aa r X i v : . [ m a t h . GN ] A ug On an exotic topology of the integers
Rezs˝o L. Lovas ∗ Istv´an Mez˝o † Abstract
We study some interesting properties of F¨urstenberg’s topology of theintegers. We show that it is metrizable, totally disconnected, and ( Z , + , · )is a topological ring with respect to this topology. As an application, weshow that any two disjoint sets of primes can be separated by arithmeticprogressions. It has been known since Euclid that there are infinitely many prime numbers.The simplest proof runs as follows: if there were a finite number of primes, say p , . . . , p k , then p p · · · p k + 1 would have at least one prime divisor differentfrom each of p , . . . , p k . Since then, many other proofs of this fact and muchstronger results on the distribution of primes have been found. See Part 1 of [1]for a delightful account of some highlights of this history.In 1955, H. F¨urstenberg found a topological proof for the fact that thereare infinitely many primes [2]. He defined a strange topology on the set Z ofintegers, and, assuming that there is only a finite number of primes, he came toa contradiction. In this topology a set is open if and only if, roughly speaking,each of its points is contained together with an infinite arithmetic progression;we present a precise definition in the next section. This topology is, of course,different from the ordinary topology of Z , i.e., the subspace topology inheritedfrom R . The latter is rather bland: every subset of Z is open.Mathematicians usually think that number theory and topology are com-pletely disjoint areas of mathematics. The construction outlined in the previousparagraph shows very vividly that this is not so: this topology establishes aconnection between these seemingly so distant branches. That is why we foundit so attractive and why we wished to study it in detail. For F¨urstenberg’stopology, we asked and answered some of the most natural questions. We alsofound an application in number theory.In section 2 we briefly recall F¨urstenberg’s proof, and we study some fur-ther interesting properties of his topology. Among others we show that it is ∗ Supported by National Science Research Foundation OTKA No. NK 81402. † Supported by National Science Research Foundation OTKA No. NK 79129.
We denote the set of natural numbers including 0 by N , and we let N ∗ := N \{ } .Similarly, Z ∗ := Z \ { } .For a ∈ Z and b ∈ N ∗ we define a + b Z := { a + bn : n ∈ Z } . If a = 0, we simply write b Z instead of 0 + b Z . We say that a set A ⊂ Z is open if, for each point a ∈ A , there is a number b ∈ N ∗ such that a + b Z ⊂ A . In otherwords, a subset of Z is open if each of its points is contained in an arithmeticprogression belonging to the set. Let T be the set of all open sets in this sense.It is easy to see that T satisfies the usual axioms for a topology, thus ( Z , T )becomes a topological space. So the arithmetic progressions form a basis forthe topology. Obviously, the basic sets a + b Z are open. More surprisingly, theyare also closed, since the complement of a + b Z is the union of other arithmeticprogressions with the same difference.How can we use this topology to show that there is an infinite number ofprimes? Suppose, indirectly, that there is only a finite number of primes, anddenote these by p , . . . , p k . Then the set C := k [ i =1 p k Z is closed, as it is a finite union of closed sets. On the other hand, all integersbut − C = Z \ {− , } . Thusits complement {− , } is open, which is clearly a contradiction, and this provesthat there are indeed infinitely many prime numbers.Now we turn to the metrizability property of the topology T . If n ∈ Z ∗ ,then we define k n k := 1max { k ∈ N ∗ : 1 | n, . . . , k | n } , i.e., then “norm” k n k is the reciprocal of the greatest natural number k withthe property that the natural numbers 1 , . . . , k are all divisors of n . Thus, forexample, k k = 1 , k k = 12 , k k = 1 , k k = 12 , k k = 1 , k k = 13 , . . . , k n ! k ≤ n , k − n k = k n k . Furthermore, we set k k := 0.Then we define the distance of the integers m and n by d ( m, n ) := k m − n k . Theorem 1.
With this distance function, ( Z , d ) becomes a metric space, andthe metric d induces the topology T .Proof. All axioms of a metric space are trivially satisfied except the triangleinequality. To prove that, first we show that k m + n k ≤ k m k + k n k if m, n ∈ Z .It is sufficient to show this when both m and n are different from 0.First suppose k m k ≤ k n k . Then the numbers 1 , , . . . , k n k are all divisors ofboth m and n , thus they are divisors of m + n as well, therefore k m + n k ≤ k n k < k m k + k n k . For k m k > k n k we get the assertion by interchanging the role of m and n . Fromthis, the triangle inequality follows easily: d ( m, n ) = k m − n k = k m − l + l − n k ≤ k m − l k + k l − n k = d ( m, l ) + d ( l, n )for any l, m, n ∈ Z .Now we show that the metric d induces T , indeed. Let A ⊂ Z be an openset with respect to the metric d . If a ∈ A , then there is a positive number r such that the open ball B ( a, r ) with center a and radius r is contained in A . Let b ∈ N ∗ be such that k b k < r . (Such a b exists: e.g., if n is such that 1 /n < r ,then take b := n !.) If a + bn is an arbitrary element of a + b Z , then d ( a, a + bn ) = k a − ( a + bn ) k = k bn k ≤ k b k < r, since the divisors of b are divisors of bn as well. This means that a + bn ∈ B ( a, r )and a + b Z ⊂ B ( a, r ) ⊂ A , thus A is open also with respect to the topology T .To see the converse, let A ∈ T . Thus, if a ∈ A , then a + b Z ⊂ A with some b ∈ N ∗ . Let r := 1 /b . If c ∈ B ( a, r ), i.e., k a − c k < r , then b | a − c , thus c = a + bn for some n ∈ Z , thus c ∈ a + b Z . We have obtained B ( a, r ) ⊂ a + b Z ⊂ A , i.e., A is open with respect to d , too.According to the previous two paragraphs, A ∈ T if and only if A is openwith respect to the metric d , which means exactly that d induces the topology T . Corollary.
