a, \frac {1}{2}(a+b),b\}$ with $a\neq b$ .

Next, we prove a result on the existence of triangles in compact sets of sufficient Yavicoli thickness, but first we need a way to categorize all triangles.

Figure 3: The triangle $T(\alpha ,\lambda )$ with vertices $x,y,z$ , largest angle at z, height $\alpha $ , and base $1$ .

Lemma 2.11 Let $\mathcal {T}$ be the vertices of any non-degenerate triangle in $\mathbb {R}^2$ . Then, there exists an $(\alpha ,\lambda )$ in

$$ \begin{align*} \mathcal{R}= \left\{ (\alpha,\lambda)\in \mathbb{R}^2 : 0< \alpha , 0\leq \lambda \leq \frac{1}{2}, \alpha^2+(1-\lambda)^2\leq 1 \right\}, \end{align*} $$

such that $\mathcal {T}$ is similar to the triangle $\mathcal {T}(\alpha ,\lambda )$ .

The lemma is immediate upon scaling, rotating, and labeling the vertices appropriately; the above inequalities are a simple consequence of the Pythagorean theorem.

Theorem 2.12 (Triangles in $\mathbb {R}^2$ )

Let $\mathcal {T}$ denote the vertices of any triangle in $\mathbb {R}^2$ , and let $\mathcal {T}(\alpha ,\lambda )$ be a triangle similar to $\mathcal {T}$ resulting from Lemma 2.11 for some $\alpha $ , $\lambda $ in $\mathcal {R}$ . Let $C\subset \mathbb {R}^2$ be a compact set generated by the system of balls $\{S_I\}$ in the Euclidean norm such that C is r-uniformly dense for some $0<r<\frac {1}{2}$ . Suppose there exists distinct first-generation children $S_{1_A}$ and $S_{1_B}$ , $1\leq 1_A<1_B\leq k_\emptyset $ , contained in $\bar {B}\left (0,\frac {1}{2}\right )$ such that $S_{1_A}$ and $S_{1_B}$ are disjoint from all other first-generation children; i.e., $S_{1_A}\cap S_i=\emptyset $ for all $i\neq 1_A$ , and $S_{1_B}\cap S_i=\emptyset $ for all $i\neq 1_B$ . Further, suppose

$$ \begin{align*}\tau\left( C,\{S_I\}\right) \geq \sqrt{\frac{\alpha^2+(1-\lambda)^2}{\alpha^2+\lambda^2}}\cdot \frac{2}{1-2r},\end{align*} $$

then C contains the vertices of a similar copy of $\mathcal {T}$ .

In other words, given any 3-point set $\mathcal {T}$ , any set C satisfying the hypotheses contains a similar copy of $\mathcal {T}$ . A key tool in the proof is the higher gap lemma due to Yavicoli (see Theorem 3.2); the hypothesis that $r\in (0,\frac 12)$ is an assumption of the Gap lemma.

For equilateral triangles, $\lambda =\frac 12$ and $\alpha = \frac {\sqrt {3}}{2}$ , and the thickness assumption is simplified so that we have the following.

Corollary 2.14 (Equilateral triangles in $\mathbb {R}^2$ )

Let $\mathcal {T}$ denote the vertices of an equilateral triangle. Let $C\subset \mathbb {R}^2$ be a compact set generated by the system of balls $\{S_I\}$ in the Euclidean norm such that C is r-uniformly dense for some $0<r<\frac {1}{2}$ . Suppose there exists first-generation children $S_{1_A}$ and $S_{1_B}$ , $1\leq 1_A<1_B\leq k_\emptyset $ , contained in $\bar {B}\left (0,\frac {1}{2}\right )$ such that $S_{1_A}$ and $S_{1_B}$ are disjoint from all other first-generation children. Further, suppose

$$ \begin{align*}\tau\left( C,\{S_I\}\right)\geq \frac{2}{1-2r},\end{align*} $$

then C contains the vertices of a similar copy of $\mathcal {T}$ .

Before, to guarantee the occurrence of a $3$ -AP, we needed $C\cap \left (\frac {A+B}{2}\right )\neq \emptyset $ for $A, B$ disjoint subsets of C. Now, to guarantee the occurrence of the vertices of an equilateral triangle, we need $C\cap \left (H(A,B)\right )\neq \emptyset $ , where $H: \mathbb {R}^2\times \mathbb {R}^2 \rightarrow \mathbb {R}^2$ is defined by $H(a,b) = \frac {a+b}{2} + \frac {\sqrt {3}}{2}(b-a)^{\perp }$ . This ensures that there’s some point $a\in A$ , $b\in B$ forming the base of our equilateral triangle and some point $c\in C\cap H(A,B)$ as the top vertex. The details are found in Section 5.3.

Remark 2.16 Theorem 2.12 offers a significant improvement over the following result of Yavicoli in the specific setting of triangles in the plane by lowering the required thickness threshold; however, for values of $\alpha $ and $\lambda $ significantly close to $0$ , Yavicoli’s result requires less thickness. Yavicoli proved that compact sets in $\mathbb {R}^d$ generated by a restricted system of balls in the infinity norm with significantly large thickness contain homothetic copies of finite sets [Reference Yavicoli40]. In particular, let $C\subset \mathbb {R}^d$ be a compact set with disjoint children. Take also constraints on the number of children $N_0$ and the radii of the children. Then, C contains a homothetic copy of every set with at most

$$ \begin{align*} N(\tau):= \left\lfloor\frac{3}{4eK_2}\frac{\tau}{\log\tau}\right\rfloor \end{align*} $$

elements where $K_2$ is a large constant dependent on $N_0$ . In fact, we can take the conservative estimate of $K_2=360,000$ which means we would need a thickness strictly greater than $10^7$ to guarantee the existence of any $3$ -point configuration.

2.3 Organization

In Section 3, we introduce systems of balls for compact sets, define r-uniformity, and introduce Yavicoli’s higher-dimensional thickness and gap lemma. We also discuss some relevant properties of this notion of thickness, including its behavior under taking subsets. In Section 6, we give some examples. Section 4 contains the proofs of the results of Section 2.1 that rely on Newhouse thickness, and the proofs of the results in Section 2.2 that rely on Yavicoli thickness appear in Section 5.

3.1 Computing the thickness of a subset

We now consider how to compute the thickness of a subset of C given the thickness of C.

Let C be a compact set with a system of balls $\{S_I\}_I$ , and let $A:= S_{1_A}\cap C$ for some $1\leq 1_A\leq k_\emptyset $ be a compact set with a system of balls $\{S_{1_A,I}\}_I$ .

While the definition of $h_I(C):=\max _{x\in S_I} \operatorname {\mathrm {dist}}(x,C)$ is used in calculating the thickness of C, when we consider the thickness of first generation subsets of the form $A=C \cap S_{1_A}$ for some $1_A$ satisfying $1\leq 1_A\leq k_\emptyset $ , we need $h_{1_A}(A):=\max _{x\in S_{1_A,I}}\operatorname {\mathrm {dist}}(x,A)$ to calculate the thickness of A:

$$ \begin{align*} \tau\left( A,\{S_{1_A,I}\}_I\right):=\inf_{n\in\mathbb{N}_0}\inf_{\ell(I)=n}\frac{\min_i \operatorname{\mathrm{rad}}(S_{1_A,I,i})}{\max_{x\in S_{1_A,I}}\operatorname{\mathrm{dist}}(x,A)}. \end{align*} $$

In the proof of Theorem 2.8, we have implicit assumptions about $\max _{x\in S_I}\operatorname {\mathrm {dist}}(x,C)$ but no assumptions about $\max _{x\in S_{1_A,I}}\operatorname {\mathrm {dist}}(x,A)$ , so we use $\max _{x\in S_{1_A}}\operatorname {\mathrm {dist}}(x,C)$ to get an upper bound on $\max _{x\in S_{1_A}}\operatorname {\mathrm {dist}}(x,A)$ in Lemma 3.4. As in (3.2), define

(3.3) $$ \begin{align} h_\emptyset(C) := \max_{x\in S_\emptyset} \operatorname{\mathrm{dist}}(x,C) \quad \text{and} \quad h_{1_A}(A) &:= \max_{x\in S_{1_A}} \operatorname{\mathrm{dist}}(x,A) \nonumber\\ &= \max_{x\in S_{1_A}} \operatorname{\mathrm{dist}}(x,S_{1_A}\cap C). \end{align} $$

Lemma 3.4 (Preliminary computation for the thickness of a subset)

Let C be a compact set in ( $\mathbb {R}^d,\operatorname {\mathrm {dist}}$ ) generated by the system of balls $\{S_I\}_I$ such that $\tau \left ( C,\{S_I\}\right )\geq 1$ . Then, for any word I, we have

$$ \begin{align*} \max_{x\in S_I} \operatorname{\mathrm{dist}}(x,S_I\cap C) \,\, \leq \,\, 2\max_{x\in S_I} \operatorname{\mathrm{dist}}(x,C). \end{align*} $$

So, if $A = S_{1_A}\cap C$ for some $1\le 1_A \le k_\emptyset $ , Lemma 3.4 implies that

$$ \begin{align*} \max_{x\in S_{1_A}} \operatorname{\mathrm{dist}}(x,S_{1_A}\cap C) \,\, \leq \,\, 2\max_{x\in S_{1_A}} \operatorname{\mathrm{dist}}(x,C) \,\, \leq \,\, 2\max_{x\in S_\emptyset} \operatorname{\mathrm{dist}}(x,C), \end{align*} $$

and it follows that

$$ \begin{align*} h_{1_A}(A) \le 2h_\emptyset(C). \end{align*} $$

Proof Fix a word I. Since $\tau \left ( C,\{S_I\}\right )\geq 1$ for all words I, it follows from the definition of thickness that

$$ \begin{align*}\min_i \operatorname{\mathrm{rad}}(S_{I,i})\geq \max_{x\in S_I}\operatorname{\mathrm{dist}}(x,C) = h_I.\end{align*} $$

In particular,

$$ \begin{align*}\operatorname{\mathrm{rad}}(S_I)\geq \min_i \operatorname{\mathrm{rad}}(S_{I,i}) \geq h_I.\end{align*} $$

This establishes that, for any $y\in S_I$ , there exists a ball of radius $h_I$ in $S_I$ containing y.

