# NOTE: Alternative proof of Infinitude of Primes

This is an alternative proof that infinite number of primes, by using an unusal topology on $$\mathbb{Z}$$

## 1. Definitions

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.

## 2. Every $$S_{a, b}$$ is closed

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.

## 3. Conclusion

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.

Date: 2023-03-07 Tue 00:00