Is $(\mathbb{Z}_p\times \mathbb{R})/\mathbb{Z}$ connected? I was reading this question The connected component of the idele class group but I am very confused about the structure of the solenoids $(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$, (where $\mathbb{Z}$ acts diagonally), more specifically why it is connected. More generally, if we replace $\widehat{\mathbb{Z}}$ by $\mathbb{Z}_p$, is the resulting quotient group $(\mathbb{Z}_p\times \mathbb{R})/\mathbb{Z}$ still connected?
Hoping if someone can give me any hints or answers!!
 A: No originality here, but I would tell the story as follows. Consider the subgroup $(\mathbb{Z}\times\mathbb{R})/\mathbb{Z}$ of $(\mathbb{Z}_p\times\mathbb{R})/\mathbb{Z}$. It is dense, because $\mathbb{Z}$ is dense in $\mathbb{Z}_p$. It is also connected, because it is isomorphic to $\mathbb{R}$ (with a coarser topology than the standard one). Therefore, in the group $(\mathbb{Z}_p\times\mathbb{R})/\mathbb{Z}$, the connected component of the identity is dense, whence it equals $(\mathbb{Z}_p\times\mathbb{R})/\mathbb{Z}$.
The same proof works for $(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$.
A: Assume that $(\Bbb{Z}_p\times\mathbb{R})/\Bbb{Z} = U\cup V$ with $U,V$ open non-empty disjoint.
$U$ contains the image of some $(a,b)$ in $\Bbb{Z}_p\times\mathbb{R}$ so it contains the image of an open $(a+p^k\Bbb{Z}_p,b+(-\epsilon,\epsilon))$ and thus it contains the image of the whole of $(a+p^k\Bbb{Z}_p,\Bbb{R})$.
Given an another $(c,d)\in (\Bbb{Z}_p\times\mathbb{R})/\Bbb{Z}$ there is an integer such that $(c+n,d+n)\in U$. So $V$ is in fact empty.
It works the same way for $(\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Z}$.
A: Hint: Connected components are always closed.