Now take any $y\in S_I$ , and observe that

$$ \begin{align*}\operatorname{\mathrm{dist}}(y,C\cap S_I)\leq \operatorname{\mathrm{dist}}(y,c'),\end{align*} $$

for any $c'\in C\cap S_I$ . We will choose $c'$ in such a way that we can bound $\max _{x\in S_I}\operatorname {\mathrm {dist}}(x,C\cap S_I)$ .

Let $\bar {B}_y$ be a closed ball of radius $h_I$ in $S_I$ containing the point y. Then, there exists some point $z \in \bar {B}_y\subset S_I$ such that

(3.4) $$ \begin{align} \operatorname{\mathrm{dist}}(y,z)\leq h_I \quad \text{and}\quad \operatorname{\mathrm{dist}}(z,\partial S_I)\geq h_I. \end{align} $$

For instance, z can be taken as the center of $\bar {B}_y$ . As a consequence of the latter inequality above combined with the definition of $h_I: = \max _{x\in S_I}\operatorname {\mathrm {dist}}(x,C)$ , there exists some $c'\in C \cap S_I$ such that

(3.5) $$ \begin{align} \operatorname{\mathrm{dist}}(z,c')\leq h_I. \end{align} $$

Now, we have

(3.6) $$ \begin{align} \operatorname{\mathrm{dist}}(y,C\cap S_I) \leq \operatorname{\mathrm{dist}}(y,c') \leq \operatorname{\mathrm{dist}}(y,z) + \operatorname{\mathrm{dist}}(z,c') \leq 2h_I. \end{align} $$

As this holds for any $y\in S_I$ , we have

$$ \begin{align*} \max_{y\in S_I}\operatorname{\mathrm{dist}}(y,C \cap S_I) \leq 2 h_I. \end{align*} $$

Now that we understand the relationship in Lemma 3.4, we can use it to calculate the relationship between the thicknesses of C and its subsets.

Lemma 3.5 (Thickness of a subset)

Let C be a compact set in $(\mathbb {R}^d,\operatorname {\mathrm {dist}})$ generated by the system of balls $\{S_I\}_I$ such that $\tau \left ( C,\{S_I\}\right )\geq 1$ . Suppose that there exists some $1\leq 1_A\leq k_\emptyset $ such that $S_{1_A}\cap S_i =\emptyset $ for all $i\neq 1_A$ , $1\le i\le k_\emptyset $ . Let $A= S_{1_A}\cap C$ . Then

$$ \begin{align*}\tau\left( A,\{S_{1_A,I}\}\right) \geq \frac{1}{2}\tau\left( C,\{S_I\}\right).\end{align*} $$

Remark 3.6 We comment on the assumption that $S_{1_A}$ and $S_{1_B}$ are disjoint from all other first-generation children. Let C be a compact set generated by $\{S_I\}_I$ and take any $S_{1_A}$ a first-generation child, not necessarily disjoint from other first-generation children. Consider the following two subsets constructed by the first-generation child $S_{1_A}$ of C: $A'$ generated by the system of balls $\{S_{1_A,I}\}_{1_A,I}$ and $A := S_{1_A}\cap C$ . We necessarily have $A'\subseteq A \subseteq C$ .

To calculate the Yavicoli thickness of a compact set E, we need: $(1)$ a system of balls that generates E and $(2)$ the value of $\max _{x\in S_{I}}\operatorname {\mathrm {dist}}(x,E)$ for all words I. In particular, we want to calculate the thickness of a subset of a compact set, so we additionally need $(1')$ a system of balls that generates the subset and relates to the system that generates C and $(2')$ the value of $\max _{x\in S_{I}}\operatorname {\mathrm {dist}}(x,E)$ compared to $\max _{x\in S_I} \operatorname {\mathrm {dist}}(x,C)$ . For A, we have $(2')$ as we can use Lemma 3.4 to obtain the estimate

$$ \begin{align*}\max_{x\in S_{1_A}}\operatorname{\mathrm{dist}}(x,A) \leq 2 \max_{x\in S_{1_A}}\operatorname{\mathrm{dist}}(x,C).\end{align*} $$

While the compact set $A'$ generated by $\{S_{1_A,I}\}$ is contained in A, if $S_{1_A}$ is not disjoint from other first-generation children it is possible we have some point $x\in S_i\cap S_{1_A}$ that is not generated by $\{S_{1_A,I}\}$ , and it becomes hard to see if A satisfies $(1')$ . For $A'$ , we have $(1')$ because $A'$ is generated by $\{S_{1_A,I}\}$ , but it does not necessarily satisfy $(2')$ as we have no way to calculate or bound $\max _{x\in S_{I}}\operatorname {\mathrm {dist}}(x,A')$ above; in general,

$$ \begin{align*}\max_{x\in S_I}\operatorname{\mathrm{dist}}(x,A')\geq \max_{x\in S_I}\operatorname{\mathrm{dist}}(x,A).\end{align*} $$

In order to guarantee the existence of a set that satisfies both $(1')$ and $(2')$ , we take $S_{1_A}$ to be disjoint from all other first-generation children. This forces $A=A'$ , so $(1')$ and $(2')$ are both satisfied. We first need a corresponding system of balls that generate the set. In the case of $A=S_{1_A}\cap C$ , the system of balls $\{S_{1_A,I}\}$ generates the set $A=S_{1_A}\cap C$ if and only if all elements of $S_{1_A}\cap C$ are generated by $\{S_{1_A,I}\}$ . This is satisfied by having $S_{1_A}$ disjoint from all other first-generation children $S_i$ where $1\leq i\leq k_\emptyset $ , $i\neq 1_A$ .

Proof Lemma 3.4 implies

$$ \begin{align*} \max_{x\in S_{I}}\operatorname{\mathrm{dist}}(x,A) := \max_{x\in S_{I}}\operatorname{\mathrm{dist}} (x,S_{1_A}\cap C) \leq 2\max_{x\in S_{I}}\operatorname{\mathrm{dist}}(x,C), \end{align*} $$

for all words I starting at $1_A$ . It follows that:

$$ \begin{align*} \tau\left( A, \{S_{1_A,I}\}\right) &= \inf_{n\geq 1} \inf_{\substack{\ell(I)=n\\ I=\{1_A,\ldots\}}} \frac{\min_i \operatorname{\mathrm{rad}}(S_{I,i})}{\max_{x\in S_{I}} \operatorname{\mathrm{dist}}(x,A)} \geq \frac{1}{2}\inf_{n\geq1} \inf_{\substack{\ell(I)=n\\I=\{1_A,\ldots\} }} \frac{\min_i \operatorname{\mathrm{rad}}(S_{I,i})}{\max_{x\in S_{I}} \operatorname{\mathrm{dist}}(x,C)} \\ &\geq \frac{1}{2}\inf_{n\geq0} \inf_{\ell(I)=n} \frac{\min_i \operatorname{\mathrm{rad}}(S_{I,i})}{\max_{x\in S_I} \operatorname{\mathrm{dist}}(x,C)} =\frac{1}{2}\tau\left( C, \{S_I\}\right), \end{align*} $$

where the first inequality follows from the estimate in Lemma 3.4, and the second inequality follows from taking the infimum over a larger set.

The content of these lemmas is significantly different from the one-dimensional case and reflect one of the technical hurdles of defining thickness in higher dimensions. In the one-dimensional setting, if $C\subset \mathbb {R}$ and $\tau $ denotes Newhouse thickness, then $\tau (A)\geq \tau (C)$ whenever $A=C\cap S$ and S is bridge. In the Appendix, we illustrate this key difference and discuss a special case in which Lemma 3.5, and hence our main results, can be improved.

4.1 Proof of Proposition 2.1

The following proof is inspired by that of Yavicoli’s [Reference Yavicoli39, Proposition 20], where the proposition is proved for $\lambda =\frac 12$ . The proof here is more involved as a number of technical hurdles arise in this more general setting.

Since thickness is invariant under scaling and translations, we may assume that $\operatorname {\mathrm {conv}}(C)=[0,1]$ . The idea is to show that $C\cap \left ( (1-\lambda )C + \lambda C \right ) \neq \emptyset $ for $\lambda \in (0,1)$ . To avoid degeneracy, we introduce disjoint subsets A and B of C, and show that $C\cap \left ( (1-\lambda )A + \lambda B \right ) \neq \emptyset $ , which will establish that there exist points $a,b \in C$ with $a\neq b$ so that

$$ \begin{align*}\{a, (1-\lambda)a + \lambda b, b\} \subset C.\end{align*} $$

A brief sketch of the proof is as follows. We observe that $t\in (1-\lambda )A+ \lambda B$ if and only if $ -(1-\lambda )A \cap \left (\lambda B-t\right ) \neq \emptyset. $ We then verify the hypotheses of the Gap Lemma and apply it to the sets $-(1-\lambda )A$ and $ \left (\lambda B-t\right )$ . A potential issue that can arise is that, for $\lambda $ small, $ \left (\lambda B-t\right )$ can be contained in a gap of $-(1-\lambda )A$ , which would violate the hypotheses of the gap lemma. To get around this obstacle, we only work with values of t and $\lambda $ that avoid this issue, mainly so that the two sets are interwoven and neither lies in the gap of the other.

Let $G=(k_1, k_2)$ denote the largest bounded gap of C. Set $A=C\cap [0,k_1]$ and $B=C\cap [k_2, 1]$ , and denote $|A| = k_1$ and $|B| = 1-k_2$ .

Set

$$ \begin{align*}m = |A|/(|A|+|B|),\end{align*} $$

$$ \begin{align*}I_\lambda = [\lambda k_2, \lambda + (1-\lambda)k_1]\end{align*} $$

and

$$ \begin{align*}\widetilde{I}_\lambda = [\lambda k_2, \lambda] \,\, \bigcup \,\, [\lambda k_2 +(1-\lambda) k_1, \lambda + (1-\lambda)k_1].\end{align*} $$

First, we use the gap lemma to establish the following claim.

Claim 4.1 For $\lambda \in (0,1)$ ,

$$ \begin{align*}\widetilde{I}_\lambda \,\, \subset \,\, (1-\lambda)A+ \lambda B \,\,\subset\,\, I_\lambda.\end{align*} $$

Proof We verify the first containment; the second containment is straightforward.

