On a Density for Sets of Integers

# On a Density for Sets of Integers

## Abstract

A relationship between the Riemann zeta function and a density on integer sets is explored. Several properties of the examined density are derived.

Keywords Number theory, probability theory, arithmetization

## 1 Introduction

Several measures for the density of sets have been discussed in the literature [1, 2, 3, 4, 5, 6]. Presumably the most employed tool for evaluating the density of sets is the asymptotic density, also referred to as natural density. The asymptotic density is expressed by

 d(A)=limn→∞∥A∩{1,2,3,…,n}∥n, (1)

provided that such a limit does exist. The symbol denotes the cardinality, and is a set of integers. Analogously, the lower and upper asymptotic densities are defined by

 d––(A) =liminfn→∞∥A∩{1,2,3,…,n}∥n, (2) ¯d(A) =limsupn→∞∥A∩{1,2,3,…,n}∥n, (3)

The asymptotic density is said to exist if and only if both the lower and upper asymptotic densities do exist and are equal.

Although the asymptotic density does not always exist, the Schnirelmann density [2, 3] is always well-defined. The Schnirelmann density is defined as

 δ(A)=infn≥1∥A∩{1,2,3,…,n}∥n. (4)

Interestingly, this density is highly sensitive to the initial elements of sequence. For instance, if , then  [7].

Another interesting tool is the logarithmic density [4, 5, 6]. Let be a set of integers. The logarithmic density of is given by

 ℓd(A)=limn→∞∑ai≤n1ailogn. (5)

In [1], Bell and Burris bring a good exposition on the Dirichlet density. The Dirichlet density is defined as the limit of the ratio between two Dirichlet series. Let be two sets. The generating series of is given by

 A(s)=∞∑n=1N(A,n)ns, (6)

where is a counting function that returns the number of elements in of norm  [1]. The Dirichlet density is then expressed by

 D(A)=lims↓αA(s)B(s), (7)

where is the generating series of and is an abscissa of convergence [1]. In [6], we also find the Dirichlet density defined as

 D(A)=lims↓1(s−1)∑n∈A1ns, (8)

whenever the limit exists. This density admits lower and upper versions, simply by replacing the above limit by and , respectively.

The aim of this work is to investigate the properties of the Dirichlet density as defined in Equation 7 in the particular case where the set is the set of natural numbers. This induces a density based on the Riemann zeta function.

## 2 A Density for Sets of Integers

In this section, we investigate the particular case of the Dirichlet function, applied for subsets of the natural numbers. In this case, taking into consideration the usual norm, where the norm of any natural number is equal to its absolute value, the counting function becomes

 N(A,n)={1,if n∈A,0,otherwise. (9)
###### Definition 1

The density of a subset is given by

 dens(A) ≜lims↓1∑∞n=1N(A,n)1ns∑∞n=1N(N,n)1ns (10) =lims↓1∑n∈A1ns∑∞n=11ns (11) =lims↓1∑n∈A1nsζ(s),s>1, (12)

if the limit exists. The quantity denotes the Riemann zeta function [8].

###### Proposition 1

The following assertions hold true:

1. (nonnegativity)

2. if and exist1 and , then exists and is equal to

Proof: Statements 1 and 2 cab be trivially checked. The additivity property can be derived as follows:

 dens(A∪B) =lims↓1∑n∈A∪B1nsζ(s) (13) =lims↓1∑n∈A1nsζ(s)+lims↓1∑n∈B1nsζ(s) (14) =dens(A)+dens(B). (15)

###### Corollary 1

The density of the null set is zero.

Proof: In fact, , since .

###### Corollary 2

.

Proof: The proof is straightforward:

 dens(B) =dens((A∩B)∪(Ac∩B)) (16) =dens(A∩B)+dens(Ac∩B) (17) =dens(A∩B)+dens(B−A), (18)

since and are disjoint.

###### Corollary 3

For every , , where is the complement of .

Proof: .

###### Proposition 2 (Monotonicity)

The discussed density is a monotone function, i.e., if , then .

Proof: We have that

 dens(B) =lims↓1∑n∈B1nsζ(s) (19) =lims↓1∑n∈A1nsζ(s)+lims↓1∑n∈B−A1nsζ(s) (20) ≥lims↓1∑n∈A1nsζ(s) (21) =dens(A). (22)

###### Proposition 3 (Finite Sets)

Every finite subset of has density zero.

