Coarse dimension and definable sets in expansions of the ordered real vector space

TL;DR

This paper investigates the properties of definable sets in expansions of the ordered real vector space, establishing conditions under which such expansions can or cannot define all bounded Borel sets, and relating these to coarse dimension and density.

Contribution

It introduces a new criterion linking coarse dimension, nowhere dense sets, and definability of bounded Borel sets in expansions of the real vector space.

Findings

If certain conditions hold, the set's intersection with integer intervals grows slower than any polynomial.

Existence of a linear map making the set dense in the real line.

If the set is nowhere dense, the expansion defines all bounded Borel subsets.

Abstract

Suppose $E \subseteq \mathbb{R}$ is nowhere dense. If $(\mathbb{R},<,+,(x \mapsto \lambda x)_{\lambda \in \mathbb{R} }, E)$ does not define every bounded Borel subset of every $\mathbb{R}^n$ then for every $s > 0$ we have $$ | \{ k \in \mathbb{Z}, -m \leq k \leq m - 1 : [k,k+1] \cap E \neq \emptyset \} | < m^s $$ for sufficiently large $m \in \mathbb{N}$. Then there is an $n \in \mathbb{N}$ and a linear $T : \mathbb{R}^n \to \mathbb{R}$ such that $T(E^n)$ is dense. It follows that if $E$ is in addition nowhere dense then $(\mathbb{R},<,+,0,(x \mapsto \lambda x)_{\lambda \in \mathbb{R}}, E)$ defines every bounded Borel subset of every $\mathbb{R}^n$.

Equations

$$|\{k\in \mathbb{Z},-m\le k\le m-1:[k,k+1]\cap E\ne \varnothing \}|\ge m^s$$

$${\dim }_{\mathrm{CM}}(Z):=\underset{r\to \infty }{\mathrm{limsup}}\frac{\mathcal{M}(1,B_n(r)\cap Z)}{\log (r)}.$$

$$\mathcal{N}(X):=\left|\left\{(k_1,\dots ,k_n)\in {\mathbb{Z}}^n:X\cap \prod _{i=1}^n[k_i,k_i+1]\ne \varnothing \right\}\right|.$$

$$K^{-1}\mathcal{M}(1,X)\le \mathcal{N}(X)\le K\mathcal{M}(1,X).$$

$${\dim }_{\mathrm{CM}}(Z)=\underset{r\to \infty }{\mathrm{limsup}}\frac{\mathcal{N}(B_n(r)\cap Z)}{\log (r)}.$$

$$\mathcal{M}({\delta }^{\prime },X)\le \mathcal{M}(\delta ,X)\le K{\left(\frac{{\delta }^{\prime }}{\delta }\right)}^n\mathcal{M}({\delta }^{\prime },X)$$

$$L^{-1}\mathcal{M}(\delta ,X)\mathcal{M}(\delta ,Y)\le \mathcal{M}(\delta ,X\times Y)\le L\mathcal{M}(\delta ,X)\mathcal{M}(\delta ,Y)$$

$$L^{-1}\mathcal{M}(\delta ,X{)}^2\le \mathcal{M}(\delta ,X^2)\le L\mathcal{M}(\delta ,X{)}^2$$

$${\dim }_{\mathrm{CM}}(X\times Y)\le {\dim }_{\mathrm{CM}}(X)+{\dim }_{\mathrm{CM}}(Y)$$

$${\dim }_{\mathrm{CM}}(X^k)=k{\dim }_{\mathrm{CM}}(X).$$

$$\frac{1}{\lambda }\Vert x-x^{\prime }\Vert -\delta \le \Vert f(x)-f(x^{\prime })\Vert \le \lambda \Vert x-x^{\prime }\Vert +\delta \text{for all}x,x^{\prime }\in X,$$

$$\Vert f(x)\Vert \ge \frac{1}{\lambda }\Vert x\Vert -\delta >r{\lambda }^{-1}-\delta .$$

$$\mathcal{M}(\lambda +2\delta ,Y(r{\lambda }^{-1}-2\delta ))\le \mathcal{M}(1,X(r))\text{for all}r>0.$$