Let $t \in \widetilde {I}_\lambda $ , and observe $t\in (1-\lambda )A+ \lambda B$ if and only if

(4.1) $$ \begin{align} -(1-\lambda)A \cap \left(\lambda B-t\right) \neq \emptyset. \end{align} $$

We verify the hypotheses of the gap lemma and apply it to the sets $-(1-\lambda )A $ and $ \left (\lambda B-t\right )$ to verify (4.1) for $t\in \widetilde {I}_\lambda $ .

First, we verify that the convex hulls, $\operatorname {\mathrm {conv}}(-(1-\lambda )A)$ and $\operatorname {\mathrm {conv}}(\lambda B-t)$ , are interwoven for $t\in \widetilde {I}_\lambda $ , where we say that two closed intervals are interwoven if they intersect and neither is contained in the interior of the other.

Observe

$$ \begin{align*}\operatorname{\mathrm{conv}}(-(1-\lambda)A) = [ - (1-\lambda) k_1, 0],\end{align*} $$

and

$$ \begin{align*}\operatorname{\mathrm{conv}}(\lambda B-t) = [\lambda k_2 -t, \lambda -t].\end{align*} $$

It follows that the convex hulls are interwoven provided that either

(4.2) $$ \begin{align} \lambda k_2 -t \le -(1-\lambda)k_1 \le \lambda -t \le 0 \end{align} $$

or

(4.3) $$ \begin{align} -(1-\lambda)k_1 \le \lambda k_2 -t \le 0 \le \lambda -t. \end{align} $$

These simplify to the conditions that, from (4.2),

$$ \begin{align*}t\in [\lambda k_2 + (1-\lambda) k_1,\,\, \lambda + (1-\lambda) k_1],\end{align*} $$

or, from (4.3), the condition that

$$ \begin{align*}t\in [\lambda k_2, \,\, \min\{ \lambda, \lambda k_2 + (1-\lambda) k_1\},\end{align*} $$

where we observe that $\min \{ \lambda , \lambda k_2+ (1-\lambda ) k_1,\} = \lambda \iff \lambda \le m.$

Taking the union, we see that the convex hulls are interwoven provided that

$$ \begin{align*}t\in \widetilde{I}_\lambda.\end{align*} $$

The interwoven condition guarantees that sets $-(1-\lambda )A$ and $(\lambda B-t)$ are not contained in each others’ gaps.

Finally, we observe that $\tau (A) = \tau ( C\cap [0, k_1]) \geq \tau (C)$ . In general, thickness may behave badly under intersections, but $\tau (C\cap [0, k_1]) \geq \tau (C)$ since G is the largest gap of C. Similarly, $\tau (B)\geq \tau (C)$ . It follows that

$$ \begin{align*}\tau(-(1-\lambda)A) \tau( \lambda B-t) \geq 1,\end{align*} $$

and the gap lemma applies.

The next step is to show that $\widetilde {I}_\lambda \cap C \neq \emptyset ,$ which will suffice to establish that $((1-\lambda )A + \lambda B) \cap C \neq \emptyset $ by the previous claim. Recall

$$ \begin{align*}\widetilde{I}_\lambda = [\lambda k_2, \lambda] \,\, \bigcup \,\, [\lambda k_2 +(1-\lambda) k_1, \lambda + (1-\lambda)k_1].\end{align*} $$

Proof We prove the claim for $\lambda \geq \frac 12$ . Then, by applying the result to $\widetilde {C} = -C+1$ , we may conclude the claim holds for any $\lambda \in (0,1)$ . Recall $G=(k_1, k_2)$ denotes the largest bounded gap of C.

Let $\lambda \geq \frac 12$ . We consider the cases when $|A| \le |B|$ and $|B| \le |A|$ separately.

Case 1: Suppose first that $|A| \le |B|$ so that $k_1 \le 1-k_2$ or

(4.4) $$ \begin{align} k_2 \le 1-k_1. \end{align} $$

Since $\tau (C)\geq 1$ , it follows that $|A| \geq |G|$ so that $k_1\geq k_2-k_1$ or that

(4.5) $$ \begin{align} k_2 \le 2k_1. \end{align} $$

Observe $k_2 \in \widetilde {I}_\lambda $ . Indeed,

(4.6) $$ \begin{align} \lambda k_2 +(1-\lambda)k_1 \le k_2 \le \lambda + (1-\lambda)k_1, \end{align} $$

where the first inequality holds trivially since such a convex combination of $k_1< k_2$ is bounded above by $k_2$ , and for the second inequality is implied, see by graphing, by (4.4) and (4.5) provided $\lambda \in [\frac 12, 1)$ .

Case 2: Suppose second that $|B| \le |A|$ so that $ 1-k_2 \le k_1$ or

(4.7) $$ \begin{align} 1-k_1 \le k_2. \end{align} $$

Since $\tau (C)\geq 1$ , it follows that $|B| \geq |G|$ so that $1-k_2\geq k_2-k_1$ or that

(4.8) $$ \begin{align} k_2 \le \frac12(1+k_1). \end{align} $$

Again, observe that $k_2 \in \widetilde {I}_\lambda $ . Indeed, as above, we must verify

(4.9) $$ \begin{align} \lambda k_2 +(1-\lambda)k_1 \le k_2 \le \lambda + (1-\lambda)k_1, \end{align} $$

where the first inequality is implied in the same way as above, and the second is implied by noting that, by (4.8), $k_2 \le \frac 12(1+k_1),$ and $\frac 12(1+k_1) \le \lambda + (1-\lambda )k_1 $ provided $\lambda \in [\frac 12, 1]$ .

4.2 Proof of Theorem 2.2

Let T be a set of three distinct vertices in $\mathbb {R}^2$ . We prove that if $C\subset \mathbb {R}$ is compact with $\tau (C)\geq 1$ , then $C\times C$ contains a similar copy of T.

If all three vertices lie on a line, the result follows from Proposition 2.1, after rotating the line to be parallel to one of the axes, if necessary. We assume then that the vertices are not collinear.

Label the vertices of $T\subset \mathbb {R}^2$ by $x=(x_1,x_2)$ , $y=(y_1,y_2)$ , and $z=(z_1,z_2)$ , with corresponding angles $\theta _1, \theta _2, \theta _3$ , with $\theta _3 \geq \theta _i$ for $i=1,2$ . Further, performing a rotation and reflection, assume that T is positioned and labeled so that x and y lie on the x-axis, and $z_2>0$ . It follows that $x_1\le z_1\le y_1$ .

Label $h = z_2$ , $b_1= (z_1-x_1)$ , $b_2= (y_1-z_1)$ , and $b= b_1+b_2$ . It follows that

(4.10) $$ \begin{align} h= \tan{\theta_1}b_1 = \tan{\theta_2}b_2. \end{align} $$

Since $\tau (C)\geq 1$ , it is a consequence of the Newhouse gap lemma that $\Delta (C)$ has non-empty interior. Further, there exists $L>0$ so that $[0,L]\subset \Delta (C)$ .

Choose $c>0$ so that $ch\le L$ and $cb\le L$ . Choose $c'\in (0,c]$ and $t\in \mathbb {R}$ so that $P = \{c'x_1+t, c'z_1+t, c'y_1+t\}\subset C$ ; such a choice is possible by Proposition 2.1. Choose $a,b\in C$ so that $b-a = c'h$ .

Now, the triangle with vertices

(4.11) $$ \begin{align} (c'x_1+t, a), (c'y_1+t, a), (c'z_1+t, b) \end{align} $$

is similar to T and each of the points in (4.11) are in C.

5.1 Setup and table of notation for the proofs of Theorems 2.8 and 2.12

Throughout, $C\subset \mathbb {R}^d$ denotes a compact set generated by the system of balls $\{S_I\}_I$ in the distance $\operatorname {\mathrm {dist}}$ generated by the norm $\|\cdot \|$ as expected: $\operatorname {\mathrm {dist}}(x,C) = \min _{y\in C}\|x-y\|$ . Because thickness is translation and scalar invariant, we assume that $C\subset \bar {B}(0,1)$ , so that $S_\emptyset = \bar {B}(0,1)$ , where $\bar {B}(x,t)= \{x \in \mathbb {R}^d: \|x\| \le t\}$ .

Further, $S_{1_A} =\bar {B}(c_{1_A}, t_{1_A})$ and $S_{1_B} =\bar {B}(c_{1_B}, t_{1_B})$ denote first-generation children and closed balls with centers $c_{1_A}, c_{1_B}$ and radii $t_{1_A}, t_{1_B}$ , respectively, to be chosen in each proof, where the radius of $S_{1_A}$ is assumed to be no more than the radius of $S_{1_B}$ :

$$ \begin{align*}t_{1_A} \le t_{1_B}.\end{align*} $$

In the proof of Theorem 2.12, we further assume that $\operatorname {\mathrm {dist}}$ is Euclidean norm $\|\cdot \|_2$ in order to guarantee that a rotated ball is still a ball in the same norm.

For convenience, we make a table of notation that will be used throughout this section, and we record some relationships between variables.

The following is an immediate consequence of the definition of thickness, Definition 3.2, applied with $n=0$ and i for $1\le i\le k_\emptyset $ , and will be used throughout:

(5.1) $$ \begin{align} \tau\left( C,\{S_I\}\right) \leq \frac{t_{i}}{h_\emptyset}. \end{align} $$

5.2 Proof of Theorem 2.8

Fix $0<r<\frac {1}{2}$ and $0<\lambda \leq \frac {1}{2}$ . Let C be a compact set in $(\mathbb {R}^d,\operatorname {\mathrm {dist}})$ generated by the system of balls $\{S_I\}_I$ such that C is r-uniformly dense and $\tau \left ( C,\{S_I\}\right ) \geq \frac {2(1-\lambda )}{\lambda (1-2r)}$ . Assume $C \subset S_\emptyset = \bar {B}(0,1)$ .

