# Elementary topological proof of twin prime conjecture.

For any function $f: \Bbb{N}^{\times} \to \Bbb{N}^{\times}$, $|f(n) - f(m)|$ is a pseudometric on $\Bbb{N}^{\times}$. When $f = \Omega$ the number of prime divisors including multiplicity, we get a topological monoid on $\Bbb{N}^{\times}$. This is because the open ball $B_{r}^{\Omega}(n) = \{m \in \Bbb{N}^{\times} : |\Omega(m) - \Omega(n)| \lt r\} = \biguplus\limits_i P^{k_i}$ where $P = \{ p \in \Bbb{N}^{\times} : p$ is prime $\}$, $P^k = \{k$-fold products of primes $\}$, and $|k_i - \Omega(n) | \lt r$. So the map $a\cdot : \Bbb{N}^{\times} \to \Bbb{N}^{\times} : x \mapsto ax$ has inverse images $(a\cdot)^{-1}(B_r^{\Omega}(n)) = \biguplus\limits_i P^{k_i - \Omega(a)} =$ some ball.

The translated pseudometrics $|f(n + t) - f(m + t)|$ generate the same topology as the $t = 0$ pseudomtric.

Let $B_r^0(n) = B_r^{\Omega}(n)$ as above. And let $B^t_r(n)$ denote translated balls. Then there is the map $B_r^0(n) \to B_r^t(n - t)$ given by $x \mapsto x - t$. Thus as long as we cut out the first $t$ elements of $\Bbb{N}^{\times}$ namely $S = \Bbb{N}^{\times} \setminus \{1, \dots, t\}$ we have a homeomorphism given by $x \mapsto x - t : S \to S$.

Proof: $$x \in B_r^0(n) \implies x - t \in B_r^t(n - t) = \\ \{m: |f(m+t) - f(n - t + t)| \lt r \} = \{m: |f(m + t) - f(n)|\lt r\}$$

So this must imply that $g(n): \Bbb{N}^{\times}\setminus \{1\} : n \to n \pm 1$ are both continuous on the topology generated by any one of the metrics $d_t (m,n) := |f(m + t) - f(n + t)|$.

Since as we've said we're in a topological semigroup, we have that $g(n) = (n-1)(n+1) = n^2 - 1$ is also continuous.

No open set in $(\Bbb{N}^{\times}\setminus \{1\}, d_0)$ is finite. This is clear since for any open ball radius $r$ and fixed $\Omega(n)$ there are an infinite number of multiples of primes $m$ that are less than $r$ away.

$P^2$ is open; it equals $B_{2}^0(p) = \{m : |\Omega(m) - 1| \lt 2 \}$ for any prime $p$.

$g^{-1}(P^2)$ cannot be finite then since $g$ is continuous. If the twin prime conjecture were false, then there would be only a finite number of $n \in \Bbb{N}^{\times}\setminus \{1\}$ such that $g(n) \in P^2$ since a twin prime occurs if and only if the number between them squared minus $1$ equals the product of the two primes; in other words, $n^2 - 1 = pq$.

Thus, since $g^{-1}(P^2)$ is indeed infinite, the twin prime conjecture must be true.

$$\square$$

Please verify this proof or identify the error(s).

This may be more of a comment or observation but I'll leave it here due to lack of rep. It seems extremely unlikely that the functions $n \mapsto n \pm 1$ are continuous with respect to the topology given by the pseudometric $\Omega$.
Since the set of primes is open, that continuity would imply that the sets of the form $(\text{prime} + 1)$ are open for any fixed $t$, or in other words that you could determine whether a number $n$ has the form $(\text{prime} + 1)$ simply by counting the number of divisors of $n$.
• @EyesOnBud The open sets in your topology are those of the form $\{n: \, \Omega(n) \in M\}$ for fixed sets $M$. The set $\{p+1: \, p \, \mathrm{prime}\}$ is not open because you can't tell whether a number is of the form $p+1$ by counting its prime divisors. Sep 2, 2017 at 6:17