Directions sets: A generalization of ratio sets

TL;DR

This paper introduces and analyzes the concept of k-directions sets, generalizing ratio sets, and characterizes their accumulation points and density properties within the unit sphere.

Contribution

It defines k-directions sets as a generalization of ratio sets, characterizes their accumulation points, and provides conditions for their density in the sphere.

Findings

Characterization of sets of accumulation points of k-directions sets.

A simple sufficient condition for density of D^k(A) in the sphere.

Extension of ratio set concepts to higher dimensions.

Abstract

For every integer $k \geq 2$ and every $A \subseteq \mathbb{N}$, we define the \emph{$k$-directions sets} of $A$ as $D^k(A) := \{{\bf a} / \|{\bf a}\| : {\bf a} \in A^k\}$ and $D^{\underline{k}}(A) := \{{\bf a} / \|{\bf a}\| : {\bf a} \in A^{\underline{k}}\}$, where $\|\cdot\|$ is the Euclidean norm and $A^{\underline{k}} := \{{\bf a} \in A^k : a_i \neq a_j \text{ for all } i \neq j\}$. Via an appropriate homeomorphism, $D^k(A)$ is a generalization of the \emph{ratio set} $R(A) := \{a / b : a,b \in A\}$, which has been studied by many authors. We study $D^k(A)$ and $D^{\underline{k}}(A)$ as subspaces of $S^{k-1} := \{{\bf x} \in [0,1]^k : \|{\bf x}\| = 1\}$. In~particular, generalizing a result of Bukor and T\'oth, we provide a characterization of the sets $X \subseteq S^{k-1}$ such that there exists $A \subseteq \mathbb{N}$ satisfying $D^{\underline{k}}(A)^\prime = X$, where $Y^\prime$ denotes the set of accumulation points of $Y$. Moreover, we provide a simple sufficient condition for $D^k(A)$ to be dense in $S^{k-1}$. We conclude leaving some questions for further research.

Equations

$$D^k(A):=\{\rho (\mathbf{a}):\mathbf{a}\in A^k\}\text{ and }{D}^{\underline{k}}(A):=\{\rho (\mathbf{a}):\mathbf{a}\in A^{\underline{k}}\},$$

$$x_i<\frac{a_{m_i}}{m}\le \frac{a_{m_i}}{a_{m_i-1}}{x}_i,$$

Full text

$\dagger\,$P. Leonetti is supported by the Austrian Science Fund (FWF), project F5512-N26

$\ddagger\,$C. Sanna is supported by a postdoctoral fellowship of INdAM and is a member of the INdAM group GNSAGA

1. Introduction

Given $A\subseteq\mathbb{N}$, its ratio set is defined as $R(A):=\{a/b:a,b\in A\}$. The study of the topological properties of $R(A)$ as a subspace of $[0,+\infty]$, especially the question of when $R(A)$ is dense in $[0,+\infty]$, is a classical topic and has been considered by many researchers [1, 3, 4, 10, 12, 13, 20, 21, 22, 23, 2, 19]. More recently, some authors have also studied $R(A)$ as a subspace of the $p$-adic numbers $\mathbb{Q}_{p}$ [6, 8, 9, 14, 15, 17].

We consider a further variation on this theme, which stems from the following easy observation: We have that $[0,+\infty]$ is homeomorphic to $S^{1}:=\{\bm{x}\in[0,1]^{2}:\|\bm{x}\|=1\}$ via the map $x\mapsto(1,x)/\|(1,x)\|$, if $x\in{[0,+\infty)}$, and $+\infty\mapsto(0,1)$. This sends $R(A)$ onto $D^{2}(A):=\{\rho(\bm{a}):\bm{a}\in A^{2}\}$, where $\rho(\bm{a}):=\bm{a}/\|\bm{a}\|$ for each $\bm{a}\neq\bm{0}$. Hence, topological questions about $R(A)$ as a subspace of $[0,+\infty]$ are equivalent to questions about $D^{2}(A)$ as a subspace of $S^{1}$. The novelty of this approach is that it can be generalized to higher dimensions. For every integer $k\geq 2$, define the $k$-directions sets of $A$ as

