Directions sets: A generalization of ratio sets
Paolo Leonetti Carlo Sanna
math.NT 11B05, 11A99


Spin polarization-scaling quantum maps and channels
    Kamil Yu. Magadov
    
===================================================




For every integer k ≥ 2 and every A ⊆ℕ, we define the k-directions sets of A as D^k(A) := {a / a : a∈ A^k} and D^k(A) := {a / a : a∈ A^k}, where · is the Euclidean norm and A^k := {a∈ A^k : a_i ≠ a_j  for all  i ≠ j}.
Via an appropriate homeomorphism, D^k(A) is a generalization of the ratio set R(A) := {a / b : a,b ∈ A}, which has been studied by many authors.
We study D^k(A) and D^k(A) as subspaces of S^k-1 := {x∈ [0,1]^k : x = 1}.
In particular, generalizing a result of Bukor and Tóth, we provide a characterization of the sets X ⊆ S^k-1 such that there exists A ⊆ℕ satisfying D^k(A)^' = X, where Y^' 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.




§ INTRODUCTION


Given A⊆ℕ, its ratio set is defined as R(A) := {a / b : a,b ∈ A}.
The study of the topological properties of R(A) as a subspace of [0, +∞], especially the question of when R(A) is dense in [0, +∞], is a classical topic and has been considered by many researchers <cit.>.
More recently, some authors have also studied R(A) as a subspace of the p-adic numbers ℚ_p <cit.>.

We consider a further variation on this theme, which stems from the following easy observation: 
We have that [0,+∞] is homeomorphic to S^1 := {x∈ [0,1]^2 : x = 1} via the map x ↦ (1, x) / (1,x), if x ∈[0,+∞), and +∞↦ (0,1). 
This sends R(A) onto D^2(A) := {ρ(a): a∈ A^2}, where ρ(a):=a / a for each a≠0. 
Hence, topological questions about R(A) as a subspace of [0, +∞] 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 ≥ 2, define the k-directions sets of A as 

    D^k(A) := {ρ(a): a∈ A^k}     and     D^k(A) := {ρ(a): a∈ A^k},

where for every set B we let B^k := {b∈ B^k : b_i ≠ b_j for all  i ≠ j} denote the set of k-tuples with pairwise distinct entries in B.
Put also S^k-1 := {x∈ [0,1]^k : x = 1}.
We shall study D^k(A) and D^k(A) as subspaces of S^k-1.

Bukor and Tóth <cit.> characterized the subsets of [0, +∞] that are equal to R(A)^' for some A ⊆ℕ, where Y^' denotes the set of accumulation points of Y.
In terms of D^2(A), via the homeomorphism [0,+∞] → S^1 mentioned above, their result is the following:


Let X ⊆ S^1.
Then there exists A ⊆ℕ such that X = D^2(A)^' if and only if the following conditions are satisfied:


  * X is closed;

  * (x_1, x_2) ∈ X implies (x_2, x_1) ∈ X;

  * if X is nonempty, then (1, 0) ∈ X.


Note that Theorem <ref> holds also if D^2(A) is replaced by D^2(A).
Indeed, D^2(A)⊆ D^2(A) ⊆ D^2(A) ∪{ρ(1,1)} and consequently D^2(A)^' = D^2(A)^'.

Our first result generalizes Theorem <ref>.
Before stating it, we need to introduce some notation.
Let x = (x_1, …, x_k) ∈ S^k-1.
For every permutation π of {1,…,k}, we put 
Also, for every I ⊆{1,…,k}, we say that I meets x if there exists j ∈ I such that x_j ≠ 0.
In such a case, we put ρ_I(x) := ρ(y), where y = (y_1, …, y_k) is defined by y_i := x_i if i ∈ I, and y_i := 0 for i ∉ I. 
(This is well defined since y≠0.)

Our first result is the following:


Let X ⊆ S^k - 1 for some integer k ≥ 2.
Then there exists A ⊆ℕ such that X = D^k(A)^' if and only if the following conditions are satisfied:


  *  X is closed;

  *  x∈ X implies π(x) ∈ X, for every permutation π of {1, …, k};

  *  x∈ X implies ρ_I(x) ∈ X, for every I ⊆{1, …, k} that meets x.



