Exercise 1.3.52* (There are infinitely many primes, third proof (topological proof))

Proof. a) We verify first that the open sets define a topology. Note that $\text{∅}$ and $\mathbb{Z}$ are open sets.

Let ${({𝒪}_i)}_{i\in I}$ be a family of open sets, and $𝒪={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{𝒪}_i$. Let $n$ be any element in $𝒪$. Then there is some $j\in I$ such that $n\in {𝒪}_j$. Since ${𝒪}_j$ is open, there is an arithmetic progression $𝒜=\{\mathit{dk}+r:k\in \mathbb{Z}\}$ such that $𝒜\subset {𝒪}_j$. Moreover ${𝒪}_j\subset 𝒪={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{𝒪}_i$, thus $n\in 𝒜\subset 𝒪$. Since this is true for every $n\in 𝒪$, $𝒪$ is an open set.

Now consider a family ${({𝒪}_i)}_{i\in I}$, where $I=[[1,m]],m\in \mathbb{N},$ is finite, and $𝒪={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{𝒪}_i$. If $n\in 𝒪$, then $n\in {𝒪}_i$ for every $i\in [[1,m]]$. Therefore there are arithmetic progressions ${𝒜}_i=\{d_{i}k+r_i:k\in \mathbb{Z}\}\subset {𝒪}_i$, such that $n\in {𝒜}_i$, for for every $i\in [[1,m]]$. Thus there is $k_i\in \mathbb{Z}$ such that

Consider the arithmetic progression

Then $n=d_1\cdots d_m\cdot 0+r\in 𝒜$. Moreover, we will show that $𝒜\subset {𝒜}_i\subset {𝒪}_i$ for every $i\in [[1,m]]$.

Let $a\in 𝒜$. Then $a=d_1\cdots d_{m}k+n$ for some $k\in \mathbb{Z}$. For every $i\in [[1,m]]$,

where $l_i={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\ne i}{d}_j+k_i\in \mathbb{Z}$. Therefore $a\in {𝒜}_i\subset {𝒪}_i$. This proves

so

This shows that ${\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{𝒪}_i$ is an open set. So the open sets define a topology.

b) Let $𝒜=\{\mathit{dk}+r^{\prime }:k\in \mathbb{Z}\}$ be an arithmetic progression. Since $\{\mathit{dk}+r:k\in \mathbb{Z}\}=\{(-d)k+r:k\in \mathbb{Z}\}$, we can assume $d>0$. If $r^{\prime }$ is the remainder of the division of $r^{\prime }$ by $d$, then $r^{\prime }=d{q}_0+r$, where $0\le r<d$, and

So we can always assume $d>0$ and $0\le r<d$ to define an arithmetic progression.

We prove that $\mathbb{Z}\setminus 𝒜$ is a (finite) union of arithmetic progressions. Let

If $j\in \mathbb{Z}\setminus 𝒜$, then the Euclidean division of $j$ by $d>0$ gives $j=\mathit{dq}+i,0\le i<d$. Moreover $i\ne r$, otherwise $j\in 𝒜$. So

Conversely, if $j\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in [[0,d-1]]\setminus \{r\}}{𝒜}_i$, then $j=\mathit{dq}+i,q\in \mathbb{Z}$ for some $i$ such that $0\le i<d,i\ne r$. Assume for contradiction that $j\in 𝒜$, then

The unicity of quotient and remainder in the Euclidean division shows that $i=r$. This is a contradiction, because $i\ne r$. So $j\in \mathbb{Z}\setminus 𝒜$. This proves

So $\mathbb{Z}\setminus 𝒜={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in [[0,d-1]]\setminus \{r\}}{𝒜}_i$ is a finite union of arithmetic progressions.

Since every element of $𝒜$ is in the arithmetic progression $𝒜$, $𝒜$ is open by definition.

Since $𝒜$ is a finite union of arithmetic progressions, it is closed. So $𝒜$ is both open and closed.

c) Write $\mathbb{P}=\{2,3,5,7,11,\dots \}$ the set of prime numbers. Let

We show that $𝒱=\{-1,1\}$.

First, $p\nmid 1,p\nmid -1$ for every prime number $p$, so $\{-1,1\}\subset 𝒱$.

Now take $n\in \mathbb{Z}\setminus \{-1,1\}$. If $n=0$, then $0\in 2\mathbb{Z}\subset 𝒰$. If $n\ne 0$ (and $n\ne 1,n\ne -1$), $n$ is divisible by some prime number $p$, so $n\in 𝒰$. This shows that $\mathbb{Z}\setminus \{-1,1\}\subset 𝒰$, therefore $𝒱=\mathbb{Z}\setminus 𝒰\subset \{-1,1\}$, so

d) $𝒱$ is not an open set. Otherwise $1\in 𝒜\subset \{-1,1\}$, where $𝒜$ is some arithmetic progression. Since $𝒜$ is infinite, and $\{-1,1\}$ finite, $𝒜\subset \{-1,1\}$ is impossible. So $𝒱$ is not open.

If there were only finitely many prime numbers, then $𝒰={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{p\in \mathbb{P}}\mathit{p\mathbb{Z}}$ would be a finite union of closed sets (by part (b)), so would be itself a closed set. Then $𝒱$ would be an open set.

This is a contradiction, so there are infinitely many prime numbers. □