[TABLE]

where for every set $B$ we let $B^{\underline{k}}:=\{\bm{b}\in B^{k}:b_{i}\neq b_{j}\text{ for all }i\neq j\}$ denote the set of $k$-tuples with pairwise distinct entries in $B$. Put also $S^{k-1}:=\{\bm{x}\in[0,1]^{k}:\|\bm{x}\|=1\}$. We shall study $D^{k}(A)$ and $D^{\underline{k}}(A)$ as subspaces of $S^{k-1}$.

Bukor and Tóth [3] characterized the subsets of $[0,+\infty]$ that are equal to $R(A)^{\prime}$ for some $A\subseteq\mathbb{N}$, where $Y^{\prime}$ denotes the set of accumulation points of $Y$. In terms of $D^{2}(A)$, via the homeomorphism $[0,+\infty]\to S^{1}$ mentioned above, their result is the following:

Theorem 1.1**.**

Let $X\subseteq S^{1}$. Then there exists $A\subseteq\mathbb{N}$ such that $X=D^{2}(A)^{\prime}$ if and only if the following conditions are satisfied:

1. (i)

$X$* is closed;* 2. (ii)

$(x_{1},x_{2})\in X$* implies $(x_{2},x_{1})\in X$;* 3. (iii)

if $X$ is nonempty, then $(1,0)\in X$.

Note that Theorem 1.1 holds also if $D^{2}(A)$ is replaced by $D^{\underline{2}}(A)$. Indeed, $D^{\underline{2}}(A)\subseteq D^{2}(A)\subseteq D^{\underline{2}}(A)\cup\{\rho(1,1)\}$ and consequently $D^{2}(A)^{\prime}=D^{\underline{2}}(A)^{\prime}$.

Our first result generalizes Theorem 1.1. Before stating it, we need to introduce some notation. Let $\bm{x}=(x_{1},\dots,x_{k})\in S^{k-1}$. For every permutation $\pi$ of $\{1,\dots,k\}$, we put $\pi(\bm{x}):=(x_{\pi(1)},\dots,x_{\pi(k)})$. Also, for every $I\subseteq\{1,\dots,k\}$, we say that $I$ meets $\bm{x}$ if there exists $j\in I$ such that $x_{j}\neq 0$. In such a case, we put $\rho_{I}(\bm{x}):=\rho(\bm{y})$, where $\bm{y}=(y_{1},\dots,y_{k})$ is defined by $y_{i}:=x_{i}$ if $i\in I$, and $y_{i}:=0$ for $i\notin I$. (This is well defined since $\bm{y}\neq\bm{0}$.)

Our first result is the following:

Theorem 1.2**.**

Let $X\subseteq S^{k-1}$ for some integer $k\geq 2$. Then there exists $A\subseteq\mathbb{N}$ such that $X=D^{\underline{k}}(A)^{\prime}$ if and only if the following conditions are satisfied:

1. (i)

$X$* is closed;* 2. (ii)

$\bm{x}\in X$* implies $\pi(\bm{x})\in X$, for every permutation $\pi$ of $\{1,\dots,k\}$;* 3. (iii)

$\bm{x}\in X$* implies $\rho_{I}(\bm{x})\in X$, for every $I\subseteq\{1,\dots,k\}$ that meets $\bm{x}$.*

Note that Theorem 1.2 is indeed a generalization of Theorem 1.1, since $\rho_{I}(\bm{x})\in\{\bm{x},(1,0),(0,1)\}$ for every $I\subseteq\{1,2\}$ that meets $\bm{x}\in S^{1}$. Furthermore, for $k\geq 3$, Theorem 1.2 is false if $D^{\underline{k}}(A)$ is replaced by $D^{k}(A)$ (see Remark 2.1 below).

