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A pinned Mattila-Sjölin type theorem for product sets

Zack Boone and Eyvindur Ari Palsson
Abstract.

We generalize a result of McDonald and Taylor which concerns the size of the tuples of edge lengths in the set C1×C2C_{1}\times C_{2} utilizing the notion of thickness. Specifically, we show that C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} compact sets with thickness satisfying τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1, then the edge lengths in C1×C2C_{1}\times C_{2} corresponding to any pinned finite tree configuration has non-empty interior. Originally proven for Cantor sets on the real line by McDonald and Taylor, we use the notion of thickness introduced by Falconer and Yavicoli which allows us to generalize the result of McDonald and Taylor to compact sets in d\mathbb{R}^{d}.

Key words and phrases:
Newhouse thickness, distance sets, point configurations
2010 Mathematics Subject Classification:
Primary 28A75, Secondary 52C10
The work of the second listed author was supported in part by Simons Foundation Grant #360560.

1. Introduction

Falconer’s distance problem conjectures a relationship between the structure of a set EE and the size of its distance set Δ(C):={|xy|:x,yC}\Delta(C):=\{|x-y|:x,y\in C\}. The notion of structure and size used by Falconer is Hausdorff dimension and Lebesgue measure denoted dimH()\dim_{H}(\cdot) and ()\mathcal{L}(\cdot) respectively. In [5], Falconer proved that if CdC\subset\mathbb{R}^{d} is compact with dimH(C)>d+12\dim_{H}(C)>\frac{d+1}{2} then (Δ(C))>0\mathcal{L}(\Delta(C))>0. The conjecture is that we can lower the threshold d+12\frac{d+1}{2} to d2\frac{d}{2}. Many improvements have been made on this threshold. When d3d\geq 3, the best known threshold is d22d1\frac{d^{2}}{2d-1} which was first achieved by Du, Guth, Ou, Wang, Wilson and Zhang in [2] when d=3d=3 and generalized to higher dimensions by Du and Zhang in [4]. Interestingly, this threshold can be improved for even dimensions. When restricting dd to even integers and for d2d\geq 2, the current threshold is d2+14\frac{d}{2}+\frac{1}{4}. For d=2d=2, this threshold was obtained by Guth, Iosevich, Ou and Wang in [10] and generalized to higher even dimensions by Du, Iosevich, Ou, Wang and Zhang in [3].

Another version of the Falconer distance problem, the pinned distance problem, has also seen much attention. Instead of looking at all possible distances from any two elements in a set, we fix one element in the set and only look at the distances the exist from this fixed point, i.e. for xCx\in C the pinned distance set is Δx(C):={|xy|:yC}\Delta_{x}(C):=\{|x-y|:y\in C\}. We can then ask what dimensional requirements are needed so that (Δx(C))>0\mathcal{L}(\Delta_{x}(C))>0. This is a stronger version of the Falconer distance problem since we can write

Δ(C)=xCΔx(C)\Delta(C)=\bigcup_{x\in C}\Delta_{x}(C)

so that if Δx(C)\Delta_{x}(C) has positive Lebesgue measure for some xCx\in C, then we automatically get that Δ(C)\Delta(C) has positive Lebesgue measure. This problem was first investigated by Peres and Schlag in [24] where they proved

dimH({xd:(Δx(E))=0})d+1dimH(E)\dim_{H}\left(\left\{x\in\mathbb{R}^{d}:\mathcal{L}(\Delta_{x}(E))=0\right\}\right)\leq d+1-\dim_{H}(E)

which, in particular, implies that if dimH(E)>d+12\dim_{H}(E)>\frac{d+1}{2} then there exists xEx\in E such that (Δx(E))>0\mathcal{L}(\Delta_{x}(E))>0. Refinements on this problem have been made, particularly when restricted to the plane. For example, Shmerkin in [25] proved that for E2E\subset\mathbb{R}^{2} one has

dimH({x2:dimH(Δx(E))<1})1\dim_{H}\left(\{x\in\mathbb{R}^{2}:\dim_{H}(\Delta_{x}(E))<1\}\right)\leq 1

when EE is a set satisfying 1<dimH(E)=dimP(E)1<\dim_{H}(E)=\dim_{P}(E), where dimP()\dim_{P}(\cdot) denotes packing dimension. Moreover, this was improved by Keleti and Shmerkin in [15] to the relaxed setting of when dimP(E)2dimH(E)1\dim_{P}(E)\leq 2\dim_{H}(E)-1.

Due to Mattila [19], one method to prove the Falconer distance problem is to show that there exists a measure μ\mu on EE such that

1(Sd1|μ^(rω)dω)2rd1𝑑r<,\int_{1}^{\infty}\left(\int_{S^{d-1}}|\widehat{\mu}(r\omega)d\omega\right)^{2}r^{d-1}dr<\infty,

and this approach has been called the Mattila integral approach. Surprisingly, the dimensional thresholds for the pinned and non-pinned versions of the distance problem match when one studies them via the Mattila integral approach. This is due to Liu in [17] where he establishes an identity for L2L^{2} integrals.

A variant of the distance problem asks to find dimensional conditions on CC such that not only will it be true that (Δ(C))>0\mathcal{L}(\Delta(C))>0 but that Δ(C)\Delta(C) will contain an interval. Let ()(\cdot)^{\circ} denote the interior of a set. This problem was also investigated by Peres and Schlag in [24] where they proved a pinned version of this problem. In particular, they showed

dimH({xd:(Δx(E))=})d+2dimH(E)\dim_{H}\left(\left\{x\in\mathbb{R}^{d}:(\Delta_{x}(E))^{\circ}=\emptyset\right\}\right)\leq d+2-\dim_{H}(E) (1)

which shows that if dimH(E)>d+22\dim_{H}(E)>\frac{d+2}{2} then one can find xEx\in E such that Δx(E)\Delta_{x}(E) contains an interval. Iosevich and Liu in [12] obtained the bound (d+1)2d1d+1d1dimH(E)\frac{(d+1)^{2}}{d-1}-\frac{d+1}{d-1}\dim_{H}(E) which improves on Equation (1) when dimH(E)>d+32\dim_{H}(E)>\frac{d+3}{2}. In the non-pinned setting, Mattila and Sjölin in [18] refined this result when they proved that if dimH(C)>d+12\dim_{H}(C)>\frac{d+1}{2} then Δ(C)\Delta(C) contains an interval, now known as the Mattila-Sjölin theorem. Finding when some kind of geometric operation on a set has non-empty interior is an active research area. For example Iosevich, Mourgoglou, and Taylor in [13] generalized the Mattila-Sjölin theorem to the case where the set ΔB(E)={xyB:x,yE}\Delta_{B}(E)=\{\|x-y\|_{B}:x,y\in E\} has non-empty interior, where B\|\cdot\|_{B} is the metric induced by the norm defined by a symmetric bounded convex body BB with a smooth boundary and everywhere non-vanishing Gaussian curvature. Improvements on this threshold have been made by Koh, Pham, and Shen in [16] for product sets, i.e. sets of the form C=E××E=EddC=E\times\cdots\times E=E^{d}\subset\mathbb{R}^{d} where EE\subset\mathbb{R} compact.

Another example of this, which serves as the basis of this paper, is Theorem 1.8 in [20] proven by McDonald and Taylor which will be stated below, and it determines the amount of edge lengths one can find in the set K1×K2K_{1}\times K_{2} when K1,K2K_{1},K_{2}\subset\mathbb{R} are Cantor sets. Instead of taking Hausdorff dimension to be the structure of the set, McDonald and Taylor take thickness, denoted by τ\tau, to be the structure of their sets. The definition of thickness is defined below.

First introduced by Newhouse in [21], he gave a definition of thickness specifically for Cantor sets on the real line. Newhouse used this notion of thickness to prove that if K1,K2K_{1},K_{2}\subset\mathbb{R} are Cantor sets such that τ(K1)τ(K2)>1\tau(K_{1})\tau(K_{2})>1 then K1K2K_{1}\cap K_{2}\neq\emptyset, a result now known as the Newhouse Gap Lemma. Problems of determining elements in distance sets can be converted into problems of determining intersections of sets, so having a notion of structure which allows one to determine when two sets intersect can be of high importance to questions related to distance sets. Falconer and Yavicoli in [6] generalized both the definition of thickness and the Newhouse Gap Lemma to arbitrary compact sets in d\mathbb{R}^{d}. Before stating the definition we describe the "gaps" of a compact set.

Definition 1.1.

Given a compact set CdC\subset\mathbb{R}^{d}, we define {Gn}n=1\{G_{n}\}_{n=1}^{\infty} to be the, at most, countably many open bounded path-connected components of CCC^{C} (the complement of CC) and EE to be the unbounded open path-connected component of CCC^{C} (except when d=1d=1 when EE consists of two unbounded intervals). We call EE the unbounded gap of CC and {Gn}n=1\{G_{n}\}_{n=1}^{\infty} the unbounded gaps of CC. Furthermore, we assume that the sequence of bounded gaps {Gn}n=1\{G_{n}\}_{n=1}^{\infty} is ordered by non-increasing diameter.

We will write dist\operatorname{dist} for the usual distance between points of non-empty subsets of d\mathbb{R}^{d} and diam\operatorname{diam} for the diameter of a non-empty subset of d\mathbb{R}^{d}.

Definition 1.2 (Thickness in d\mathbb{R}^{d}).

The thickness of CC is

τ(C):=infndist(Gn,1in1GiE)diam(Gn),\tau(C):=\inf_{n\in\mathbb{N}}\frac{\operatorname{dist}\left(G_{n},\bigcup_{1\leq i\leq n-1}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})},

provided that EE is not the only path-connected component of CCC^{C}. When the only complementary path-connected component is EE, we define

τ(C)={+if C0if C=\tau(C)=\begin{cases}+\infty&\text{if $C^{\circ}\neq\emptyset$}\\ 0&\text{if $C^{\circ}=\emptyset$}\end{cases}

We say CC is thick if τ(C)>0\tau(C)>0.

When restricted to \mathbb{R}, this definition of thickness is equivalent to one given by Newhouse. With this definition in mind, the kinds of sets that we are considering are ones that look like they have holes which are been punched out of them. Another way of putting it is that the sets we consider need to have some kind of border. Thus, one drawback to this definition is that there are sets which are of importance in geometric measure theory but do not fit into the framework of Definition 1.2. One example is the four corner Cantor set (and its generalizations) because it does not have any bounded gaps. Therefore it should be noted that recently, Yavicoli in [27] gave a different definition of thickness and proved a type of gap lemma which did allow her to study sets such as the four corner Cantor set. Nevertheless, Falconer and Yavicoli in [6] used the definition of thickness given above to give a generalized version of the Newhouse Gap Lemma which is stated as follows:

Theorem 1.3 (Gap lemma in d\mathbb{R}^{d}).

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} be compact such that neither of them is contained in a gap of the other and τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Then C1C2C_{1}\cap C_{2}\neq\emptyset.

Before stating the result of McDonald and Taylor, we provide some preliminary definitions on graphs and specifically, trees.

Definition 1.4 (Graphs).

A (finite) graph is a pair G=(V,E)G=(V,E) where VV is a (finite) set and EE is a set of 2-element subsets of VV. If {i,j}E\{i,j\}\in E we say ii and jj are adjacent and write iji\sim j.

Definition 1.5 (Chain and tree graphs).

The k-chain is the graph on the vertex set {1,,k+1}\{1,\cdots,k+1\} with iji\sim j if and only if |ij|=1|i-j|=1. A tree is a connected, acyclic graph; equivalently, a tree is a graph in which any two vertices are connected by exactly one path. If TT is a tree, the leaves of TT are the vertices which are adjacent to exactly one other vertex of TT.

Definition 1.6 (G distance sets).

Let GG be a graph on the vertex set {1,,k+1}\{1,\cdots,k+1\} with mm edges, and let \sim denote the adjacency relation on GG. Define the GG distance set of BB to be

ΔG(B):={(|xixj|)ij:x1,,xk+1B,xixj},\Delta_{G}(B):=\left\{\left(|x^{i}-x^{j}|\right)_{i\sim j}:x^{1},\cdots,x^{k+1}\in B,x^{i}\neq x^{j}\right\},

where (ai,j)ij(a_{i,j})_{i\sim j} denotes a vector in m\mathbb{R}^{m} with coordinates indexed by the edges of GG.

The usual definition of the distance set is a 11-chain except with the element 0 removed, i.e. if we make GG a 11-chain then Δ(B)=ΔG(B){0}\Delta(B)=\Delta_{G}(B)\cup\{0\}. The element 0 is removed from GG distance sets because we require non-degeneracy conditions in some of the later proofs. McDonald and Taylor in [20] used Newhouse’s definition to prove the following theorem:

Theorem 1.7 (Theorem 1.8 in [20]).

Let K1,K2K_{1},K_{2}\subset\mathbb{R} be Cantor sets satisfying τ(K1)τ(K2)>1\tau(K_{1})\tau(K_{2})>1. For any finite tree TT, the set ΔT(K1×K2)\Delta_{T}(K_{1}\times K_{2}) has non-empty interior.

The goal of this paper is to extend the above theorem to compact sets in d\mathbb{R}^{d} satisfying the same thickness condition, and also extending the work to a pinned tree distance set. This theorem uses and extends the work from Simon and Taylor in [26]. In that paper, they also prove a Mattila-Sjölin type theorem by considering thickness instead of Hausdorff dimension. In particular, Corollary 2.12 in [26] states that for K1,K2K_{1},K_{2}\subset\mathbb{R} Cantor, if τ(K1)τ(K2)>1\tau(K_{1})\tau(K_{2})>1 then

(Δx(K1×K2)).\left(\Delta_{x}(K_{1}\times K_{2})\right)^{\circ}\neq\emptyset.

