Harry Furstenberg is an American-Israeli mathematician and a laurea...

A basis for a topology on $\mathbb{Z}$ is a collection B of subsets...

The author is defining a subset U of Z to be open if for every $m \...

We say a set is closed if its complement is open. It's not difficul...

Harry Furstenberg is an American-Israeli mathematician and a laureate of the Wolf Prize in Mathematics.
In 1955, when Harry was still a 20 years-old undergraduate student at Yeshiva University he gained attention for coming up with this innovative topological proof of the infinitude of prime numbers.
![](https://upload.wikimedia.org/wikipedia/commons/a/a7/Harry_Furstenberg.jpeg)
We say a set is closed if its complement is open. It's not difficult to see that the complement of $N_{7,3}$ is $N_{8,3} \cup N_{9,3}$. By definition since $N_{7,3}$ is open its complement is closed but at the same time by axiom #2 we have that $N_{8,3} \cup N_{9,3}$ is open - $N_{a,b}$ is open and closed at the same time!
The author is defining a subset U of Z to be open if for every $m \in U$, there is a set of the form
$$
N_{a,b} = \{a+nb: n \in \mathbb{Z}\}, \ \ b>0
$$
such that $m \in N_{a,b} \subset U$. Here's an example of what $N_{1,2}$ looks like
![](https://i.imgur.com/6tv2viq.png)
Now that we've defined our open sets, we need to make sure they satisfy the axioms of topology:
#1 both the empty set and $\mathbb{Z}$ are open
#2 a union of open sets is also open
#3 a finite intersection of open sets is open
For axiom #1, by definition, the empty set is open; $\mathbb{Z}$ is just the sequence $N_{1,0}$, and so is open as well.
For axiom #2, it's not difficult to see that $N_{a,b} \cap N_{c,d} = N_{a+c,b+d}$
For axiom #3, we note that $N_{a,b} \cup N_{a,c} = N_{a,lcm(b,c)}$ where $lcm(b,c)$ is the least common multiple of b and c.
How can $\mathbb{Z}$ be $N_{1,0}$ if $b > 0$ by definition? Is it $N_{1,1}$?
Yes, that's a typo :) - $N_{1,1}$
A basis for a topology on $\mathbb{Z}$ is a collection B of subsets of $\mathbb{Z}$ (called basis elements) satisfying the following properties.
1. For each x in $\mathbb{Z}$, there is at least one basis element B containing x.
2. If x belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing x such that $B_3 \subset B_1 \cap B_2$.
It is not difficult to see that 1. is verified because for any x in $\mathbb{Z}$, we can always have arithmetic progressions that include x.
To prove 2. let's suppose $a_{1}$ and $a_{2}$ are the differences between successive terms in the sequences $B_{1}$ and $B_{2}$. If $a$ is the least common multiple of $a_{1}$ and $a_{2}$, then $B_{3}=\{\ldots,n-a, n, n+a, n+2a,\ldots\}$.
We have now proven that the sequence of arithmetic progressions forms a basis for a topology on $\mathbb{Z}$!