Now we turn our attention to the question of when $D^{k}(A)$ is dense in $S^{k-1}$. First, we have the following easy proposition.

Proposition 1.3**.**

Let $k\geq 2$ be an integer and fix $A\subseteq\mathbb{N}$. We have that $D^{k}(A)$ is dense in $S^{k-1}$ if and only if $D^{\underline{k}}(A)$ is dense in $S^{k-1}$.

Proof.

On the one hand, since $D^{\underline{k}}(A)\subseteq D^{k}(A)$, if $D^{\underline{k}}(A)$ is dense in $S^{k-1}$ then $D^{k}(A)$ is dense in $S^{k-1}$. On the other hand, suppose that $D^{k}(A)$ is dense in $S^{k-1}$. Then, for every $\bm{x}\in S^{k-1}\cap\mathbb{R}^{\underline{k}}$, there exists $\bm{a}^{(n)}\in A^{k}$ such that $\rho(\bm{a}^{(n)})\to\bm{x}$. Consequently, for all sufficiently large $n$ we have $\bm{a}^{(n)}\in A^{\underline{k}}$. This implies that $D^{\underline{k}}(A)$ is dense in $S^{k-1}\cap\mathbb{R}^{\underline{k}}$. Since $S^{k-1}\cap\mathbb{R}^{\underline{k}}$ is dense in $S^{k-1}$, we get that $D^{\underline{k}}(A)$ is dense in $S^{k-1}$, as desired. ∎

The next result shows that if $D^{k}(A)$ is dense in $S^{k-1}$, for some integer $k\geq 3$ and $A\subseteq\mathbb{N}$, then $D^{k-1}(A)$ is dense in $S^{k-2}$, but the opposite implication is false.

Theorem 1.4**.**

Let $k\geq 3$ be an integer. On the one hand, if $D^{k}(A)$ is dense in $S^{k-1}$, for some $A\subseteq\mathbb{N}$, then $D^{k-1}(A)$ is dense in $S^{k-2}$. On the other hand, there exists $A\subseteq\mathbb{N}$ such that $D^{k}(A)$ is not dense in $S^{k-1}$ but $D^{k-1}(A)$ is dense in $S^{k-2}$.

We also provide a simple sufficient condition for $D^{k}(A)$ to be dense in $S^{k-1}$.

Theorem 1.5**.**

Let $A\subseteq\mathbb{N}$. If there exists an increasing sequence $a_{n}\in A$ such that $a_{n-1}/a_{n}\to 1$, then $D^{k}(A)$ is dense in $S^{k-1}$ for every integer $k\geq 2$.

The case $k=2$ of Theorem 1.5 was proved by Starni [19] (hereafter, we tacitly express all the results about $R(A)$ in terms of $D^{2}(A)$), who also showed that the condition is sufficient but not necessary.

Let $\mathbb{P}$ be the set of prime numbers. It is known that $D^{2}(\mathbb{P})$ is dense in $S^{1}$ [13, 19] (see also [7, 18, 5, 16] for similar results in number fields). Let $p_{n}$ be the $n$th prime number. As a consequence of the Prime Number Theorem, we have that $p_{n}\sim n\log n$ [11, Theorem 8]. Hence, $p_{n-1}/p_{n}\to 1$ and thus Theorem 1.5 yields the following:

Corollary 1.6**.**

$D^{k}(\mathbb{P})$* is dense in $S^{k-1}$, for every integer $k\geq 2$.*

We leave the following questions to the interested readers:

Question 1.1*.*

What is a simple characterization of the sets $X\subseteq S^{k-1}$, $k\geq 2$, such that there exists $A\subseteq\mathbb{N}$ satisfying $X=D^{k}(A)^{\prime}$ ?

