Hausdorff space not completely Hausdorff

On the set + of strictly positive integers, let a and b be two different integers b0 and consider the set


such set is the infiniteMathworldPlanetmath arithmetic progression of positive integers with differencePlanetmathPlanetmath b and containing a. The collectionMathworldPlanetmath of all S(a,b) sets is a basis for a topologyMathworldPlanetmathPlanetmath on +. We will use a coarserPlanetmathPlanetmath topology induced by the following basis:


The collection 𝔹 is basis for a topology on +

We first prove such collection is a basis. Suppose xS(a,b)S(c,d). By Euclid’s algorithm we have S(a,b)=S(x,b) and S(c,d)=S(x,d) and


besides, since gcd(x,b)=1 and gcd(x,d)=1 then gcd(x,bd)=1 so x and bd are coprimesMathworldPlanetmath and S(x,bd)𝔹. This concludes the proof that 𝔹 is indeed a basis for a topology on +.

The topology on + induced by 𝔹 is Hausdorff

Let m,n integers two different integers. We need to show that there are open disjoint neighborhoodsMathworldPlanetmathPlanetmath Um and Un such that mUm and nUn, but it suffices to show the existence of disjoint basic open sets containing m and n.

Taking d=|m-n|, we can find an integer t such that t>d and such that gcd(m,t)=gcd(n,t)=1. A way to accomplish this is to take any multipleMathworldPlanetmath of mn greater than d and add 1.

The basic open sets S(m,t) and S(n,t) are disjoint, because they have common elements if and only if the diophantine equationMathworldPlanetmath m+tx=n+ty has solutions. But it cannot have since t(x-y)=n-m implies that t divides n-m but t>|n-m| makes it impossible.

We conclude that S(m,t)S(n,t)= and this means that + becomes a Hausdorff space with the given topology.

Some properties of S(a,b)¯

We need to determine first some facts about S(a,b)¯. in order to take an example, consider S(3,5) first. Notice that if we had considered the former topology (where in S(a,b), a and b didn’t have to be coprime) the complement of S(3,5) would have been S(4,5)S(5,5)S(6,5)S(7,5) which is open, and so S(3,5) would have been closed. In general, in the finer topology, all basic sets were both open and closed. However, this is not true in our coarser topology (for instance S(5,5) is not open).

The key fact to prove + is not a completely Hausdorff space is: given any S(a,b), then b+={n+:b divides n} is a subset of S(a,b)¯.

Indeed, any basic open set containing bk is of the form S(bk,t) with t,bk coprimes. This means gcd(t,b)=1. Now S(bk,t) and S(a,b) have common terms if an only if bk+tx=a+by for some integers x,y. But that diophantine equation can be rewritten as


and it always has solutions because 1=gcd(t,b) divides a-bk.

This also proves S(a,b)S(a,b)¯, because b is not in S(a,b) but it is on the closureMathworldPlanetmathPlanetmath.

The topology on + induced by 𝔹 is not completely Hausdorff

We will use the closed-neighborhood sense for completely HausdorffPlanetmathPlanetmath, which will also imply the topology is not completely Hausdorff in the functionalPlanetmathPlanetmathPlanetmath sense.

Let m,n different positive integers. Since 𝔹 is a basis, for any two disjoint neighborhoods Um,Un we can find basic sets S(m,a) and S(n,b) such that


and thus


But then g=ab is both a multiple of a and b so it must be in S(m,a)¯ and S(n,b)¯. This means


and thus Um¯Un¯.

This proves the topology under consideration is not completely Hausdorff (under both usual meanings).

Title Hausdorff space not completely Hausdorff
Canonical name HausdorffSpaceNotCompletelyHausdorff
Date of creation 2013-03-22 14:16:05
Last modified on 2013-03-22 14:16:05
Owner drini (3)
Last modified by drini (3)
Numerical id 21
Author drini (3)
Entry type Example
Classification msc 54D10
Synonym T2 space not T212
Synonym example of a Hausdorff space that is not completely Hausdorff
Related topic CompletelyHausdorff
Related topic SeparationAxioms
Related topic FrechetSpace
Related topic RegularSpace
Related topic FurstenbergsProofOfTheInfinitudeOfPrimes
Related topic SeparationAxioms
Related topic T2Space