Our proof is motivated by the following key observation. If we were to take two disjoint subsets A, B of C and show that

$$ \begin{align*}\left(\lambda A+ (1-\lambda) B\right) \cap C\neq\emptyset,\end{align*} $$

then there would exists some element $x\in \lambda A+(1-\lambda ) B$ of the form $x=\lambda a+(1-\lambda )b$ for some $a\in A$ and $b\in B$ and $x \in C$ , with $a\neq b$ . Thus, C would contain the $3$ -point convex combination $\{a, \lambda a+(1-\lambda )b,b\}$ . We proceed with this plan in place.

Set $A:= S_{1_A}\cap C$ and $B:=S_{1_B}\cap C$ , where $1\leq 1_A<1_B\leq k_\emptyset $ , and $S_{1_A}$ , $S_{1_B}$ are disjoint first-generation children that are disjoint from all other children. Observe that our choice of A and B imply

(5.2) $$ \begin{align} t_{1_A}\leq t_{1_B}. \end{align} $$

We express A as the compact set generated by $\{S_{1_A,I}\}_I$ . We express B similarly.

As in (3.2), we define

(5.3) $$ \begin{align} h_\emptyset&:=h_\emptyset(C) = \max_{x\in S_\emptyset} \operatorname{\mathrm{dist}}(x,C), \quad h_{1_A}:= h_{1_A}(A) = \max_{x\in S_{1_A}} \operatorname{\mathrm{dist}}(x,A), \quad \text{and} \\ h_{1_B} &:= h_{1_B}(B)= \max_{x\in S_{1_B}}\operatorname{\mathrm{dist}}(x,B).\nonumber \end{align} $$

Recall, it is a consequence of Lemma 3.4 that

(5.4) $$ \begin{align} h_{1_A} \le 2h_\emptyset \,\,\, \text{ and } \,\,\, h_{1_B} \le 2h_\emptyset. \end{align} $$

We now prove a key lemma, which states that the set $\lambda A+(1-\lambda )B$ contains a disk.

Lemma 5.1 The set $\lambda A+(1-\lambda )B$ contains the closed ball

$$ \begin{align*} D & := \bar{B}\left(\lambda c_{1_A} + (1-\lambda) c_{1_B}, t_D\right), \end{align*} $$

where $t_D:= \lambda (1-2r)t_{1_A}+(1-\lambda )t_{1_B}-(1-\lambda )h_{1_B}$ .

Proof of Lemma 5.1

To prove the lemma, we verify the following implications:

(5.5) $$ \begin{align} t \in D & \Rightarrow \left((1-\lambda)S_{1_B}-t\right) \cap \bar{B}\left(-\lambda c_{1_A},t_D-(1-\lambda)t_{1_B}\right)\neq \emptyset \end{align} $$

(5.6) $$ \begin{align} & \Rightarrow \left((1-\lambda)B-t\right) \cap (1-2r)\cdot \left( -\lambda S_{1_A}\right) \neq \emptyset \end{align} $$

(5.7) $$ \begin{align} & \Rightarrow \left( (1-\lambda)B-t\right)\cap\left( -\lambda A\right)\neq \emptyset. \end{align} $$

Since $ \left ((1-\lambda ) B-t\right )\cap (-\lambda A) \neq \emptyset $ if and only if $t\in \lambda A+(1-\lambda ) B$ , this will complete the proof of the lemma.

The first two implications are purely geometric and follow from simple algebraic manipulations. The final implication utilizes the Gap Lemma and relies on Lemma 3.5.

Verifying implication (5.5): First, observe that the radius $t_D-(1-\lambda )t_{1_B}$ is in fact nonnegative. Combining our assumed lower bound on $\tau \left ( C,\{S_I\}\right )$ with the upper bound in (5.1):

(5.8) $$ \begin{align} \frac{2(1-\lambda)}{\lambda(1-2r)}\leq \tau\left( C,\{S_I\}\right) \leq \frac{t_{1_A}}{h_\emptyset}, \end{align} $$

which implies

$$ \begin{align*} 2(1-\lambda)h_\emptyset \leq \lambda (1-2r) t_{1_A}. \end{align*} $$

By (5.4), we know $h_{1_B}\leq 2h_\emptyset $ which means

$$ \begin{align*} (1-\lambda)h_{1_B} \leq \lambda (1-2r) t_{1_A}, \end{align*} $$

so

$$ \begin{align*} t_D-(1-\lambda)t_{1_B}\geq 0. \end{align*} $$

Second, let $t \in D$ and write $t = \lambda c_{1_A} + (1-\lambda ) c_{1_B} + x$ for some

$$ \begin{align*} \|x\| \leq t_D. \end{align*} $$

Recalling $(1-\lambda )S_{1_B}-t = \bar {B}\left ( (1-\lambda )c_{1_B}-t,(1-\lambda )t_{1_B} \right )$ , we wish to show that

$$ \begin{align*} \bar{B}\left( (1-\lambda)c_{1_B}-t,(1-\lambda)t_{1_B} \right) \cap \bar{B}\left(-\lambda c_{1_A},t_D-(1-\lambda)t_{1_B}\right)\neq \emptyset. \end{align*} $$

By the definition of t, this holds if and only if

$$ \begin{align*}\bar{B}\left( -\lambda c_{1_A}-x,(1-\lambda)t_{1_B} \right) \cap \bar{B}\left(-\lambda c_{1_A},t_D-(1-\lambda)t_{1_B}\right) \neq \emptyset,\end{align*} $$

which, shifting everything by $\lambda c_{1_A}$ , holds if and only if

$$ \begin{align*}\bar{B}\left( -x,(1-\lambda)t_{1_B} \right) \cap \bar{B} \left( \vec{0}, t_D-(1-\lambda)t_{1_B} \right) \neq \emptyset,\end{align*} $$

which is true since $\| x\| \le t_D$ .

Verifying implication (5.6): Let $t\in D$ . By (5.5), there exists a

$$ \begin{align*}z\in \left((1-\lambda)S_{1_B}-t\right) \cap \bar{B}\left(-\lambda c_{1_A},t_D-(1-\lambda)t_{1_B}\right).\end{align*} $$

Since $z \in \left ((1-\lambda )S_{1_B}-t\right )$ , we know by definition of $h_{1_B}$ that there exists $y \in (1-\lambda )B-t$ such that

$$ \begin{align*}\|z-y\|\leq (1-\lambda)h_{1_B}.\end{align*} $$

Since $z \in \bar {B}\left (-\lambda c_{1_A},t_D-(1-\lambda )t_{1_B}\right )$ , we know

$$ \begin{align*} \|y-(-\lambda c_{1_A})\| &\leq \|y-z\| + \|z - (-\lambda c_{1_A})\| \\ &\leq (1-\lambda)h_{1_B} + \left( t_D-(1-\lambda)t_{1_B}\right) = \lambda (1-2r)t_{1_A}. \end{align*} $$

Thus, $y\in \left ( (1-2r)\cdot (-\lambda S_{1_A})\right ) \cap \left ( (1-\lambda )B-t\right )$ , and it follows that $\left ( (1-2r)\cdot (-\lambda S_{1_A})\right ) \cap \left ( (1-\lambda )B-t\right )\neq \emptyset $ .

Verifying implication (5.7): Implication (5.7) will follow from an application of the Gap Lemma (Theorem 3.2) applied to the sets $(1-\lambda )B-t$ and $-\lambda A$ , and we need only verify that the hypotheses hold.

First, we calculate the thickness of A and B. By Lemma 3.5,

$$ \begin{align*}\tau\left( A,\{S_{1_A,I}\}\right) \geq \frac{1}{2} \tau\left( C,\{S_I\}\right).\end{align*} $$

Because $\tau \left ( C,\{S_I\}\right ) \geq \frac {2(1-\lambda )}{\lambda (1-2r)}$ ,

$$ \begin{align*}\tau\left( A,\{S_{1_A,I}\}\right) \geq \frac{1-\lambda}{\lambda(1-2r)}.\end{align*} $$

We similarly get $\tau \left ( B,\{S_{1_B,I}\}\right ) \geq \frac {1-\lambda }{\lambda (1-2r)}$ , and since thickness is translation and scalar invariant, we verify (i) of the Gap Lemma for $0\leq \lambda \leq \frac {1}{2}$ as follows:

$$ \begin{align*} \tau\left( -\lambda A,\{-\lambda S_{1_A,I}\}\right) \,\tau\left((1-\lambda)B-t, \{(1-\lambda)S_{1_B,I}-t\}\right) &= \tau\left( A,\{S_{1_A,I}\}\right)\,\tau\left( B,\{S_{1_B,I}\}\right) \\ &\geq \frac{(1-\lambda)^2}{\lambda^2(1-2r)^2} \geq \frac{1}{(1-2r)^2}. \end{align*} $$

Next, by (5.6), we have a t value such that $\left ( (1-2r)\cdot (-\lambda S_{1_A})\right ) \cap \left ( (1-\lambda )B-t\right )\neq \emptyset $ , and (ii) is satisfied.

Next, by assumption (5.2), $t_{1_B}\geq t_{1_A}$ . Hence, $\operatorname {\mathrm {rad}}(S_{1_B}) \geq \operatorname {\mathrm {rad}}(S_{1_A})> r\, \operatorname {\mathrm {rad}}(S_{1_A})$ . Moreover, $(1-\lambda )\geq \lambda $ as $0\leq \lambda \leq \frac {1}{2}$ , and we conclude that

$$ \begin{align*}(1-\lambda)\operatorname{\mathrm{rad}}(S_{1_B})> \lambda r\, \operatorname{\mathrm{rad}}(S_{1_A}) ,\end{align*} $$

so that $\operatorname {\mathrm {rad}} ((1-\lambda )S_{1_B})> r \operatorname {\mathrm {rad}}( \lambda S_{1_A})$ and part (iii) of the Gap Lemma holds.

Lastly, observe that A and B inherit r-uniformity from C and r-uniformity is translation and scalar invariant; hence, (iv) of the Gap Lemma is satisfied.

Because all assumptions of the Gap Lemma hold, we conclude that $(-\lambda A)\cap \left ( (1-\lambda ) B-t\right )\neq \emptyset $ . This concludes implication (5.7).