Proof: Let be a finite set. For , it yields

 ∑n∈A1ns≤∑n∈A1n=constant<∞. (23)

Thus,

 dens(A)=lims↓1∑n∈A1nsζ(s)≤lims↓1constantζ(s)=0. (24)

As a consequence, sets of nonzero density must be infinite.

###### Corollary 4

The density of a singleton is zero.

###### Proposition 4 (Union)

Let and be two sets of integers. Then the density of is given by

 dens(A∪B)=dens(A)+dens(B)−dens(A∩B). (25)

Proof: Observe that , and . Then, it follows directly from the properties of the density measure that

 dens(A∪B) =dens(A∪(B−A)) (26) =dens(A)+dens(B−A) (27) =dens(A)+dens(B)−dens(A∩B). (28)

###### Proposition 5 (Heavy Tail)

Let . If and , then

 dens(A)=dens(A2). (29)

Proof: Observe that and and are disjoint. Therefore, . Since is a finite set, .

Now consider the following operation . This can be interpreted as a dilation operation on the elements of .

In [9, 5, 4], Erdös et al. examined the density of the set of multiples , showing the existence of a logarithmic density equal to its lower asymptotic density. Herein we investigate further this matter, evaluating the density of sets of multiples.

###### Proposition 6 (Dilation)

Let be a set, such as . Then

 dens(m⊗A)=1mdens(A). (30)

Proof: This result follows directly from the definition of the discussed density:

 dens(m⊗A) =lims↓1∑n∈A1(mn)sζ(s) (31) =lims↓11ms∑n∈A1nsζ(s) (32) =1mdens(A). (33)

Let . This process is called a translation of by units [10, p.49]. Our aim is to show that the discussed density is translation invariant, i.e., , . Before that we need the following lemma.

###### Lemma 1 (Unitary Translation)

Let be a set, such as . Then

 dens(A⊕1)=dens(A). (34)

Proof: Let . We can split into two disjoint sets as shown below:

 A ={a1,a2,…,aN−1,aN,aN+1,…} (35) ={a1,a2,…,aN−1}∪{aN,aN+1,…} (36) =AN1∪AN2, (37)

where is a finite set and is a ‘tail’ set starting at the element . By the Heavy Tail property, we have that . For the same reason, , .

Note that, for , if , then the following inequalities hold:

 ∑n∈AN21(kn)sζ(s)≤∑n∈AN21(n+1)sζ(s)≤∑n∈AN21nsζ(s). (38)

In other words, for the above inequality to be valid, we must have

 k≥n+1n,∀n∈AN2. (39)

For instance, let . Consequently, we establish that

 ∑n∈AN21(aN+1aNn)sζ(s)≤∑n∈AN21(n+1)sζ(s)≤∑n∈AN21nsζ(s). (40)

Taking the limit as , the above upper bound becomes

 lims↓1∑n∈AN21nsζ(s)=dens(AN2)=dens(A). (41)

Now examining the lower bound and using the dilation property, we have that

 lims↓1∑n∈AN21(aN+1aNn)sζ(s) (42) =aNaN+1dens(AN2) (43) =aNaN+1dens(A). (44)

Since , such that

 aNaN+1dens(A)>dens(A)−ϵ, (45)

and as

 lims↓1∑n∈AN21(n+1)sζ(s)=dens(AN2⊕1)=dens(A⊕1), (46)

it follows that we have that

 dens(A)−ϵ≤dens(A⊕1)≤dens(A). (47)

Therefore, letting , it follows that

 dens(A⊕1)=dens(A). (48)

###### Proposition 7 (Translation Invariance)

Let be a set, such as . Then

 dens(A⊕m)=dens(A), (49)

where is a positive integer.

Proof: The proof follows by finite induction. We have already proven that . Therefore, we have that

 dens(A⊕(m+1))=dens((A⊕m)⊕1) (50) =dens(A⊕m)=dens(A). (51)

One possible application for a density on a set of natural numbers is to interpret it as the chance of choosing a natural number in when all natural numbers are equally likely to be chosen. Interestingly, as we show next the above measure of uncertainty does not obey all the axioms of Kolmogorov since it is not -additive. Additionally, we show that it is impossible to define a finite -additive translation invariant measure on . This result emphasizes an important point that there are reasonable measures of uncertainty that do not satisfy the formal standard definition of a probability measure.

###### Theorem 1

There is no -additive measure defined on the measurable space such that:

1. ; and

2. is translation invariant.