Extensions to pinned tree distance sets for Cantor sets in \mathbb{R} can also be concluded from the work of McDonald and Taylor, which is discussed below.

A key ingredient in proving statements about non-empty interior of tree distance sets is having access to a strong statement about pinned distance sets, which is presented next and is adapted and extended from Lemma 3.5 in [20].

Theorem 1.8.

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} be compacts sets satisfying τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Then for any x0C1×C2x^{0}\in C_{1}\times C_{2} there exists an open neighborhood SS about x0x^{0} such that

xSΔx(C1×C2)\bigcap_{x\in S}\Delta_{x}(C_{1}\times C_{2})

has non-empty interior.

The proof of this theorem is given in Section 5. We note that this result not only states existence of such an x0x^{0}, but rather we can take any x0C1×C2x^{0}\in C_{1}\times C_{2} and obtain this result. Furthermore, this theorem gives a Mattila-Sjölin type result in the pinned distance setting. Indeed, since we are taking a neighborhood around x0x^{0}, we immediately obtain the following corollary:

Corollary 1.9.

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} be compacts sets satisfying τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Then for any xC1×C2x\in C_{1}\times C_{2} we have

(Δx(C1×C2)).\left(\Delta_{x}(C_{1}\times C_{2})\right)^{\circ}\neq\emptyset.

In Section 2 we construct sets C1×C2C_{1}\times C_{2}, where C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d}, for which the above theorem can be applied and such that the dimension of C1×C2C_{1}\times C_{2} approaches 2d12d-1. Then note that when d=1d=1 (so C1×C22)C_{1}\times C_{2}\subset\mathbb{R}^{2}), the threshold of 2d12d-1 turns into 11, which matches the threshold in the Falconer distance conjecture and both the pinned Falconer problem as well as the Mattila-Sjölin variant. This gives some evidence to the conjecture that we should expect the same threshold for the pinned Mattila-Sjölin question. We also point out that, since we are looking at the product of sets, the threshold that comes from Equation (1) is 2d+22\frac{2d+2}{2}. Thus, we can find a sequence of sets with dimension converging to 2d12d-1, which falls below 2d+22\frac{2d+2}{2} for d=1d=1 and matches it when d=2d=2. However, for d3d\geq 3, the threshold 2d+22\frac{2d+2}{2} is lower than 2d12d-1 (and thus, better).

Before stating the main result, we fix some notation. Let TkT^{k} denote a tree with kk edges, i.e. a tree with k+1k+1 vertices. Then we define the pinned tree distance set with kk edges by

ΔTxk(C):={(|xixj|)ij:x1,,xk+1C,xixj,x=xi for some i}.\Delta_{T_{x}^{k}}(C):=\{(|x^{i}-x^{j}|)_{i\sim j}:x^{1},\cdots,x^{k+1}\in C,x^{i}\neq x^{j},x=x^{i}\text{ for some }i\}.

Using the above theorem, we can extend the work of McDonald and Taylor to the following, which is the main result of this paper:

Theorem 1.10.

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} be compact sets such that τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Then for any finite pinned tree TxkT_{x}^{k} of k+1k+1 vertices, the set ΔTxk(C1×C2)\Delta_{T_{x}^{k}}(C_{1}\times C_{2}) has non-empty interior.

Note that the requirements for Theorem 1.10 are no more stringent than the requirements for Theorem 1.8. Thus, we again can construct a wide range of sets of the form C1×C22dC_{1}\times C_{2}\in\mathbb{R}^{2d} of with dimension going down to dimension 2d12d-1.

To conclude the introduction, we note that there have been numerous other results on establishing the existence of configurations of a set from the perspective of Hausdorff dimension. For example, Palsson and Romero Acosta in [23] establish that when dimH(E)>23d+1\dim_{H}(E)>\frac{2}{3}d+1 for d4d\geq 4, then the set of congruence classes of triangles formed by triples of points in EE will have non-empty interior, with further improvements in low dimensions in a very recent preprint [22].

As another variant of the problem, we can describe distances corresponding to a tree as a function Φ\Phi, which is a vector-valued function describing the distances between a set of points. There has also been work when allowing Φ\Phi to describe more general operations, and determining when the resulting set under Φ\Phi has non-empty interior. This was done by Greenleaf, Iosevich, and Taylor in [7] where they considered a suitable class of functions Φ:d×dk\Phi:\mathbb{R}^{d}\times\mathbb{R}^{d}\rightarrow\mathbb{R}^{k} to establish the resulting 2-point configuration set ΔΦ(E):={Φ(x,y):x,yE}\Delta_{\Phi}(E):=\{\Phi(x,y):x,y\in E\} has non-empty interior given lower bounds on dimH(E)\dim_{H}(E) which depend on Φ\Phi. In [9], the same authors built upon this work to establish similar results when considering more general k-point configurations and in their most recent paper [8] they can even handle triangles matching the result in [23].

The proof of Theorem 1.10 will be done by induction on kk (the number of edges in the tree), and heavily relying on Theorem 1.8. Proving Theorem 1.8 will largely follow the same strategy as is done in [20], however there are clear difficulties that arise when going to higher dimensions, and these issues will be discussed throughout this document


2. Application to Sierpiński carpets

Here, we give an example of a sequence of sets, which will be Sierpiński carpets, for which we can apply Corollary 1.10 and for which the dimension of the sets in this sequence converges to 2d12d-1. For a more detailed discussion and construction of these sets see [6]. For simplicity, we will assume each function is the iterated function system of the following Sierpiński carpets will have the same contraction ratios. That is, let n3n\in\mathbb{N}_{\geq 3} be an odd number and set

D={𝕚=(i1,,id):1ikn with (i1,,id)((n+1)/2,,(n+1)/2)}.D=\{\mathbb{i}=(i_{1},\cdots,i_{d}):1\leq i_{k}\leq n\text{ with }(i_{1},\cdots,i_{d})\neq((n+1)/2,\cdots,(n+1)/2)\}.

Then the family of maps {f𝕚:𝕚D}\{f_{\mathbb{i}}:\mathbb{i}\in D\}, where f𝕚:ddf_{\mathbb{i}}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} and is defined by

f𝕚(x1,,xd)=(x1+i11n,,xd+id1n),f_{\mathbb{i}}(x_{1},\cdots,x_{d})=\left(\frac{x_{1}+i_{1}-1}{n},\cdots,\frac{x_{d}+i_{d}-1}{n}\right),

forms an iterated function system, so that there exists a unique and non-empty compact CdC\subset\mathbb{R}^{d} such that C=𝕚Df𝕚(C)C=\bigcup_{\mathbb{i}\in D}f_{\mathbb{i}}(C). Furthermore, dimH(C)=s\dim_{H}(C)=s where ss is the unique number satisfying #D(1n)s=1\#D\left(\frac{1}{n}\right)^{s}=1. For verification of these facts, see [11]. This means dimH(C)=log(#D)log(n)\dim_{H}(C)=\frac{\log(\#D)}{\log(n)}. The set CC corresponds to initially breaking up the unit cube [0,1]d[0,1]^{d} into ndn^{d} boxes and removing the middle one, then iteratively performing this process on the remaining boxes. Since we are only removing the middle one, we have #D=nd1\#D=n^{d}-1. Thus, dimH(C)=log(nd1)log(n)\dim_{H}(C)=\frac{\log(n^{d}-1)}{\log(n)}. Furthermore, in [6] it is shown that τ(C)=n12d.\tau(C)=\frac{n-1}{2\sqrt{d}}.

Another kind of Sierpiński carpet that we can calculate the Haudsorff dimension and thickness of is one in which instead of just removing the middle box, we only keep the "outer boundary" of boxes. That is, for n5n\geq 5 and nn odd, we remove the middle (n2)d(n-2)^{d} squares and keep the boxes on the outer edge, then perform this process iteratively on the remaining squares. For this set, say CC, we have dimH(C)=log(nd(n2)d)log(n)\dim_{H}(C)=\frac{\log(n^{d}-(n-2)^{d})}{\log(n)}. By a similar discussion as found in [6], the thickness is then

τ(C)=1nd(n2n)2=1(n2)d.\tau(C)=\frac{\frac{1}{n}}{\sqrt{d\left(\frac{n-2}{n}\right)^{2}}}=\frac{1}{(n-2)\sqrt{d}}.

Thus, if we take CnC_{n} to be a Sierpiński carpet of the first kind described, with contraction ratio 1/n1/n, and CmC_{m} to be a Sierpiński carpet of the second kind described, with contraction ratio 1/m1/m. Then if 1<τ(Cn)τ(Cm)1<\tau(C_{n})\tau(C_{m}), this yields 2d<n1m22d<\frac{n-1}{m-2}. If nn is fixed, this inequality is satisfied, for example, by taking mm to be largest odd integer such that m<n1+4d2dm<\frac{n-1+4d}{2d}. So an nn\rightarrow\infty, we choose mnm_{n} to a sequence of the largest odd numbers such that mn<n1+4d2dm_{n}<\frac{n-1+4d}{2d}. Then dimH(Cn)d\dim_{H}(C_{n})\rightarrow d and dimH(Cmn)d1\dim_{H}(C_{m_{n}})\rightarrow d-1 as nn\rightarrow\infty. Thus, for any xCn×Cmnx\in C_{n}\times C_{m_{n}} and any kk\in\mathbb{N}, the set ΔTxk(Cn×Cmn)\Delta_{T_{x}^{k}}(C_{n}\times C_{m_{n}}) has non-empty interior and

dimH(Cn×Cmn)=dimH(Cn)+dimH(Cmn)2d1\dim_{H}(C_{n}\times C_{m_{n}})=\dim_{H}(C_{n})+\dim_{H}(C_{m_{n}})\rightarrow 2d-1

as nn\rightarrow\infty.

3. Limitations of the Techniques

First, note that Theorem 1.10 immediately establishes the following result:

Corollary 3.1.

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} be compacts sets such that τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Then for any finite tree TT, the set ΔT(C1×C2)\Delta_{T}(C_{1}\times C_{2}) has non-empty interior.

One can then ask how low can we go, in terms of Hausdorff dimension, to obtain the result of the above theorem? Falconer and Yavicoli in [6] provide two estimates on this subject. The first of these estimates uses the following definitions:

K1:=2d(24d)dlog(16d)112dandK2:=((24d)d(1+4d2)112d)2K_{1}:=\frac{2d(24\sqrt{d})^{d}\log\left(16\sqrt{d}\right)}{1-\frac{1}{2^{d}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K_{2}:=\left(\frac{(24\sqrt{d})^{d}(1+4^{d}2)}{1-\frac{1}{2^{d}}}\right)^{2}

where dd is being taken as in d\mathbb{R}^{d}.

Theorem 3.2 (Corollary 19 in [6]).

Let CC be a compact set in d\mathbb{R}^{d} with positive thickness τ\tau (so diam(C)<+\operatorname{diam}(C)<+\infty and there is a ball BB such that BEB\cap E\neq\emptyset where EE is the unbounded gap of CC). If there exists c(0,d)c\in(0,d) such that τ(C)c1K2βc(1βdc)\tau(C)^{-c}\leq\frac{1}{K_{2}}\beta^{c}(1-\beta^{d-c}) for β:=min{14,diam(B)diam(C)}\beta:=\min\{\frac{1}{4},\frac{\operatorname{diam}(B)}{\operatorname{diam}(C)}\} then

dimH(C)dimH(BC)dK1τ(C)dβd|log(β)|>0.\dim_{H}(C)\geq\dim_{H}(B\cap C)\geq d-K_{1}\frac{\tau(C)^{-d}}{\beta^{d}|\log(\beta)|}>0.
Theorem 3.3 (Proposition 21 in [6]).

Let C0dC_{0}\subset\mathbb{R}^{d} be a proper compact convex set, and let C=C0\i=1GiC=C_{0}\backslash\bigcup_{i=1}^{\infty}G_{i} where {Gi}i\{G_{i}\}_{i} are open convex gaps. Then

dimH(C)d1+log2log(2+1/τ(C)).\dim_{H}(C)\geq d-1+\frac{\log 2}{\log(2+1/\tau(C))}.

For CdC\subset\mathbb{R}^{d}, since dimH(C×C)2dimH(C)\dim_{H}(C\times C)\geq 2\dim_{H}(C) the lower estimates on dimH(C×C)\dim_{H}(C\times C) will be of the form 2(dK1τ(C)dβd|log(β)|)2\left(d-K_{1}\frac{\tau(C)^{-d}}{\beta^{d}|\log(\beta)|}\right) and 2(d1+log2log(1+1/τ(C)))2\left(d-1+\frac{\log 2}{\log(1+1/\tau(C))}\right). For large thickness, the first estimate is lower (and therefore better since we want to know how small we can make the dimension) and for small thickness, the second estimate is lower (and therefore better). But when considering compact sets CC such τ(C)\tau(C) is close to 11, this restricts us to the second estimate. This is because to fulfill the requirement τ(C)c1K2βc(1βdc)1K2\tau(C)^{-c}\leq\frac{1}{K_{2}}\beta^{c}(1-\beta^{d-c})\leq\frac{1}{K_{2}}, since K2K_{2} is very large, we need to τ(C)\tau(C) to be much larger than 11. Thus, for τ(C)>1\tau(C)>1 and observing the second estimate, we have

dimH(C×C)2(d1+log2log(2+1/τ(C)))>2(d1+log2log3).\dim_{H}(C\times C)\geq 2\left(d-1+\frac{\log 2}{\log(2+1/\tau(C))}\right)>2\left(d-1+\frac{\log 2}{\log 3}\right).