Note that Theorem <ref> is indeed a generalization of Theorem <ref>, since ρ_I(x) ∈{x, (1, 0),(0, 1)} for every I ⊆{1,2} that meets x∈ S^1.
Furthermore, for k ≥ 3, Theorem <ref> is false if D^k(A) is replaced by D^k(A) (see Remark <ref> 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.


Let k ≥ 2 be an integer and fix A ⊆ℕ.
We have that D^k(A) is dense in S^k-1 if and only if D^k(A) is dense in S^k-1.


On the one hand, since D^k(A) ⊆ D^k(A), if D^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 x∈ S^k-1∩ℝ^k, there exists a^(n)∈ A^k such that ρ(a^(n)) →x.
Consequently, for all sufficiently large n we have a^(n)∈ A^k.
This implies that D^k(A) is dense in S^k-1∩ℝ^k.
Since S^k-1∩ℝ^k is dense in S^k-1, we get that D^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 ≥ 3 and A ⊆ℕ, then D^k-1(A) is dense in S^k - 2, but the opposite implication is false.


Let k ≥ 3 be an integer.
On the one hand, if D^k(A) is dense in S^k-1, for some A ⊆ℕ, then D^k-1(A) is dense in S^k-2.
On the other hand, there exists A ⊆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.


Let A ⊆ℕ.
If there exists an increasing sequence a_n ∈ A such that , then D^k(A) is dense in S^k-1 for every integer k ≥ 2.


The case k = 2 of Theorem <ref> was proved by Starni <cit.> (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 ℙ be the set of prime numbers.
It is known that D^2(ℙ) is dense in S^1 <cit.> (see also <cit.> for similar results in number fields).
Let p_n be the nth prime number.
As a consequence of the Prime Number Theorem, we have that p_n ∼ n log n <cit.>.
Hence, p_n-1 / p_n → 1 and thus Theorem <ref> yields the following:


D^k(ℙ) is dense in S^k-1, for every integer k ≥ 2.


We leave the following questions to the interested readers:


What is a simple characterization of the sets X ⊆ S^k-1, k ≥ 2, such that there exists A ⊆ℕ satisfying X = D^k(A)^' ?



Strauch and Tóth <cit.> proved that if A ⊆ℕ has lower asymptotic density at least 1/2, then D^2(A) is dense in S^1.
Moreover, they showed that for every δ∈[0, 1/2) there exists some A ⊆ℕ with lower asymptotic density equal to δ and such that D^2(A) is not dense in S^1.
How can these results be generalized to D^k(A) with k ≥ 3 ?



Bukor, Šalát, and Tóth <cit.> proved that 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 ≥ 3 ?





 §.§ Notation

We use ℕ to denote the set of positive integers.
We write vectors in bold and we use subscripts to denote their components, so that x = (x_1, …, x_k).
Also, we put x := √(x_1^2 + ⋯ + x_k^2) for the Euclidean norm of x.
If X is a subset of a topological space T, then X^' denotes the set of accumulation points of X.
Given a sequence x^(n)∈ T, we write x^(n) x to mean that x^(n)→ x as n → +∞ and x^(n)≠ x for infinitely many n.



§ PROOF OF THEOREM <REF>


Only If Part. Suppose that X = D^k(A)^' for some A ⊆ℕ.
We shall prove that X satisfies <ref>–<ref>.
Clearly, X is closed, since it is a set of accumulation points.
Hence, <ref> holds.
Pick x∈ X.
Then there exists a sequence a^(n)∈ A^k such that ρ(a^(n)) x.
In particular, this implies that a^(n)→ +∞ and that A is infinite. 
Let π be a permutation of {1, …, k}.
Setting b^(n) := π(a^(n)), it follows easily that b^(n)∈ A^k and ρ(b^(n)) π(x).
Consequently, π(x) ∈ X and <ref> holds.
Finally, assume that I ⊆{1, …, k} meets x.
Up to passing to a subsequence of a^(n), we can assume that each sequence a_i^(n), with i ∈{1, …, k}, is nondecreasing.
Recalling that A is infinite, this implies that we can fix k - #I distinct c_i ∈ A, with i ∈{1, …, k}∖ I, such that d^(n)∈ A^k for every sufficiently large n ∈ℕ, where d^(n)∈ℕ^k is defined by d_i^(n) := a_i^(n) if i ∈ I, and d_i^(n) := c_i if i ∉ I.
Since I meets x, there exists j ∈ I such that x_j ≠ 0, which in turn implies that a_j^(n)→ +∞ and consequently d^(n)→ +∞.
At this point, it follows easily that ρ(d^(n)) ρ_I(x).
Hence, ρ_I(x) ∈ X and <ref> holds too.