Next, we show that the closed ball D contains an element of C whenever $\lambda \in [0,\frac {1}{2}]$ , where we recall that

$$ \begin{align*}D= \bar{B}\left(\lambda c_{1_A} + (1-\lambda) c_{1_B}, t_D\right)\end{align*} $$

and

$$ \begin{align*}t_D= \lambda(1-2r)t_{1_A}+(1-\lambda)t_{1_B}-(1-\lambda)h_{1_B}.\end{align*} $$

Lemma 5.2 Let $\lambda \in [0,\frac {1}{2}]$ . Then

$$ \begin{align*}D \cap C \neq \emptyset.\end{align*} $$

Proof Observe $ D \subset S_\emptyset $ . This is true by Lemma 5.1 because $S_\emptyset $ is a convex set and the elements of D are constructed by taking convex combinations of elements in A and B.

Before proceeding, recall (5.4) implies $ h_{1_B} \le 2h_\emptyset $ where $h_\emptyset $ , $h_{1_B}$ are defined in (5.3). Also, recall from (5.8) that

$$ \begin{align*} \frac{2(1-\lambda)}{\lambda(1-2r)}\leq \tau(C) \leq \frac{t_{1_A}}{h_\emptyset}. \end{align*} $$

We first show that the radius of D is greater than $h_\emptyset := \max _{x\in S_\emptyset } \operatorname {\mathrm {dist}}(x,C)$ . Indeed,

$$ \begin{align*} t_D &=\lambda(1-2r)t_{1_A}+(1-\lambda)t_{1_B}-(1-\lambda)h_{1_B} \\ &\geq \lambda(1-2r)t_{1_A}+(1-\lambda)t_{1_A}-(1-\lambda)h_{1_B} &\text{because } t_{1_A} \leq t_{1_B} \text{ by choice of } A, B \\ &= (1-2\lambda r) t_{1_A} -(1-\lambda) h_{1_B} & \\ &\geq (1-2\lambda r)t_{1_A} -2(1-\lambda) h_\emptyset &\text{ by (5.4) and since } \lambda \leq 1-\lambda \\ &= h_\emptyset\left((1-2\lambda r)\frac{t_{1_A}}{h_\emptyset} - 2(1-\lambda)\right) \\ &\geq h_\emptyset\left( (1-2\lambda r)\frac{2(1-\lambda)}{\lambda(1-2r)} - 2(1-\lambda)\right) &\text{ by (5.8)} \\ &= 2 h_\emptyset\left( \frac{ (1-\lambda)^2}{\lambda(1-2r)}\right) \\ &\geq h_\emptyset, \end{align*} $$

where the last inequality follows from $2\frac {(1-\lambda )^2}{\lambda (1-2r)}\geq 1$ for $0< \lambda \leq \frac {1}{2}$ . Thus, $D\subset S_\emptyset $ is a ball of radius larger than $h_\emptyset $ , so there exists some $c\in C$ such that $c\in D$ .

Combining Lemmas 5.1 and 5.2, the disc D is contained in $\lambda A+(1-\lambda )B$ , and since $D \cap C$ is not empty, then there is an element of C in $\lambda A+(1-\lambda )B$ for each $\lambda \in [0,\frac {1}{2}]$ .

In the following proof, as above, we use the notation and definitions from Section 3.

5.3 Proof of Theorem 2.12

Let $C\subset \mathbb {R}^d$ be a compact set generated by a system of balls $\{S_I\}_I$ in the Euclidean norm $\|\cdot \|_2$ . Suppose additionally that C is r-uniformly dense for some $0<r<\frac {1}{2}$ , and without loss of generality assume that $S_\emptyset =\bar {B}(0,1)$ .

Let $\mathcal {T}$ be any triangle. By Lemma 2.11, we know there exists some $\mathcal {T}(\alpha ,\lambda )$ , determined by a fixed $\alpha $ , $\lambda $ in $\mathcal {R}$ , similar to $\mathcal {T}$ . We show that C contains a similar copy of $\mathcal {T}(\alpha ,\lambda )$ when

(5.9) $$ \begin{align} \sqrt{\frac{\alpha^2+(1-\lambda)^2}{\alpha^2+\lambda^2}}\cdot \frac{2}{1-2r} \le \tau\left( C,\{S_I\}\right). \end{align} $$

The key idea of the proof is as follows. Consider the function

$$ \begin{align*}H:\mathbb{R}^2\times\mathbb{R}^2\rightarrow \mathbb{R}^2\end{align*} $$

defined by

$$ \begin{align*} (x,y) \mapsto \lambda x + (1-\lambda)y + \alpha(y-x)^\perp \end{align*} $$

where $(y-x)^\perp = (-x_2+y_2,x_1-y_1)$ . This function takes as input base vertices x and y, makes the convex combination $\lambda x+(1-\lambda )y$ , and sums it with an element of the perp space to output the third vertex $z:=H(x,y)$ of a triangle similar to $\mathcal {T}(\alpha ,\lambda )$ . So, if we had $A\subset C$ and $B\subset C$ disjoint such that $H(A,B)\cap C \neq \emptyset $ , then there would exist distinct points $x=(x_1,x_2)\in A$ and $y=(y_1,y_2)\in B$ forming the base of a triangle similar to $\mathcal {T}(\alpha ,\lambda )$ with the top vertex at the point

$$ \begin{align*}t=(t_1,t_2):=\left( \lambda x_1 +(1-\lambda) y_1, \lambda x_2+(1-\lambda) y_2\right) + \alpha \left( -x_2+y_2, x_1-y_1\right),\end{align*} $$

in C.

Instead of working directly with $H(A,B)$ , we consider the set $f(A)-g(B)$ , where the functions f and g are defined by identifying the above coordinates into two equations and rearranging them as shown below:

$$ \begin{align*} \lambda x_1 -\alpha x_2 -t_1 = -(1-\lambda)y_1 +\alpha y_2 \end{align*} $$

and

$$ \begin{align*} \alpha x_1 +\lambda x_2 - t_2 = \alpha y_1 -(1-\lambda) y_2. \end{align*} $$

Then, we can combine the x-coordinates and define the function

(5.10) $$ \begin{align} f(x_1,x_2):= \left( \lambda x_1-\alpha x_2,\alpha x_1+\lambda x_2\right), \end{align} $$

and similarly combine the y-coordinates and define the function

(5.11) $$ \begin{align} g(y_1,y_2):= \left(-(1-\lambda)y_1-\alpha y_2, \alpha y_1-(1-\lambda)y_2\right). \end{align} $$

Now,

$$ \begin{align*}t\in H(A,B) \text{ if and only if } t\in f(A) - g(B),\end{align*} $$

and it suffices to show that $\left ( f(A) - g(B) \right ) \cap C\neq \emptyset $ for disjoint subsets A and B of C.

To show that $\left ( f(A) - g(B) \right ) \cap C\neq \emptyset $ , we demonstrate that $f(A) -g(B)$ contains a ball D that, in turn, contains a point $c\in C$ . We break the proof into steps.

Step 1. Analyze the functions f and g: Since f is a linear operator on each variable, it can be interpreted as a $2\times 2$ matrix as follows:

$$ \begin{align*} f (x,y) = \begin{pmatrix} \lambda & -\alpha \\ \alpha & \lambda \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}. \end{align*} $$

Such a matrix can be re-written to be a scalar times a rotation matrix:

(5.12) $$ \begin{align} \begin{pmatrix} \lambda & -\alpha \\ \alpha & \lambda \end{pmatrix} = \begin{pmatrix} s_f & 0 \\ 0 & s_f \end{pmatrix} \begin{pmatrix} \cos(\theta_f) & -\sin (\theta_f) \\ \sin(\theta_f) & \cos(\theta_f) \end{pmatrix} :=s_fR_f, \end{align} $$

where

(5.13) $$ \begin{align} s_f = \sqrt{\alpha^2+\lambda^2}, \quad \cos(\theta_f) = \frac{\lambda}{s_f}, \quad \sin(\theta_f) = \frac{\alpha}{s_f}, \end{align} $$

and $\theta _f = \arctan \left (\frac {\alpha }{\lambda }\right )$ .

Similarly, for g, we can write

$$ \begin{align*} g= \begin{pmatrix} -(1-\lambda) & -\alpha \\ \alpha & -(1-\lambda) \end{pmatrix} = \begin{pmatrix} s_g & 0\\ 0 & s_g \end{pmatrix} \begin{pmatrix} \cos(\theta_g) & -\sin (\theta_g) \\ \sin(\theta_g) & \cos(\theta_g) \end{pmatrix} :=s_gR_g, \end{align*} $$

where

(5.14) $$ \begin{align} s_g = \sqrt{\alpha^2+(1-\lambda)^2},\quad\cos(\theta_g) = \frac{-(1-\lambda)}{s_g}, \quad\sin(\theta_g) = \frac{\alpha}{s_g}, \end{align} $$

and $\theta _g = \arctan \left ( \frac {-\alpha }{1-\lambda }\right )+\pi $ .

Now, the assumed lower bound on thickness in (5.9) can be rephrased as

(5.15) $$ \begin{align} \frac{s_g}{s_f}\cdot \frac{2}{1-2r} \le \tau\left( C,\{S_I\}\right). \end{align} $$

Since $0\leq \lambda \leq \frac {1}{2}$ , we note that $ s_f\leq s_g.$

Step 2. Choose disjoint subsets A and B of C: By assumption, there exist closed balls that are first-generation children $S_{1_A}$ and $S_{1_B}$ , $1\leq 1_A<1_B\leq k_\emptyset $ , contained in $\bar {B}\left (0,\frac {1}{2}\right )$ such that $S_{1_A}$ and $S_{1_B}$ are disjoint from all other children. This implies $t_{1_A}\leq t_{1_B}$ , where $t_{1_A}$ , $t_{1_B}$ are the radii of $S_{1_A}$ , $S_{1_B}$ , respectively. Set

$$ \begin{align*}A:= S_{1_A}\cap C \,\,\,\text{ and }\,\,\, B:= S_{1_B}\cap C.\end{align*} $$

Step 3. Determine the thickness of $f(A)$ and $g(B)$ : Recall that C is a compact set constructed by a system of balls $\{S_I\}$ using the Euclidean norm such that $S_\emptyset = \bar {B}(0,1)$ and there exists two first-generation children $S_{1_A}$ , $S_{1_B}$ that are disjoint from all other children; i.e., $S_{1_A}\cap S_i=\emptyset $ for all $1\leq i\leq k_\emptyset $ , $i\neq 1_A$ and similarly for $S_{1_B}$ . Consequently, by applying Lemma 3.5, we know

$$ \begin{align*} \tau\left( A,\{S_{1_A,I}\}\right) \geq \frac{1}{2}\tau\left( C,\{S_I\}\right) \quad\text{and}\quad \tau\left( B,\{S_{1_B,I}\}\right) \geq \frac{1}{2}\tau\left( C, \{S_I\}\right). \end{align*} $$

Moreover, when we take any ball $S_I=\bar {B}( c_I,t_I)$ and apply the function f to it we get $f(S_I)=\bar {B}\left ( s_fR_fc_I, s_ft_I \right )$ which is a scaled rotation of $S_I$ , so it is still a ball in the Euclidean norm, where $R_f$ and $s_f$ are defined in (5.12) and (5.13).

