Constructing topology on $\Bbb{Z}$

Question

Fix an infinite subset $A$ of $\mathbb Z$ whose complement $\mathbb{Z}\setminus A$ is also infinite. Construct a topology on $\mathbb{Z}$ in which:

(a) $A$ is open

(b) Singletons are never open (i.e $\forall n \in \mathbb{Z}$, $\{n\}$ is not open).

(c) For any pair of distinct integers $m$ and $n$, there are disjoint open sets $U$ and $V$ such that $m \in U$ and $n \in V$.

My solution: I know this is wrong but I don't understand why, so please be as detailed as possible. I think the Furstenberg topology satisfies all these conditions. Let $\mathcal{B}$ be the collection of all arithmetic progressions on $\mathbb Z$, then an open set $U \in \mathcal{B}$ will be of the form $Z(m,b) = \{mx + b: x \in \mathbb{Z}\}$, where $m$ and $b\in \mathbb{Z}$ and $m\ne 0$.

1. Let $U$ be an open set of $T_{\text furst}$ then $\forall x \in U \ \exists \ Z(m,x) \subseteq U$

2. Every non-empty open set in in $T_{\text furst}$ is infinite.

Then let $A$ be a basic open set of $\mathcal{B}$, then $A$ is infinite and $\mathbb{Z} \setminus A$ is open (can be written as a union of arithmetic progressions) and hence, infinite by (2). Sigletons are not open by (1) and if we describe $\mathbb{Z}$ as $\bigcup_{r=0}^{|m|-1}Z(m,r)$ then it is a union of pairwise disjoint sets which satisfies part c. of the question.

Answer 1

The problem is that you’re supposed to give a construction that works for any infinite $A\subseteq\Bbb Z$ whose complement is also infinite. Your idea works fine if $A$ and $\Bbb Z\setminus A$ happen to be unions of arithmetic progressions, but it doesn’t work if, for instance, $A$ is the set of non-negative integers.

HINT: There is a bijection $h:\Bbb Z\to\Bbb Q$ such that $h[A]=\Bbb Q\cap(0,1)$, and you can use $h$ to define a topology on $\Bbb Z$ that has the desired properties.

Answer 2

Note that not all open sets of the Fürstenberg topology are arithmetic progressions (for example, singletons are closed). A subset of $\Bbb Z$ is Fürstenberg-open iff it is the union of arithmetic sequences. But with that correction, your proof can easily be mended, and it can also easily adapted to an arbitrary infinite and co-infinite subset $A$: Just pick a bijection of $A$ and $2\Bbb Z$ as well as a bijection of $\Bbb Z\setminus A$ and $2\Bbb Z+1$, for example.

So let us treat the case $A=2\Bbb Z$ and the Fürstenberg topology. Then indeed

• $A$ is open, because $2\Bbb Z$ is an arithmetic progression
• $\{n\}$ is not open, because any non-empty open set contains an rithmetic progression and hence is infinite
• Given distinct $m,n$, we have $m\in U:=m+2(m-n)\Bbb Z$ and $n\in V:=n+ 2(m-n)\Bbb Z$ and $U\cap V=\emptyset$.