Question 1.2*.*

Strauch and Tóth [20] proved that if $A\subseteq\mathbb{N}$ has lower asymptotic density at least $1/2$, then $D^{2}(A)$ is dense in $S^{1}$. Moreover, they showed that for every $\delta\in{[0,1/2)}$ there exists some $A\subseteq\mathbb{N}$ with lower asymptotic density equal to $\delta$ and such that $D^{2}(A)$ is not dense in $S^{1}$. How can these results be generalized to $D^{k}(A)$ with $k\geq 3$ ?

Question 1.3*.*

Bukor, Šalát, and Tóth [4] proved that $\mathbb{N}$ can be partitioned into three sets $A$, $B$, $C$, such that none of $D^{2}(A)$, $D^{2}(B)$, $D^{2}(C)$ is dense in $S^{1}$. Moreover, they showed that such a partition is impossible using only two sets. How can these results be generalized to $D^{k}(A)$ with $k\geq 3$ ?

Notation

We use $\mathbb{N}$ to denote the set of positive integers. We write vectors in bold and we use subscripts to denote their components, so that $\bm{x}=(x_{1},\dots,x_{k})$. Also, we put $\|\bm{x}\|:=\sqrt{x_{1}^{2}+\cdots+x_{k}^{2}}$ for the Euclidean norm of $\bm{x}$. If $X$ is a subset of a topological space $T$, then $X^{\prime}$ denotes the set of accumulation points of $X$. Given a sequence $x^{(n)}\in T$, we write x^{(n)}\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}x to mean that $x^{(n)}\to x$ as $n\to+\infty$ and $x^{(n)}\neq x$ for infinitely many $n$.

2. Proof of Theorem 1.2

Only If Part. Suppose that $X=D^{\underline{k}}(A)^{\prime}$ for some $A\subseteq\mathbb{N}$. We shall prove that $X$ satisfies (i)–(iii). Clearly, $X$ is closed, since it is a set of accumulation points. Hence, (i) holds. Pick $\bm{x}\in X$. Then there exists a sequence $\bm{a}^{(n)}\in A^{\underline{k}}$ such that \rho(\bm{a}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\bm{x}. In particular, this implies that $\|\bm{a}^{(n)}\|\to+\infty$ and that $A$ is infinite. Let $\pi$ be a permutation of $\{1,\dots,k\}$. Setting $\bm{b}^{(n)}:=\pi(\bm{a}^{(n)})$, it follows easily that $\bm{b}^{(n)}\in A^{\underline{k}}$ and \rho(\bm{b}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\pi(\bm{x}). Consequently, $\pi(\bm{x})\in X$ and (ii) holds. Finally, assume that $I\subseteq\{1,\dots,k\}$ meets $\bm{x}$. Up to passing to a subsequence of $\bm{a}^{(n)}$, we can assume that each sequence $a_{i}^{(n)}$, with $i\in\{1,\dots,k\}$, is nondecreasing. Recalling that $A$ is infinite, this implies that we can fix $k-\#I$ distinct $c_{i}\in A$, with $i\in\{1,\dots,k\}\setminus I$, such that $\bm{d}^{(n)}\in A^{\underline{k}}$ for every sufficiently large $n\in\mathbb{N}$, where $\bm{d}^{(n)}\in\mathbb{N}^{k}$ is defined by $d_{i}^{(n)}:=a_{i}^{(n)}$ if $i\in I$, and $d_{i}^{(n)}:=c_{i}$ if $i\notin I$. Since $I$ meets $\bm{x}$, there exists $j\in I$ such that $x_{j}\neq 0$, which in turn implies that $a_{j}^{(n)}\to+\infty$ and consequently $\|\bm{d}^{(n)}\|\to+\infty$. At this point, it follows easily that \rho(\bm{d}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\rho_{I}(\bm{x}). Hence, $\rho_{I}(\bm{x})\in X$ and (iii) holds too.