$$L\mathcal{M}(1,Y(r{\lambda }^{-1}-2\delta ))\le \mathcal{M}(\lambda +2\delta ,Y(r{\lambda }^{-1}-2\delta ))$$

$$L\mathcal{M}(1,Y(r{\lambda }^{-1}-2\delta ))\le \mathcal{M}(1,X(r)).$$

$$C_{s,\epsilon }^u:=\{tv:t\in \mathbb{R},|t|>s,v\in \mathbb{S},\Vert v-u\Vert <\epsilon \}$$

$$C_{s,\epsilon }^u=\{(x,y)\in {\mathbb{R}}^2:|y|>\lambda |x|\text{and}\Vert (x,y)\Vert >s\}$$

$$\Vert (x,y)-(x^{\prime },y^{\prime })\Vert <s\text{or}|y-y^{\prime }|\le \lambda |x-x^{\prime }|.$$

$$\Vert (x,y)-(x^{\prime },y^{\prime })\Vert \le |x-x^{\prime }|+|y-y^{\prime }|\le (1+\lambda )|x-x^{\prime }|$$

$$\frac{1}{1+\lambda }\Vert (x,y)-(x^{\prime },y^{\prime })\Vert \le |x-x^{\prime }|.$$

$$\Vert (x,y)-(x^{\prime },y^{\prime })\Vert -s<|x-x^{\prime }|.$$

$$\frac{1}{1+\lambda }\Vert (x,x^{\prime })-(y,y^{\prime })\Vert -s\le |x-x^{\prime }|.$$

$$C_{s,\epsilon }^{u,+}:=\{tv:t\in \mathbb{R},t>s,v\in \mathbb{S},\Vert v-u\Vert <\epsilon \}$$

$$\{u\in {\mathbb{S}}^{+}:T_u(F)\cap I\ne \varnothing \}$$

$$D=\bigcup _{q\in I}\{r\in \mathbb{H}:\angle pou<\angle pqr<\angle pov\}=\bigcup _{q\in I}\bigcup _{w\in J}{T}_w^{-1}(\{q\})=\bigcup _{w\in J}{T}_w^{-1}(I).$$

$$\angle pou<\angle po{z}_1<\angle po{z}_2<\angle pov.$$

$$W:=\{z\in \mathbb{H}:\Vert z\Vert \ge t,\angle po{z}_1<\angle poz<\angle po{z}_2\}\subseteq D.$$

$$S(x,y,x^{\prime },y^{\prime })=T_u(x-x^{\prime },y-y^{\prime })\text{for all}x,y,x^{\prime },y^{\prime }\in \mathbb{R}.$$

$${\dim }_{\mathrm{CM}}T(F^2)={\dim }_{\mathrm{CM}}(F^2)=2{\dim }_{\mathrm{CM}}(F)>s.$$

$${\pi }_k(x_1,\dots ,x_n)=x_k\text{for all}(x_1,\dots ,x_n)\in {\mathbb{R}}^n.$$

Full text

[TABLE]

for all sufficiently large $m\in\mathbb{N}$. Then there is an $n\in\mathbb{N}$ and a linear $T:\mathbb{R}^{n}\to\mathbb{R}$ such that $T(E^{n})$ is dense. It follows that if $E$ is in addition nowhere dense then $(\mathbb{R},<,+,0,(x\mapsto\lambda x)_{\lambda\in\mathbb{R}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$.

1. Introduction

Let $X\subseteq\mathbb{R}^{n}$ be bounded and $Z\subseteq\mathbb{R}^{n}$. Given a positive $\delta\in\mathbb{R}$ we let $\mathscr{M}(\delta,X)$ be the minimum number of open $\delta$-balls required to cover $X$. Equivalently $\mathscr{M}(\delta,X)$ is the minimal cardinality of a subset $S$ of $X$ such that every $x\in X$ lies within distance $\delta$ of some element of $S$. Let $B_{n}(p,r)$ be the open ball in $\mathbb{R}^{n}$ with center $p$ and radius $r>0$ and let $B_{n}(r)=B_{n}(0,r)$. We define the coarse Minkowski dimension of $Z$ to be

