Exercise solved topology I do not understand the resolution of this exercise
For any pair of integers $a, b \in \mathbb{R}$ , with $b > 0$, let’s write $N_{a,b}= \{ a+kb |  \in \mathbb{Z} \} $.  Prove the following facts:

*

*arithmetic progressions $\beta = \{ N_{a,b} | a, b   \in \mathbb{Z},>0 \} $ form a basis for a topology $\tau$ on $\mathbb{Z}$

*every $N_{a,b}$ is both open and closed in $\tau$ ;

*call $P = \{ 2, 3, . . . \} \subset \mathbb{N}$ the set of primes. Then

$\mathbb{Z} − \{ −1, 1 \} = \cup \{ N_{0,p} | p \in P \}$
Therefore if $P$ were finite $\{−1, 1 \}$ would be open in $\tau$ .
Solutions
We need to show that the family of progressions $N_{a,b}$ satisfies this Theorem (but
this is an immediate consequence of the formulas) (Why was the theorem not used?)
(Theorem
Let $X$ be a set and $\mathcal{B} \subset \mathcal{P} (X)$ a family of subsets. There exists a
topology on $X$ for which $\mathcal{B}$ is a basis if and only if two conditions hold:

*

*$X = \cup \{ B | B \in \mathcal{B} \}$

*for any pair $A, B \in \mathcal{B}$ and any point $x \in A \cap B$ there exists $C \in \mathcal{B}$ such that $x \in C \subset A \cap B$)

$N_{0,1} = \mathbb{Z}$,  $N_{a,b} \cup N_{c,d} = \cup \{ N_{s,bd} | s \in N_{a,b} \cap N_{c,d} \}$ (
(I do not understand this)
As $N_{a,b}$ is the complement in $\mathbb{Z}$of the open union
$N_{a+1,b} \cup N{a+2,b} \cup . . . \cup N_{a+b−1,b}$ , (I do not understand this)
the open set $N_{a,b}$ is also closed. Note that any non-empty open set contains at least
one arithmetic progression, and hence it must be infinite.
Thanks
 A: $N_{0,1} = \Bbb Z$ is clear, and shows that condition 1 of the basis theorem is satisfied, even one basic element already covers the whole space.
To show 2, the statement you gave is not enough. Let $x \in N_{a,b} \cap N_{c,d}$.
So $x= a+k_1b = c + k_2d$ for some $k_1,k_2 \in \Bbb Z$. Now let $e:=(b,d)$ be the gcd of $b$ and $d$, it's then clear that $x \in N_{x,e} \subseteq N_{a,b}\cap N_{c,d}$ so that 2 is satisfied. (this is an improvement on the wrong argument you cited.)
So the $N_{a,b}$ form a base. If $x \in \Bbb Z$ and $b>0$, the clases modulo $b$ divide $\Bbb Z$ into $b$ many disjoint classes which are $N_{0,b}, N_{1,b}, \ldots N_{b-1,b}$, covering $\Bbb Z$, depending on whether the remainder modulo $b$ is $0, \ldots, b-1$.
So all sets $N_{a,b}$ are also closed as the complement is the union of the other $b-2$ sets in this partition, and thus open.
The final remark that $$\Bbb Z\setminus \{-1,1\} = \bigcup_{p \in P} N_{0,p}$$
follows trivially from the fact "any integer unequal to $1$ or $-1$ has a prime divisor".
If $P$ were finite, that union would be closed as a finite union of closed sets and $\{-1,1\}$ would be open (and finite), a contradiction, as all open sets are unions of basic open sets, and these are all infinite. So finite open sets cannot exists (except $\emptyset$)