So for τ(C)\tau(C) close to 11, we have the Mattila-Sjölin type result of Theorem 3.1 where we allow the dimension of our set to get down to 2(d1+log2log3)2\left(d-1+\frac{\log 2}{\log 3}\right). Note that this lower bound is worse than what is found in the Mattila-Sjölin theorem, which in this case would be 2d+12\frac{2d+1}{2}, with the only improvement on the bound being the case when d=1d=1. However, there has been work done on establishing chain and tree configurations from the point of view of Hausdorff dimension rather than thickness. Bennett, Iosevich, and Taylor in [1] were able to establish the existence of finite chain configurations with the Mattila-Sjölin threshold of d+12\frac{d+1}{2}, i.e. the assumption they were placing on sets was that EdE\subset\mathbb{R}^{d} for d2d\geq 2 and dimH(E)>d+12\dim_{H}(E)>\frac{d+1}{2}, which is enough to guarantee the existence of chain configurations. Building on this work, Iosevich and Taylor in [14] generalized the result to establish the existence of tree configurations using the Mattila-Sjölin threshold.

4. Discussion of the distance function

We now introduce our distance function, gx,t:ddg_{x,t}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}. For a fixed xdx\in\mathbb{R}^{d} this function will have the property that for C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} we will have tΔx(C1×C2)t\in\Delta_{x}(C_{1}\times C_{2}) provided that C2gx,t(C1)C_{2}\cap g_{x,t}(C_{1})\neq\emptyset. As pointed out in the introduction, problems of distance sets can be turned into problems of finding when sets intersect. This is one variant of that idea. Now fix (x,t)2d×[0,)(x,t)\in\mathbb{R}^{2d}\times[0,\infty) and for i{1,,d}i\in\{1,\cdots,d\} define g~i:d\widetilde{g}^{i}:\mathbb{R}^{d}\rightarrow\mathbb{R} by

g~i(y1,,yd)\displaystyle\widetilde{g}^{i}(y_{1},\cdots,y_{d}) :=xi+d+1dt20(x1y1)20(xi1yi1)2(xiyi)20(xi+1yi+1)20(xdyd)2\displaystyle:=x_{i+d}+\sqrt{\begin{aligned} \frac{1}{d}t^{2}-&0\cdot(x_{1}-y_{1})^{2}-\cdots-0\cdot(x_{i-1}-y_{i-1})^{2}\\ &-(x_{i}-y_{i})^{2}-0\cdot(x_{i+1}-y_{i+1})^{2}-\cdots-0\cdot(x_{d}-y_{d})^{2}\end{aligned}}
=xi+d+1dt2(xiyi)2=:gi(yi).\displaystyle=x_{i+d}+\sqrt{\frac{1}{d}t^{2}-(x_{i}-y_{i})^{2}}=:g^{i}(y_{i}).

We do this because if we set yi+d=gi(yi)y_{i+d}=g^{i}(y_{i}) then 1dt2=(yi+dxi+d)2+(yixi)2\frac{1}{d}t^{2}=(y_{i+d}-x_{i+d})^{2}+(y_{i}-x_{i})^{2} for all ii, which then means that by summing each of these terms we will have

i=12d(yixi)2=t2\sum_{i=1}^{2d}(y_{i}-x_{i})^{2}=t^{2}

or in other words, the distance between (y1,,y2d)(y_{1},\cdots,y_{2d}) and (x1,,x2d)(x_{1},\cdots,x_{2d}) is tt. Finally defined gx,t:ddg_{x,t}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} by

gx,t(y1,,yd)=(g~1(y1,,yd),,g~d(y1,,yd))=(g1(y1),,gd(yd)).g_{x,t}(y_{1},\cdots,y_{d})=(\widetilde{g}^{1}(y_{1},\cdots,y_{d}),\cdots,\widetilde{g}^{d}(y_{1},\cdots,y_{d}))=(g^{1}(y_{1}),\cdots,g^{d}(y_{d})).

Then we do indeed obtain the property that tΔx(C1×C2)t\in\Delta_{x}(C_{1}\times C_{2}) provided C2gx,t(C1)C_{2}\cap g_{x,t}(C_{1})\neq\emptyset.

Now we list an interesting property of gx,tg_{x,t}. From here on out, gx,tg_{x,t}^{\prime} will denote the Jacobian matrix of gx,tg_{x,t}. Then for z=(z1,,zd)dz=(z_{1},\cdots,z_{d})\in\mathbb{R}^{d} we have

gx,t(z)=[dg1dz10000dg2dz200000dgddzd]=[x1z11dt2(z1x1)20000x2z21dt2(z2x2)200000xdzd1dt2(zdxd)2].g_{x,t}^{\prime}(z)=\begin{bmatrix}\frac{dg^{1}}{dz_{1}}&0&0&\cdots&0\\ 0&\frac{dg^{2}}{dz_{2}}&0&\cdots&0\\ \vdots&&\ddots&&\vdots\\ 0&0&\cdots&0&\frac{dg^{d}}{dz_{d}}\end{bmatrix}=\begin{bmatrix}\frac{x_{1}-z_{1}}{\sqrt{\frac{1}{d}t^{2}-(z_{1}-x_{1})^{2}}}&0&0&\cdots&0\\ 0&\frac{x_{2}-z_{2}}{\sqrt{\frac{1}{d}t^{2}-(z_{2}-x_{2})^{2}}}&0&\cdots&0\\ \vdots&&\ddots&&\vdots\\ 0&0&\cdots&0&\frac{x_{d}-z_{d}}{\sqrt{\frac{1}{d}t^{2}-(z_{d}-x_{d})^{2}}}\end{bmatrix}.

Thus, gx,t(z)g^{\prime}_{x,t}(z) is a diagonal matrix. Furthermore, the norm that we will be working with for gx,t(z)g^{\prime}_{x,t}(z) is the operator norm which is the square root of the largest eigenvalue of gx,t(z)Tgx,t(z)g^{\prime}_{x,t}(z)^{T}g^{\prime}_{x,t}(z) where TT denotes the transpose. Then because gx,t(z)g^{\prime}_{x,t}(z) is diagonal, we have gx,t(z)T=gx,t(z)g^{\prime}_{x,t}(z)^{T}=g^{\prime}_{x,t}(z) and therefore

gx,t(z)Tgx,t(z)=[(x1z1)21dt2(z1x1)20000(x2z2)21dt2(z2x2)200000(xdzd)21dt2(zdxd)2]g^{\prime}_{x,t}(z)^{T}g^{\prime}_{x,t}(z)=\begin{bmatrix}\frac{(x_{1}-z_{1})^{2}}{\frac{1}{d}t^{2}-(z_{1}-x_{1})^{2}}&0&0&\cdots&0\\ 0&\frac{(x_{2}-z_{2})^{2}}{\frac{1}{d}t^{2}-(z_{2}-x_{2})^{2}}&0&\cdots&0\\ \vdots&&\ddots&&\vdots\\ 0&0&\cdots&0&\frac{(x_{d}-z_{d})^{2}}{\frac{1}{d}t^{2}-(z_{d}-x_{d})^{2}}\end{bmatrix}

so that

gx,t(z)op=max{|x1z1|1dt2(z1x1)2,,|xdzd|1dt2(zdxd)2}=max{|dg1dz1|,,|dgddzd|}.\|g^{\prime}_{x,t}(z)\|_{\text{op}}=\max\left\{\frac{|x_{1}-z_{1}|}{\sqrt{\frac{1}{d}t^{2}-(z_{1}-x_{1})^{2}}},\cdots,\frac{|x_{d}-z_{d}|}{\sqrt{\frac{1}{d}t^{2}-(z_{d}-x_{d})^{2}}}\right\}=\max\left\{\left|\frac{dg^{1}}{dz_{1}}\right|,\cdots,\left|\frac{dg^{d}}{dz_{d}}\right|\right\}. (2)

This estimate will come in later. The function gx,tg_{x,t} satisfies another useful property.

Lemma 4.1 (Mean Value Theorem for gx,tg_{x,t}).

Let a,bda,b\in\mathbb{R}^{d} be given such that ai<bia_{i}<b_{i} for all i{1,,d}i\in\{1,\cdots,d\} and gx,tg_{x,t} is defined on [a1,b1]××[ad,bd][a_{1},b_{1}]\times\cdots\times[a_{d},b_{d}]. Then there exists zi(ai,bi)z_{i}\in(a_{i},b_{i}) such that

(ba)gx,t(z)=gx,t(b)gx,t(a)(b-a)g_{x,t}^{{}^{\prime}}(z)=g_{x,t}(b)-g_{x,t}(a)

where z=(z1,,zd)z=(z_{1},\cdots,z_{d}).

Proof.

Since each gig^{i} is a mapping from \mathbb{R} to \mathbb{R} and will be differentiable in [ai,bi][a_{i},b_{i}], by the one-dimensional mean value theorem, there exists zi(ai,bi)z_{i}\in(a_{i},b_{i}) such that (biai)dgidzi=gi(bi)gi(ai).(b_{i}-a_{i})\frac{dg^{i}}{dz_{i}}=g^{i}(b_{i})-g^{i}(a_{i}). So

(ba)gx,t(z)=[b1a1bdad][dg1dz10000dgddzd]=[(b1a1)dg1dz1(bdad)dg2dzd]\displaystyle(b-a)g_{x,t}^{{}^{\prime}}(z)=\begin{bmatrix}b_{1}-a_{1}\\ \vdots\\ b_{d}-a_{d}\end{bmatrix}\begin{bmatrix}\frac{dg^{1}}{dz_{1}}&0&\cdots&0\\ \vdots&\ddots&&\vdots\\ 0&0&\cdots&\frac{dg^{d}}{dz_{d}}\end{bmatrix}=\begin{bmatrix}(b_{1}-a_{1})\frac{dg^{1}}{dz_{1}}\\ \vdots\\ (b_{d}-a_{d})\frac{dg^{2}}{dz_{d}}\end{bmatrix} =[g1(b1)g1(a1)gd(bd)gd(ad)]\displaystyle=\begin{bmatrix}g^{1}(b_{1})-g^{1}(a_{1})\\ \vdots\\ g^{d}(b_{d})-g^{d}(a_{d})\end{bmatrix}
=[g1(b1)gd(bd)][g1(a1)gd(ad)]\displaystyle=\begin{bmatrix}g^{1}(b_{1})\\ \vdots\\ g^{d}(b_{d})\end{bmatrix}-\begin{bmatrix}g^{1}(a_{1})\\ \vdots\\ g^{d}(a_{d})\end{bmatrix}
=gx,t(b)gx,t(a).\displaystyle=g_{x,t}(b)-g_{x,t}(a).

5. Proof of Theorem 1.10

We start this section with a lemma that allows us to write τ\tau in a, sometimes, more convenient way.

Lemma 5.1.

Let CdC\subset\mathbb{R}^{d} be compact with bounded gaps {Gn}n=1\{G_{n}\}_{n=1}^{\infty}. Set Λn:={in:diam(Gn)diam(Gi)}\Lambda_{n}:=\{i\neq n:\operatorname{diam}(G_{n})\leq\operatorname{diam}(G_{i})\}. Then we can also write τ\tau as

τ(C)=infndist(Gn,iΛnGiE)diam(Gn).\tau(C)=\inf_{n\in\mathbb{N}}\frac{\operatorname{dist}\left(G_{n},\bigcup_{i\in\Lambda_{n}}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}.
Proof.

To see this, assume that we cannot. Set

τ^(C):=infndist(Gn,iΛnGiE)diam(Gn).\widehat{\tau}(C):=\inf_{n\in\mathbb{N}}\frac{\operatorname{dist}\left(G_{n},\bigcup_{i\in\Lambda_{n}}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}.

First note that since diam(Gn)diam(Gn+1)\operatorname{diam}(G_{n})\geq\operatorname{diam}(G_{n+1}) we have τ^(C)τ(C)\widehat{\tau}(C)\leq\tau(C) as we could only be increasing the number gaps in our analysis using Λn\Lambda_{n} (as would be the case if we had diam(Gi)=diam(Gn)\operatorname{diam}(G_{i})=\operatorname{diam}(G_{n}) with i>ni>n). So therefore we may assume for contradiction that τ^(C)<τ(C)\widehat{\tau}(C)<\tau(C). Thus, there exists η1>0\eta_{1}>0 such that τ^(C)+η1=τ(C)\widehat{\tau}(C)+\eta_{1}=\tau(C). Since τ^\widehat{\tau} is an infimum, by properties of infimums we can find NN\in\mathbb{N} such that

dist(GN,iΛNGiE)diam(GN)<τ^(C)+η1=τ(C).\frac{\operatorname{dist}\left(G_{N},\bigcup_{i\in\Lambda_{N}}G_{i}\cup E\right)}{\operatorname{diam}(G_{N})}<\widehat{\tau}(C)+\eta_{1}=\tau(C).

Also note that this means

dist(GN,iΛNGiE)diam(GN)<τ(C)dist(GN,1iN1GiE)diam(GN)\frac{\operatorname{dist}\left(G_{N},\bigcup_{i\in\Lambda_{N}}G_{i}\cup E\right)}{\operatorname{diam}(G_{N})}<\tau(C)\leq\frac{\operatorname{dist}\left(G_{N},\bigcup_{1\leq i\leq N-1}G_{i}\cup E\right)}{\operatorname{diam}(G_{N})}

and therefore, the gap with the smallest distance is not achieved by EE since EE is present in both the LHS and RHS of the above inequality. Now set η2>0\eta_{2}>0 such that

dist(GN,iΛNGiE)diam(GN)+η2=τ(C).\frac{\operatorname{dist}\left(G_{N},\bigcup_{i\in\Lambda_{N}}G_{i}\cup E\right)}{\operatorname{diam}(G_{N})}+\eta_{2}=\tau(C).

