I know this question has been asked before (here and here) still I believe this is not a duplicated question.
I am trying to learn a little bit of topology following this course, and I came across the following problem. Given an infinite subset $A \subset \Bbb{Z}$ with its complement $A^c = \Bbb{Z}/A$ also infinite, construct a topology with the following properties:
- $A$ is open
- Singletons are not open
- For any two elements $n,m\in \Bbb{Z}$, there exists two disjoint open sets $U,V\in \mathcal{T}$ with $n\in U$, $m\in V$.
After a few minutes, I came up with the following construction, but I'm not entirely sure if it is valid, so I appreciate it if anyone can give me some feedback and tell me if it is correct, or where it fails. I know there are probably simpler and more elegant ways to do it, and while you are welcome to share them with me, it's not what I am looking for here. My construction goes as follows:
Let $\mathcal{B}_2 = \{A, A^c\}$. Now, given $\mathcal{B}_n$ let's construct $\mathcal{B}_{n+1}$ as follows: For each $B\in \mathcal{B}_n$, split it into two infinite complementary subsets, i.e. $B = B_1 \cup B_2$ with $B_1 \cap B_2 = \emptyset$ and both $B_1$ and $B_2$ having infinitely many elements. Furthermore, given any order in $\Bbb{Z}$, let $x,y\in B$ be the first two elements, the splitting must separate them, i.e. $x\in B_1, y\in B_2$. Then the set $\mathcal{B}_{n+1}$ is defined as the set of all these splittings.
Clearly, $\mathcal{B}_{n+1}$ is a base of $\Bbb{Z}$ since it covers it, and all the elements of $\mathcal{B}_{n+1}$ are disjoint. Furthermore, the topology generated by $\mathcal{B}_{n}$ contains all the previous basis $\mathcal{B}_k$ $k\leq n$
The topology we are looking for is the topology $\mathcal{T}$ generated by $\mathcal{B}_{\infty}$. $A$ is open because $A\in \mathcal{B}_{2} \subset \mathcal{T}$. Singletons are not open since any element of $\mathcal{B}_{\infty}$ has infinitely many elements, and therefore any open set must have infinitely many elements. Finally, given two integers $n,m$ with positions $i_n$ and $i_m$ in whatever order we chose for $\Bbb{Z}$, then property 3 is ensured in the topology generated by $\mathcal{B}_{k}$ as long as $k > \max\{i_n, i_m\}$.
My main concern is about $\mathcal{B}_{\infty}$, I'm not really sure if it even makes sense to talk about such a set, and even if it makes sense, I'm not sure if the properties of $\mathcal{B}_n$ are inherited by $\mathcal{B}_{\infty}$ (for example, all elements in $\mathcal{B}_n$ are infinite, but I'm not sure this would remain true for $\mathcal{B}_{\infty}$)
EDIT: After thinking a little bit more, I thought that maybe it was better to define $\mathcal{B}_{n+1}$ as the set $\mathcal{B}_{n}$ extended with all the splittings. Then I think I can define $$\mathcal{B}_{\infty} = \bigcup_n \mathcal{B}_{n}$$ is this approach better?