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It is relatively easy to see that $\mathbb{Q}$ (diagonally embedded) is dense in $\mathbb{A}_\mathrm{fin} = \hat{\prod}^{Z_p} Q_p$ (the 'finite adeles where the restricted product is only taken for the finite places $p$) so it cannot be a closed subset. My question is: can one see that the diagonal embedding of whole $\mathbb{Q}$ is not closed in the full adeles $\mathbb{R} \times \mathbb{A}_\mathrm{fin}$? Does somebody know what the closure is?

Thanks in advance,

Fabian Werner

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    Of course you are right... lazy me, thanks for the improvement :D2012-09-05

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"Recall" the following:

  1. $\mathbb{Q}$ is discrete in $\mathbb{A} = \mathbb{R}\times\mathbb{A}_{\text{fin}}$.
  2. $\mathbb{A}$ is Hausdorff.
  3. Every discrete subgroup of a Hausdorff group is closed.
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    If anyone is interested: a proof of 3. can be found here: http://groupprops.subwiki.org/wiki/Discrete_subgroup_implies_closed and the construction of the symmetric square root can be found here: math.uh.edu/~vern/grouprepn.pdf, prop. 5.7.2012-09-06