If Part. Suppose that $X\subseteq S^{k-1}$ satisfies (i)–(iii). We shall prove that there exists $A\subseteq\mathbb{N}$ such that $X=D^{\underline{k}}(A)^{\prime}$. Since $X$ is a closed subset of $S^{k-1}$, we have that $X$ has a countable dense subset, say $Y:=\{\bm{y}^{(m)}:m\in\mathbb{N}\}$.

Claim* 1**.*

There exists a sequence $\bm{c}^{(m)}$ such that:

1. (c1)

$\bm{c}^{(m)}\in\mathbb{N}^{\underline{k}}$ for every $m\in\mathbb{N}$; 2. (c2)

$m\mapsto\rho(\bm{c}^{(m)})$ is an injection; 3. (c3)

\big{|}\frac{1}{m!}c_{i}^{(m)}-y_{i}^{(m)}\big{|}\to 0, for every $i\in\{1,\dots,k\}$; 4. (c4)

\big{\|}\rho(\bm{c}^{(m)})-\bm{y}^{(m)}\big{\|}\to 0.

Proof.

For every $m\in\mathbb{N}$ and $i\in\{1,\dots,k\}$, we define $c_{i}^{(m)}:=\lfloor m!\,y_{i}^{(m)}\rfloor+s_{i}^{(m)}+t^{(m)}$, where $\bm{s}^{(m)}\in\mathbb{N}^{k}$ and $t^{(m)}\in\mathbb{N}$ will be chosen later. For each $m\in\mathbb{N}$, it is easy to see that we can choose $\bm{s}^{(m)}\in\{1,\dots,k\}^{k}$ such that $\bm{c}^{(m)}\in\mathbb{N}^{\underline{k}}$. (Note that this property does not depend on $t^{(m)}$.) We make this choice so that (c1) holds. Now note that for every fixed $u,v\in\mathbb{R}^{+}$, with $u\neq v$, the function $\mathbb{R}^{+}\to\mathbb{R}:t\mapsto\frac{u+t}{v+t}$ is injective. Therefore, for each $m\in\mathbb{N}$ we can choose $t^{(m)}\in\{1,\dots,m\}$ such that $c_{1}^{(m)}/c_{2}^{(m)}\neq c_{1}^{(\ell)}/c_{2}^{(\ell)}$ for every positive integer $\ell<m$. In turn, this choice implies that (c2) holds. At this point, both (c3) and (c4) follow easily. ∎

Define $A:=\bigcup_{i=1}^{k}A_{i}$, where $A_{i}:=\{c_{i}^{(m)}:m\in\mathbb{N}\}$ for every $i\in\{1,\dots,k\}$. We claim that $X=D^{\underline{k}}(A)^{\prime}$.

First, let us prove that $X\subseteq D^{\underline{k}}(A)^{\prime}$. Pick some $\bm{x}\in X$. Since $Y$ is a dense subset of $X$, there exists an increasing sequence of positive integers $(m_{n})_{n\in\mathbb{N}}$ such that $\bm{y}^{(m_{n})}\to\bm{x}$. By the definition of $A$ and by (c1), we have that $\bm{c}^{(m_{n})}\in A^{\underline{k}}$. Moreover, (c2) and (c4) imply that \rho(\bm{c}^{(m_{n})})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\bm{x}. Hence, we have $\bm{x}\in D^{\underline{k}}(A)^{\prime}$, as desired.