Further, any subset $E= C\cap S_i$ under f will still be an r-uniform subset of thickness $\tau \left ( f(E)\right ) =\tau (E)$ as thickness is rotation, translation, and scalar invariant. A similar result is obtained for the function g. Thus, we conclude that $f(A)$ and $g(B)$ are generated by the system of balls $\{f\left ( S_{1_A,I}\right )\}_I$ and $\{g\left ( S_{1_B,I}\right )\}_I$ , respectively, which are both r-uniformly dense and have thickness given by

(5.16) $$ \begin{align} \tau\left( f(A),\{f(S_{1_A,I})\}\right) &= \tau\left( A, \{S_{1_A,I}\}\right)\geq \frac{1}{2}\tau\left( C,\{S_I\}\right),\quad \text{and} \\ \tau\left( g(B),\{g(S_{1_B,I}\}\right) &= \tau\left( B, \{S_{1_B,I}\}\right) \geq \frac{1}{2}\tau\left( C,\{S_I\}\right). \nonumber \end{align} $$

Step 4. Apply the Gap Lemma to show that $f(A)- g(B)$ contains a disc: We have now arrived at the heart of the argument in which the Gap Lemma is used, but we must first make some geometric observations and verify the hypotheses of the lemma.

We briefly recall that $r\in (0,\frac 12)$ is the uniformity constant, $s_f$ and $s_g$ are the scaling factors defined in (5.13) and (5.14), and $t_{1_A}$ and $t_{1_B}$ are the radii of $S_{1_A}$ and $S_{1_B}$ , respectively. Also, $h_\emptyset = \max _{x\in S_\emptyset } \operatorname {\mathrm {dist}}(x,C)$ , $h_{1_B}=\max _{x\in S_{1_B}}\operatorname {\mathrm {dist}}(x,B)$ were defined in (5.3) and satisfy $h_{1_B} \le 2h_\emptyset $ from (5.4).

Lemma 5.3 The set $f(A)-g(B)$ contains the disc

$$ \begin{align*}D:=\bar{B}\left( f(c_{1_A})-g(c_{1_B}), t_D\right),\end{align*} $$

where $t_D:=(1-2r)s_ft_{1_A}+s_gt_{1_B}-s_gh_{1_B}$ .

Proof To prove the lemma, we verify the following implications:

(5.17) $$ \begin{align} t\in D &\Rightarrow g(S_{1_B}) \cap \bar{B}\left( f(c_{1_A})-t,t_D - s_gt_{1_B} \right)\neq\emptyset \end{align} $$

(5.18) $$ \begin{align} &\Rightarrow g(B) \cap (1-2r)\cdot \left( f(S_{1_A})-t\right) \neq \emptyset \end{align} $$

(5.19) $$ \begin{align} &\Rightarrow \ g(B) \cap \left( f(A)-t\right) \neq \emptyset. \end{align} $$

Since $g(B)\cap \left ( f(A)-t\right ) \neq \emptyset $ if and only if $t\in f(A)-g(B)$ , verifying these implications will complete the proof of the lemma.

The first two implications are purely geometric and follow from simple algebraic manipulations. The final implication utilizes the Gap Lemma and relies on Lemma 3.5. Let $t\in D$ .

Verifying implication (5.17): First, we verify that $t_D - s_gt_{1_B} = (1-2r)s_ft_{1_A}-s_gh_{1_B}$ is nonnegative. Combining the lower bound in (5.15) with the upper bound in (5.1), we have

(5.20) $$ \begin{align} \frac{s_g}{s_f}\cdot\frac{2}{1-2r}\leq \tau\left( C,\{S_I\}\right) \leq \frac{t_{1_A}}{h_\emptyset}, \end{align} $$

which implies

$$ \begin{align*}2s_gh_\emptyset \leq (1-2r)s_ft_{1_A}.\end{align*} $$

By (5.4), we know $h_{1_B} \leq 2h_\emptyset $ , and combining this with the previous line implies that

$$ \begin{align*}s_gh_{1_B} \leq (1-2r)s_ft_{1_A},\end{align*} $$

so that $t_D - s_gt_{1_B}$ is nonnegative.

Moving on, $t\in D$ implies that

(5.21) $$ \begin{align} t=f(c_{1_A})-g(c_{1_B}) +x \end{align} $$

for some $\|x\|_2\leq t_D$ . Recall that $g(S_{1_B}) = \bar {B}\left ( g(c_{1_B}), s_gt_{1_B} \right )$ . We wish to show

$$ \begin{align*}\bar{B}\left( g(c_{1_B}), s_gt_{1_B}\right) \cap \bar{B}\left( f(c_{1_A})-t, t_D - s_gt_{1_B}\right) \neq\emptyset.\end{align*} $$

Substituting (5.21) for t, this holds if and only if

$$ \begin{align*}\bar{B}\left( g(c_{1_B}), s_gt_{1_B}\right) \cap \bar{B}\left( g(c_{1_B})-x, t_D - s_gt_{1_B}\right) \neq\emptyset.\end{align*} $$

Shifting everything by $g(c_{1_B})$ , this holds if and only if

$$ \begin{align*}\bar{B}\left( \vec{0}, s_gt_{1_B}\right) \cap \bar{B}\left( -x,t_D-s_gt_{1_B}\right) \neq\emptyset,\end{align*} $$

which is true since $\|x\|_2\leq t_D$ .

Verifying implication (5.18): Let $t\in D$ , and assume $ g(S_{1_B}) \cap \bar {B}\left ( f(c_{1_A})-t,t_D - s_gt_{1_B} \right )\neq \emptyset $ . (5.17) Let

$$ \begin{align*}z\in g(S_{1_B}) \cap \bar{B}\left( f(c_{1_A})-t, t_D - s_gt_{1_B} \right).\end{align*} $$

Since $z\in g(S_{1_B})$ , we know by definition of $h_{1_B}$ that there exists $y \in g(B)$ such that

$$ \begin{align*}\|y-z\|_2\leq s_g h_{1_B}.\end{align*} $$

Because $z\in \bar {B}\left ( f(c_{1_A})-t,t_D - s_gt_{1_B} \right )$ , we know

$$ \begin{align*} \|y-\left( f(c_{1_A})-t\right)\|_2 &\leq \|y-z\|_2 + \|z-\left( f(c_{1_A})-t\right)\|_2 \\&\leq s_gh_{1_B} + \left( t_D - s_gt_{1_B}\right) \\&= s_gh_{1_B} + \left( (1-2r)s_ft_{1_A}-s_gh_{1_B} \right) \\&= (1-2r)s_ft_{1_A} \\&< s_f t_{1_A}. \end{align*} $$

Recalling that $f(S_{1_A}) = B(s_f R_f c_{1_A}, s_f t_{1_A})$ , we conclude that $y \in g(B) \cap \left ( f(S_{1_A})-t \right )$ , so that $g(B) \cap \left ( f(S_{1_A})-t \right )\neq \emptyset $ .

Verifying implication (5.19): Implication (5.19) follows from applying the Gap Lemma (Theorem 3.2) to the sets $f(A)-t$ and $g(B)$ for $t\in D$ , and we need only verify that the hypotheses hold.

First, using the inequalities in (5.16) and (5.20), we have

$$ \begin{align*} \tau\left( f(A),\{f(S_{1_A,I})\}\right) \tau\left( g(B),\{g(S_{1_B,I})\}\right) \geq \frac{s_g^2}{s_f^2}\cdot \frac{1}{(1-2r)^2} \geq \frac{1}{(1-2r)^2}, \end{align*} $$

for $\alpha $ , $\lambda $ in $\mathcal {R}$ , which verifies (i) of the Gap Lemma.

By implication (5.18), we have $g(B) \cap (1-2r)\cdot \left ( f(S_{1_A})-t\right ) \neq \emptyset $ for $t\in D$ , which is hypothesis (ii) of the Gap Lemma.

By assumption, $\operatorname {\mathrm {rad}}(S_{1_B})\geq \operatorname {\mathrm {rad}}(S_{1_A})$ , which implies $\operatorname {\mathrm {rad}}\left ( f(S_{1_B}) \right )\geq r \, \operatorname {\mathrm {rad}}\left ( g(S_{1_A})\right )$ , and (iii) of the Gap Lemma holds.

Lastly, $f(A)$ and $g(B)$ inherit r-uniformity from C as r-uniformity is translation, rotation, and scalar invariant; hence, (iv) of the Gap Lemma is satisfied.

Because all assumptions of the Gap Lemma hold, $\left ( f(A) - t\right ) \cap g(B)\neq \emptyset $ for $t\in D$ . This concludes implication (5.19).

Step 5. Show D contains an element of C: Recall as in Lemma 5.3 that

$$ \begin{align*}D = \bar{B}\left( f(c_{1_A})-g(c_{1_B}), t_D\right)\end{align*} $$

and

$$ \begin{align*}t_D=(1-2r)s_ft_{1_A}+s_gt_{1_B}-s_gh_{1_B}.\end{align*} $$

Lemma 5.4 Let $\alpha $ and $\lambda $ be elements of $\mathcal {R}$ . Then,

$$ \begin{align*}D\cap C\neq \emptyset.\end{align*} $$

Proof We will show that the center of the disc D lies inside the closed disc $S_\emptyset =\bar {B}(0,1)$ , and the radius $t_D$ is larger than $2h_\emptyset $ , so D contains a disc of radius $h_\emptyset $ inside $S_\emptyset $ . Since this disc ball is contained in $S_\emptyset $ , it must contain a point in C by definition of $h_\emptyset $ . From this, we conclude that D contains a point in C.

