Hausdorff and non-discrete topology on $mathbb{Z}$

Construct a topology $mathfrak{T}$ on $mathbb{Z}$ such that $mathbb{Z}$ is Hausdorff and non-discrete with respect to $mathfrak{T}$.

$textbf{My idea}$ : We know that $mathbb{Q}$ is Hausdorff and non-discrete with respect to the topology inherited from $mathbb{R}$. So we should use this fact in a following way.

Let $pi:mathbb{Q}tomathbb{Z}$ be any $textit{onto}$ function. We define a topology $mathfrak{T}$ on $mathbb{Z}$ such that $Usubsetmathbb{Z}$ is open in $mathbb{Z}$ if and only if $pi^{-1}(U)$ is open in $mathbb{Q}$. In other words, we define the largest topology on $mathbb{Z}$ such that $pi$ becomes continuous. Now it is a matter of choosing an appropriate onto map $pi$. I am having a difficulty in choosing such $pi$.

I have tried the integer part function as $pi$, and the resultant topology $mathfrak{T}$ on $mathbb{Z}$ is indeed $textit{non-discrete}$ but the $textit{Hausdorffness}$ is not clear. Can anyone please suggest me some other functions as $pi$ ? Also I would be really grateful if someone suggests me an entire new way to define a topology $mathfrak{T}$ on $mathbb{Z}$.

Mathematics Asked by UDAY PATEL on November 18, 2021

3 Answers

3 Answers

It seems to me the simplest solution is to find a countable space $X$ that is Hausdorff and non-discrete, and then endow a topology on $mathbb{Z}$ through a bijection with $X$.

Explicitly: suppose $X$ is countable and Hausdorff and non-discrete. Let $varphi: mathbb{Z}rightarrow X$ be a bijection. Define a topology on $mathbb{Z}$ by $$text{A set }mathcal{U}subseteqmathbb{Z}text{ is open } iff phi(mathcal{U}) text{ is open in }X.$$

You can readily check that this defines a topology $tau$ on $mathbb{Z}$ such that $(mathbb{Z},tau)$ is homeomorphic to $X$. In particular, $(mathbb{Z},tau)$ is Hausdorff and non-discrete.

Now the problem is reduced to finding a space $X$ that's countable and Hausdorff and non-discrete. And you already stated $mathbb{Q}$ is one such space. Another easy example is a ${0}cup{frac{1}{n}}_n$ with the subspace topology inherited from $mathbb{R}$.

Answered by JKEG on November 18, 2021

I'm really beginning with topology so if I make any mistakes please correct me.

A general surjective map $pi colon mathbb{Q} to mathbb{Z}$ will not guarantee that the obtained topology is Hausdorff.

The map $pi colon mathbb{Q} to mathbb{Z}$ is the integer part map if given $q in mathbb{Q}$, $pi(q) = [q]$ where $q - 1 < [q] leq q$ and $[q] in mathbb{Z}$.

Given $z in mathbb{Z}$, $pi^{-1}(z) = [z,z+1) cap mathbb{Q}$. If $A subset mathbb{Z}$ is a subset, we know that $pi^{-1}(A) = cup_{a in A}pi^{-1}(a) =$ $cup_{a in A} [a , a+1) cap mathbb{Q}$ and so $pi^{-1}(A)$ is open if, and only if, $cup_{a in A} [a , a+1) cap mathbb{Q}$ is open in $mathbb{Q}$ with the induced topology from $mathbb{R}$.

If the set $A$ is a finite set, then it has a minimal element $alpha$, and as it is minimal in $A$, for sure it will be minimal in $cup_{a in A} [a , a+1)$, hence $cup_{a in A} [a , a+1)$ is not open, so $pi^{-1}(A)$ is not open which implies $A$ not open in $mathbb{Z}$.

If $A$ in infinite, then it is open if, and only if, all the elements of $A$ are consecutive numbers, cause if there are $a,b in A$ such that $a < b$ and $a neq b-1$, then $b$ is a problematic point for the openess of $cup_{a in A} [a , a+1)$, as the ball $B(b; frac12)$ is not contained in the set. Hence, all the elements must be consecutives, so $A$ is infinite and not lower bounded.

We can conclude, then, that the topology you're looking at is $tau = {A subset mathbb{Z} mid exists z in mathbb{Z}, A = (-infty, z) cap mathbb{Z}}$.

Answered by Jefferson Johnes on November 18, 2021

Perhaps the simplest solution is to let $tau$ be the topology on $Bbb Z$ generated by the following base:

$$big{{n}:ninBbb Zsetminus{0}big}cup{cup(leftarrow,-n]cup{0}cup[m,to):n,minBbb Z^+};.$$

That is, each point of $Bbb Zsetminus{0}$ is isolated, and nbhds of $0$ contain a tail of the positive integers and a tail of the negative integers.

For your original idea I’d not even bother with an explicit $pi$: $Bbb Q$ is countably infinite, so there is a bijection between it and $Bbb Z$, and you can use that to define a topology on $Bbb Z$ making it homeomorphic to $Bbb Q$. That’s sufficient unless you absolutely need an explicit description of the topology on $Bbb Z$.

Answered by Brian M. Scott on November 18, 2021

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