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Let $\phi : \mathbb{Q}_{>0} \to \mathbb{Z}$ be the group morphism defined by $\phi(p) = p$ for $p$ a prime number.

It follows that $\phi(1)=0$, $\phi(a.b) = \phi(a)+\phi(b)$, $\phi(a^{-1}) = -a$ and in general:
$$\phi(\prod_i p_i^{n_i}) = \sum_i n_i p_i$$ with $p_i$ a prime number and $n_i \in \mathbb{Z}$.

Let $\mathcal{K} = ker (\phi) $, a subgroup of $\mathbb{Q}_{>0}$.
Then, $\mathcal{K}= \{ r \in \mathbb{Q}_{>0} \vert r=\prod_i p_i^{n_i} \text{ and } \sum_i n_i p_i = 0 \}$

Question: Is $\mathcal{K}$ a dense subset of $\mathbb{R}_{ +}$ ?

Sebastien Palcoux
  • 5,797

2 Answers2

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Yes, $\mathcal{K}$ is dense. If we assume as known that the closed subgroups of $(\mathbb{R},+)$ are

  • $\{0\}$,
  • $\mathbb{Z}\cdot g$ for some $g\neq 0$, and
  • $\mathbb{R}$,

since the logarithm is an isomorphism of topological groups between $(\mathbb{R}_{> 0}, \cdot)$ and $(\mathbb{R},+)$, it suffices to see that $\mathcal{K}$ is not cyclic. Since $\frac{8}{9} \in \mathcal{K}$, the kernel is not trivial. It is not cyclic, since for every $\frac{m}{n}\in \mathcal{K}$, we can find two primes $p,q$ dividing neither $m$ nor $n$, and $\frac{p^q}{q^p}$ is an element of the kernel that is not a power of $\frac{m}{n}$.

Hence $\overline{\log\mathcal{K}} = \mathbb{R}$, which means $\overline{\mathcal{K}} = \mathbb{R}_+$.

Daniel Fischer
  • 206,697
2

I like Daniel Fischer's highbrow proof, but here's a direct one using a great theorem:

Because $\mathcal K$ is multiplicatively closed (and closed under inverses), it suffices to prove that there is an element of $\mathcal K$ in every interval of the form $(1-\varepsilon,1)$.

Green and Tao proved that for every $k\ge3$, there are arithmetic progressions $a,a+d,\dots,a+(k-1)d$ consisting entirely of primes. Letting $p_1<p_2<p_3$ be the largest three of those $k$ primes, set $r=\frac{p_1p_3}{p_2^2}$. It's now easy to show that $r \in\mathcal K$ and $1 > r > 1-\frac1{(k-2)^2}$.

Greg Martin
  • 78,820