Because dist\operatorname{dist} is an infimum property, because the minimal distance to GNG_{N} is not achieved by GiG_{i} with 1iN11\leq i\leq N-1, and because we are always assuming diam(Gn)diam(Gn+1)\operatorname{diam}(G_{n})\geq\operatorname{diam}(G_{n+1}), we can find MΛNM\in\Lambda_{N} such that M>NM>N, diam(GM)=diam(GN)\operatorname{diam}(G_{M})=\operatorname{diam}(G_{N}), and

dist(GN,GM)diam(GM)<dist(GN,iΛNGiE)diam(GN)+η2=τ(C).\frac{\operatorname{dist}(G_{N},G_{M})}{\operatorname{diam}(G_{M})}<\frac{\operatorname{dist}\left(G_{N},\bigcup_{i\in\Lambda_{N}}G_{i}\cup E\right)}{\operatorname{diam}(G_{N})}+\eta_{2}=\tau(C).

But since M>NM>N and diam(GM)=diam(GN)\operatorname{diam}(G_{M})=\operatorname{diam}(G_{N}) this implies

dist(GN,GM)diam(GN)=dist(GM,GN)diam(GM)dist(GM,1iM1GiE)diam(GM)\frac{\operatorname{dist}(G_{N},G_{M})}{\operatorname{diam}(G_{N})}=\frac{\operatorname{dist}(G_{M},G_{N})}{\operatorname{diam}(G_{M})}\geq\frac{\operatorname{dist}\left(G_{M},\bigcup_{1\leq i\leq M-1}G_{i}\cup E\right)}{\operatorname{diam}(G_{M})}

which means

dist(GM,1iM1GiE)diam(GM)<τ(C).\frac{\operatorname{dist}\left(G_{M},\bigcup_{1\leq i\leq M-1}G_{i}\cup E\right)}{\operatorname{diam}(G_{M})}<\tau(C).

However this is a contradiction since τ\tau is an infimum over all such elements and thus, we can write

τ(C)=infndist(Gn,1in1GiE)diam(Gn)=infndist(Gn,iΛnGiE)diam(Gn).\tau(C)=\inf_{n}\frac{\operatorname{dist}\left(G_{n},\bigcup_{1\leq i\leq n-1}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}=\inf_{n}\frac{\operatorname{dist}\left(G_{n},\bigcup_{i\in\Lambda_{n}}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}.

Throughout the proof of Theorem 1.8 and Lemma 5.11, it will be convenient for our pinned point to be the origin and for us to be able work with boundary points of gaps of C1C_{1} and C2C_{2} to exist on the diagonal. This seems to be, initially, a highly restrictive condition to put on our sets C1C_{1} and C2C_{2}. Fortunately though, we can do this by utilizing the fact that affine transformations are isometries and that thickness is preserved under affine transformations. The next lemma characterizes this shift.

Lemma 5.2.

Let C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d} with τ(C1)τ(C2)>1\tau(C_{1})\tau(C_{2})>1. Let x0C1×C2x^{0}\in C_{1}\times C_{2} be given and let u1C1u^{1}\in C_{1}, u2C2u^{2}\in C_{2} be such that they are boundary points of some gap (could also be the unbounded gaps) of C1C_{1} and C2C_{2} respectively. Then there exists affine transformations S1:ddS_{1}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} and S2:ddS_{2}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} with the following properties:

  1. (1)

    x00x^{0}\mapsto\vec{0} under S1S_{1} and S2S_{2}

  2. (2)

    u1w1=(w11,,wd1)u^{1}\mapsto w^{1}=(w_{1}^{1},\cdots,w_{d}^{1}) under S1S_{1} such that wi1=w11w_{i}^{1}=w_{1}^{1} for all ii and u2w2=(w12,,wd2)u^{2}\mapsto w^{2}=(w_{1}^{2},\cdots,w_{d}^{2}) under S2S_{2} such that wi2=w12w_{i}^{2}=w_{1}^{2} for all ii

  3. (3)

    τ(S1(C1))=τ(C1)\tau(S_{1}(C_{1}))=\tau(C_{1}) and τ(S2(C2))=τ(C2)\tau(S_{2}(C_{2}))=\tau(C_{2})

  4. (4)

    Δ0(S1(C1)×S2(C2))=Δx0(C1×C2)\Delta_{\vec{0}}(S_{1}(C_{1})\times S_{2}(C_{2}))=\Delta_{x^{0}}(C_{1}\times C_{2}).

Proof.

Let x0,1:=(x10,,xd0)x^{0,1}:=(x_{1}^{0},\cdots,x_{d}^{0}) and x0,2:=(xd+10,,x2d0)x^{0,2}:=(x_{d+1}^{0},\cdots,x_{2d}^{0}). Find z1,z2dz^{1},z^{2}\in\mathbb{R}^{d} such that x0,1+z1=0x^{0,1}+z^{1}=\vec{0} and x0,2+z2=0x^{0,2}+z^{2}=\vec{0}. Let SO(d)SO(d) denote the group of dd-dimensional rotations. Then there exists A1,A2SO(d)A_{1},A_{2}\in SO(d) such that A1(u1+z1)=w1=(w11,,wd1)A_{1}(u^{1}+z^{1})=w^{1}=(w_{1}^{1},\cdots,w_{d}^{1}) such that wi1=w11w_{i}^{1}=w_{1}^{1} for all i{1,,d}i\in\{1,\cdots,d\} and A2(u2+z2)=w2=(w12,wd2)A_{2}(u^{2}+z^{2})=w^{2}=(w_{1}^{2},\cdots w_{d}^{2}) such that wi2=w12w_{i}^{2}=w_{1}^{2} for all ii, i.e. we are rotating the points ui+ziu^{i}+z^{i} so that they are on the diagonal. Furthermore, elements in SO(d)SO(d) are linear. So for any xdx\in\mathbb{R}^{d} we have that the function Si:ddS_{i}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d} defined by

Si(x)=Ai(x+zi)=Aix+AiziS_{i}(x)=A_{i}(x+z^{i})=A_{i}x+A_{i}z^{i}

which shows that SiS_{i} is an affine transformation. Therefore, since thickness is preserved under affine transformations we get τ(Si(Ci))=τ(Ci)\tau(S_{i}(C_{i}))=\tau(C_{i}). Furthermore, both translations by ziz^{i} and rotation by AiA_{i} are isometries which means that SiS_{i} is an isometry. Thus, for tΔ0(S1(C1)×S2(C2))t\in\Delta_{\vec{0}}(S_{1}(C_{1})\times S_{2}(C_{2})) we can find y1C1y^{1}\in C_{1} and y2C2y^{2}\in C_{2} such that

t2\displaystyle t^{2} =[dist(S1(x0,1),S1(y1))]2+[dist(S2(x0,2),S2(y2))]2\displaystyle=\left[\operatorname{dist}(S_{1}(x^{0,1}),S_{1}(y^{1}))\right]^{2}+\left[\operatorname{dist}(S_{2}(x^{0,2}),S_{2}(y^{2}))\right]^{2}
=[dist(x0,1,y1)]2+[dist(x0,2,y2)]2\displaystyle=\left[\operatorname{dist}(x^{0,1},y^{1})\right]^{2}+\left[\operatorname{dist}(x^{0,2},y^{2})\right]^{2}
=i=1d(xi0yi1)2+j=d+12d(xj0yj2)2\displaystyle=\sum_{i=1}^{d}(x_{i}^{0}-y_{i}^{1})^{2}+\sum_{j=d+1}^{2d}(x_{j}^{0}-y_{j}^{2})^{2}

and therefore, tΔx0(C1×C2).t\in\Delta_{x^{0}}(C_{1}\times C_{2}). Since equality held throughout, we can say Δx0(C1×C2)=Δ0(S1(C1)×S2(C2))\Delta_{x^{0}}(C_{1}\times C_{2})=\Delta_{\vec{0}}(S_{1}(C_{1})\times S_{2}(C_{2})). ∎

Since we are now working in d\mathbb{R}^{d} we now to deal with more complicated sets and in particular, more complicated boundaries. But the next series of facts, which is presented without proof, allows us nicely characterize how the boundary and diameter of a set is affected by gx,tg_{x,t}.

Fact 5.3.

Suppose ff is continuous and HH is open where f(H)=Hf(H)=H^{\prime} are open sets with HH having compact closure. Then f(H)=Hf(\partial H)=\partial H^{\prime}.

In our language, this will mean gx,t(H)=gx,t(H)g_{x,t}(\partial H)=\partial g_{x,t}(H) where HH is a gap of CC.

Fact 5.4.

If HH is open, then diam(H)=diam(H)\operatorname{diam}(H)=\operatorname{diam}(\partial H).

Combining these two facts gives us the next fact.

Fact 5.5.

Since gx,tg_{x,t} is continuous we have

diam(gx,t(H))=diam(gx,t(H))=diam(gx,t(H)).\operatorname{diam}(g_{x,t}(H))=\operatorname{diam}(\partial g_{x,t}(H))=\operatorname{diam}(g_{x,t}(\partial H)).

Thus, when trying to characterize the diameter of some gap under gx,tg_{x,t}, it suffices to look how the boundary is affected by gx,tg_{x,t}. Then next part of this discussion characterizes how thickness is changed under gx,tg_{x,t}.

As pointed out in Section 4, we will have that tΔx(C1×C2)t\in\Delta_{x}(C_{1}\times C_{2}) provided C2gx,t(C1)C_{2}\cap g_{x,t}(C_{1})\neq\emptyset. So to use the Gap Lemma we need to understand the thickness of the set gx,t(C1)g_{x,t}(C_{1}). To do this, we introduce a notion of ε\varepsilon-thickness that gives us some breathing room to work with in later calculations. The following definition is adapted from Definition 3.2 in [20].

Definition 5.6.

Let CdC\in\mathbb{R}^{d} be a compact set with bounded gaps {Gn}n=1\{G_{n}\}_{n=1}^{\infty}. Let unGnu_{n}\in\partial G_{n} and define Hε,n:={in:(1ε)diam(Gn)<diam(Gi)}.H_{\varepsilon,n}:=\{i\neq n:(1-\varepsilon)\operatorname{diam}(G_{n})<\operatorname{diam}(G_{i})\}. Then the ε\varepsilon-thickness of CC at unu_{n} is defined to be

τε(C,un):=dist(un,iHε,nGiE)diam(Gn)\tau_{\varepsilon}(C,u_{n}):=\frac{\operatorname{dist}\left(u_{n},\bigcup_{i\in H_{\varepsilon,n}}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}

and the ε\varepsilon-thickness of CC is

τε(C)=infninfunτε(C,un)\tau_{\varepsilon}(C)=\inf_{n\in\mathbb{N}}\inf_{u_{n}}\tau_{\varepsilon}(C,u_{n})

the infimum being taken over all boundary points of all gaps.

This definition leads to two important properties, which we quickly prove.

Proposition 5.7.

Let CdC\subset\mathbb{R}^{d} be compact. Then

  • (i)

    If ε1<ε2\varepsilon_{1}<\varepsilon_{2} then τε2(C)τε1(C)\tau_{\varepsilon_{2}}(C)\leq\tau_{\varepsilon_{1}}(C)

  • (ii)

    τε(C)τ(C)\tau_{\varepsilon}(C)\rightarrow\tau(C) as ε0\varepsilon\rightarrow 0.

Proof.

For the first claim, if (1ε1)diam(Gn)<diam(Gi)(1-\varepsilon_{1})\operatorname{diam}(G_{n})<\operatorname{diam}(G_{i}) then

(1ε2)diam(Gn)<(1ε1)diam(Gn)<diam(Gi).(1-\varepsilon_{2})\operatorname{diam}(G_{n})<(1-\varepsilon_{1})\operatorname{diam}(G_{n})<\operatorname{diam}(G_{i}).

So if iHε1,ni\in H_{\varepsilon_{1},n} then iHε2,ni\in H_{\varepsilon_{2},n}. Therefore, we include possibly more gaps into consideration with indices in the set Hε2,nH_{\varepsilon_{2},n} and thus, the distance to GnG_{n} can only shrink. So τε2(C)τε1(C).\tau_{\varepsilon_{2}}(C)\leq\tau_{\varepsilon_{1}}(C).

Now for the second claim, we can rewrite τ(C)\tau(C) as is done in Lemma 5.1. Then note, in terms of lim sup\limsup and lim inf\liminf sets, we have limmH1/m,n=Λn\lim_{m\rightarrow\infty}H_{1/m,n}=\Lambda_{n} and thus, as ε0\varepsilon\rightarrow 0 we get Hε,nΛnH_{\varepsilon,n}\rightarrow\Lambda_{n} which shows τε(C)τ(C)\tau_{\varepsilon}(C)\rightarrow\tau(C). ∎

The next two lemmas will tell us that the thickness of CC under gx,tg_{x,t} either is not changed at all or only changes slightly when restricting our viewpoint to a small section of CC. We break up the two lemmas by whether or not CC has any bounded gaps.

Lemma 5.8.

Let CdC\subset\mathbb{R}^{d} compact with τ(C)>0\tau(C)>0 and assume that CC has no bounded gaps. Then there exists a rectangle (hyperrectangle in d\mathbb{R}^{d}) RCR\subset C such that τ(gx~,t(R))=τ(C)\tau(g_{\widetilde{x},t}(R))=\tau(C) where x~\widetilde{x} is being taken from a neighborhood with center xx.

