# Hausdorff space not completely Hausdorff

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

$$S(a,b)=\{a+kb\in {\mathbb{Z}}^{+}:k\in \mathbb{Z}\}$$ |

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 ${\mathbb{Z}}^{+}$. We will use a coarser^{} topology induced by the following basis:

$$\mathbb{B}=\{S(a,b):\mathrm{gcd}(a,b)=1\}$$ |

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

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

## The topology on ${\mathbb{Z}}^{+}$ induced by $\mathbb{B}$ 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
$\mathrm{gcd}(m,t)=\mathrm{gcd}(n,t)=1$. 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 $S(m,t)\cap S(n,t)=\mathrm{\varnothing}$ and this means that ${\mathbb{Z}}^{+}$ 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 ${\mathbb{Z}}^{+}$ is not a completely Hausdorff space is: given any $S(a,b)$, then $b{\mathbb{Z}}^{+}=\{n\in {\mathbb{Z}}^{+}:b\text{divides}n\}$ 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 $\mathrm{gcd}(t,b)=1$. 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 $1=\mathrm{gcd}(t,b)$ divides $a-bk$.

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

## The topology on ${\mathbb{Z}}^{+}$ induced by $\mathbb{B}$ 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 $\mathbb{B}$ 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

$$m\in S(m,a)\subseteq {U}_{m},n\in S(n,b)\subseteq {U}_{n}$$ |

and thus

$$S(m,a)\cap S(n,b)=\mathrm{\varnothing}.$$ |

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

$$\overline{S(m,a)}\cap \overline{S(n,b)}\ne \mathrm{\varnothing}$$ |

and thus $\overline{{U}_{m}}\cap \overline{{U}_{n}}\ne \mathrm{\varnothing}$.

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 ${T}_{2\u2064\frac{1}{2}}$ |

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 |