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[parent] Hausdorff space not completely Hausdorff (Example)

On the set $\mathbbmss{Z}^+$ of strictly positive integers, let $a$ and $b$ be two different integers $b\neq0$ and consider the set

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

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:
\begin{displaymath}\mathbbmss{B}=\{S(a,b): \gcd(a,b)=1\}\end{displaymath}

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
\begin{displaymath} x\in S(x,bd)\subset S(x,d)\cap S(c,d) \end{displaymath}

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 $d=\vert m-n\vert$, 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 $t>\vert n-m\vert$ 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

\begin{displaymath}tx - by = a-bk\end{displaymath}

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

\begin{displaymath}m\in S(m,a)\subseteq U_m,\qquad n\in S(n,b)\subseteq U_n\end{displaymath}

and thus
\begin{displaymath}S(m,a)\cap S(n,b)=\emptyset.\end{displaymath}

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

\begin{displaymath}\overline{S(m,a)}\cap\overline{S(n,b)}\neq\emptyset\end{displaymath}

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).



"Hausdorff space not completely Hausdorff" is owned by drini. [ full author list (3) | owner history (2) ]
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See Also: completely Hausdorff, separation axioms, Fréchet space, regular space, Fürstenberg's proof of the infinitude of primes, separation axioms, Hausdorff space

Other names:  $T_2$ space not $T_{2\frac12}$, example of a Hausdorff space that is not completely Hausdorff

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Cross-references: functional, closure, terms, subset, completely Hausdorff space, basic sets, finer, closed, complement, order, Hausdorff space, divides, implies, solutions, Diophantine equation, multiple, open sets, neighborhoods, disjoint, open, proof, coprimes, Euclid's algorithm, induced, coarser, topology, basis, collection, difference, arithmetic progression, infinite, integers, positive, strictly
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This is version 18 of Hausdorff space not completely Hausdorff, born on 2004-03-17, modified 2005-05-10.
Object id is 5718, canonical name is HausdorffSpaceNotCompletelyHausdorff.
Accessed 3376 times total.

Classification:
AMS MSC: 54D10 (General topology :: Fairly general properties :: Lower separation axioms )
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