$p$-adic valuations of infinite products

Question

There are several ways to argue that the $q$-adic valuation on $\mathbb Q$ can be extended to a valuation $\mathsf v_q$ on the field $\mathbb R$ of real numbers.

I know that such a valuation cannot be explicitely given. Nevertheless I suspect that it is possible to compute very special values.

I summarize two questions here, because they are related. Denote by $\mathbb P$ the set of prime numbers.

1. Let $(a_p)_{p \in \mathbb P}$ be a sequence of rational numbers such that $\prod_{p \in \mathbb P} a_p$ is convergent (wrt euclidean topology). Let $q \in \mathbb P$ and assume that $\sum_{p \in \mathbb P} \mathsf v_q(a_p)$ is also convergent. Does it follow that $\mathsf v_q(\prod_{p \in \mathbb P} a_p) = \sum_{p \in \mathbb P} \mathsf v_q(a_p)$?

An obvious proof attempt is to show that $\mathsf v_q(\prod_{p >n} a_p)$ tends to $0$ as $n$ goes to infinity. This would of course be easy if $\mathsf v_p$ would be continuous with respect to the euclidean topology.

2. The understanding of 1. could lead to constructions of real numbers with prescribed $q$-adic valuations. For instance, does there exists $x \in \mathbb R$ such that $\mathsf v_q(x) = 1$ for all $q \in \mathbb P$?

Thank you in advance for your help!

Answer 1

$\sum_{p}(v_{q}(a_{p}))$ is given to be convergent. Now, as the discrete valuation takes discrete values, preferably maybe from $Z$, the above series is convergent $\iff$ it is eventually zero. Which means question 1 boils down to finite case for the same problem, to which the answer is yes. Which I believe leads to a vacuous answer for question $2$.

Answer 2

For a sequence $(a_n)\in \Bbb{Q}$ then $\prod_{n\ge 1} a_n$ is the limit of a sequence, namely $(\prod_{n\le N} a_n)_{N\ge 1}$, and the limit if it exists (often it doesn't) is not the same at all in $\Bbb{R}$ and in $\Bbb{Q}_p$. The convergence of one is not related to the other, you can have that $\prod_{n\ge 1} a_n$ converges to $1$ in $\Bbb{R}$ and to $p$ in $\Bbb{Q}_p$.

If $\prod_{n\ge 1} a_n$ converges to $b$ in $\Bbb{Q}_p$ then $v_p(b)=\sum_{n\ge 1} v_p(a_n)$.

So if $\prod_{n\ge 1} a_n$ converges to $1$ in $\Bbb{R}$ and to $p$ in $\Bbb{Q}_p$ then $v_p(1)=0$ is not related to $\sum_{n\ge 1} v_p(a_n)=v_p(b)=1$.