How can $\mathbb{Z}$ be $N_{1,0}$ if $b > 0$ by definition? Is it $N_{1,1}$? 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! 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.  This proof is featured in "Proofs from the Book" by Martin Aigner, Günter M. Ziegler. This book is supposed to present some of the most elegant proofs. Its title is a reference to a story by Erdos about a book where God has written the most beautiful proofs. 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 $\mathbb{Z}$! 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  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.