Proof: Suppose that is translation invariant. Then every singleton set must have the same measure. Let be any natural numbers. Since , we have

 μ({ω2})=μ({ω1}⊕(ω2−ω1))=μ({ω1}),

where the last equality follows from translation invariance. Let . If , then . Thus is not -additive. If , then , which also implies that is not -additive.

###### Proposition 8 (Criterion for Zero Density)

Let . If   converges, then .

Proof: It follows directly from the definition of .

###### Definition 2 (Sparse Set)

A zero density set is said to be a sparse set.

As matter of fact, any criterion that ensure the convergence of can be taken into consideration. In the next result, we utilize the Ratio Test for convergence of series.

###### Corollary 5

Let . If  , then .

Proof: Observe that for

 ∑n∈A1ns<∑n∈A1n=∞∑n=11an. (52)

According to the Ratio Test [11, p.68], the series on the right-hand side of the above inequality converges whenever

 limn→∞1an+11an=limn→∞anan+1<1. (53)

Thus, by series dominance, it follows that also converges. And finally, this implies that the density of must be zero.

###### Lemma 2 (Powers of the Set Elements)

Let and be an integer. The density of is zero.

Proof: Notice that, for and ,

 ∞∑n=11(amn)s<∞∑n=11amn<∞∑n=11nm<∞ (54)

Thus, it follows from the preceding discussion that .

###### Corollary 6

The density of the set of perfect squares is zero.

###### Corollary 7 (Geometric Progressions)

Let , where and are positive integers. Then

 dens(G)=0. (55)

Proof: Notice that, for ,

 ∞∑n=11(arn)s=1as∞∑n=11rns<1as∞∑n=1(1r)n<∞ (56)

Thus, it also follows from the preceding discussion that .

Let be an arithmetic progression, where . For example, , , etc. Regarding the cardinality of these sets, we have that .

###### Proposition 9 (Arithmetic Progressions)

For a fixed integer , the density of the set is given by

 dens(Mp)=1p. (57)

Proof: Note that . Thus, according to Proposition 6, we have that

 dens(Mp)=dens(p⊗N)=1pdens(N)=1p. (58)

Consequently, we have, for instance, , , etc.

One can derive a physical interpretation for the density of . Consider discrete-time signals characterized by a sequence of discrete-time impulses associated to the sets . The signals are built according to a binary function that returns one if is a multiple of , and zero otherwise. For example, we can have and , associated to and , as depicted in Figure 1. In terms of signal analysis, has some correspondence to the average value of .

## 3 Density of Particular Sets

In this section, some special sets have their density examined. We focus our attention in three notable sets: (i) set of prime numbers, (ii) Fibonacci sequence, and (iii) set of square-free integers.

### 3.1 Set of Prime Numbers

Let be the set of prime numbers.

###### Proposition 10

The prime numbers set is sparse.

Proof: Utilizing the definition of the prime zeta function, , we have that . Additionally observe that for . Taking in account that

 lnζ(s)=∞∑k=1ζ′(ks)k, (59)

for we have that

 lnζ(1+ϵ) =∞∑k=1ζ′(k(1+ϵ))k (60) =ζ′(1+ϵ)+∞∑k=2ζ′(k(1+ϵ))k. (61)

Now consider the following inequalities:

 0≤ζ′(1+ϵ)ζ(1+ϵ) =lnζ(1+ϵ)−∑∞k=2ζ′(k(1+ϵ))kζ(1+ϵ) (62) =lnζ(1+ϵ)ζ(1+ϵ)−∞∑k=2ζ′(k(1+ϵ))kζ(1+ϵ). (63) ≤lnζ(1+ϵ)ζ(1+ϵ). (64)

As , we have that , therefore

 limϵ→0lnζ(1+ϵ)ζ(1+ϵ)=0, (65)

because . Therefore, we can apply the Squeeze Theorem once more, and find that

 limϵ→0ζ′(1+ϵ)ζ(1+ϵ)=0. (66)

Since the density is zero, it indicates that the associated discrete-time signal , as shown in Figure 2, has null average value.

### 3.2 Fibonacci Sequence

The Fibonacci sequence constitutes another interesting subject of investigation. Fibonacci numbers are constructed according to the following recursive equation , , for , where denotes the nth Fibonacci number [12, p.160]. This procedure results in the Fibonacci set .

###### Proposition 11

The Fibonacci set is sparse.