We proceed by first analyzing the center and radius of D.

The center of D is $f(c_{1_A})-g(c_{1_B})$ . A consequence of the choice of the sets $S_{1_A}$ and $S_{1_B}$ is that it sufficiently minimizes the distance between $f(c_{1_A})$ and $g(c_{1_B})$ . Recall that, by assumption, $S_{1_A}$ , $S_{1_B}$ are both contained in $\bar {B}\left (0,\frac {1}{2}\right )$ . We can actually take a larger – though uglier – ball, and in this proof, we will suppose that $S_{1_A}$ , $S_{1_B}$ are contained inside the ball $\bar {B}\left (0,\frac {1}{2}+t_1-\frac {h_\emptyset x}{2s_f}\right )$ , where $x=\max \left \{1-\frac {2r}{1-2r},0\right \}$ .

Note: $\bar {B}\left (0,\frac {1}{2}\right )\subset \bar {B}\left ( 0,\frac {1}{2}+t_1-\frac {h_\emptyset x}{2s_f}\right )$ . This can be seen by combining (5.15) and (5.1):

$$ \begin{align*} \frac{s_g}{s_f}\frac{2}{1-2r} &\leq \frac{t_1}{h_\emptyset}. \end{align*} $$

Rearranging then gives

$$ \begin{align*} h_\emptyset \frac{s_g}{s_f}\frac{2}{1-2r} &\leq t_1. \end{align*} $$

Because

$$ \begin{align*} x &<1, & \frac{1}{2} &< s_g, & 1 &< \frac{2}{1-2r}, \end{align*} $$

we can combine the above inequalities to see

$$ \begin{align*} h_\emptyset \frac{x}{2s_f} < h_\emptyset \frac{s_g}{s_f}\frac{2}{1-2r} < t_1. \end{align*} $$

Returning to our analysis of the center $f(c_{1_A})-g(c_{1_B})$ , observe that the centers of $c_{1_A}$ , $c_{1_B}$ of $S_{1_A}$ , $S_{1_B}$ satisfy

$$ \begin{align*}\|c_{1_A}\|\leq \frac{1}{2}-\frac{h_\emptyset x}{2s_f} \quad \text{and}\quad \|c_{1_B}\|\leq \frac{1}{2}-\frac{h_\emptyset x}{2s_f}.\end{align*} $$

Because f, respectively g, rotates and scales by $s_f\leq 1$ , respectively $s_g\leq 1$ , we know

$$ \begin{align*}\|f(c_{1_A})\|\leq \frac{1}{2}s_f -\frac{h_\emptyset x}{2} \quad \text{and}\quad \|g(c_{1_B})\|\leq \frac{1}{2}s_g-\frac{h_\emptyset xs_g}{2s_f} \leq \frac{1}{2}s_g-\frac{h_\emptyset x}{2}.\end{align*} $$

Thus,

(5.22) $$ \begin{align} \|f(c_{1_A})-g(c_{1_B})\|\leq \frac{1}{2}(s_f+s_g) - h_\emptyset x \leq 1-h_\emptyset x. \end{align} $$

where the last inequality is from maximizing $s_f+s_g = \sqrt {\alpha ^2+\lambda ^2}+\sqrt {\alpha ^2+(1-\lambda )^2}$ on $\bar {\mathcal {R}}$ , and the center of D, $f(c_{1_A})-g(c_{1_B})$ , is contained in $S_\emptyset $ .

Next, we analyze the radius of D. Observe that

(5.23) $$ \begin{align} t_D &= (1-2r)s_ft_{1_A} + s_gt_{1_B}-s_gh_{1_B} & \nonumber\\ & \geq \left( (1-2r)s_f + s_g \right) t_1 -2s_gh_\emptyset & t_1\leq t_{1_A},t_{1_B}, \text{ and } h_{1_B}\le 2h_\emptyset \text{ by (5.4)} \nonumber\\ &= h_\emptyset\left[ \left( (1-2r)s_f+s_g \right)\frac{t_1}{h_\emptyset} -2s_g \right] \nonumber\\ &\geq h_\emptyset\left[ \left( (1-2r)s_f + s_g\right) \frac{s_g}{s_f}\frac{2}{(1-2r)}-2s_g\right] & \text{ applying (5.20)} \nonumber\\ &= h_\emptyset \frac{s_g^2}{s_f} \frac{2}{(1-2r)} \nonumber\\ &= 2h_\emptyset \frac{s_g^2}{s_f} + 2h_\emptyset \frac{s_g^2}{s_f} \frac{2r}{(1-2r)}. & \end{align} $$

Claim: $\frac {s_g^2}{s_f} = \frac {\alpha ^2+(1-\lambda )^2}{\sqrt {\alpha ^2+\lambda ^2}}$ is minimized when $\alpha =0$ , $\lambda =\frac {1}{2}$ in $\bar {\mathcal {R}}$ with minimum value $\frac {1}{2}$ .

Then, (5.23) becomes

(5.24) $$ \begin{align} \qquad\qquad\quad\qquad t_D &\geq 2h_\emptyset \frac{s_g^2}{s_f} + 2h_\emptyset \frac{s_g^2}{s_f} \frac{2r}{(1-2r)} &\nonumber \\ &\geq h_\emptyset + h_\emptyset \frac{2r}{1-2r} \\ &> h_\emptyset. \nonumber \end{align} $$

Now if $D\subseteq S_\emptyset $ , then D is itself a ball of radius larger than $h_\emptyset $ by (5.24), so $D\subset S_\emptyset $ contains a point $c\in C$ .

If $D\not \subseteq S_\emptyset $ , then it must be the case that $|f(c_{1_A}) - g(c_{1_B}) + t_D|>1$ , and we will use the lower bound on the radius (5.24) and upper bound on the norm of the center (5.22) below.

If

$$ \begin{align*} \left|\|f(c_{1_A})-g(c_{1_B})\|_2 - t_D\right| \le 1- 2h_\emptyset, \end{align*} $$

then it follows that the disk $D=\bar {B}\left ( f(c_{1_A})-g(c_{1_B}),t_D\right )$ intersects $S_\emptyset =\bar {B}(0,1)$ in such a way that the intersection contains a ball of radius $h_\emptyset $ .

Hence, it remains to show that

(5.25) $$ \begin{align} 1+t_D -2h_\emptyset- \|f(c_{1_A})-g(c_{1_B})\|_2 \geq 0. \end{align} $$

Indeed,

$$ \begin{align*} 1+t_D -2h_\emptyset- \|f(c_{1_A})-g(c_{1_B})\|_2 &\geq 1+\left( h_\emptyset + h_\emptyset\frac{2r}{1-2r}\right) -2h_\emptyset - \left(1-h_\emptyset x\right) \\ &= h_\emptyset \left( \frac{2r}{1-2r} - 1\right) +h_\emptyset x \\ &\geq 0, \end{align*} $$

because $0<r<\frac {1}{2}$ and $x = \max \left \{1-\frac {2r}{1-2r},0\right \}$ . This is where our choice of x in the radius comes from.

Thus, (5.25) is confirmed, and we conclude $D\cap S_\emptyset $ contains a ball of radius $h_\emptyset $ . This ball of radius $h_\emptyset $ is contained in $S_\emptyset $ , so it must contain a point in C by definition of $h_\emptyset $ . Therefore, D contains a point in C.

Combining Lemmas 5.3 and 5.4, the disk D is contained in $f(A)-g(B)$ and $D\cap C\neq \emptyset $ . This implies that there is an element of c in $f(A)-g(B)$ .

6.1 Convex combinations in $\mathbb {R}^d$

As Yavicoli illustrated in [Reference Yavicoli40], compact sets can be constructed using a system of balls, including self-similar sets where each generation of children are equally spaced in a grid. For such an example, the existence of an arithmetic progression is immediate regardless of the thickness as there will be three children in a row (or column) all containing the exact same points through self-similarity.

In what follows, we provide an example of a compact set using the infinity norm $\|\cdot \|_\infty $ which contains a $3$ -term arithmetic progression that is not obvious. Note, one can modify this example to work with other norms by re-arranging the norm balls to their respective optimal packing shape, such as a hexagonal packing arrangement for $\|\cdot \|_2$ ; this may introduce constants that could alter the stated thickness condition.

We first construct a self-similar compact set C, and then we introduce randomness to the construction. Let $S_\emptyset = \bar {B}(0,1)$ be a ball in the infinity norm. Let $n^2$ be the number of children in each generation and $\rho $ be the fixed radius for all of the first-generation children. We take the $n^2$ children to be equidistant in an $n\times n$ grid, where the children in a generation are all distance d apart from each other and distance $d/2$ away from the boundary of $\bar {B}(0,1)$ , as shown in Figure 4.

Figure 4: Parent square $S_\emptyset $ and first-generation children of radius $\rho $ for self-similar compact set C in the infinity norm $\|\cdot \|_\infty $ .

Note that we must have

$$ \begin{align*}2\rho n+nd=2\end{align*} $$

because $S_\emptyset =B(0,1)$ . A compact set such as this can be described by an iterated function system $f_{i}(x) = \rho x + t_{i}$ where the $t_{i}$ are the equidistributed centers of each child for $1\leq i\leq n^2$ . By construction, $f_i\left ( \bar {B}(0,1)\right ) \subset \bar {B}(0,1)$ for all $1\leq i\leq n^2$ . We label these sets $S_{i_1\cdots i_{j}} = f_{i_1}\cdots f_{i_j}(\bar {B}(0,1))$ .

As previously mentioned, such a self-similar set has “obvious” $3$ -term arithmetic progressions and is a trivial illustration of our proof. However, we can introduce randomness to this IFS to make the existence of a $3$ -term arithmetic progression nontrivial.

