[NOTE] Alternative proof of Infinitude of Primes

1. The topology is on \(\mathbb{Z}\)
2. and define \(S_{a, b} = \{ an + b \mid n \in \mathbb{Z} \}\)

It's easy to check \(S_{a, b}\) is a subset of \(\mathbb{Z}\). The conditions of open are

1. \(S_{a, b}\) is open iff \(a \in \mathbb{Z} \setminus \{0\}\) and \(b \in \mathbb{Z}\)
2. \(\varnothing\) is open

and you should check above definitions truly cover \(\mathbb{Z}\) is open.

If \(a = 0\), in this case \(S_{a, b} = S_{0, b} = \{ b \}\). Therefore, \(S_{0, b}\)'s complement is \(\mathbb{Z} \setminus \{b\}\), that can be re-expressed to

\[ \bigcup_{\forall x \ge 2} S_{x, b-1} \cup S_{x, b+1} \]

since for all \(x \ge 2\), \(S_{x, b-1}\) and \(S_{x, b+1}\) are open, the complement is open, and \(S_{0, b} = \{ b \}\) is closed.

If \(|a| = 1\), the complement of \(S_{1, b}\) is \(\varnothing\), and hence \(S_{1, b}\) is closed.

If \(|a| > 1\), the complement is the union of every \(S_{a, b + y}\) where \(0 < y < |a|\). Since every set in the complement is open, \(S_{a, b}\) is closed.

Now, we have enough information to say that all \(S_{a, b}\) are closed.

Because every integer has a prime factor except \(1\) and \(-1\), so we can say they are the union of all primes can cover, that can be written as

Note: You can say \(z \in S_{p, 0}\) if \(z\) has \(p\) as a prime factor of it.

\[ \mathbb{Z} \setminus \{1, -1\} = \bigcup_{p \; \text{prime}} S_{p, 0} \]

the left hand side \(\mathbb{Z} \setminus \{1, -1\}\) is open, and hence we must have infinitely many closed set on the right! Because the union of a finite collection of closed sets is closed. Therefore, we can have conclusion there are infinitely many primes.