[TABLE]

It is easy to see that the coarse Minkowski dimension of $Z$ is bounded above by $n$ and the coarse Minkowski dimension of a bounded set is zero. An application of the first claim of Fact 2.1 below shows that replacing one with any fixed real number $\delta>0$ does not change the coarse Minkowski dimension. A simple computation shows that $\dim_{\mathrm{CM}}(Z)$ is the infimum of the set of positive $s\in\mathbb{R}$ such that $\mathscr{M}(1,B_{n}(r)\cap Z)<r^{s}$ for all sufficiently large $r>0$.

We define

[TABLE]

It is well-known and easy to see that there is a real number $K>0$ depending only $n$ such that

[TABLE]

So

[TABLE]

Our main geometric result is Theorem 1.1.

Theorem 1.1**.**

Suppose $E\subseteq\mathbb{R}$. If $\dim_{\mathrm{CM}}(E)>0$ then $T(E^{n})$ is dense for some $n\in\mathbb{N}$ and linear $T:\mathbb{R}^{n}\to\mathbb{R}$. Equivalently, if there is a positive $s\in\mathbb{R}$ such that $\mathscr{N}(B_{1}(r)\cap E)\geq r^{s}$ for all sufficiently large $r\in\mathbb{R}$ then there exist $n\in\mathbb{N}$ and linear $T:\mathbb{R}^{n}\to\mathbb{R}$ such that $T(E^{n})$ is dense.

The converse implication to Theorem 1.1 does not hold. Let $D=\{2^{n},2^{n}+n:n\in\mathbb{N}\}$. A simple computation shows that $D$ has coarse Minkowski dimension zero. Let $S:\mathbb{R}^{4}\to\mathbb{R}$ be given by $S(x_{1},x_{2},x_{3},x_{4})=(x_{1}-x_{2})+\alpha(x_{3}-x_{4})$ for a fixed $\alpha\in\mathbb{R}\setminus\mathbb{Q}$. Then $S(D^{4})$ is dense.

Theorem 1.1 is motivated by an application to logic that we now describe. Let $\mathbb{R}_{\mathrm{Vec}}$ be the ordered vector space $(\mathbb{R},<,+,0,(x\mapsto\lambda x)_{\lambda\in\mathbb{R}})$ of real numbers. For any subset $E$ of $\mathbb{R}$ let $(\mathbb{R}_{\mathrm{Vec}},E)$ be the expansion of $\mathbb{R}_{\mathrm{Vec}}$ by a unary predicate defining $E$. When we say that a subset of $\mathbb{R}^{n}$ is definable in a first order expansion of $(\mathbb{R},<,+,0)$ such as $(\mathbb{R}_{\mathrm{Vec}},E)$ we mean that it is first order definable possibly with parameters from $\mathbb{R}$.

Hieronymi and Tychonievich [6] show that $(\mathbb{R}_{\mathrm{Vec}},\mathbb{Z})$ defines all bounded Borel subsets of all $\mathbb{R}^{n}$. In contrast, it follows from [8, 9] that every subset of $\mathbb{R}^{n}$ definable in $(\mathbb{R},<,+,0,\mathbb{Z})$ is a finite union of locally closed sets.

The theorem of Hieronymi and Tychonievich is a special case of Theorem 1.2. Theorem 1.2 also follows from a more general theorem of Fornasiero, Hieronymi, and Walsberg [2, Theorem 7.3, Corollary 7.5]. We let $\mathrm{Cl}(E)$ be the closure of $E\subseteq\mathbb{R}$ and $\mathrm{Bd}(E)$ be the boundary of $E$. Recall that the boundary of a subset of $\mathbb{R}$ is always closed.