If Part. Suppose that X ⊆ S^k-1 satisfies <ref>–<ref>.
We shall prove that there exists A ⊆ℕ such that X = D^k(A)^'.
Since X is a closed subset of S^k-1, we have that X has a countable dense subset, say Y := {y^(m) : m ∈ℕ}.

There exists a sequence c^(m) such that:


  *  c^(m)∈ℕ^k for every m ∈ℕ;

  *   m ↦ρ(c^(m)) is an injection;

  *   |1/m!c_i^(m) - y_i^(m)| → 0, for every i ∈{1, …, k};

  *  ρ(c^(m)) - y^(m)→ 0.



For every m ∈ℕ and i ∈{1, …, k}, we define c_i^(m) := ⌊ m! y_i^(m)⌋ + s_i^(m) + t^(m), where s^(m)∈ℕ^k and t^(m)∈ℕ will be chosen later.
For each m ∈ℕ, it is easy to see that we can choose s^(m)∈{1, …, k}^k such that c^(m)∈ℕ^k. (Note that this property does not depend on t^(m).)
We make this choice so that <ref> holds.
Now note that for every fixed u,v ∈ℝ^+, with u ≠ v, the function ℝ^+ →ℝ : t ↦u+t/v+t is injective.
Therefore, for each m ∈ℕ we can choose t^(m)∈{1, …, m} such that c_1^(m) / c_2^(m)≠ c_1^(ℓ) / c_2^(ℓ) for every positive integer ℓ < m.
In turn, this choice implies that <ref> holds.
At this point, both <ref> and <ref> follow easily.


Define A := ⋃_i=1^k A_i, where A_i := {c_i^(m) : m ∈ℕ} for every i ∈{1,…,k}.
We claim that X = D^k(A)^'.



First, let us prove that X ⊆ D^k(A)^'.
Pick some x∈ X.
Since Y is a dense subset of X, there exists an increasing sequence of positive integers (m_n)_n ∈ℕ such that y^(m_n)→x.
By the definition of A and by <ref>, we have that c^(m_n)∈ A^k.
Moreover, <ref> and <ref> imply that ρ(c^(m_n)) x.
Hence, we have x∈ D^k(A)^', as desired.



Now let us prove that D^k(A)^'⊆ X.
Pick x∈ D^k(A)^'.
Then there exists a sequence a^(n)∈ A^k such that ρ(a^(n)) x.
Up to passing to a subsequence, we can assume that there exist some j_1, …, j_k ∈{1,…,k} such that a^(n)∈ A_j_1×⋯× A_j_k for every n ∈ℕ.
In turn, this implies that there exists a sequence m^(n)∈ℕ^k such that a_i^(n) = c_j_i^(m_i^(n)) for every n ∈ℕ and i ∈{1, …, k}.
Thanks to <ref>, without loss of generality, we can reorder the entries of a^(n).
Hence, up to reordering and up to passing to a subsequence, we can assume that there exists h ∈{1, …, k} such that y_j_1^(m_1^(n)),…,y_j_h^(m_h^(n))≠ 0 and y_j_h+1^(m_h+1^(n)) = ⋯ = y_j_k^(m_k^(n)) = 0 for every n ∈ℕ.
Similarly, again up to reordering and up to passing to a subsequence, we can assume that there exists ℓ∈{1, …, h} such that m_1^(n) = ⋯ = m_ℓ^(n) > m_ℓ+1^(n)≥⋯≥ m_h^(n) for every n ∈ℕ.
In particular, since a^(n)∈ A^k for every n ∈ℕ, we get that j_1, …, j_ℓ are pairwise distinct.
Let π be any permutation of {1, …, k} such that π(i) = j_i for all i ∈ I := {1, …, ℓ}.
Note that I meets π(y^(m_1^(n))) for every n ∈ℕ.
Put z^(n) := ρ_I(π(y^(m_1^(n)))) for every n ∈ℕ.
Hence, by <ref> and <ref> we have that z^(n)∈ X for every n ∈ℕ.
Thanks to <ref>, we have that |1m_1^(n)! a_i^(n) - y_j_i^(m_1^(n))| → 0 for each i ∈ I, and 1m_1^(n)! a_i^(n)→ 0 for each i ∈{1, …, k}∖ I, as n → +∞.
As a consequence, ρ(a^(n)) - z^(n)→ 0, which in turn implies that z^(n)→x.
Finally, since X is closed by <ref>, we obtain that x∈ X, as desired.

