A proof for $\widehat{\Bbb Z_{p^\infty}}\cong Z_p$ According to wikipedia, the Pontryagin dual of a Prüfer group is isomorphic to a group of p-adic integers.
Where can I find a proof for it on the internet?
 A: The Pruefer group ${\mathbb Z}_{p^{\infty}}$ is the direct limit of the sequence $${\mathbb Z}/p{\mathbb Z}\hookrightarrow {\mathbb Z}/p^2{\mathbb Z}\hookrightarrow ...$$
Applying $(-)^{\wedge} = \text{Hom}_{\text{cnt}}(-,{\mathbb S}^1)$ shows that $\left({\mathbb Z}_{p^\infty}\right)^{\wedge}$ is the inverse limit of the Pontryagin duals of ${\mathbb Z}/p^k {\mathbb Z}$. Finite groups are self-dual, so you end up with the $p$-adic integers.
This is a bit sloppy, since, as Alex pointed out, one has to be careful in which categories to talk about direct and inverse limits. See the comment below.
Addendum
Unfolding the definitions, you get the following explicit pairing: If $x := \sum\limits_{n=0}^{\infty} a_n p^n\in{\mathbb Z}_p$ is a $p$-adic integer and $y\in{\mathbb Z}[\frac{1}{p}]/{\mathbb Z}={\mathbb Z}_{p^{\infty}}$ is an element of the Pruefer $p$-group, then $\langle x,y\rangle := xy\in {\mathbb Z}_{p^\infty}\subset{\mathbb Q}/{\mathbb Z}\subset{\mathbb S}^1$ is the value of the character ${\mathbb Z}_{p^\infty}\to{\mathbb S}^1$ associated to $x$ when evaluated at $y$. Here the product $xy$ makes sense since multiples of $y$ by sufficiently high $p$-powers vanish in ${\mathbb Q}/{\mathbb Z}$.
A: Since $\mathbb{Z}(p^\infty)$ is torsion, the image of a character $\mathbb{Z}(p^\infty)\to\mathbb{T}=\mathbb{R}/\mathbb{Z}$ is contained in the torsion part of $\mathbb{T}$, which is $\mathbb{Q}/\mathbb{Z}$. Since $\mathbb{Z}_{p^\infty}$ is a $p$-group, the image is further contained in the $p$-component, which is exactly $\mathbb{Z}_{p^\infty}$. So
$$
\operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{T})=
\operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{Z}(p^\infty)).
$$
The topology on this group is still the one of pointwise convergence.
Now write $\mathbb{Z}(p^\infty)= \operatorname{colim}\limits_{n\in\mathbb{N}}\mathbb{Z}(p^n)$, so
$$
\operatorname{Hom}(\mathbb{Z}(p^\infty),\mathbb{Z}(p^\infty))\cong
\lim_{n\in\mathbb{N}}\operatorname{Hom}(\mathbb{Z}(p^n),\mathbb{Z}(p^\infty))\cong
\lim_{n\in\mathbb{N}}\operatorname{Hom}(\mathbb{Z}(p^n),\mathbb{Z}(p^n))
$$
and it's just a matter of checking that this is the required isomorphism.