A sequence ( a n ) n ∈ N in Z converges to in the topology T if andonly if, for every k ∈ N ∗ , there is a number N ∈ N such that n ≥ N implies k | a n .Proof. The statement follows from the fact that each neighborhood of 0 containsa set of the form k Z .Thus a sequence converges to 0 if and only if any positive k is a divisor ofall members whose index is sufficiently large. A typical sequence converging to0 in our topology is ( n !) n ∈ N . This too shows that our topology is different from3he ordinary topology of Z , since in the latter one a sequence can converge to 0only if all members but finitely many are zero.This has a strange consequence as well. Namely, let a := 1 , and a n := ( n + 1)! − n ! = n · n ! if n ≥ . Then P ∞ n =0 a n = 0 with respect to the topology T , although every member ofthis series is a positive integer.Recall that a topological space is said to be totally disconnected if each ofits maximal connected subsets consists of one single point. Theorem 2.
The topological space ( Z , T ) is totally disconnected.Proof. We show that if a set A ⊂ Z contains at least two different elements a and b , then it cannot be connected. Let k be a nonzero integer which is not adivisor of b − a . Then A ∩ { a + nk : n ∈ Z } and A ∩ { a + 1 + nk, . . . , a + k − nk : n ∈ Z } are nonempty disjoint open subsets of A whose union is A , thus A is not con-nected.A topological ring is a ring which is a topological space at the same time suchthat the ring operations are continuous with respect to the product topology,and so is the additive inversion. Theorem 3.
The set Z is a topological ring with respect to the usual additionand multiplication and the topology T .Proof. The additive inversion n ∈ Z
7→ − n is obviously continuous. To provethe continuity of addition, by translation invariance, it is enough to show thatit is continuous at the point (0 , ε >
0, and k a k , k b k < ε , then k a + b k < ε by the properties of the norm.To prove the continuity of multiplication, we show that a n b n → ab if a n → a and b n → b . Indeed, if k ∈ N ∗ , then there exist N , N ∈ N such that n ≥ N implies k | a n − a , and n ≥ N implies k | b n − b . If N := max { N , N } , and n ≥ N , then a n ≡ a, b n ≡ b (mod k ) , therefore a n b n ≡ ab (mod k ) , i.e., k | a n b n − ab . As an application, we show that any two (possibly infinite) disjoint sets of primescan be separated by arithmetic progressions. So, for example, let A = { p , p , . . . } and B = { p , p , . . . } , p i is the i th positive prime number.Then there are two sets, say A and B , obtained as unions of arithmeticprogressions, such that A ⊂ A , B ⊂ B and A ∩ B = ∅ . More precisely, the next theorem is valid.
Theorem 4.
Let A = { p , p , p , . . . } and B = { q , q , q , . . . } two disjoint sets of positive primes. Then there are numbers a , a , a , . . . ∈ N ∗ and b , b , b , . . . ∈ N ∗ such that A ⊂ { p + k a , p + k a , p + k a , . . . : k i ∈ Z } ,B ⊂ { q + k b , q + k b , q + k b , . . . : k i ∈ Z } , and p i + k i a i = q j + k j b j for any i, j ∈ N and k i , k j ∈ Z . We remark that the crucial point in the proof is a theorem of Urysohn[3, 2.5.7]: a topological space with a countable basis is metrizable if and only ifthe singletons are closed and any two disjoint closed sets can be separated bydisjoint open sets.
Proof.
Consider the set e Z := { n ∈ Z : n ≥ } with the topology induced by T . The sets A and B are closed subsets of e Z ,since, e.g., e Z \ A = e Z ∩ [ n ∈ e Z \ A n Z , and similarly for b . Since e Z is a metrizable topological space, two disjoint closedsets can be separated by disjoint open sets, which we may assume to be of theforms e Z ∩ U and e Z ∩ V , where U = { p + k a , p + k a , p + k a , . . . : k i ∈ Z } ,V = { q + k b , q + k b , q + k b , . . . : k i ∈ Z } . Note that, a priori, we do not know that U and V are disjoint, we only knowthat e Z ∩ U ∩ V = ∅ . If, however, we had p i + k i a i = q j + k j b j for some i, j ∈ N and k i , k j ∈ Z , then this Diophantine equation would haveinfinitely many solutions for k i and k j , and substituting some of these into theequation would render both sides ≥
2, which would contradict e Z ∩ U ∩ V = ∅ .This completes the proof that U ∩ V = ∅ .5t is worth noting that the Diophantine equation a i k i − b j k j = q j − p i has a solution if and only if ( a i , b j ), the greatest common divisor of a i and b j , isa divisor of p i − q j . Therefore, the condition obtained on a i and b j is equivalentto ( a i , b j ) ∤ p i − q j for any i, j ∈ N . Note
Having completed this paper, we realized that in November 2009 another metricinducing this topology had been proposed, as far as we make it out, by Jim Ferry[4]. In a way which is similar to ours, he defines a kind of norm k n k := X k ∈ N ∗ ,k ∤ n − k for n ∈ Z , and then he lets d ( m, n ) := k m − n k for m, n ∈ Z . He provesthat this is indeed a metric, but he leaves it to heuristic judgement whether itinduces F¨urstenberg’s topology. References [1] M. Aigner, G. M. Ziegler,
Proofs from the Book , Springer, Berlin, 2002.[2] H. F¨urstenberg, On the infinitude of primes,
Amer. Math. Monthly (1955)353.[3] A. N. Kolmogorov, S. V. Fomin, Elements of the Theory of Functions andFunctional Analysis , Dover Publications, Mineola, NY, 1999.[4] Science Forum Index,