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

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