Topology on the idele group. For the rational number field, the adele ring $A_{\Bbb Q}$ is defined. On the other hand, its multiplicative subgroup $I_{\Bbb Q}$ is also defined.
$A_{\Bbb Q}$ is endowed with the topology as its subgroups $n{\widehat {\Bbb Z}}$ over all integers $n$ make the fundamental neighbourhoods.
$I_{\Bbb Q}$ is endowed with the fundamental neighbourhoods as the intersection with the subsets $(x, y) \in A_{\Bbb Q} × A_{\Bbb Q}$ s.t. $xy = 1$.
Q. Why is thus defined topology on $I_{\Bbb Q}$ stronger than the induced topology from $A_{\Bbb Q}$?
 A: If you're asking for a proof that the topology on $I_\Bbb{Q}$ is stronger than the induced topology from the inclusion $I_\Bbb{Q}\subseteq A_\Bbb{Q},$ note that for any topological ring $R,$ the topology induced by the embedding
\begin{align*}
\iota: R^\times&\to R\times R\\
r&\mapsto (r,r^{-1})
\end{align*}
is at least as fine as the topology on $R$: the projection maps $\pi_1,\pi_2 : R\times R\to R$ are continuous. Then for any open $U\subseteq R,$ we have $(\pi_1\circ\iota)^{-1}(U) = U\cap R^\times,$ which is open by continuity of $\pi_1$ and by definition of the topology on $R^\times.$
For a specific instance of the failure of inversion to be continuous, see example 23.1 here. You can use this to construct an example of an open in $I_\Bbb{Q}$ which is not the restriction of an open in $A_\Bbb{Q}$.
If you're asking why the topology is defined in this way, we want $I_\Bbb{Q}$ to be a topological group. But $I_\Bbb{Q}$ with the subspace topology induced from $I_\Bbb{Q}\subseteq A_\Bbb{Q}$ is not a topological group -- inversion is not continuous! Instead, we use the subspace topology induced by the embedding (of groups!) $\mathrm{GL}_1(R)\to\mathrm{GL}_2(R)$ which sends $r\mapsto \begin{pmatrix}r&0\\0&r^{-1}\end{pmatrix}.$ This is equivalent to the topology induced by the map $\iota : R^\times\to R\times R$ I wrote above, thinking of $\mathrm{GL}_2(R)$ as the subspace
$$
\mathrm{GL}_2(R) = \{(a,b,c,d,y)\in R^5\mid (ad - bc)y= 1\}\subseteq R^5
$$
(Notice that the topology on $\mathrm{GL}_2(R)$ is the subspace topology induced by this inclusion!)
Since the map $\mathrm{GL}_1(R)\to\mathrm{GL}_2(R)$ lands in the subspace $\{(a,0,0,d,1)\in R^5\},$ we identify this subspace with $R\times R$ and there is no harm in using the map $\iota : R^\times\to R\times R$ to define the topology on $R^\times.$
