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...

Discussion

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 $\mathbb{Z}$! 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. 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) How can $\mathbb{Z}$ be $N_{1,0}$ if $b > 0$ by definition? Is it $N_{1,1}$? Yes, that's a typo :) - $N_{1,1}$ 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!