On page 68 of Nancy Childress' book "Class Field Theory", it says "We want to put a topology on $J_F$ that will make it a locally compact topological group." ($J_F$ is the group of ideles of $F$), which explains the need for the restricted product. My question is why do we want to do make it a locally compact topological group?
It will be great if you can make it as simple as possible as I do not have much background on algebraic topology, but I do know a little about metric spaces and topological spaces. Thank you in advance.
One way to think about the construction of the adeles, and especially their topology, is by analogy with the the $p$-adic numbers.
The $p$-adic integers $\mathbb Z_p$ are compact. To get to $\mathbb Q_p$, we then invert $p$, and this is a discrete process, in the sense that we have inclusions $$\mathbb Z_p \subset \frac{1}{p} \mathbb Z_p \subset \frac{1}{p^2} \mathbb Z_p \subset \cdots \subset \frac{1}{p^n} \mathbb Z_p \subset \cdots,$$ with each set being both open and closed in the next.
Now the finite adeles are similar, except we don't have just one prime.
First we have $\widehat{\mathbb Z},$ which is the projective limit $\mathbb Z/N$ over all positive integers $N$ (rather than just taking the limit over powers of the fixed prime $p$); by the Chinese Remainder Theorem, it can also be written as a product over all $p$ of $\mathbb Z_p$.
Now we want to invert all the non-zero integers, so we form the various abelian groups $$\frac{1}{N} \widehat{\mathbb Z},$$ with $N$ running over all positive integers.
Instead of forming a linearly ordered set (as they do if $N$ is restricted to powers of a single prime $p$), these are ordered in a more complicated way ($\frac{1}{M}\widehat{\mathbb Z}$ is contained in $\frac{1}{N}\widehat{\mathbb Z}$ if and only if $M$ divides $N$), but just as in the $p$-adic case, each of these inclusions is an open and closed embedding, and we take the union of all of them.
This exactly gives the ring of finite adeles, with its correct topology.
To get the full ring of adeles, just take the product with $\mathbb R$.
So rather than thinking about (and getting confused by) the restricted product, think about the adeles as being analogous to the passage from $\mathbb Z_p$ to $\mathbb Q_p$, except that we have every prime, so we go from $\widehat{\mathbb Z}$ to $\mathbb A$ by inverting all the primes. (And then add a copy of $\mathbb R$, for the infinite place.)