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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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I changed $\mathbb{A}_{fin}$ to $\mathbb{A}_\mathrm{fin}$, which I believe is standard. (Code: \mathbb{A}_{fin} changed to \mathbb{A}_\mathrm{fin}) If letters are not set in \mathrm{} or \text{} or the like, then the italicization and spacing is what is appropriate for juxtaposed variables. – Michael Hardy Sep 5 '12 at 15:08
Of course you are right... lazy me, thanks for the improvement :D – Fabian Werner Sep 5 '12 at 15:10
up vote 3 down vote accepted

"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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erm... wait... isnt the set $\{ 1/n : n \in \mathbb{N} \}$ discrete in $\mathbb{R}$ but not closed? – Fabian Werner Sep 5 '12 at 15:36
Ah, of course you are right. What I'm thinking of is restricted to subgroups. I'll edit. – Sean Eberhard Sep 5 '12 at 15:39
The (surprisingly complicated) proof of 3 has been looked at before on MSE. See…. – Sean Eberhard Sep 5 '12 at 16:00
I found it on the internet already, thanks for pointing out the direction :D unbelievable, $\mathbb{A}$ and $\mathbb{A}_{\text{fin}}$ look so close, yet they are so far apart... – Fabian Werner Sep 5 '12 at 16:05
If anyone is interested: a proof of 3. can be found here: and the construction of the symmetric square root can be found here:, prop. 5.7. – Fabian Werner Sep 6 '12 at 8:10

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