Computing Pontryagin Duals I would like to know how to compute the Pontryagin dual of a certain class of continua which are (compact and therefore locally compact) abelian topological groups, in particular, $P$-adic solenoids, $\Sigma_P$, where $P = (p_1, p_2, p_3, \ldots)$ is a sequence of prime numbers.  In one reference, it states that the Pontryagin dual of the diadic solenoid, $\Sigma_2$ (where each prime number in the sequence is 2), is isomorphic to the additive discrete group they call $Q_2 = \big\{\frac{k}{2^n} \mid k \in \mathbb{Z} \text{ and } n \geq 0\big\}$.  It is clear to me how the Pontryagin dual of the circle group, $\mathbb{T} = S^1$, is simply $\mathbb{Z}$, but how is the Pontryagin duality of these solenoids computed?  
My conjecture is that for any $P$-adic solenoid where $P = (p_1, p_2, p_3, \ldots)$, the Pontryagin dual is isomorphic to the additive discrete group,
$$Q_P = \Bigg\{\frac{k}{\displaystyle \prod_{j \in F}p_{n_j}^{m_j}} \mathrel{\Bigg|} \forall k \in \mathbb{Z}, \forall F \subset \mathbb{N} \text{ such that } |F| < \infty, \forall m_j \in \mathbb{N}\Bigg\};$$
that is, $Q_P$ is the group of all rational numbers with numerators ranging over the integers and denominators being arbitrary finite products of arbitrary powers of members of $P$.  Note that $Q_P$ is an additive group since $0 \in Q_P$, each member clearly has an inverse being its negative, it is closed under addition since the sum of two members will have integer numerators and denominators of the given form, and associativity obviously holds.  
If my conjecture is true, then I'm not sure how to prove that it is.  Obviously, I need to show that the group of all characters of $\Sigma_P$ is isomorphic to $Q_P$, but I'm not sure how to do this.  How is it done with $\Sigma_2$ and $Q_2$? Any input would be appreciated.  Sorry for such a long post. 
 A: In general, let $S$ be a set of primes. Write $S^{-1} \mathbb{Z}$ (the localization) for the rational numbers whose denominators are divisible only by primes in $S$. I claim that the Pontryagin dual of $S^{-1} \mathbb{Z}$ is the $S$-adic solenoid; it follows by Pontryagin duality that the Pontryagin dual of the $S$-adic solenoid is $S^{-1} \mathbb{Z}$.
To see this, write $S^{-1} \mathbb{Z}$ as a filtered colimit (substitute "direct limit" if you're more comfortable with that terminology, although I hate it because direct limits are not limits) of copies of $\mathbb{Z}$. The diagram you're taking a colimit over is easiest to describe in the case that $S$ consists of a single prime $p$: then it's given by
$$\mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{p} \mathbb{Z} \dots$$
corresponding to the sequence of inclusions
$$\mathbb{Z} \subset \frac{1}{p} \mathbb{Z} \subset \frac{1}{p^2} \mathbb{Z} \dots$$
of subgroups of $\mathbb{Q}$. If $S$ consists of two primes $\{ p, q \}$ then there are various diagrams we could choose; one way to describe a diagram that works is to consider the subgroups of $S^{-1} \mathbb{Z}$ corresponding to rational numbers with denominator at most $n$, and then keep increasing $n$. Another is to consider a diagram with a more complicated shape given by repeatedly multiplying by both $p$ and $q$. And similarly for larger $S$. There are many possible choices of diagram because they are cofinal in each other. 
In any case, Pontryagin duality sends all of these filtered colimits of copies of $\mathbb{Z}$ to cofiltered limits of copies of $S^1$ (substitute "inverse limit" if you're more comfortable with that terminology), which are (depending on your definitions) the diagrams whose limits are the $S$-adic solenoids. 
Pontryagin duality in fact sends all limits to colimits and all colimits to limits, since it's a contravariant equivalence, and this observation is enough to compute many other Pontryagin duals as well. For example, you can compute that the Pontryagin dual of $\mathbb{Q}/\mathbb{Z}$ is the profinite integers. 