Theorem 1.2**.**

Suppose that $E\subseteq\mathbb{R}$ is not dense and co-dense in any nonempty open interval. Then the following are equivalent:

1. (1)

$(\mathbb{R}_{\mathrm{Vec}},E)$* does not define every bounded Borel subset of every $\mathbb{R}^{n}$,* 2. (2)

Every subset of $\mathbb{R}$ definable in $(\mathbb{R}_{\mathrm{Vec}},E)$ either has interior or is nowhere dense, 3. (3)

$T(\mathrm{Bd}(E)^{n})$* is nowhere dense for every linear $T:\mathbb{R}^{n}\to\mathbb{R}$.*

The implication $(3)\Rightarrow(2)$ is a corollary of a result of Friedman and Miller [3]. The implication $(1)\Rightarrow(3)$ is a corollary of the main theorem of [6]. Note that $\mathrm{Bd}(E)$ is nowhere dense as $E$ is not dense and co-dense in any open interval. If $E$ is bounded then $(3)$ above is equivalent to a natural geometric condition on $E$. This equivalence, observed in [2, Theorem 7.3], is an easy consequence of the famous Marstrand projection theorem (see Mattila [7, Chapter 9]) and the classical theorem of Steinhaus that $Z-Z:=\{z-z^{\prime}:z,z^{\prime}\in Z\}$ has interior whenever $Z\subseteq\mathbb{R}^{n}$ has positive $n$-dimensional Lebesgue measure.

Fact 1.3**.**

Suppose $F\subseteq\mathbb{R}$ is bounded. Then $T(F^{n})$ is nowhere dense for every linear $T:\mathbb{R}^{n}\to\mathbb{R}$ if and only if $\mathrm{Cl}(F)^{n}$ has Hausdorff dimension zero for all $n\in\mathbb{N}$.

Fact 1.3 does not hold for unbounded subsets of $\mathbb{R}$. The set of integers, like any countable set, has Hausdorff dimension zero, and $T(\mathbb{Z}^{2})$ is dense for any linear $T:\mathbb{R}^{2}\to\mathbb{R}$ of the form $T(x,y)=x+\alpha y$ with $\alpha\in\mathbb{R}\setminus\mathbb{Q}$. Combining Theorem 1.1 and Theorem 1.2 we obtain the following.

Theorem 1.4**.**

Suppose $E\subseteq\mathbb{R}$ is not dense and co-dense in any nonempty open interval. If $\mathrm{Bd}(E)$ has positive coarse Minkowski dimension then $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$. In particular if $E$ is nowhere dense and has positive coarse Minkowski dimension then $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$.

Note that $\mathbb{Z}$ has coarse Minkowski dimension one so Theorem 1.4 generalizes the result of Hieronymi and Tychonievich described above. There are subsets $E$ of $\mathbb{R}$ with coarse Minkowski dimension zero such that $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$ such as $\{2^{n},2^{n}+n:n\in\mathbb{N}\}$ (see the comment after Theorem 1.1). Theorem 1.4 fails without the assumption that $E$ is not dense and co-dense in any nonempty open interval. Block-Gorman, Hieronymi, and Kaplan [4] show that every closed subset of $\mathbb{R}^{n}$ definable in $(\mathbb{R}_{\mathrm{Vec}},\mathbb{Q})$ is already definable in $\mathbb{R}_{\mathrm{Vec}}$ and $\mathrm{Bd}(\mathbb{Q})=\mathbb{R}$ has coarse Minkowski dimension one.

The present paper is part of the broader study of the metric geometry of definable sets in first order structures expanding $(\mathbb{R},<,+,0)$, see [1, 2, 5]. Fornsiero, Hieronymi, and Miller [1] show that if $E\subseteq\mathbb{R}$ is nowhere dense and has positive Minkowski dimension then $(\mathbb{R},<,+,\cdot,0,1,E)$ defines every Borel subset of every $\mathbb{R}^{n}$. This statement fails over $\mathbb{R}_{\mathrm{Vec}}$, as $D=\{\frac{1}{n}:n\in\mathbb{N},n\geq 1\}$ has Minkowski dimension one and Fact 1.3 and Theorem 1.2 together imply that every subset of $\mathbb{R}$ definable in $(\mathbb{R}_{\mathrm{Vec}},D)$ either has interior or is nowhere dense. It is shown in [2] that if $E\subseteq\mathbb{R}^{n}$ is closed and the topological dimension of $E$ is strictly less than the Hausdorff dimension of $E$ then $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$.

