The interval [0,1] with additional modulo 1 The interval $[0, 1]$ is an abelian group with addition modulo $1$.
Let $H$ be a proper subgroup of $[0, 1]$, which is closed as subset of $[0,1]$.
Show $H$ is finite.
I assumed $H$ is infinite: since $H$ is closed, all limit points of $H$ are in $H$.
Then let $x$ be in $[0, 1]\setminus H=S$ ($S$ is open since $H$ is closed) so there exists an open set $U_x$ which contains $x$ in it and $U_x \subset S$.
Let $\{x_n \}_n$ be a sequence converging to $x$. Thus for all $\epsilon>0$ there exists a $N$ natural number so that for all $n>N$ $|x_n-x|<\epsilon$.
For some $\epsilon_1 >0$ $B(x,\epsilon_1)$ is contained in $U_x$, then if I prove that $x$ is an limit point of $H$ then I can say $H$ is dense (because I express a generic element of $[0, 1]$ as belonging to $H$ or as limit point of $H$) then closure of $H$ will equal to $[0, 1]$ and this would contradict $H$ being proper subgroup.
But I couldn't find a way to show that.

Hint : Consider the function $\pi : \mathbb{R} \rightarrow [0,1)$ defined for all $x \in \mathbb{R}$ by $\pi(x)=x-\lfloor x \rfloor$. Then :

*

*Show that $\pi^{-1}(H)$ is a subgroup of $\mathbb{R}$.


*Deduce that either $\pi^{-1}(H)=a\mathbb{Z}$ for a real number $a$, either $\pi^{-1}(H)$ is dense in $\mathbb{R}$.


*Show that the fact that $H$ is closed in $[0,1)$ implies that $\pi^{-1}(H)$ cannot be dense, so it has to be of the form $a \mathbb{Z}$.


*Deduce that $H$ is finite.