The proof is complete.


We note that for k ≥ 3 the statement of Theorem <ref> is false if D^k(A) is replaced by D^k(A).
In fact, fix an integer k ≥ 3 and let X be the subset of S^k-1 containing all the permutations of η:=ρ(1,√(2),0,…,0) and ρ(1,0,…,0) (and nothing else).
It follows by Theorem <ref> that there exists A⊆ℕ such that X = D^k(A)^'.
For the sake of contradiction, let us suppose that there exists B⊆ℕ such that X = D^k(B)^'. 
Since η∈ X, there exists a sequence b^(n)∈ B^k such that ρ(b^(n)) η. 
Let c^(n)∈ℕ^k be the sequence defined by c_i^(n)=b_1^(n) if i ≠ 2, and c_i^(n) := b_2^(n) if i = 2.
We obtain that c^(n)∈ B^k and ρ(c^(n)) θ, where θ := ρ(1,√(2),1,…,1).
(Here we have used that η_1 / η_2 is irrational and consequently ρ(c^(n)) ≠θ.)
Therefore, θ∈ D^k(B)^' = X, which is a contradiction.




§ PROOF OF THEOREM <REF>


Let k ≥ 3 be an integer and let A ⊆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 x∈ S^k-2, let f_k(x) ∈ S^k - 1 be defined by f_k(x) := ρ(x_1, …, x_k- 1, 0).
Since D^k(A) is dense in S^k-1, we have that there exists a sequence a^(n)∈ A^k such that ρ(a^(n)) → f_k(x).
In turn, this implies that ρ(b^(n)) →x, where b^(n)∈ A^k-1 is defined by b_i^(n) := a_i^(n) for i ∈{1, …, k - 1}.
Hence, D^k-1(A) is dense in S^k-2, as desired.

Now given an integer k ≥ 3, we shall prove that there exists A ⊆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 := {x∈ S^k - 1 : x_i = 0  for some  i}.
Clearly, X satisfies conditions <ref>–<ref> of Theorem <ref>, and consequently there exists A ⊆ℕ such that D^k(A)^' = X.
Therefore, D^k(A) is not dense in S^k-1 and, in light of Proposition <ref>, D^k(A) is not dense in S^k-1 as well.
Finally, for every x∈ S^k - 2 we have f_k(x) ∈ X, and the same reasonings of the previous paragraph show that D^k - 1(A) is dense in S^k - 2.



§ PROOF OF THEOREM <REF>


Suppose that there exists an increasing sequence a_n ∈ A such that a_n-1 / a_n → 1.
Fix an integer k ≥ 2 and pick x∈ S^k-1 with x_1, …, x_k > 0.
Clearly, for every integer m ≥ a_1  / min{x_1, …, x_k} there exist integers m_1, …, m_k ≥ 2 such that a_m_i - 1≤ m x_i < a_m_i for each i ∈{1, …, k}.
Hence, for every i ∈{1, …, k}, we have that

    x_i < a_m_i/m≤a_m_i/a_m_i-1  x_i ,

which, since m_i → +∞ as m → +∞, yields that a_m_i / m → x_i as m → +∞.
Putting a^(m) := (a_m_1, …, a_m_k), it follows that ρ(a^(m)) →x.
Therefore, D^k(A) is dense in S^k-1, as claimed.

amsplain