As a closed subset of $\mathbb{R}$ has topological dimension zero if it is nowhere dense and topological dimension one if it has interior, Theorem 1.4 shows that if $E\subseteq\mathbb{R}$ is closed and the topological dimension of $E$ is strictly less than the coarse Minkowski dimension of $E$ then $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$. It is natural to conjecture that if $E\subseteq\mathbb{R}^{n}$ is closed and the topological dimension of $E$ is strictly less then the coarse Minkowski dimension of $E$ then $(\mathbb{R}_{\mathrm{Vec}},E)$ defines every bounded Borel subset of every $\mathbb{R}^{n}$. In Theorem 4.1 we will show as a corollary to Theorem 1.4 that if $Z\subseteq\mathbb{R}^{n}$ is closed and has topological dimension zero and positive coarse Minkowksi dimension then $(\mathbb{R}_{\mathrm{Vec}},Z)$ defines every bounded Borel subset of $\mathbb{R}^{n}$.

Acknowledgements

I thank the referee for many improvements and Philipp Hieronymi for useful discussions.

2. Metric Notions

We recall two useful facts about $\mathscr{M}(\delta,X)$ and $\mathscr{N}(X)$, both of which are easy to see. One can find more information about these invariants in Yomdin and Comte [10, Chapter 2] and many other places.

Fact 2.1**.**

Let $n\in\mathbb{N}$. There are $K,L>0$ such that for all bounded $X,Y\subseteq\mathbb{R}^{n}$ and $0<\delta<\delta^{\prime}$

[TABLE]

and

[TABLE]

In particular

[TABLE]

for all bounded $X\subseteq\mathbb{R}^{n}$.

The proof of the fact below is a straightforward computation that is essentially the same as the proof of the analogous fact for Minkowski dimension. We leave the proof to the reader.

Fact 2.2**.**

For any $X\subseteq\mathbb{R}^{n},Y\subseteq\mathbb{R}^{m}$ and $k\in\mathbb{N}$ we have

[TABLE]

and

[TABLE]

Suppose that $X\subseteq\mathbb{R}^{n}$, $Y\subseteq\mathbb{R}^{m}$, $f$ is a map $X\to Y$, and $\lambda,\delta>0$. Then $f$ is a $(\lambda,\delta)$-quasi-isometry if

[TABLE]

and if for every $y\in Y$ we have $\|f(x)-y\|<\delta$ for some $x\in X$. We say that $f$ is a quasi-isometry if it is a $(\lambda,\delta)$-quasi-isometry for some $\lambda,\delta>0$. It is well-known and easy to see that if there is a quasi-isometry $X\to Y$ then there is also a quasi-isometry $Y\to X$. A map $g:X\to\mathbb{R}^{n}$ is a quasi-isometric embedding if it yields a quasi-isometry $X\to g(X)$.

Lemma 2.3**.**

Suppose $X\subseteq\mathbb{R}^{n}$, $Y\subseteq\mathbb{R}^{m},0\in X,0\in Y$, and $f:X\to Y$ is a quasi-isometry such that $f(0)=0$. Then $X$ and $Y$ have the same coarse Minkowski dimension.

Lemma 2.3 holds without the assumptions that $0\in X,0\in Y$, and $f(0)=0$. We do not prove this more general result to avoid technicalities.

Proof.

We show that $\dim_{\mathrm{CM}}(Y)\leq\dim_{\mathrm{CM}}(X)$. As there is a quasi-isometry $Y\to X$ that also maps [math] to [math] the same argument yields the other inequality. Fix $\lambda,\delta>0$ such that $f$ is a $(\lambda,\delta)$-quasi-isometry.

Fix $r>0$. Let $X(r)=B_{n}(0,r)\cap X$ and $Y(r)=B_{m}(0,r)\cap Y$. Let $\{B_{n}(p_{i},1)\}_{i=1}^{k}$ be a minimal covering of $X(r)$ by balls with radius $1$. Then $\{f(B_{n}(p_{i},1))\}_{i=1}^{k}$ covers $f(X(r))$. Let $q_{i}=f(p_{i})$ for all $i$. As $f$ is a $(\lambda,\delta)$-quasi-isometry we see that $f(B_{n}(p_{i},1))$ is contained in $B_{m}(q_{i},\lambda+\delta)$ for all $i$. So $\{B_{m}(q_{i},\lambda+\delta)\}_{i=1}^{k}$ covers $f(X(r))$.

