Verify that every arithmetic progression in $\mathbb Z$ is clopen (i.e., both open and closed) with respect to the evenly-spaced topology

Question

A (bi-infinite) arithmetic progression in $\mathbb Z$ is a set of the form $A(a, b)$ := {$a \mathbb Z + b$} = {$an + b : n ∈ \mathbb Z$}, where $a, b ∈ \mathbb Z$ and $a \not= 0$.

(a) Verify that the collection of all the arithmetic progressions in Z is a basis for a topology on Z.

1. Every $a \in \mathbb Z$ is in some base element because $\mathbb Z = A(0,1)$, so one base element suffices.

2. For any two elements $B_1=A(a_1,b_1)$ and $B_2=A(a_2,b_2)$ and any $x∈B_1∩B_2$ we must find some $B_3$ containing $x$ sitting inside this intersection. We know that $x=a_1+n_1b_1=a_2+n_2b_2$ for some $n_1,n_2∈\mathbb Z$. But then $x∈A(x,b_1b_2)⊆B_1∩B_2$, as e.g. for any $n∈\mathbb Z$: $x+nb_1b_2=a_1+n_1b_1+nb_1b_2=a_1+(n_1+b_2n)b_1∈A(a_1,b_1)$ and likewise for $A(a_2,b_2)$.

(b) The topology generated by the arithmetic progressions is called the evenly-spaced topology or Furstenberg’s topology. Verify that every arithmetic progression in $\mathbb Z$ is clopen (i.e., both open and closed) with respect to the evenly-spaced topology.

I don't know how to verify this. Any help please?

Answer 1

Note that $x\in\mathbb{Z}$ belongs to $A(a,b)$ if and only if $x\equiv b\text{ mod }a$.

With this you can easily verify that for any $a\in\mathbb{N}$ and any $n,m\in\{0,1,\ldots,a-1\}$, $n\neq m$ we have $A(a,n)\cap A(a,m)=\emptyset$ and

$$\mathbb{Z}=\bigcup_{n=0}^{a-1} A(a,n)$$

Meaning $\{A(a,0), A(a,1),\ldots, A(a,a-1)\}$ is a partitioning of $\mathbb{Z}$. Thus if every $A(a,b)$ is open then so is its complement.