Proof.

First note that since gx,tg_{x,t} is continuous, the image of any compact set is also compact. So it is valid for us to talk about the thickness of the image of a compact set under gx,tg_{x,t}. Now because we are assuming CC has no bounded gaps, by definition this implies either τ(C)=0\tau(C)=0 or τ(C)=+\tau(C)=+\infty and since τ(C)>0\tau(C)>0 we have that τ(C)=+\tau(C)=+\infty and that CC has non-empty interior. So there exists a ball BCB\subset C and a (filled-in) rectangle RBCR\subset B\subset C. Then recall that each gig^{i} only takes in one coordinate for their argument. So because of this and because each gig^{i} is decreasing and continuous, gx,t(R)g_{x,t}(R) is a filled-in rectangle and therefore, τ(gx,t(R))=+\tau(g_{x,t}(R))=+\infty. The same argument that will be given below in the proof of Lemma 5.11 also shows that we can have τ(gx~,t(R))=+\tau(g_{\widetilde{x},t}(R))=+\infty where x~\widetilde{x} is being taken from a neighborhood of xx. Thus, in the case when CC has no bounded gaps, we simply restrict to this trivial case where thickness is in fact preserved. ∎

The calculations in the rest of this document will be more convenient if we can work with a specific type of boundary point.

Definition 5.9.

Let CdC\subset\mathbb{R}^{d} be compact with HH begin some gap of CC. Then we call uu a right endpoint of HH provided that for any ε>0\varepsilon>0 we have u+(ε,,ε)Hu+(\varepsilon,\cdots,\varepsilon)\notin H.

Remark 5.10.

Intuitively, a right endpoint of a gap HH is a boundary point which sits at the "top right" of HH. Clearly every gap will have a right endpoint.

Now we state a key lemma which tells us how thickness of CC is affected under continuously differentiable mappings where CC has at least one bounded gap. Furthermore, in the next lemma we will take x=(x1,,x2d)2dx=(x_{1},\cdots,x_{2d})\in\mathbb{R}^{2d} such that x1=xix_{1}=x_{i} for all i{1,,d}i\in\{1,\cdots,d\}. But this is not a concern for us because by Lemma 5.2, we can always rotate and translate our sets so that we are in this case. This lemma is a higher dimensional version of Lemma 3.4 in [20].

Lemma 5.11 (Thickness is nearly preserved).

Let CdC\subset\mathbb{R}^{d} compact with τ(C)>0\tau(C)>0 and let uu be a right endpoint of a bounded gap of CC. Find x2dx\in\mathbb{R}^{2d} such that x1=xix_{1}=x_{i} for all i{1,,d}i\in\{1,\cdots,d\} and for which uu is in the domain of gx,tg_{x,t} and such that dgidui0\frac{dg^{i}}{du_{i}}\neq 0 for all i{1,,d}i\in\{1,\cdots,d\}. Then for every ε>0\varepsilon>0, there exists δ>0\delta>0 such that

τ(gx~,t(C([u1,u1+δ]××[ud,ud+δ])))>τ2εε2(C)(1ε)\tau\left(g_{\widetilde{x},t}(C\cap([u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta]))\right)>\tau_{2\varepsilon-\varepsilon^{2}}(C)(1-\varepsilon)

where x~\widetilde{x} is being taken from a neighborhood with center xx.

Proof.

Let ε>0\varepsilon>0 be given. Note that since x1=xix_{1}=x_{i} for i{1,,d}i\in\{1,\cdots,d\}, this means

dgidzi=x1zi1dt2(x1zi)2.\frac{dg^{i}}{dz_{i}}=\frac{x_{1}-z_{i}}{\sqrt{\frac{1}{d}t^{2}-(x_{1}-z_{i})^{2}}}.

Define f:f:\mathbb{R}\rightarrow\mathbb{R} by f(z)=x1z1dt2(x1z)2f(z)=\frac{x_{1}-z}{\sqrt{\frac{1}{d}t^{2}-(x_{1}-z)^{2}}}. Then ff is continuous and for zdz\in\mathbb{R}^{d} we have f(zi)=dgi/dzif(z_{i})=dg^{i}/dz_{i} for i{1,,d}i\in\{1,\cdots,d\}. Furthermore, since each dgi/duidg^{i}/du_{i} is nonzero, ff will also be nonzero in a neighborhood of uu. With this and because ff is continuous, there exists δ~>0\widetilde{\delta}>0 such that for any z[u1,u1+δ]××[ud,ud+δ]z\in[u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta] where δ=δ~/d\delta=\widetilde{\delta}/\sqrt{d}, we have

||dgi/dzi||dgj/dzj|1|=||f(zi)||f(zj)|1|<ε\left|\frac{|dg^{i}/dz_{i}|}{|dg^{j}/dz_{j}|}-1\right|=\left|\frac{|f(z_{i})|}{|f(z_{j})|}-1\right|<\varepsilon (3)

for any i,j{1,,d}i,j\in\{1,\cdots,d\} since this means |zizj|dδ=δ~|z_{i}-z_{j}|\leq\sqrt{d}\cdot\delta=\widetilde{\delta} which invokes continuity of ff. We can also do this for a neighborhood of xx values. Indeed, take z[u1+δ2d,u1+(2d1)δ2d]××[ud+δ2d,ud+(2d1)δ2d]z\in\left[u_{1}+\frac{\delta}{2d},u_{1}+\frac{(2d-1)\delta}{2d}\right]\times\cdots\times\left[u_{d}+\frac{\delta}{2d},u_{d}+\frac{(2d-1)\delta}{2d}\right] and αi[δ2d,δ2d]\alpha_{i}\in\left[-\frac{\delta}{2d},\frac{\delta}{2d}\right] for i{1,,d}i\in\{1,\cdots,d\}. Then by setting x~=(x1+α1,,xd+αd)\widetilde{x}=(x_{1}+\alpha_{1},\cdots,x_{d}+\alpha_{d}) we will have that

x~izi1dt2(x~izi)2=x1(ziαi)1dt2(x1(ziαi))2=dgid(ziαi)\frac{\widetilde{x}_{i}-z_{i}}{\sqrt{\frac{1}{d}t^{2}-(\widetilde{x}_{i}-z_{i})^{2}}}=\frac{x_{1}-(z_{i}-\alpha_{i})}{\sqrt{\frac{1}{d}t^{2}-(x_{1}-(z_{i}-\alpha_{i}))^{2}}}=\frac{dg^{i}}{d(z_{i}-\alpha_{i})}

and z(α1,,αd)[u1,u1+δ]××[ud,ud+δ]z-(\alpha_{1},\cdots,\alpha_{d})\in[u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta]. So again if we take z[u1+δ2d,u1+(2d1)δ2d]××[ud+δ2d,ud+(2d1)δ2d]z\in\left[u_{1}+\frac{\delta}{2d},u_{1}+\frac{(2d-1)\delta}{2d}\right]\times\cdots\times\left[u_{d}+\frac{\delta}{2d},u_{d}+\frac{(2d-1)\delta}{2d}\right] and allowing x~\widetilde{x} to have the form (x1+α1,,xd+αd)(x_{1}+\alpha_{1},\cdots,x_{d}+\alpha_{d}), this will still invoke continuity of ff so that the ratio |dgi/d(ziαi)||dgj/d(zjαj)|\frac{|dg^{i}/d(z_{i}-\alpha_{i})|}{|dg^{j}/d(z_{j}-\alpha_{j})|} is still close to 11 as this is functionally the same as allowing z[u1,u1+δ]××[ud,ud+δ]z\in[u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta], but now this allows us to let the xx-values we are observing to be taken from an entire neighborhood. Explicitly, the neighborhood of xx-values is the set x+([δ2d,δ2d]××[δ2d,δ2d])x+\left(\left[-\frac{\delta}{2d},\frac{\delta}{2d}\right]\times\cdots\times\left[-\frac{\delta}{2d},\frac{\delta}{2d}\right]\right) which also shows that the center of this neighborhood is xx. From here on out, we will abuse notation and use the term dgi/dzidg^{i}/dz_{i} when we really mean dgi/d(ziαi)dg^{i}/d(z_{i}-\alpha_{i}) and we will use gx,tg_{x,t} when we really mean gx+α,tg_{x+\alpha,t} for αi[δ2d,δ2d]\alpha_{i}\in\left[-\frac{\delta}{2d},\frac{\delta}{2d}\right] and α=(α1,,αd)\alpha=(\alpha_{1},\cdots,\alpha_{d}).

Before going further, we make a quick note about ff being well-defined. Since we are assuming uu is in the domain of gx,tg_{x,t}, we have that each gi(ui)=xi+d+1dt2(xiui)2g^{i}(u_{i})=x_{i+d}+\sqrt{\frac{1}{d}t^{2}-(x_{i}-u_{i})^{2}} is well-defined. So then clearly we can find an interval II such that for any ηiIi\eta_{i}\in I_{i}, the function gi(ui+ηi)g^{i}(u_{i}+\eta_{i}) is also well-defined. Thus, gx,tg_{x,t} is well-defined for an entire neighborhood, say NN, for which uNu\in N. This further implies that each ui+ηiu_{i}+\eta_{i} is also in the domain of ff where ηiIi\eta_{i}\in I_{i}. Here, we are assuming without loss of generality that NN intersects the set [u1,u1+δ]××[ud,ud+δ][u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta].

Resuming, note that any gap in A1:=gx,t(C([u1,u1+δ]××[ud,ud+δ]))A_{1}:=g_{x,t}(C\cap([u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta])) is the image of a gap in A2:=C([u1,u1+δ]××[ud,ud+δ])A_{2}:=C\cap([u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta]). In particular, all gaps in A1A_{1} will be of the form gx,t(Gn)g_{x,t}(G_{n}) where GnG_{n} is a gap in A2A_{2}. After possibly reordering the original gaps, we then produce a new list of gaps of A1A_{1}, say {gx,t(Gn)}n=1\{g_{x,t}(G_{n})\}_{n=1}^{\infty}, with diam(gx,t(Gn))diam(gx,t(Gn+1))\operatorname{diam}(g_{x,t}(G_{n}))\geq\operatorname{diam}(g_{x,t}(G_{n+1})). By Fact 5.5, diam(gx,t(Gn))=diam(gx,t(Gn))=diam(gx,t(Gn))\operatorname{diam}(g_{x,t}(G_{n}))=\operatorname{diam}(\partial g_{x,t}(G_{n}))=\operatorname{diam}(g_{x,t}(\partial G_{n})). Thus, for ε\varepsilon-thickness we can take unGnu_{n}\in\partial G_{n}, for valid GnG_{n}, and have

τε(A1,gx,t(un))=dist(gx,t(un),iHε,ngx,t(Gi)E)diam(gx,t(Gn))\tau_{\varepsilon}(A_{1},g_{x,t}(u_{n}))=\frac{\operatorname{dist}(g_{x,t}(u_{n}),\bigcup_{i\in H_{\varepsilon,n}}g_{x,t}(G_{i})\cup E)}{\operatorname{diam}(g_{x,t}(G_{n}))}

where EE now denotes the unbounded gap of A1A_{1} and Hε,n={in:(1ε)diam(gx,t(Gn))<diam(gx,t(Gi))}H_{\varepsilon,n}=\{i\neq n:(1-\varepsilon)\operatorname{diam}(g_{x,t}(G_{n}))<\operatorname{diam}(g_{x,t}(G_{i}))\}. Now let vnGnv_{n}\in\partial G_{n} such that gx,t(Gn)g_{x,t}(G_{n}) is a gap in A1A_{1}. Note that since we are assuming that we have thick compact sets, then the thickness of neither is equal to zero. So we always get nonzero values in the numerator for the expression of τε(A1,g(vn))\tau_{\varepsilon}(A_{1},g(v_{n})). So we can find some element, say vjGjv_{j}\in\partial G_{j} where GjG_{j} could also be the unbounded gap of A1A_{1} such that (1ε)diam(gx,t(Gn))<diam(gx,t(Gj))(1-\varepsilon)\operatorname{diam}(g_{x,t}(G_{n}))<\operatorname{diam}(g_{x,t}(G_{j})) and dist(gx,t(vn),iHε,ngx,t(Gi)E)=gx,t(vn)gx,t(vj)\operatorname{dist}(g_{x,t}(v_{n}),\bigcup_{i\in H_{\varepsilon,n}}g_{x,t}(G_{i})\cup E)=\|g_{x,t}(v_{n})-g_{x,t}(v_{j})\|. Set A3:=[u1,u1+δ]××[ud,ud+δ]A_{3}:=[u_{1},u_{1}+\delta]\times\cdots\times[u_{d},u_{d}+\delta]. By Lemma 4.1, there exists zA3z\in A_{3} such that gx,t(vn)gx,t(vj)=gx,t(z)(vnvj)\|g_{x,t}(v_{n})-g_{x,t}(v_{j})\|=\|g_{x,t}^{{}^{\prime}}(z)(v_{n}-v_{j})\|. Furthermore, diam(gx,t(Gn))=gx,t(an)gx,t(bn)\operatorname{diam}(g_{x,t}(G_{n}))=\|g_{x,t}(a_{n})-g_{x,t}(b_{n})\| for some an,bnGna_{n},b_{n}\in\partial G_{n}. Again by Lemma 4.1 we have that there exists some wA3w\in A_{3} such that gx,t(an)gx,t(bn)=gx,t(w)(anbn).\|g_{x,t}(a_{n})-g_{x,t}(b_{n})\|=\|g_{x,t}^{{}^{\prime}}(w)(a_{n}-b_{n})\|. This gives

τε(A1,gx,t(vn))\displaystyle\tau_{\varepsilon}(A_{1},g_{x,t}(v_{n})) =gx,t(z)(vnvj)diam(gx,t(Gn))\displaystyle=\frac{\|g_{x,t}^{{}^{\prime}}(z)(v_{n}-v_{j})\|}{\operatorname{diam}(g_{x,t}(G_{n}))}
=gx,t(z)(vnvj)gx,t(w)(anbn)\displaystyle=\frac{\|g_{x,t}^{{}^{\prime}}(z)(v_{n}-v_{j})\|}{\|g_{x,t}^{{}^{\prime}}(w)(a_{n}-b_{n})\|}
gx,t(z)(vnvj)gx,t(w)opanbn.\displaystyle\geq\frac{\|g_{x,t}^{{}^{\prime}}(z)(v_{n}-v_{j})\|}{\|g_{x,t}^{{}^{\prime}}(w)\|_{op}\|a_{n}-b_{n}\|}.