We now show that every point in $Y(r\lambda^{-1}-2\delta)$ lies within distance $\delta$ of $f(X(r))$. Fix $y\in Y(r\lambda^{-1}-2\delta)$. As $f$ is a $(\lambda,\delta)$-quasi-isometry there is $x\in X$ such that $\|f(x)-y\|<\delta$. Suppose $\|x\|>r$. Then as $f(0)=0$ we have

[TABLE]

As $\|f(x)-y\|<\delta$ the triangle inequality yields $\|y\|>r\lambda^{-1}-2\delta$. Contradiction.

Combining the previous paragraphs we see that $\{B_{m}(q_{i},\lambda+2\delta)\}_{i=1}^{k}$ covers $Y(r\lambda^{-1}-2\delta)$. Thus

[TABLE]

Applying the first claim of Fact 2.1 we obtain a constant $L>0$ depending only on $m$ such that

[TABLE]

hence

[TABLE]

Taking logarithms of of both sides of the expression above, dividing both sides by $\log(r)$, and taking the limit as $r\to\infty$ we see that $\dim_{\mathrm{CM}}(Y)\leq\dim_{\mathrm{CM}}(X)$. ∎

3. Proof of Theorem 1.1

Let $\mathbb{S}$ be the unit circle in $\mathbb{R}^{2}$. Given $u\in\mathbb{S}$ we let $T_{u}:\mathbb{R}^{2}\to\mathbb{R}$ be the orthogonal projection parallel to $u$, i.e., $T_{u}$ is the orthogonal projection such that $T_{u}(x)=T_{u}(y)$ if and only if $x-y=tu$ for some $t\in\mathbb{R}$. For our purposes a double wedge around $u\in\mathbb{S}$ is a subset of $\mathbb{R}^{2}$ of the form

[TABLE]

for some $s,\varepsilon>0$.

Lemma 3.1**.**

Let $F$ be a nonempty subset of $\mathbb{R}^{2}$ and $u\in\mathbb{S}$. If $F-F=\{x-y:x,y\in F\}$ is disjoint from some double wedge around $u$ then the restriction of $T_{u}$ to $F$ is a quasi-isometric embedding $F\to\mathbb{R}$.

Lemma 3.1 is a quasi-isometric version of a well-known fact from geometric measure theory: if $F$ is a nonempty subset of $\mathbb{R}^{2}$ such that $F-F$ is disjoint from a double wedge of the form $C^{u}_{\varepsilon,0}$ then the restriction of $T_{u}$ to $F$ is a bilipschitz embedding $F\to\mathbb{R}$. This fact is applied in [1, 5].

Proof.

Suppose that $F-F$ is disjoint from $C^{u}_{s,\varepsilon}$. As $T_{u}$ is an othogonal projection we have $\|T_{u}(x)-T_{u}(x^{\prime})\|\leq\|x-x^{\prime}\|$ for all $x,x^{\prime}\in\mathbb{R}^{2}$, so it suffices to obtain a lower bound on $\|T_{u}(X)-T_{u}(x^{\prime})\|$ of the appropriate form.

After making a change of coordinates if necessary we suppose $u=(0,1)$ so that $T_{u}(x,y)=x$ for all $(x,y)\in\mathbb{R}^{2}$. Then we have

[TABLE]

for some $\lambda>0$ depending only on $\varepsilon$. Thus, if $(x,y)\in F-F$ then either $\|(x,y)\|<s$ or $|y|\leq\lambda|x|$. Equivalently, for all $(x,y),(x^{\prime},y^{\prime})\in F$ we either have

[TABLE]

In the latter case we have

[TABLE]

hence

[TABLE]

In the first case we have

[TABLE]

So for all $(x,y),(x^{\prime},x^{\prime})\in F$ we have

[TABLE]

So the restriction of $T_{u}$ to $F$ is a quasi-isometric embedding $F\to\mathbb{R}^{2}$. ∎

We let $\mathbb{H}$ be the upper half plane $\{(x,y)\in\mathbb{R}^{2}:y>0\}$ and let $\mathbb{S}^{+}=\mathbb{S}\cap\mathbb{H}$. A wedge in $\mathbb{H}$ around $u\in\mathbb{S}^{+}$ is a set of the form

[TABLE]

such that $C^{u,+}_{s,\varepsilon}\subseteq\mathbb{H}$.