Now let us prove that $D^{\underline{k}}(A)^{\prime}\subseteq X$. Pick $\bm{x}\in D^{\underline{k}}(A)^{\prime}$. Then there exists a sequence $\bm{a}^{(n)}\in A^{\underline{k}}$ such that \rho(\bm{a}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\bm{x}. Up to passing to a subsequence, we can assume that there exist some $j_{1},\dots,j_{k}\in\{1,\dots,k\}$ such that $\bm{a}^{(n)}\in A_{j_{1}}\times\cdots\times A_{j_{k}}$ for every $n\in\mathbb{N}$. In turn, this implies that there exists a sequence $\bm{m}^{(n)}\in\mathbb{N}^{k}$ such that $a_{i}^{(n)}=c_{j_{i}}^{(m_{i}^{(n)})}$ for every $n\in\mathbb{N}$ and $i\in\{1,\dots,k\}$. Thanks to (ii), without loss of generality, we can reorder the entries of $\bm{a}^{(n)}$. Hence, up to reordering and up to passing to a subsequence, we can assume that there exists $h\in\{1,\dots,k\}$ such that $y_{j_{1}}^{(m_{1}^{(n)})},\ldots,y_{j_{h}}^{(m_{h}^{(n)})}\neq 0$ and $y_{j_{h+1}}^{(m_{h+1}^{(n)})}=\cdots=y_{j_{k}}^{(m_{k}^{(n)})}=0$ for every $n\in\mathbb{N}$. Similarly, again up to reordering and up to passing to a subsequence, we can assume that there exists $\ell\in\{1,\dots,h\}$ such that $m_{1}^{(n)}=\cdots=m_{\ell}^{(n)}>m_{\ell+1}^{(n)}\geq\cdots\geq m_{h}^{(n)}$ for every $n\in\mathbb{N}$. In particular, since $\bm{a}^{(n)}\in A^{\underline{k}}$ for every $n\in\mathbb{N}$, we get that $j_{1},\dots,j_{\ell}$ are pairwise distinct. Let $\pi$ be any permutation of $\{1,\dots,k\}$ such that $\pi(i)=j_{i}$ for all $i\in I:=\{1,\dots,\ell\}$. Note that $I$ meets $\pi(\bm{y}^{(m_{1}^{(n)})})$ for every $n\in\mathbb{N}$. Put $\bm{z}^{(n)}:=\rho_{I}(\pi(\bm{y}^{(m_{1}^{(n)})}))$ for every $n\in\mathbb{N}$. Hence, by (ii) and (iii) we have that $\bm{z}^{(n)}\in X$ for every $n\in\mathbb{N}$. Thanks to (c3), we have that \big{|}\tfrac{1}{m_{1}^{(n)}!}a_{i}^{(n)}-y_{j_{i}}^{(m_{1}^{(n)})}\big{|}\to 0 for each $i\in I$, and $\tfrac{1}{m_{1}^{(n)}!}a_{i}^{(n)}\to 0$ for each $i\in\{1,\ldots,k\}\setminus I$, as $n\to+\infty$. As a consequence, \big{\|}\rho(\bm{a}^{(n)})-\bm{z}^{(n)}\big{\|}\to 0, which in turn implies that $\bm{z}^{(n)}\to\bm{x}$. Finally, since $X$ is closed by (i), we obtain that $\bm{x}\in X$, as desired.

The proof is complete.

Remark 2.1*.*

We note that for $k\geq 3$ the statement of Theorem 1.2 is false if $D^{\underline{k}}(A)$ is replaced by $D^{k}(A)$. In fact, fix an integer $k\geq 3$ and let $X$ be the subset of $S^{k-1}$ containing all the permutations of $\bm{\eta}:=\rho(1,\sqrt{2},0,\ldots,0)$ and $\rho(1,0,\ldots,0)$ (and nothing else). It follows by Theorem 1.2 that there exists $A\subseteq\mathbb{N}$ such that $X=D^{\underline{k}}(A)^{\prime}$. For the sake of contradiction, let us suppose that there exists $B\subseteq\mathbb{N}$ such that $X=D^{k}(B)^{\prime}$. Since $\bm{\eta}\in X$, there exists a sequence $\bm{b}^{(n)}\in B^{k}$ such that \rho(\bm{b}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\bm{\eta}. Let $\bm{c}^{(n)}\in\mathbb{N}^{k}$ be the sequence defined by $c_{i}^{(n)}=b_{1}^{(n)}$ if $i\neq 2$, and $c_{i}^{(n)}:=b_{2}^{(n)}$ if $i=2$. We obtain that $\bm{c}^{(n)}\in B^{k}$ and \rho(\bm{c}^{(n)})\mathrel{\ooalign{\rightarrow\cr\kern 3.87495pt\raise 3.01385pt\hbox{\scalebox{0.5}[0.5]{\bullet}}\cr}}\bm{\theta}, where $\bm{\theta}:=\rho(1,\sqrt{2},1,\ldots,1)$. (Here we have used that $\eta_{1}/\eta_{2}$ is irrational and consequently $\rho(\bm{c}^{(n)})\neq\bm{\theta}$.) Therefore, $\bm{\theta}\in D^{k}(B)^{\prime}=X$, which is a contradiction.

