Why is the restricted direct product topology on the idele group stronger than the topology induced by the adele group? I am confused by the saying that the restricted direct product topology on the idele group is stronger than the topology induced by the adele group. And perhaps this is elaborated by the following example

Let $p_n$ be the $n$-th positive prime in $\mathbb{Z}$, and let $\alpha^{n}=(\alpha^{(n)}_v)\in\mathbb{A}_\mathbb{Q}$
  with  $\alpha^{(n)}_v=p_n$ if $v=p_n$ and $\alpha^{(n)}_v=1$ if $v\neq p_n$. The result is a sequence
  $\{\alpha^{n}\}$ of ideles in $\mathbb{I}_\mathbb{Q}$. Then this sequence converges to the idele $(1)_v$ in
  the topology of the adeles but not converges in the topology of the ideles.

Any comment is appreciated!
 A: First of all, the reason why we define the ideles and the adeles in the first place is so that we can study number theory using a local-global principle. Two consequences of the definition using the restricted product:


*

*A character $\chi$ is trivial on all but finitely many components, and so we can write $\chi=\otimes'\chi_\nu$; conversely, a family of characters $\{\chi_\nu:\mathbb{Q}_\nu\to\mathbb{C}^\times\}$ such that finitely many are trivial allows you to define a character as above.

*Similarly, we can construct a measure on the adeles (or the ideles) by using (normalized) measures on each component.


Now, because the ideles form a subset of the adeles, there are two candidates for the topology, and they are not equivalent, as your example shows. Basically, for the sequence constructed, $\alpha_{p_n}^{(n)}=p_n$ is not a unit (although the adele $\alpha^n$ is an idele), and so a small enough neighbourhood of the identity in $\mathbb{I}_\mathbb{Q}$ won't contain your sequence. Hence, the idele topology is stronger than the relative topology.
A: Yes, the topology on the ideles is finer than the subspace topology. For your example,
In the ideles, consider the neighborhood of $1$ which is just the product of $\mathbb{Z}_p^\times$ for all $p$. Your sequence never enters this neighborhood, hence can't converge to $1$.
In the topology coming from the adeles, that's not a neighborhood of $1$. Any neighborhood of 1 is a translate by 1 of a neighborhood of 0, so contains a translate by 1 of a basic open neighborhood of 0. A basic open neighborhood of 0 in the adeles looks like $p^{n_p}\mathbb{Z}_p$ at finitely many spots and $\mathbb{Z}_p$ in all the others (the $n_p$s can be negative), so a basic open neighborhood of 1 in the subspace topology consists of ideles is given by the set of ideles which are in $1 + p^{n_p}\mathbb{Z}_p$ in those finitely many slots and $1 + \mathbb{Z}_p = \mathbb{Z}_p$ in all the others. Your sequence clearly eventually lands in any such neighborhood, as soon as $n$ is larger than the index of any of the special slots.