Lemma 3.2**.**

Suppose $F\subseteq\mathbb{H}$ intersects every wedge in $\mathbb{H}$. Then there is a $u\in\mathbb{S}^{+}$ such that $T_{u}(F)$ is dense.

The reader may find that drawing a few pictures greatly assists in comprehending the proof of Lemma 3.2. We let $p=(-1,0)$ and $o=(0,0)$. Note that if $z\in\mathbb{H}$, $q$ is a positive real number, and $u\in\mathbb{S}^{+}$, then $T_{u}(z)=q$ if and only if $\angle pou=\angle pqz$.

Proof.

We show that the set of $u\in\mathbb{S}^{+}$ such that $T_{u}(F)$ is dense in $\mathbb{R}$ is comeager in $\mathbb{S}^{+}$. It suffices to show that

[TABLE]

is open and dense in $\mathbb{S}^{+}$ for every nonempty open interval $I$ with rational endpoints. Fix a nonempty open interval $I=(q_{1},q_{2})$ with rational endpoints. We suppose that $q_{1},q_{2}>0$ for the sake of simplicity, the more general case follows by trivial modifications of our argument. The map $T:\mathbb{S}^{+}\times\mathbb{R}^{2}\to\mathbb{R}$ given by $T(u,x)=T_{u}(x)$ is continuous. Thus if $T_{u}(x)\in I$ then $T_{v}(x)\in I$ for all $v\in\mathbb{S}^{+}$ sufficiently close to $u$. It follows that the set of $u$ such that $T_{u}(F)\cap I\neq\emptyset$ is open in $\mathbb{S}^{+}$.

It now suffices to show that the set of $w\in\mathbb{S}^{+}$ such that $T_{w}(F)\cap I\neq\emptyset$ is dense in $\mathbb{S}^{+}$. Fix $u,v\in\mathbb{S}^{+}$ such that $\angle pou<\angle pov$ and let $J$ be the set of $w\in\mathbb{S}^{+}$ such that $\angle pou<\angle pow<\angle pov$. We show there is a $w\in J$ such that $T_{w}(F)\cap I\neq\emptyset$. Let $r_{1},r_{2}\in\mathbb{H}$ be such that $\angle pq_{1}r_{1}=\angle pou$ and $\angle pq_{2}r_{2}=\angle pov$. Let $D$ be the set of points in $\mathbb{H}$ that lie in between the rays $\overrightarrow{q_{1}r_{1}}$ and $\overrightarrow{q_{2}r_{2}}$. It is easy to see that

[TABLE]

It therefore suffices to show that $D$ intersects $F$. Let $z_{1},z_{2}\in\mathbb{S}^{+}$ be such that

[TABLE]

As $\angle pq_{1}r_{1}<\angle poz_{1}<\angle poz_{2}<\angle pq_{2}r_{2}$, we see that every element of $\overrightarrow{oz_{1}}$ or $\overrightarrow{oz_{2}}$ sufficiently far from the origin lies in $D$. It follows that there is a $t>0$ such that

[TABLE]

Then $W$ is a wedge in $\mathbb{H}$ and so contains an element of $F$. Thus $D$ contains an element of $F$. ∎

Lemma 3.3**.**

Suppose $E\subseteq\mathbb{R}$. Then one of the following holds:

1. (1)

there is a $u\in\mathbb{S}$ such that the restriction of $T_{u}$ to $E^{2}$ is a quasi-isometric embedding $E^{2}\to\mathbb{R}$, 2. (2)

there is a linear $S:\mathbb{R}^{4}\to\mathbb{R}$ such that $S(E^{4})$ is dense.