Proof: It is known that the sum of the reciprocals of the Fibonacci numbers converges to a constant (reciprocal Fibonacci constant), whose value is approximately  [13]. Consequently, for , we have that

 dens(F)=lims↓1∑∞n=21u[n]sζ(s) (67)

### 3.3 Set of Square-free Integers

The set of square-free numbers can be defined as , where is the Möbius function, which is given by

 μ(n)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1,if n=1,0,if p2|n, for some prime number p,(−1)r,if n is the product of distinctprime numbers. (69)

Thus the first elements of are given by

###### Proposition 12

The density of the set of square-free integers is .

Proof: It is known [14] that , thus it follows easily that:

 dens(S) =lims↓1∑n∈S1nsζ(s) (70) =lims↓1∑∞n=1|μ(n)|nsζ(s) (71) =lims↓1ζ(s)ζ(2s)ζ(s)=lims↓11ζ(2s) (72) =1ζ(2)=6π2. (73)

## 4 Computational Results

###### Definition 3

Let be a set of positive integers. The truncated set is defined by

 AT=A∩{1,2,3,…,T−1,T}, (74)

where is a positive integer.

In other words, contains the elements of which are no greater than .

The concept of truncated sets can furnish computational approximations for the numerical value of some densities. Consider, for example, the quantity

 d′(AT)≜∥AT∥∥NT∥. (75)

This expression can be interpreted in a frequentist way as the ratio between favorable cases and possible cases. Clearly, the asymptotic density can be expressed in terms of :

 d(A)=limT→∞d′(AT). (76)

In a similar fashion, we can consider a version of the discussed density for truncated sets as defined below. First, observe that discussed density can be expressed as the following double limit

 dens(A)=lims↓1limT→∞∑n∈AT1nsT∑n=11ns. (77)

Restricted to the class of sets for which the above limits can have their order interchanged, we find that

 dens(A) =limT→∞lims↓1∑n∈AT1ns∑Tn=11ns (78) =limT→∞⎛⎝∑n∈AT1n∑Tn=11n⎞⎠. (79)
###### Definition 4 (Approximate Density)

The approximate density for a truncated set is given by

 dens′(AT)=∑n∈AT1n∑Tn=11n. (80)

Again, whenever the limit order of Equation 77 can be interchanged, we have that

 dens(A)=limT→∞dens′(AT). (81)

The quantity furnishes a computationally feasible way to investigate the behavior of .

In the following, we obtain computational approximations for the asymptotic density and the discussed density. The approximate densities are numerically evaluated as increases in the range from 1 to 1000. Now we investigate (i) arithmetic progressions, (ii) the set of prime numbers, and (iii) the Fibonacci sequence.

### 4.1 Arithmetic Progressions

Considering an arithmetic progressions , we have that:

 d(Mq)=limT→∞d′(MTq)=1q, (82)

and

 dens(Mq)=limT→∞dens′(MTq). (83)

Both quantities limits converge to the same quantity . Intuitively, we have that the chance of “selecting” a multiple of among all integers is . This is exactly the density of . Figure 3 shows the result of computational calculations of and for .

### 4.2 Prime numbers sets

Consider the truncated set of prime numbers . One may easily compute both and for any finite .

An alternative path for the estimation of , where the function represents the number of primes that do not exceed , is to utilize the approximation for given by the Prime Number Theorem [12, 15, p.336]:

 π(T)≈li(T), (84)

where denotes the logarithmic integral [16]. The curves displayed in Figure 4 correspond to the calculation of , and . As expected, all curves decay to zero.

### 4.3 Fibonacci Sequence

Now consider the truncated Fibonacci set . The result of calculating

 d′(FT)=∥FT∥∥NT∥ (85)

and

 dens′(FT)=∑n∈FT1n∑Tn=11n, (86)

are shown in Figure 5. Both curves tend to zero as grows. This fact is expected since we have already verified that . The small convergence rate of is typical of problems that deal with the harmonic series.

## 5 Conclusion

A density for infinite sets of integers was investigated. It was shown that the suggested density shares several properties of usual probability. A computational simulation of the proposed approximate density for finite sets was addressed. An open topic is the characterization of sets for which the limiting value of the approximate density equals their density.

### Footnotes

1. For ease of exposition, in the following results, we assume that the densities of the relevant sets always do exist.

### References

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5. H. Davenport and P. Erdös, “On sequences of positive integers,” Journal of the Indian Mathematical Society, vol. 15, pp. 19–24, 1951.
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