## § Topological proof of infinitude of primes

We take the topological proof and try to view it from the topology as
semidecidability perspective.
- Choose a basis for the topology as the basic open sets $S(a, b) = \{ a n + b : n \in \mathbb Z \} = \{ mathbb Z + b \}$. This set is indeed semi-decidable. Given a number $k$, I can check if $(k - b) % a == 0$. So this is our basic decidability test.
- By definition, $\emptyset$ is open, and $\mathbb Z = S(1, 0)$. Thus it is a valid basis for the topology. Generate a topology from this. So we are composing machines that can check in parallel if for
*some * $i$, $(k - b[i]) % a[i] == 0$ for some index. - The basis $S(a, b)$ is clopen, hence the theory is decidable.
- Every number other than the units $\{+1, -1\}$ is a multiple of a prime.
- Hence, $\mathbb Z \setminus \{ -1, +1 \} = \cup_{p \text{prime}} S(p, 0)$.
- Since there a finite number of primes [for contradiction ], the right hand side must be must be closed.
- The complement of $\mathbb Z \setminus \{ -1, +1 \}$ is $\{ -1, +1 \}$. This set cannot be open, because it cannot be written as the union of sets of the form $\{ a n + b \}$: any such union would have infinitely many elements. Hence, $\mathbb Z \setminus \{ -1, +1 \}$ cannot be closed.