3. Proof of Theorem 1.4

Let $k\geq 3$ be an integer and let $A\subseteq\mathbb{N}$. Suppose that $D^{k}(A)$ is dense in $S^{k-1}$. We shall prove that $D^{k-1}(A)$ is dense in $S^{k-2}$. For every $\bm{x}\in S^{k-2}$, let $f_{k}(\bm{x})\in S^{k-1}$ be defined by $f_{k}(\bm{x}):=\rho(x_{1},\dots,x_{k-1},0)$. Since $D^{k}(A)$ is dense in $S^{k-1}$, we have that there exists a sequence $\bm{a}^{(n)}\in A^{k}$ such that $\rho(\bm{a}^{(n)})\to f_{k}(\bm{x})$. In turn, this implies that $\rho(\bm{b}^{(n)})\to\bm{x}$, where $\bm{b}^{(n)}\in A^{k-1}$ is defined by $b_{i}^{(n)}:=a_{i}^{(n)}$ for $i\in\{1,\dots,k-1\}$. Hence, $D^{k-1}(A)$ is dense in $S^{k-2}$, as desired.

Now given an integer $k\geq 3$, we shall prove that there exists $A\subseteq\mathbb{N}$ such that $D^{k-1}(A)$ is dense in $S^{k-2}$, but $D^{k}(A)$ is not dense in $S^{k-1}$. Let $X:=\{\bm{x}\in S^{k-1}:x_{i}=0\text{ for some }i\}$. Clearly, $X$ satisfies conditions (i)–(iii) of Theorem 1.2, and consequently there exists $A\subseteq\mathbb{N}$ such that $D^{\underline{k}}(A)^{\prime}=X$. Therefore, $D^{\underline{k}}(A)$ is not dense in $S^{k-1}$ and, in light of Proposition 1.3, $D^{k}(A)$ is not dense in $S^{k-1}$ as well. Finally, for every $\bm{x}\in S^{k-2}$ we have $f_{k}(\bm{x})\in X$, and the same reasonings of the previous paragraph show that $D^{k-1}(A)$ is dense in $S^{k-2}$.

4. Proof of Theorem 1.5

Suppose that there exists an increasing sequence $a_{n}\in A$ such that $a_{n-1}/a_{n}\to 1$. Fix an integer $k\geq 2$ and pick $\bm{x}\in S^{k-1}$ with $x_{1},\dots,x_{k}>0$. Clearly, for every integer $m\geq a_{1}\,/\min\{x_{1},\dots,x_{k}\}$ there exist integers $m_{1},\dots,m_{k}\geq 2$ such that $a_{m_{i}-1}\leq mx_{i}<a_{m_{i}}$ for each $i\in\{1,\dots,k\}$. Hence, for every $i\in\{1,\dots,k\}$, we have that

[TABLE]

which, since $m_{i}\to+\infty$ as $m\to+\infty$, yields that $a_{m_{i}}/m\to x_{i}$ as $m\to+\infty$. Putting $\bm{a}^{(m)}:=(a_{m_{1}},\dots,a_{m_{k}})$, it follows that $\rho(\bm{a}^{(m)})\to\bm{x}$. Therefore, $D^{k}(A)$ is dense in $S^{k-1}$, as claimed.