To get a further lower estimate on this quantity, we can assume without loss of generality that |dg2dz2|=min{|dg1dz1|,,|dgddzd|}\left|\frac{dg^{2}}{dz_{2}}\right|=\min\left\{\left|\frac{dg^{1}}{dz_{1}}\right|,\cdots,\left|\frac{dg^{d}}{dz_{d}}\right|\right\}. Then

gx,t(z)(vnvj)\displaystyle\|g_{x,t}^{{}^{\prime}}(z)(v_{n}-v_{j})\| =[dg1dz10000dgddzd][vn,1vj,1vn,dvj,d]\displaystyle=\left\|\begin{bmatrix}\frac{dg^{1}}{dz_{1}}&0&\cdots&0\\ \vdots&\ddots&&\vdots\\ 0&0&\cdots&\frac{dg^{d}}{dz_{d}}\end{bmatrix}\begin{bmatrix}v_{n,1}-v_{j,1}\\ \vdots\\ v_{n,d}-v_{j,d}\end{bmatrix}\right\|
=(dg1dz1)2(vn,1vj,1)2++(dgddzd)2(vn,dvj,d)2\displaystyle=\sqrt{\left(\frac{dg^{1}}{dz_{1}}\right)^{2}\left(v_{n,1}-v_{j,1}\right)^{2}+\cdots+\left(\frac{dg^{d}}{dz_{d}}\right)^{2}\left(v_{n,d}-v_{j,d}\right)^{2}}
(dg2dz2)2(vn,1vj,1)2++(dg2dz2)2(vn,dvj,d)2\displaystyle\geq\sqrt{\left(\frac{dg^{2}}{dz_{2}}\right)^{2}\left(v_{n,1}-v_{j,1}\right)^{2}+\cdots+\left(\frac{dg^{2}}{dz_{2}}\right)^{2}\left(v_{n,d}-v_{j,d}\right)^{2}}
=|dg2dz2|vnvj.\displaystyle=\left|\frac{dg^{2}}{dz_{2}}\right|\|v_{n}-v_{j}\|.

Then

τε(A1,gx,t(vn))\displaystyle\tau_{\varepsilon}(A_{1},g_{x,t}(v_{n})) |dg2dz2|vnvjgx,t(w)opanbn\displaystyle\geq\frac{\left|\frac{dg^{2}}{dz_{2}}\right|\|v_{n}-v_{j}\|}{\|g_{x,t}^{{}^{\prime}}(w)\|_{op}\|a_{n}-b_{n}\|}
|dg2dz2|vnvjgx,t(w)opdiam(Gn).\displaystyle\geq\frac{\left|\frac{dg^{2}}{dz_{2}}\right|\|v_{n}-v_{j}\|}{\|g_{x,t}^{{}^{\prime}}(w)\|_{op}\operatorname{diam}(G_{n})}.

Using Equation (2), without loss of generality we can assume gx,t(w)op=|dg1dw1|.\|g_{x,t}^{{}^{\prime}}(w)\|_{op}=\left|\frac{dg^{1}}{dw_{1}}\right|. By Equation (3) we have

|dg2dz2||dg1dw1|>1ε\frac{\left|\frac{dg^{2}}{dz_{2}}\right|}{\left|\frac{dg^{1}}{dw_{1}}\right|}>1-\varepsilon

which yields

τε(A1,gx,t(vn))\displaystyle\tau_{\varepsilon}(A_{1},g_{x,t}(v_{n})) |dg2dz2|vnvj|dg1dw1|diam(Gn)\displaystyle\geq\frac{\left|\frac{dg^{2}}{dz_{2}}\right|\|v_{n}-v_{j}\|}{\left|\frac{dg^{1}}{dw_{1}}\right|\operatorname{diam}(G_{n})}
>(1ε)vnvjdiam(Gn).\displaystyle>(1-\varepsilon)\frac{\|v_{n}-v_{j}\|}{\operatorname{diam}(G_{n})}.

Note that vnv_{n} and vjv_{j} were chosen such that gx,t(vn)gx,t(vj)=dist(gx,t(vn),iHε,ngx,t(Gi)gx,t(E))\|g_{x,t}(v_{n})-g_{x,t}(v_{j})\|=\operatorname{dist}\left(g_{x,t}(v_{n}),\bigcup_{i\in H_{\varepsilon,n}}g_{x,t}(G_{i})\cup g_{x,t}(E)\right) and recall that since jHε,nj\in H_{\varepsilon,n} we have (1ε)diam(gx,t(Gn))<diam(gx,t(Gj))(1-\varepsilon)\operatorname{diam}(g_{x,t}(G_{n}))<\operatorname{diam}(g_{x,t}(G_{j})). Note that jj being in Hε,nH_{\varepsilon,n} refers to the ordering that was placed on {gx,t(Gn)}n=1\{g_{x,t}(G_{n})\}_{n=1}^{\infty}, not necessarily the ordering that we placed on {Gn}n=1\{G_{n}\}_{n=1}^{\infty}. We will now try to find η>0\eta>0 such that jHη,nj\in H_{\eta,n} where this now refers to the ordering that we placed on {Gn}n=1\{G_{n}\}_{n=1}^{\infty} which will provide a lower estimate on vnvj\|v_{n}-v_{j}\|. Let α1,β1Gn\alpha^{1},\beta^{1}\in\partial G_{n} such that diam(Gn)=α1β1\operatorname{diam}(G_{n})=\|\alpha^{1}-\beta^{1}\| and α2,β2Gj\alpha^{2},\beta^{2}\in\partial G_{j} such that diam(gx,t(Gj))=gx,t(α2)gx,t(β2)\operatorname{diam}(g_{x,t}(G_{j}))=\|g_{x,t}(\alpha^{2})-g_{x,t}(\beta^{2})\|. Then

(1ε)gx,t(α1)gx,t(β1)(1ε)gx,t(an)gx,t(bn)<gx,t(α2)gx,t(β2).(1-\varepsilon)\|g_{x,t}(\alpha^{1})-g_{x,t}(\beta^{1})\|\leq(1-\varepsilon)\|g_{x,t}(a_{n})-g_{x,t}(b_{n})\|<\|g_{x,t}(\alpha^{2})-g_{x,t}(\beta^{2})\|.

As before, we can find ρ1,ρ2A3\rho^{1},\rho^{2}\in A_{3} such that gx,t(α1)gx,t(β1)=gx,t(ρ1)(α1β1)\|g_{x,t}(\alpha^{1})-g_{x,t}(\beta^{1})\|=\|g_{x,t}^{{}^{\prime}}(\rho^{1})(\alpha^{1}-\beta^{1})\| and gx,t(α2)gx,t(β2)=gx,t(ρ2)(α2β2)\|g_{x,t}(\alpha^{2})-g_{x,t}(\beta^{2})\|=\|g_{x,t}^{{}^{\prime}}(\rho^{2})(\alpha^{2}-\beta^{2})\|. By a similar calculations as above, we can assume without loss of generality that

gx,t(ρ1)(α1β1)|dg1dρ11|α1β1\|g_{x,t}^{{}^{\prime}}(\rho^{1})(\alpha^{1}-\beta^{1})\|\geq\left|\frac{dg^{1}}{d\rho_{1}^{1}}\right|\|\alpha^{1}-\beta^{1}\|

and

gx,t(ρ2)(α2β2)gx,t(ρ2)opα2β2=|dg2dρ22|α2β2.\|g_{x,t}^{{}^{\prime}}(\rho^{2})(\alpha^{2}-\beta^{2})\|\leq\|g_{x,t}^{{}^{\prime}}(\rho^{2})\|_{op}\|\alpha^{2}-\beta^{2}\|=\left|\frac{dg^{2}}{d\rho_{2}^{2}}\right|\|\alpha^{2}-\beta^{2}\|.

Along with Equation (3) this implies

diam(Gj)\displaystyle\operatorname{diam}(G_{j}) α2β2\displaystyle\geq\|\alpha^{2}-\beta^{2}\|
>(1ε)|dg1dρ11||dg2dρ22|α1β1\displaystyle>(1-\varepsilon)\frac{\left|\frac{dg^{1}}{d\rho_{1}^{1}}\right|}{\left|\frac{dg^{2}}{d\rho_{2}^{2}}\right|}\|\alpha^{1}-\beta^{1}\|
>(1ε)(1ε)α1β1\displaystyle>(1-\varepsilon)(1-\varepsilon)\|\alpha^{1}-\beta^{1}\|
=(1(2εε2))diam(Gn).\displaystyle=(1-(2\varepsilon-\varepsilon^{2}))\operatorname{diam}(G_{n}).

Therefore, jH2εε2,nj\in H_{2\varepsilon-\varepsilon^{2},n} where this refers to the ordering placed on {Gn}n=1\{G_{n}\}_{n=1}^{\infty}. So

τε(A1,gx,t(vn))\displaystyle\tau_{\varepsilon}(A_{1},g_{x,t}(v_{n})) (1ε)vnvjdiam(Gn)\displaystyle\geq(1-\varepsilon)\frac{\|v_{n}-v_{j}\|}{\operatorname{diam}(G_{n})}
(1ε)dist(vn,iH2εε2,nGiE)diam(Gn)\displaystyle\geq(1-\varepsilon)\frac{\operatorname{dist}\left(v_{n},\bigcup_{i\in H_{2\varepsilon-\varepsilon^{2},n}}G_{i}\cup E\right)}{\operatorname{diam}(G_{n})}
=(1ε)τ2εε2(C,vn).\displaystyle=(1-\varepsilon)\tau_{2\varepsilon-\varepsilon^{2}}(C,v_{n}).

Taking infimums produces

τ(A1)τε(A1)>τ2εε2(C)(1ε)\tau(A_{1})\geq\tau_{\varepsilon}(A_{1})>\tau_{2\varepsilon-\varepsilon^{2}}(C)(1-\varepsilon)

which finishes the proof. ∎

This lemma ensures that we can find a lower bound on τ(gx,t(C))\tau(g_{x,t}(C)) in terms of CC. Thus, even though it can be difficult to see how CC, globally, is affected by gx,tg_{x,t}, we can restrict CC in a local manner and use the above lemma to ensure that the thickness of CC does not change much under gx,tg_{x,t}.

In the proof of Theorem 1.8, we will require our pinned points to not share a coordinate and Lemma 5.13 ensures that we can do this. The proof relies on the fact that unions of hyperplanes have zero thickness, which we now present.

Fact 5.12.

Let CdC\subset\mathbb{R}^{d} be compact such that Ci=1nJiC\subseteq\bigcup_{i=1}^{n}J_{i} where each JiJ_{i} is a bounded hyperplane. Then τ(C)=0\tau(C)=0.

Proof.

Let C1C_{1} denote the part of CC that is a subset of J1J_{1}, i.e. C1=CJ1C_{1}=C\cap J_{1}. Then since C1C_{1} can only be removing bounded gaps of CC we will have τ(C)τ(C1)\tau(C)\leq\tau(C_{1}). Furthermore, the only path connected component of C1CC_{1}^{C} is its unbounded gap and because hyperplanes have empty interior, this implies C1C_{1} has empty interior as well. By the definition of thickness, this yields τ(C1)=0\tau(C_{1})=0 showing the claim. ∎

Lemma 5.13.

Assume τ(C1),τ(C2)>0\tau(C_{1}),\tau(C_{2})>0 where C1,C2dC_{1},C_{2}\subset\mathbb{R}^{d}. Then given any x1C1×C2x^{1}\in C_{1}\times C_{2} we can always find x2,,xk+1C1×C2x^{2},\cdots,x^{k+1}\in C_{1}\times C_{2} such that every element in the set {xi1,xi2,,xin}\{x_{i}^{1},x_{i}^{2},\cdots,x_{i}^{n}\} does not share a coordinate for all i{1,,2d}i\in\{1,\cdots,2d\}. Thus, in particular we have xijxikx_{i}^{j}\neq x_{i}^{k} for all i{1,,2d}i\in\{1,\cdots,2d\} and j,k{1,,n}j,k\in\{1,\cdots,n\}, i.e. none of x1,,xk+1x^{1},\cdots,x^{k+1} share a coordiante.

Proof.

