What is following product: $$ G = \prod_{n \in \mathbb{N}} \mathbb{Z}_{p^n} $$ where $p$ is a prime and $\mathbb{Z}_{p^n}$ is usual cyclic group of order $p^n$.
Obviously it is not isomorphic to the direct coproduct. It has both torsion end infinite-order elements and it has uncountable cardinality. So my first thought was unit circle or some kind of factorization of $p$-adic numbers. Is it some kind of metric completion?
$G$ carries a natural topology: consider all $\mathbb{Z}_{p^n}$ as discrete, then $G$ endowed with the product topology becomes a compact, totally disconnected space. The group operations are continuous, so this is a compact group, actually a profinite group, or even more precise, an abelian pro-$p$ group. The direct sum $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}$ is a dense subgroup (consisting of the elements of $G$ with only finitely many entries $\ne 0$). Since the topology is also metric, $G$ can be viewed as metric completion of $\bigoplus_{n\in \mathbb{N}}\mathbb{Z}_{p^n}$. The group of $p$-adic integers is a closed subgroup of $G$, namely the topological closure of the subgroup generated by $(1,1,\dots)$.