# Is there a way to show that this topology is compact?

Define the norm $\|x\| = \min\{\sum_i |k_i| : k_i \in \Bbb{Z}, \sum_i (k_i z_i) = x\}$ where $z = \{z_1, z_2, \dots \}$ is a set with two at least two coprime elements. This norm defines a topology on $\Bbb{Z}$ turning it into a topological ring. My goal is to show that this ring is profinite at least when $z =$ the primes.

I can do that if I can show that $(\Bbb{Z}, \|\cdot\|)$ is Hausdorff and compact according to this nLab article.

According to a recent Terry Tao result, the odd numbers are all expressible as a sum of $c$ primes where $3 \leq c \leq 5$. (Is that correct?)

Thus the open balls $\overline{B_c}(0) = \{x \in \Bbb{Z} : \|x\| \leq c \}$ (they're open as we can take $c' = c + 1$ and use $\lt$ instead), are such that

$\{0\} = \overline{B_0}(0) \subset \overline{B_1}(0) \subset \dots \overline{B_5}(0) = \Bbb{Z}$. (This might need to go up to $6$ here instead, not sure...)

Question: Doesn't $\{0\} = \overline{B_{\epsilon}}(0)$ for any $0 \lt \epsilon \lt 1$ mean that $\{0\}$ is a closed set and so our space is Hausdorff?

This is according to S. Warner's Topological Rings book (Theorem 3.4), since $\Bbb{Z}$ is in particular a topological group. I just want to be sure that $\{0\}$ was indeed shown to be closed easily.

We also have that $B_c(x) = B_c(0) + x$ for any $x \in \Bbb{Z}$.

Compactness.

According to a generalization of the Heine-Borel theorem, I merely need to show that $\Bbb{Z}$ is totally bounded and that it is Cauchy complete in the norm $\|\cdot \|$.

Clearly $\Bbb{Z}$ is Cauchy complete under this norm since as $0\lt \epsilon \lt 1$ approaches zero, we're left with one and only one point, which is in $\Bbb{Z}$ already.

• Harald Helfgott proved in 2013 that every odd number greater than $5$ may be written as the sum of three primes (the weak Goldbach conjecture). It is not true that $1$ and $3$ can be written as the sum of $3$, $4$ or $5$ primes. – John Gowers Sep 7 '17 at 22:21

The topology defined by your norm is the discrete topology: $\|x\|\geq 1$ for all $x\neq 0$ and so any open ball of radius less than $1$ is a singleton, and so any singleton is an open set. In particular, it is not compact, so it is not profinite.