We prove this by induction. Since we are always given x1x^{1}, our induction starts at n=2n=2. Thus, we will find x2C1×C2x^{2}\in C_{1}\times C_{2} so that each xi1xi2x_{i}^{1}\neq x_{i}^{2} for i{1,,2d}i\in\{1,\cdots,2d\}. Assume for contradiction that we cannot do this. Define

Hi:={(x11+α1,,xi11+αi1,xi1,xi+11+αi+1,,x2d1+α2d):diam(C1×C2)αjdiam(C1×C2}H_{i}:=\{(x_{1}^{1}+\alpha_{1},\cdots,x_{i-1}^{1}+\alpha_{i-1},x_{i}^{1},x_{i+1}^{1}+\alpha_{i+1},\cdots,x_{2d}^{1}+\alpha_{2d}):-\operatorname{diam}(C_{1}\times C_{2})\leq\alpha_{j}\leq\operatorname{diam}(C_{1}\times C_{2}\}

for all i{1,,2d}i\in\{1,\cdots,2d\}, i.e. HiH_{i} is a bounded hyperplane with one of the coordinates fixed. Note that our contradiction assumption means that for all xC1×C2x\in C_{1}\times C_{2}, we have that xx and x1x^{1} must share some coordinate. Therefore, xHix\in H_{i} for some ii which further implies that C1×C21k2dHkC_{1}\times C_{2}\subseteq\bigcup_{1\leq k\leq 2d}H_{k}. But if there exists α1,,α2d\{0}\alpha_{1},\cdots,\alpha_{2d}\in\mathbb{R}\backslash\{0\} such that (x11+α1,,xd1+αd)C1(x_{1}^{1}+\alpha_{1},\cdots,x_{d}^{1}+\alpha_{d})\in C_{1} and (xd+11+αd+1,x2d1+α2d)C2(x_{d+1}^{1}+\alpha_{d+1},\cdots x_{2d}^{1}+\alpha_{2d})\in C_{2} then (x11+α1,,x2d1+α2d)C1×C2(x_{1}^{1}+\alpha_{1},\cdots,x_{2d}^{1}+\alpha_{2d})\in C_{1}\times C_{2} and (x11+α1,,x2d1+α2d)1k2dHk(x_{1}^{1}+\alpha_{1},\cdots,x_{2d}^{1}+\alpha_{2d})\notin\bigcup_{1\leq k\leq 2d}H_{k} which would give us the desired contradiction. Clearly we can do this. Indeed, if we couldn’t then by defining

Ji:={(x11+β1,,xi11+βi1,xi1,xi+11βi+1,,xd1+βd):diam(C1)βjdiam(C1)}J_{i}:=\{(x_{1}^{1}+\beta_{1},\cdots,x_{i-1}^{1}+\beta_{i-1},x_{i}^{1},x_{i+1}^{1}\beta_{i+1},\cdots,x_{d}^{1}+\beta_{d}):-\operatorname{diam}(C_{1})\leq\beta_{j}\leq\operatorname{diam}(C_{1})\}

we would have C11kdJkC_{1}\subseteq\bigcup_{1\leq k\leq d}J_{k} and that each JkJ_{k} is a bounded hyperplane. So by Fact 5.12 this means τ(C1)=0\tau(C_{1})=0 which gives us a contradiction. We similarly come to the same conclusion for C2C_{2}. This proves the base case.

Now assume the claim is true for all nkn\leq k and consider the k+1k+1 case. Again assume for contradiction that we cannot find xk+1C1×C2x^{k+1}\in C_{1}\times C_{2} such that xik+1xijx_{i}^{k+1}\neq x_{i}^{j} for i{1,,2d}i\in\{1,\cdots,2d\} and j{1,,k}j\in\{1,\cdots,k\}. We proceed similarly as in the base case. Since the points x2,,xkx^{2},\cdots,x^{k} are generated in terms of x1x^{1}, we can write xj=(x11+η1j,,x2dj+η2dj)x^{j}=(x_{1}^{1}+\eta_{1}^{j},\cdots,x_{2d}^{j}+\eta_{2d}^{j}) for diam(C1×C2)ηijdiam(C1×C2).-\operatorname{diam}(C_{1}\times C_{2})\leq\eta_{i}^{j}\leq\operatorname{diam}(C_{1}\times C_{2}). Then define

Hij:={(x11+η1j+α1j,,xi11+ηi1j+αi1j,xi1+ηi,xi+11+ηi+1j+αi+1j,,x2d1+η2dj+α2dj:diam(C1×C2)αijdiam(C1×C2)}.\displaystyle H_{i}^{j}:=\left\{\begin{aligned} (x_{1}^{1}+\eta_{1}^{j}+\alpha_{1}^{j}&,\cdots,x_{i-1}^{1}+\eta_{i-1}^{j}+\alpha_{i-1}^{j},x_{i}^{1}+\eta_{i},x_{i+1}^{1}+\eta_{i+1}^{j}+\alpha_{i+1}^{j},\cdots,x_{2d}^{1}+\eta_{2d}^{j}+\alpha_{2d}^{j}\\ &:-\operatorname{diam}(C_{1}\times C_{2})\leq\alpha_{i}^{j}\leq\operatorname{diam}(C_{1}\times C_{2})\end{aligned}\right\}.

Again by our contradiction assumption we have C1×C21jk1i2dHijC_{1}\times C_{2}\subseteq\bigcup_{1\leq j\leq k}\bigcup_{1\leq i\leq 2d}H_{i}^{j}. Similarly, if there exists α1,,α2d\{0}\alpha_{1},\cdots,\alpha_{2d}\in\mathbb{R}\backslash\{0\} and some jj for which (x11+η1j+α1,,xd1+ηdj+αd)C1(x_{1}^{1}+\eta_{1}^{j}+\alpha_{1},\cdots,x_{d}^{1}+\eta_{d}^{j}+\alpha_{d})\in C_{1} and (xd+11+ηd+1j+αd+1,,x2d1+η2dj+α2d)C2(x_{d+1}^{1}+\eta_{d+1}^{j}+\alpha_{d+1},\cdots,x_{2d}^{1}+\eta_{2d}^{j}+\alpha_{2d})\in C_{2} then (x11+η1j+α1,,x2d1+η2dj+α2d)C1×C2(x_{1}^{1}+\eta_{1}^{j}+\alpha_{1},\cdots,x_{2d}^{1}+\eta_{2d}^{j}+\alpha_{2d})\in C_{1}\times C_{2} and (x11+η1j+α1,,x2d1+η2dj+α2d)1jk1i2dHij(x_{1}^{1}+\eta_{1}^{j}+\alpha_{1},\cdots,x_{2d}^{1}+\eta_{2d}^{j}+\alpha_{2d})\notin\bigcup_{1\leq j\leq k}\bigcup_{1\leq i\leq 2d}H_{i}^{j} which would give us the desired conclusion. If we couldn’t do this, then for all jj, by defining

Jij:={(x11+η1j+β1j,,xi11+ηi1j+βi1j,xi1+ηi,xi+11+ηi+1j+βi+1j,,x2d1+η2dj+β2dj:diam(C1×C2)βijdiam(C1×C2)}\displaystyle J_{i}^{j}:=\left\{\begin{aligned} (x_{1}^{1}+\eta_{1}^{j}+\beta_{1}^{j}&,\cdots,x_{i-1}^{1}+\eta_{i-1}^{j}+\beta_{i-1}^{j},x_{i}^{1}+\eta_{i},x_{i+1}^{1}+\eta_{i+1}^{j}+\beta_{i+1}^{j},\cdots,x_{2d}^{1}+\eta_{2d}^{j}+\beta_{2d}^{j}\\ &:-\operatorname{diam}(C_{1}\times C_{2})\leq\beta_{i}^{j}\leq\operatorname{diam}(C_{1}\times C_{2})\end{aligned}\right\}

we would have C11jk1idJijC_{1}\subseteq\bigcup_{1\leq j\leq k}\bigcup_{1\leq i\leq d}J_{i}^{j} and by the same argument as before, this would τ(C1)=0\tau(C_{1})=0, a contradiction. Then the same argument works for C2C_{2} which gives us the desired conclusion. ∎

This lemma allows us to generate a set of points which will be useful in the proof of Theorem 1.8. The next lemma helps to give a criterion that will describe the convex hull of a small subset of a compact set. In particular, this criterion will be used in the proof of Theorem 1.8.

Lemma 5.14.

Let CdC\subset\mathbb{R}^{d} compact such that τ(C)>0\tau(C)>0. Let uCu\in C be a right endpoint of some gap of CC. Then we can find δ=(δ1,,δd)\delta=(\delta_{1},\cdots,\delta_{d}) such that u+δCu+\delta\in C where δi>0\delta_{i}>0. Furthermore, for any η>0\eta>0 we can, possibly, shrink δ\delta so that |δ|<|η||\delta|<|\eta| and we can have δ1=δi\delta_{1}=\delta_{i} for all i{1,,d}i\in\{1,\cdots,d\}.

Proof.

Assume for contradiction that we cannot find δ\delta such that u+δCu+\delta\in C. So now let η>0\eta>0 be given and restrict δ\delta such that |δ|<|η||\delta|<|\eta|. Let GnG_{n} denote the gap for which uGnu\in\partial G_{n}. Define

Lu:={u(1t)+t(u+δ):t(0,1)},L_{u}:=\{u(1-t)+t(u+\delta):t\in(0,1)\},

i.e. LuL_{u} is the line segment connecting uu and u+δu+\delta but not including uu and u+δu+\delta. Recall that since uu is a right endpoint of GnG_{n}, we have that for any ε>0\varepsilon>0, u+(ε,,ε)Gnu+(\varepsilon,\cdots,\varepsilon)\notin G_{n}, so we can take δ\delta such that δ1=δi\delta_{1}=\delta_{i} for all i{1,d}i\in\{1,\cdots d\}. Furthermore, by the contradiction assumption we have LuC=L_{u}\cap C=\emptyset and therefore there exists some index set Λ\Lambda such that LuiΛGiL_{u}\subset\bigcup_{i\in\Lambda}G_{i} where the GiG_{i} are gaps of CC and such that LuGiL_{u}\cap G_{i}\neq\emptyset. We claim there exists a single gap of CC, say GiG_{i}, such that LuGiL_{u}\subset G_{i}. To see this, assume this is not true. Thus, we have that LuGiL_{u}\cap G_{i}\neq\emptyset and GiGj=G_{i}\cap G_{j}=\emptyset for all i,jΛi,j\in\Lambda. This implies that LuL_{u} is a disconnected space which is a contradiction. Therefore, LuGiL_{u}\subset G_{i} for some gap GiG_{i} of CC. But this implies dist(Gi,Gn)=0\operatorname{dist}(G_{i},G_{n})=0. Without loss of generality, assume diam(Gi)diam(Gn)\operatorname{diam}(G_{i})\leq\operatorname{diam}(G_{n}). Then

0<τ(C)dist(Gi,jΛiGjE)diam(Gi)dist(Gi,Gn)diam(Gi)=00<\tau(C)\leq\frac{\operatorname{dist}\left(G_{i},\bigcup_{j\in\Lambda_{i}}G_{j}\cup E\right)}{\operatorname{diam}(G_{i})}\leq\frac{\operatorname{dist}(G_{i},G_{n})}{\operatorname{diam}(G_{i})}=0

a contradiction. Note that Λi\Lambda_{i} is being defined as in Lemma 5.1. ∎

Checking that a compact set is not contained in a gap of another compact set is tricky (which is a hypothesis of Theorem 1.3), so the next lemma and theorem gives a sufficient and easier criterion to work with. Let conv()\operatorname{conv}(\cdot) denote the convex hull of a set.

Lemma 5.15.

Let CdC\subset\mathbb{R}^{d} be a compact set such that τ(C)>0\tau(C)>0. Let U=conv(C)U=\operatorname{conv}(C)^{\circ}. Then UU is non-empty.

Proof.

Assume for contradiction that U=U=\emptyset. Then note that conv(C)\operatorname{conv}(C) has no bounded gaps. Then since U=U=\emptyset, by definition of thickness, this implies τ(conv(C))=0\tau(\operatorname{conv}(C))=0. But note that τ(C)τ(conv(C))\tau(C)\leq\tau(\operatorname{conv}(C)) since conv(C)\operatorname{conv}(C) has simply removed all of the bounded gaps of CC which then means

0<τ(C)τ(conv(C))=00<\tau(C)\leq\tau(\operatorname{conv}(C))=0

a contradiction. ∎

Theorem 5.16.

Let C1C_{1} and C2C_{2} be compact sets in d\mathbb{R}^{d} with τ(C1),τ(C2)>0\tau(C_{1}),\tau(C_{2})>0 and such that their convex hulls are linked. That is, by setting U1=conv(C1)U_{1}=\operatorname{conv}(C_{1})^{\circ} and U2=conv(C2)U_{2}=\operatorname{conv}(C_{2})^{\circ}, we have that UVU\cap V\neq\emptyset, (U)\V\left(\partial U\right)\backslash V\neq\emptyset, and (V)\U\left(\partial V\right)\backslash U\neq\emptyset. Then neither C1C_{1} or C2C_{2} is contained in a gap of the other.

Proof.

Set U1=conv(C1)U_{1}=\operatorname{conv}(C_{1})^{\circ} and U2=conv(C2)U_{2}=\operatorname{conv}(C_{2})^{\circ}. By Lemma 5.15 both U1U_{1} and U2U_{2} are non-empty. Assume for contradiction that C1C_{1} is contained in a gap of C2C_{2}, say G2G^{2}. So C1G2C_{1}\subseteq G^{2}. Note that C1G2C_{1}\subset G^{2} since G2G^{2} is open and C1C_{1} is closed. Then conv(C1)conv(G2)\operatorname{conv}(C_{1})\subseteq\operatorname{conv}(G^{2}). Note that since G2G^{2} is open, conv(G2)\operatorname{conv}(G^{2}) is open. Thus,

U3:=conv(G2)=conv(G2).U_{3}:=\operatorname{conv}(G^{2})^{\circ}=\operatorname{conv}(G^{2}).

