# To prove that there are infinitely many prime numbers using topology ### Proof

Let $$\tau$$ denote that collection of $$S(a,b)$$. We show $$\tau$$ is topology. $$\varnothing \in \tau$$ is automatic. Next, since $$\mathbb{Z} = \bigcup \{ n \}$$ and $$\{ n \} = S(1,0)$$, then it is in $$\tau$$. Now, take a collection of $$\{ S(a,b) \}_{a,b \in \mathbb{Z}}$$. we need prove $$\bigcup_{a,b} S(a,b) \in \tau$$. Isnt this automatic by definition?

Finally, if $$S(a_1,b_1)$$ and $$S(a_2,b_2)$$ are two arithmetic progressions, then

$$S(a_1,b_1 ) \cap S(a_2, b_2) = \{ a_1 n + b_1 \} \cap \{ a_2 n + b_2 \}$$

By choosing $$n$$, I think it is possible to write this intersection as union of elements of the form $$\{ a_3 k + b_3 \}$$ but, I am unable to do this rigorously. But I know it is possible by choosing $$n$$ appropriately..

## (b)

If $$x \in \bigcup_p S(p,0)$$ then $$x$$ lies in some $$S(p,0)$$, that is $$x = pn$$for some $$p$$. Since $$p \neq 1,-1$$, then $$x \in \mathbb{Z} \setminus \{-1,1\}$$. Im stuck on the other inclusion. I mean it seems intutitively obvious, Im having hard time writing it rigorously.

finally, assume we have only finite number of primes. Notice that $$\mathbb{Z} \setminus S(p,0) = \bigcup_{q \neq p} S(q,0)$$ which is open so $$S(p,0)$$ is closed.

The complement of $$\mathbb{Z} \setminus \{-1,1\}$$ is $$\{-1,1\}$$ which is not open since set if finite... I havent used the fact that there are finitely many primes... where did I make a mistake?

• There may be a relevant result of Von Furstumberg. – coffeemath Jun 6 at 6:08
• The other inclusion is just the fact that any $x \neq 1,-1$ has a non-trivial prime factor. You need to use too that all $S(a,b)$ are closed too, and then a finite union of $S(p,0)$ is closed and $\{-1,1\}$ would be open. – Henno Brandsma Jun 6 at 7:07
• $\mathbb{Z} \setminus S(p,0) = \bigcup_{q \neq p} S(q,0)$ is false, as $1$ is in the left hand side, but not in the right. Better use $\mathbb{Z} \setminus S(p,0)=\bigcup_{i=1}^{p-1} S(p,i)$ e.g., based on the remainder modulo $p$. – Henno Brandsma Jun 6 at 8:43
(a) If a set $$\tau$$ of subsets of a set $$X$$ is defined as consisting of any union of elements of a set $$B\subset\mathcal P(X)$$, then it is automatic that the union of a family of elements of $$\tau$$ belongs to $$\tau$$ too.
And if $$U_1,U_2\in\tau$$ (now I mean the $$\tau$$ of your specific problem) and if $$x\in U_1\cap U_2$$, then there are integers $$a_1$$ and $$a_2$$ such that $$S(a_1,x)\subset U_1$$ and that $$S(a_2,x)\subset U_2$$. But then $$S(\operatorname{lcm}(a_1,a_2),x)\subset U_1\cap U_2$$. So, $$U_1\cap U_2$$ can be written as an union of sets of the form $$S(a,b)$$ and therefore it belongs to $$\tau$$.
(b) If $$k\in\Bbb Z\setminus\{1,-1\}$$, then there is som number $$p$$ such that $$p\mid k$$ and therefor $$k\in S(p,0)$$. And, if $$l\in\Bbb Z$$ is such that $$l\in S(p,0)$$ for some prime number $$p$$; then $$l\ne\pm1$$; in other words, $$l\in\Bbb Z\setminus\{1,-1\}$$.
Now, note that $$S(p,0)^\complement=S(p,1)\cup S(p,2)\cup\ldots\cup S(p,p-1)$$ ans that therefore $$S(p,0)$$ is a closed. If there were only finitely many primes, then $$\bigcup_pS(p,0)$$ would be a closed set too, and therefore its complement would be an open set. But the complement is $$\{1,-1\}$$ which is not open, since the only finite element of $$\tau$$ is $$\emptyset$$.