We modify our previous construction by starting with $S_\emptyset = \bar {B}(0,1)$ and let $f_{i}^k = \rho x + \tilde {t}_{i}^k$ , where $\tilde {t}_{i}^k = t_{i} + u_{i}^k$ such that $|u_{i}^k|< \frac {d}{2}$ is random for all $1\leq i\leq n^2$ and $k\geq 1$ . Even with the added randomness, we see that C is $(2\rho +d)$ -uniformly dense in Figure 5.

Figure 5: Parent square $S_\emptyset $ and first-generation children of radius $\rho $ for randomly perturbed self-similar compact set C in the infinity norm $\|\cdot \|_\infty $ .

Then, by construction

$$ \begin{align*}\max_{x\in S_\emptyset} \operatorname{\mathrm{dist}}\left( x,\bigcup_{i} S_i^k \right)\leq d,\end{align*} $$

because each $S_i^k$ is a maximum distance d apart. As this is repeated at each level but scaled to $\rho $ , in general, we have

$$ \begin{align*}\max_{x\in S_I^K} \operatorname{\mathrm{dist}}\left( x,\bigcup_{i} S_{I,i}^{K,k} \right)\leq \rho^{\ell(I)} d.\end{align*} $$

Consequently,

$$ \begin{align*}h_I(C) \leq d\rho^{\ell(I)} + d\rho^{\ell(I)+1}+ d\rho^{\ell(I)+2}+\cdots = \frac{d\rho^{\ell(I)}}{1-\rho}.\end{align*} $$

By construction,

$$ \begin{align*}\min_i \operatorname{\mathrm{rad}} (S_{I,i}^{K,k}) = \rho ^{\ell(I)+1}.\end{align*} $$

This gives a lower bound on the thickness of our compact set:

(6.1) $$ \begin{align} \tau\left( C,\{S_I\}\right)\geq \frac{\rho(1-\rho)}{d}. \end{align} $$

Corollary 2.9 then gives the existence of $3$ APs in these compact sets C when $0 < 2\rho + d < \frac {1}{2}$ and $\tau \left ( C,\{S_I\}\right ) \geq \frac {2}{1-4\rho -2d}$ . In particular, we can take $n=10$ , $\rho =0.095$ , and $d=0.01$ . Then, by inequality (6.1),

$$ \begin{align*}\tau\left( C,\{S_I\}\right) \geq \frac{0.095(1-0.095)}{0.01} = 8.5975,\end{align*} $$

which is larger than the Corollary 2.9 requirement of

$$ \begin{align*}\frac{2}{1-4\rho-2d} = \frac{2}{1-4\cdot 0.095 -2\cdot 0.1} = 3.\bar{3}.\end{align*} $$

Thus, C contains a $3$ -term arithmetic progression. In fact, we can apply Theorem 2.8 to see that C contains a homothetic copy of all convex combinations of the form $\{a,\lambda a+(1-\lambda )b, b\}$ for $\lambda \in [0.27938814,0.5]$ .

Other n, $\rho $ , and d values can be chosen to construct a different C which also contain $3$ -term arithmetic progressions or convex combinations.

Additional examples can be constructed using Yavicoli’s method [Reference Yavicoli40, Section 4.1, Lemma 7].

6.2 Triangles in $\mathbb {R}^2$

We now construct a compact set $C\in \mathbb {R}^2$ using the Euclidean norm to which we can apply Theorem 2.12 and Corollary 2.14 to get the existence of nondegenerate $3$ -point configurations. Theorem 2.8 and Corollary 2.9 will also apply and give the existence of linear configurations.

We begin by taking the best-known packing of $55$ congruent circles inside the circle $S_\emptyset =\bar {B}(0,1)$ , as determined by [Reference Graham, Lubachevsky, Nurmela and Östergård13] and illustrated in Figure 6.

Figure 6: Best known packing [Reference Graham, Lubachevsky, Nurmela and Östergård13] of $55$ congruent circles in $\bar {B}(0,1)$ .

Observe that these circles, call them $S_1, S_2, \cdots S_{55}$ , are in a hexagonal packing arrangement, the most dense packing arrangement for circles. This forces all the congruent circles, which will become our first-generation children, to have radii $\rho \approx 0.12179$ . At this moment, notice that

$$ \begin{align*} \max_{x\in \bar{B}(0,1)}\operatorname{\mathrm{dist}}\left( x,\cup_1^{55} S_i\right)> \rho, \end{align*} $$

and this would cause our thickness to be less than or equal to $1$ . Hence, we add $30$ additional congruent circles $S_{56},\ldots ,S_{85}$ around the edges, shown in Figure 7. This provides the better bound

(6.2) $$ \begin{align} \max_{x\in \bar{B}(0,1)}\operatorname{\mathrm{dist}}\left( x,\cup_1^{85} S_i\right) = \frac{2-\sqrt{3}}{\sqrt{3}}\rho. \end{align} $$

Figure 7: Best known packing of $55$ congruent circles in $\bar {B}(0,1)$ with congruent circles added to minimize $\max _{x\in \bar {B}(0,1)}\operatorname {\mathrm {dist}}(x,C)$ .

Using this structure, we can construct the compact set $C\subset \mathbb {R}^d$ by translating, scaling by $\rho $ , (and optionally rotating) a copy of Figure 7 into each $S_i$ and repeating at every level of the construction. The resulting compact set C generated by $\{S_I\}_I$ is $\frac {2+\sqrt {3}}{\sqrt {3}}\rho $ -uniformly dense, or $0.26243$ -uniformly dense. Using self-similarity with inequality (6.2) gives

$$ \begin{align*} \max_{x\in S_I}\operatorname{\mathrm{dist}}(x,C) \leq \frac{2-\sqrt{3}}{\sqrt{3}} \frac{\rho^{\ell(I)}}{1+\rho}. \end{align*} $$

Because

$$ \begin{align*} \min_i\operatorname{\mathrm{rad}}(S_{I,i}) = \rho^{\ell(I)+1}, \end{align*} $$

we know

$$ \begin{align*} \tau\left( C,\{S_I\}\right) \geq \frac{\rho(1+\rho)}{\frac{2-\sqrt{3}}{\sqrt{3}} \rho}\approx 7.25137. \end{align*} $$

While this result establishes the existence of compact sets in $\mathbb {R}^2$ using the Euclidean norm of sufficient thickness, it does not satisfy the requirement in Theorem 2.12 that there are two first-generation children that are disjoint from the others. We remedy this by constructing a new compact set $\tilde {C}$ from the compact set C generated by $\{S_I\}_I$ by taking two first-generation balls $S_{1_A}$ , $S_{1_B}$ in $\bar {B}(0,\frac {1}{2})$ , as illustrated in Figure 8, and scaling them, and all their children, by a factor of $\gamma $ for $0<\gamma <1$ ; e.g., $\gamma \cdot S_{1_A} = \bar {B}(c_{1_A},\gamma t_{1_A})$ .

Figure 8: Modifying construction so two congruent circles in $\bar {B}(0,\frac {1}{2})$ are disjoint.

We then construct the compact set $\tilde {C}$ as before with the new generating system of balls $\{\tilde {S}_I\}$ . This then shifts inequality (6.2) to become

$$ \begin{align*} \max_{x\in\bar{B}(0,1)}\operatorname{\mathrm{dist}}\left( x,\cup_1^{81}\tilde{S}_i\right) \leq \frac{2-\sqrt{3}}{\sqrt{3}}\rho +(1-\gamma)\rho = \left(\frac{2}{\sqrt{3}}-\gamma\right)\rho. \end{align*} $$

For words $I\neq \emptyset $ and not starting with $1_A$ or $1_B$ , we still have

$$ \begin{align*} \max_{x\in \tilde{S}_I}\operatorname{\mathrm{dist}}(x,\tilde{C}) \leq \frac{2-\sqrt{3}}{\sqrt{3}} \frac{\rho^{\ell(I)}}{1+\rho} \quad\text{and}\quad \min_i\operatorname{\mathrm{rad}}(\tilde{S}_{I,i}) = \rho^{\ell(I)+1}. \end{align*} $$

However, for words $I\neq \emptyset $ starting with $1_A$ or $1_B$ ,

$$ \begin{align*} \max_{x\in \tilde{S}_I}\operatorname{\mathrm{dist}}(x,\tilde{C}) \leq \gamma \frac{2-\sqrt{3}}{\sqrt{3}} \frac{\rho^{\ell(I)}}{1+\rho} \quad\text{and}\quad \min_i\operatorname{\mathrm{rad}}(\tilde{S}_{I,i})=\gamma\rho^{\ell(I)+1}, \end{align*} $$

and lastly for $I=\emptyset $ ,

$$ \begin{align*} \max_{x\in \tilde{S}_\emptyset}\operatorname{\mathrm{dist}}(x,\tilde{C}) \leq \left(1-\gamma\right)\rho + \frac{2-\sqrt{3}}{\sqrt{3}} \frac{\rho}{1+\rho} \quad\text{and}\quad \min_i \operatorname{\mathrm{rad}}(\tilde{S}_i) = \gamma\rho. \end{align*} $$

Consequently, the thickness of $\tilde {C}$ generated by this system is

$$ \begin{align*} \tau\left( \tilde{C},\{\tilde{S}_I\}\right) \geq \frac{\gamma\rho}{\left(1-\gamma\right)\rho + \frac{2-\sqrt{3}}{\sqrt{3}} \frac{\rho}{1+\rho}}. \end{align*} $$

In the case of $\gamma =0.99999$ , $\tau (\tilde {C},\{\tilde {S}_I\}) \geq 7.25077$ , so

$$ \begin{align*} \tau(\tilde{C},\{\tilde{S}_I\}) \geq \sqrt{ \frac{\alpha^2+(1-\lambda)^2}{\alpha^2+\lambda^2} }\cdot\frac{2}{1-2\cdot 0.262421} \end{align*} $$

for all $(\alpha ,\lambda )\in \mathcal {R}\cap (0,\sqrt {3}/2]\times [3/10,1/2]$ . Thus, by Theorem 2.12, $\tilde {C}$ contains a similar triangle to $\mathcal {T}(\alpha ,\lambda )$ for all $(\alpha ,\lambda )\in \mathcal {R}\cap (0,\sqrt {3}/2]\times [3/10,1/2]$ . In particular, $\tilde {C}$ contains a similar copy of an equilateral triangle.