I want to consider the (topological) group of 'finite' ideles: If
$$\mathbb{A}_\text{fin} = \widehat{\prod}^{\mathbf{Z}_p}_p \mathbf{Q}_p$$
(the 'hat' indicates the so-called restricted product topology) then
$$\mathbb{I}_\text{fin} := \mathbb{A}_\text{fin}^\times$$
The topology is such that it becomes homeomorphic to (i.e. we can also take this as a definition)
$$\mathbb{I}_\text{fin} = \widehat{\prod}^{\mathbf{Z}_p^\times}_p \mathbf{Q}_p^\times$$
My question:
What is the closure of $\mathbb{Q}^\times$ which is considered to be diagonally embedded into $\mathbb{I}_\text{fin}$?
I know already that it is not whole $\mathbb{I}_\text{fin}$ as a direct computation shows that there is no $q \in \mathbb{Q}^\times$ in the neighborhood
$$ U_{1/2}(2) \times U_{1/3}(1) \times \prod_{p \geq 5} \mathbf{Z}_p^\times$$ of the element $$(2,1,1,1,1,...)$$ Thanks in advance,
FW
Note that $\mathbb{Q}^\times \cap \prod_p \mathbb{Z}_p^\times = \{\pm 1\}$. In particular, $(\prod_p \mathbb{Z}_p^\times) \setminus \{-1\}$ is an open neighbourhood of $1$ which meets $\mathbb{Q}^\times$ only at $1$. This shows that $\mathbb{Q}^\times$ is discrete in $\mathbb{A}_\text{fin}^\times$.