Note that U1conv(C1)\partial U_{1}\subseteq\partial\operatorname{conv}(C_{1}). Then if conv(C1)U3\partial\operatorname{conv}(C_{1})\subset U_{3} we’re done since this would imply

(U1)\U2(conv(C1))\U2(conv(C1))\U3=\emptyset\neq(\partial U_{1})\backslash U_{2}\subseteq\left(\partial\operatorname{conv}(C_{1})\right)\backslash U_{2}\subseteq\left(\partial\operatorname{conv}(C_{1})\right)\backslash U_{3}=\emptyset

which yields a contradiction. So now assume conv(C1)U3\partial\operatorname{conv}(C_{1})\supseteq U_{3}. To see that this implies these sets are equal, take xconv(C1)x\in\partial\operatorname{conv}(C_{1}). Then since C1C_{1} is closed, conv(C1)\operatorname{conv}(C_{1}) is also closed and thus we have conv(C1)conv(C1)\partial\operatorname{conv}(C_{1})\subseteq\operatorname{conv}(C_{1}) and therefore, xconv(C1)x\in\operatorname{conv}(C_{1}). But

xconv(C1)conv(G2)=conv(G2)=U3x\in\operatorname{conv}(C_{1})\subseteq\operatorname{conv}(G^{2})=\operatorname{conv}(G^{2})^{\circ}=U_{3}

and therefore, U3=conv(C1)U_{3}=\partial\operatorname{conv}(C_{1}) which is a contradiction since U3U_{3} is open and conv(C1)\partial\operatorname{conv}(C_{1}) is closed. This gives the desired conclusion. ∎

Thus, rather than checking that the two sets are not contained in a gap of the other, we will use the above criterion. We can now finally prove the main ingredient needed which is Theorem 1.8.

Proof of Theorem 1.8.

Let u1u^{1} and u2u^{2} be right endpoints of some gap (not necessarily bounded gaps) of C1C_{1} and C2C_{2} respectively. By Lemma 5.13 we may assume (u1,u2)(u^{1},u^{2}) lives to the upper right of x0x^{0}, i.e. that for i{1,,d}i\in\{1,\cdots,d\} and j{d+1,,2d}j\in\{d+1,\cdots,2d\} we have xi0<ui1x_{i}^{0}<u_{i}^{1} and xj0<uj2x_{j}^{0}<u_{j}^{2}. We assume this so that we are in the setting of Lemma 5.11. By Lemma 5.14 we find δjd\delta^{j}\in\mathbb{R}^{d} such that uj,uj+δjCju^{j},u^{j}+\delta^{j}\in C_{j} for which δ1j=δij\delta_{1}^{j}=\delta_{i}^{j} for all ii. So set Cj~:=Cj([u1i,u1i+δ1i]××[udi,udi+δ1i])\widetilde{C_{j}}:=C_{j}\cap\left([u_{1}^{i},u_{1}^{i}+\delta_{1}^{i}]\times\cdots\times[u_{d}^{i},u_{d}^{i}+\delta_{1}^{i}]\right). In particular we can also choose δi\delta^{i} small enough so that C1~\widetilde{C_{1}} is in the domain of gx,tg_{x,t} and for which τ(C2~)τ(gx,t(C1~))>1\tau(\widetilde{C_{2}})\tau(g_{x,t}(\widetilde{C_{1}}))>1 by Lemma 5.14 and Lemma 5.11. Note that we are invoking Lemma 5.11 for a neighborhood of xx values centered at x0x^{0}. Recall that by Lemma 5.2, we can translate and rotate our set so that we may also assume u1u^{1} and u2u^{2} lie on the diagonal, i.e. ui1=u11u_{i}^{1}=u_{1}^{1} and ui2=u12u_{i}^{2}=u_{1}^{2} for all i{1,d}i\in\{1,\cdots d\}, and that x0=0x^{0}=\vec{0}. To use the Gap Lemma we find (x,t)(x,t) such that C2~\widetilde{C_{2}} and gx,t(C1~)g_{x,t}(\widetilde{C_{1}}) are linked where xx is coming from a neighborhood around x0x^{0}. The parameters xx and tt are important for this proof so let gx,tig_{x,t}^{i} denote gig^{i}. Note that each gx,tig_{x,t}^{i} is decreasing in its argument.

(u12,u22+δ22)(u_{1}^{2},u_{2}^{2}+\delta_{2}^{2})(u12+δ12,u22+δ22)(u_{1}^{2}+\delta_{1}^{2},u_{2}^{2}+\delta_{2}^{2})(u12+δ12,u22)(u_{1}^{2}+\delta_{1}^{2},u_{2}^{2})(u12,u22)(u_{1}^{2},u_{2}^{2})(gx,t1(u11+δ11),gx,t2(u21))(g_{x,t}^{1}(u_{1}^{1}+\delta_{1}^{1}),g_{x,t}^{2}(u_{2}^{1}))(gx,t1(u11),gx,t2(u21))(g_{x,t}^{1}(u_{1}^{1}),g_{x,t}^{2}(u_{2}^{1}))(gx,t1(u11+δ11),gx,t2(u21+δ21))(g_{x,t}^{1}(u_{1}^{1}+\delta_{1}^{1}),g_{x,t}^{2}(u_{2}^{1}+\delta_{2}^{1}))(gx,t1(u11),gx,t2(u21+δ21))(g_{x,t}^{1}(u_{1}^{1}),g_{x,t}^{2}(u_{2}^{1}+\delta_{2}^{1}))
Figure 1. How we want the conditions of UU to look like in 2\mathbb{R}^{2}. The (filled in) rectangles represent the convex hulls of C2~\widetilde{C_{2}} and gx,t(C1~)g_{x,t}(\widetilde{C_{1}}) and thus, their convex hulls will be linked provided that they are in this type of configuration.

So consider the set

U:={(x,t)2d×:gx,t1(u11+δ11)<u12<gx,t1(u11)<u12+δ12,,gx,td(ud1+δd1)<ud2<gx,td(ud1)<ud2+δd2}.U:=\{(x,t)\in\mathbb{R}^{2d}\times\mathbb{R}:g_{x,t}^{1}(u_{1}^{1}+\delta_{1}^{1})<u_{1}^{2}<g_{x,t}^{1}(u_{1}^{1})<u_{1}^{2}+\delta_{1}^{2},\cdots,g_{x,t}^{d}(u_{d}^{1}+\delta_{d}^{1})<u_{d}^{2}<g_{x,t}^{d}(u_{d}^{1})<u_{d}^{2}+\delta_{d}^{2}\}.

Again, note that δ1j=δij\delta_{1}^{j}=\delta_{i}^{j}. We simply use this definition of UU in terms of δij\delta_{i}^{j} to make UU look symmetric. Then if (x,t)U(x,t)\in U, we have that C2~\widetilde{C_{2}} and gx,t(C1~)g_{x,t}(\widetilde{C_{1}}) are linked because their convex hulls will be rectangles in d\mathbb{R}^{d} and for (x,t)U(x,t)\in U, this guarantees that these rectangles intersect each other and exist outside of each other. Then by Theorem 5.16 and Theorem 1.3, we will get that C2~gx,t(C1~)\widetilde{C_{2}}\cap g_{x,t}(\widetilde{C_{1}})\neq\emptyset. Note this means we are using Lemma 5.14.

We will show that UU is a non-empty open set containing a point of the form (x0,t)U(x^{0},t)\in U. The claim then follows because we can take open neighborhoods S,TS,T of x0,tx^{0},t respectively such that

TxSΔx(C1×C2)T\subset\bigcap_{x\in S}\Delta_{x}(C_{1}\times C_{2})

where SS is some subset of the neighborhood found in Lemma 5.11. To see that UU is non-empty, set 1d(t1)2=(x10u11)2+(xd+10u12)2\frac{1}{d}(t_{1})^{2}=(x_{1}^{0}-u_{1}^{1})^{2}+(x_{d+1}^{0}-u_{1}^{2})^{2}. By construction, we have gx0,t11(u11)=u12g_{x^{0},t_{1}}^{1}(u_{1}^{1})=u_{1}^{2}. Since each gx,tig_{x,t}^{i} is increasing in tt, for any t>t1t>t_{1} we will have gx0,t1(u11)>u12g_{x^{0},t}^{1}(u_{1}^{1})>u_{1}^{2}. Also, gx,t(z)g_{x,t}(z) is a continuous function of tt so that when tt is sufficiently close to t1t_{1} we will get gx0,t1(u11+δ11)<u12g_{x^{0},t}^{1}(u_{1}^{1}+\delta_{1}^{1})<u_{1}^{2} and gx0,t1(u11)<u12+δ12g_{x^{0},t}^{1}(u_{1}^{1})<u_{1}^{2}+\delta_{1}^{2}. This satisfies the first condition of UU.

For the rest of the conditions of UU, set 1d(ti)2=(xi0ui1)2+(xi+d0ui2)2\frac{1}{d}(t_{i})^{2}=(x_{i}^{0}-u_{i}^{1})^{2}+(x_{i+d}^{0}-u_{i}^{2})^{2}. But by assumption, since x0=0x^{0}=\vec{0}, ui1=u11u_{i}^{1}=u_{1}^{1}, and ui2=u12u_{i}^{2}=u_{1}^{2} this yields

1d(ti)2=(xi0ui1)2+(xi+d0ui2)2=(u11)2+(u12)2=(x10u11)2+(xd+10u12)2=1d(t1)2.\frac{1}{d}(t_{i})^{2}=(x_{i}^{0}-u_{i}^{1})^{2}+(x_{i+d}^{0}-u_{i}^{2})^{2}=(u_{1}^{1})^{2}+(u_{1}^{2})^{2}=(x_{1}^{0}-u_{1}^{1})^{2}+(x_{d+1}^{0}-u_{1}^{2})^{2}=\frac{1}{d}(t_{1})^{2}.

So by the same argument as above, the same tt which worked for gx0,t1g_{x^{0},t}^{1} will work for gx0,tig_{x^{0},t}^{i} as well. This satisfies the remaining conditions of UU and therefore there exists a point of the form (x0,t)U(x^{0},t)\in U. To see that UU is open, note that for a fixed input zz, each gx,ti(z)g_{x,t}^{i}(z) is continuous in (x,t)(x,t) and therefore, UU is open. This satisfies the claim. ∎

Finally, we conclude this paper with the proof of Theorem 1.10 which we show by induction.

Proof of Theorem 1.10.

To start the induction, we first let k=1k=1 and let x0C1×C2x^{0}\in C_{1}\times C_{2} denote our fixed point. Then observe

ΔTx1(C1×C2)=Δx(C1×C2)\{0}.\Delta_{T_{x}^{1}}(C_{1}\times C_{2})=\Delta_{x}(C_{1}\times C_{2})\backslash\{0\}.

Thus, by Theorem 1.8 there exists and neighborhood SS about xx such that the set

x~SΔx~(C1×C2)\{0}Δx(C1×C2)\{0}\bigcap_{\widetilde{x}\in S}\Delta_{\widetilde{x}}(C_{1}\times C_{2})\backslash\{0\}\subset\Delta_{x}(C_{1}\times C_{2})\backslash\{0\}

has non-empty interior. This proves the base case.

Now take k>1k>1. Then define ΔTxk~(E)\Delta_{\widetilde{T_{x}^{k}}}(E) to be the pinned tree distance set of k+1k+1 vertices where xx is labeled as the first leaf, the first coordinate is always a distance from the fixed point xx to the unique vertex which forms an edge with xx, and the rest of the coordinates behave in the same way that the set ΔTk1(E)\Delta_{T^{k-1}}(E) behaves. In set form,

ΔTxk~(E)={(|xxn|,|xixj|)ij:x2,,xk+1E,xixj,x is a leaf,xn unique vertex forming edge with x}\Delta_{\widetilde{T_{x}^{k}}}(E)=\left\{(|x-x^{n}|,|x^{i}-x^{j}|)_{i\sim j}:\begin{aligned} &x^{2},\cdots,x^{k+1}\in E,x^{i}\neq x^{j},x\text{ is a leaf},\\ &x^{n}\text{ unique vertex forming edge with }x\end{aligned}\right\}

where (|xxn|,|xixj|)ijk(|x-x^{n}|,|x^{i}-x^{j}|)_{i\sim j}\in\mathbb{R}^{k}. Note that

ΔTxk(C1×C2)ΔTxk~(C1×C2)=(Δx(C1×C2)\{0})×ΔTk1(C1×C2).\Delta_{T_{x}^{k}}(C_{1}\times C_{2})\supset\Delta_{\widetilde{T_{x}^{k}}}(C_{1}\times C_{2})=\left(\Delta_{x}(C_{1}\times C_{2})\backslash\{0\}\right)\times\Delta_{T^{k-1}}(C_{1}\times C_{2}).

We have already seen that Δx(C1×C2)\{0}\Delta_{x}(C_{1}\times C_{2})\backslash\{0\} has non-empty interior. Furthermore,

ΔTk1(C1×C2)=x~C1×C2ΔTx~k1(C1×C2)\Delta_{T^{k-1}}(C_{1}\times C_{2})=\bigcup_{\widetilde{x}\in C_{1}\times C_{2}}\Delta_{T_{\widetilde{x}}^{k-1}}(C_{1}\times C_{2})

and by the inductive hypothesis, ΔTx~k1(C1×C2)\Delta_{T_{\widetilde{x}}^{k-1}}(C_{1}\times C_{2}) has non-empty interior for any x~C1×C2\widetilde{x}\in C_{1}\times C_{2} which implies ΔTk1(C1×C2)\Delta_{T^{k-1}}(C_{1}\times C_{2}) has non-empty interior, and thus, the claim is satisfied. ∎

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