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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.

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If you did not know you are asking about the subgroups of the circle $S^1$ in the complex plane $\mathbb{C}$. The exponential $e^{ 2\pi i}\colon [0, 1]_{/\mathbb{Z}}\rightarrow S^1$ defines a group isomorphism that is actually an homeomorphism of spaces.

The problem is widely explained in other questions, for example this one should contain all you need: Subgroup of the unit circle under complex multiplication

Anyway I see a problem in your proof: basically you are not using the fact that $H$ is a subgroup of $S^1$. To get that the $x$ is limiting point of $H$ you have to use this, otherwise you would show that every infinite subset of $S^1$ is dense (clearly false). You could use an argument similar to the one suggested by Seirios in the question I linked: by taking the preimage along the projection get a subgroup of $\mathbb{R}$ and consider the infimum of its positive elements.

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  • $\begingroup$ I didnt know. Thank you $\endgroup$ – Zikzak Nov 6 '18 at 11:03

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