# 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.

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.
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.