Proof.

Consider $E^{2}-E^{2}\subseteq\mathbb{R}^{2}$. If $E^{2}-E^{2}$ is disjoint from a double wedge in $\mathbb{R}^{2}$ then Lemma 3.2 shows that some $T_{u}$ quasi-isometrically embeds $E^{2}$ into $\mathbb{R}$.

Suppose $E^{2}-E^{2}$ intersects every double wedge in $\mathbb{R}^{2}$. Note that if $(x,y)\in E^{2}-E^{2}$ then $(-x,-y)$ is also an element of $E^{2}-E^{2}$. It is easy to see that this implies that $E^{2}-E^{2}$ intersects every wedge in $\mathbb{H}$. Applying Lemma 3.3 we fix a $u\in\mathbb{S}$ such that $T_{u}(E^{2}-E^{2})$ is dense. Let $S:\mathbb{R}^{4}\to\mathbb{R}$ be the linear function given by

[TABLE]

Then $S(E^{4})$ is dense. ∎

We now prove Theorem 1.1.

Proof.

Suppose towards a contradiction that $E\subseteq\mathbb{R}$ has positive coarse Minkowski dimension and $T(E^{n})$ is not dense for every $n\in\mathbb{N}$ and linear $T:\mathbb{R}^{n}\to\mathbb{R}$. We may suppose that $0\in E$. Let $\mathcal{S}$ be the collection of sets of the form $T(E^{n})$ for linear $T:\mathbb{R}^{n}\to\mathbb{R}$. It is easy to see that if $F\in\mathcal{S}$ and $T:\mathbb{R}^{n}\to\mathbb{R}$ is linear then $T(F^{n})$ is also in $\mathcal{S}$. We let $s$ be the supremum of the coarse Minkowski dimensions of members of $\mathcal{S}$. Every element of $\mathcal{S}$ has coarse Minkowski dimension $\leq 1$, so $s$ exists and $s\leq 1$. As $\dim_{\mathrm{CM}}(E)>0$ we have $s>0$. Let $F\in\mathcal{S}$ be such that $\dim_{\mathrm{CM}}(F)>\frac{1}{2}s$. An application of Lemma 3.3 yields a linear $T:\mathbb{R}^{2}\to\mathbb{R}$ such that the restriction of $T$ to $F^{2}$ is a quasi-isometric embedding $F^{2}\to\mathbb{R}$. Lemma 2.3 and Fact 2.2 together show that

[TABLE]

But $T(F^{2})\in\mathcal{S}$, contradiction. ∎

4. A corollary in $\mathbb{R}^{n}$

We prove a higher dimensional version of the second claim of Theorem 1.4. (Recall that a closed subset of $\mathbb{R}^{n}$ has topological dimension zero if and only if it is nowhere dense.)

Theorem 4.1**.**

Suppose $Z$ is a closed subset of $\mathbb{R}^{n}$ with topological dimension zero. If $Z$ has positive coarse Minkowski dimension then $(\mathbb{R}_{\mathrm{Vec}},Z)$ defines all bounded Borel subsets of all $\mathbb{R}^{n}$.

Proof.

We suppose that $(\mathbb{R}_{\mathrm{Vec}},Z)$ does not define all bounded Borel subsets of all $\mathbb{R}^{n}$ and show that $\dim_{\mathrm{CM}}(Z)=0$. Given $1\leq k\leq n$ we let $\pi_{k}:\mathbb{R}^{n}\to\mathbb{R}$ be given by

[TABLE]

An application of [2, Theorem D, Theorem E] shows that $\pi_{k}(Z)$ is nowhere dense for all $1\leq k\leq n$. Theorem 1.4 shows that $\dim_{\mathrm{CM}}\pi_{k}(Z)=0$ for all $1\leq k\leq n$. As $Z$ is a subset of $\pi_{1}(Z)\times\ldots\times\pi_{n}(Z)$ repeated application of Fact 2.2 shows that $\dim_{\mathrm{CM}}(Z)=0$. ∎