# Hausdorff space not completely Hausdorff

On the set $\mathbbmss{Z}^{+}$ of strictly positive integers, let $a$ and $b$ be two different integers $b\neq 0$ and consider the set

 $S(a,b)=\{a+kb\in\mathbbmss{Z}^{+}\colon k\in\mathbbmss{Z}\}$

such set is the infinite arithmetic progression of positive integers with difference $b$ and containing $a$. The collection of all $S(a,b)$ sets is a basis for a topology on $\mathbbmss{Z}^{+}$. We will use a coarser topology induced by the following basis:

 $\mathbbmss{B}=\{S(a,b):\gcd(a,b)=1\}$

## The collection $\mathbbmss{B}$ is basis for a topology on $\mathbbmss{Z}^{+}$

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

 $x\in S(x,bd)\subset S(x,d)\cap S(c,d)$

besides, since $\gcd(x,b)=1$ and $\gcd(x,d)=1$ then $\gcd(x,bd)=1$ so $x$ and $bd$ are coprimes and $S(x,bd)\in\mathbbmss{B}$. This concludes the proof that $\mathbbmss{B}$ is indeed a basis for a topology on $\mathbbmss{Z}^{+}$.

## The topology on $\mathbbmss{Z}^{+}$ induced by $\mathbbmss{B}$ is Hausdorff

Let $m,n$ integers two different integers. We need to show that there are open disjoint neighborhoods $U_{m}$ and $U_{n}$ such that $m\in U_{m}$ and $n\in U_{n}$, 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 multiple 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 equation $m+tx=n+ty$ 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 $S(m,t)\cap S(n,t)=\emptyset$ and this means that $\mathbbmss{Z}^{+}$ becomes a Hausdorff space with the given topology.

## Some properties of $\overline{S(a,b)}$

We need to determine first some facts about $\overline{S(a,b)}$. 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 $S(4,5)\cup S(5,5)\cup S(6,5)\cup S(7,5)$ 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 $\mathbbmss{Z}^{+}$ is not a completely Hausdorff space is: given any $S(a,b)$, then $b\mathbbmss{Z}^{+}=\{n\in\mathbbmss{Z}^{+}:b\mbox{ divides }n\}$ is a subset of $\overline{S(a,b)}$.

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

 $tx-by=a-bk$

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

This also proves $S(a,b)\neq\overline{S(a,b)}$, because $b$ is not in $S(a,b)$ but it is on the closure.

## The topology on $\mathbbmss{Z}^{+}$ induced by $\mathbbmss{B}$ is not completely Hausdorff

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

Let $m,n$ different positive integers. Since $\mathbbmss{B}$ is a basis, for any two disjoint neighborhoods $U_{m},U_{n}$ we can find basic sets $S(m,a)$ and $S(n,b)$ such that

 $m\in S(m,a)\subseteq U_{m},\qquad n\in S(n,b)\subseteq U_{n}$

and thus

 $S(m,a)\cap S(n,b)=\emptyset.$

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

 $\overline{S(m,a)}\cap\overline{S(n,b)}\neq\emptyset$

and thus $\overline{U_{m}}\cap\overline{U_{n}}\neq\emptyset$.

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 $T_{2}$ space not $T_{2\frac{1}{2}}$ 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