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I'm trying to prove that infinite subgrups of $\mathbb{S}^1 $ are dense. I've proved that it suffices to show that $\mathbb{Z}\alpha +\mathbb{Z}$ is dense in $\mathbb{R} $ (with $\alpha $ irrational). I've found a way to prove the later statement by showing that $\mathbb{Z}\alpha +\mathbb{Z}$ has not minimum elements. In this (Show that there are only two types of subgroups in R , either Discrete or Dense? )question someone prove it by assuming that a minimum element $x$ of the set $\mathbb{Z}\alpha +\mathbb{Z}$ must verify $nx=1 $ and $mx=\alpha $ for some $m,n\in \mathbb{Z} $.

I was trying to prove his assumption but i can't. I need some help from you. Thanks :)

Natalio
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  • If $\mathbb S^1$ is compact and the subgroup is not finite there must be a convergent sequence of subgroup elements, from which you can derive the rest – kimchi lover Nov 12 '17 at 17:36

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You mean a minimum positive element of $\Bbb Z\alpha+\Bbb Z$.

If there is such an element $x$, then every element of the group $\Bbb Z\alpha+\Bbb Z$ must be an integer multiple of it, since otherwise one would get an element $y$ in the group with $0<y<x$. In particular $1$ must be an integer multiple of $x$.

Angina Seng
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