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 $\pi:\mathbb{Q}\to\mathbb{Z}$ be any $\textit{onto}$ function. We define a topology $\mathfrak{T}$ on $\mathbb{Z}$ such that $U\subset\mathbb{Z}$ 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 $\mathbb{Z}$.
 A: Perhaps the simplest solution is to let $\tau$ be the topology on $\Bbb Z$ generated by the following base:
$$\big\{\{n\}:n\in\Bbb Z\setminus\{0\}\big\}\cup\{\cup(\leftarrow,-n]\cup\{0\}\cup[m,\to):n,m\in\Bbb Z^+\}\;.$$
That is, each point of $\Bbb Z\setminus\{0\}$ 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 $\Bbb Z$.
A: 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 $\tau = \{A \subset \mathbb{Z} \mid \exists z \in \mathbb{Z}, A = (-\infty, z) \cap \mathbb{Z}\}$.
A: 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
$$\text{A set }\mathcal{U}\subseteq\mathbb{Z}\text{ is open } \iff \phi(\mathcal{U}) \text{ is open in }X.$$
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 $\mathbb{R}$.
