If $H$ is a subgroup of $\mathbb{R}$ and there is $h\in H$, $0I'm trying to prove that if $H$ is a subgroup of $(\mathbb{R},+)$ and for every $\epsilon >0$ there is $h\in H$ with $0<h<\epsilon$ then $H$ is dense in $\mathbb{R}$.
Let $\epsilon >0$ and $x\in\mathbb{R}$. So I want to see that there is some $h'\in H$ in $(x-\epsilon,x+\epsilon)$. Since we can find any nonzero $h$ as small as we want, I think we will find some integer $n$ such that $nh\in (x-\epsilon,x+\epsilon)$. And we use that $nh\in H$.
However, I don't know how to do this formally. Would anyone give me a hint?
Thank you.
 A: I'd do it slightly differently, I hope you like it.
We have to prove for every open interval $(a,b)$ there is $h\in H$ with $h\in (a,b)$.
Take $h\in H$ with $0<h<b-a$.
Notice that if $n<\frac{b}{h}$ then $nh<b$.
Let $m$ be the largest integer with $n<\frac{b}{h}$ (it exists because it is a subset of $\mathbb Z$ bounded above).
Suppose that $mh\leq a$, then $(m+1)h=mh+h\leq a+h<a+(b-a)=b$.
Contradicting the maximality of $m$.
We conclude $mh>a$ and $mh<b$. So $mh\in (a,b)$ and $mh\in H_\blacksquare$
A: If you are familiar with the Archimedean Property of the Reals and the Well-Ordering Property of the Naturals, then there is a fairly straightforward approach you can take.
First, apply the hypothesis to find $h\in H$ such that $$0<h<\epsilon.$$ Next, if $|x-\epsilon|\le|x+\epsilon|,$ take the least natural number $n$ such that $(n+1)h>x+\epsilon$; otherwise, take the least $n$ such that $-(n+1)h<x-\epsilon.$ Finally, show that $nh\in(x-\epsilon,x+\epsilon)$ or $-nh\in(x-\epsilon,x+\epsilon),$ whichever is appropriate.
A: Take $c>0, d\in H, 0<d<c$, write $x=nd+e, 0<e<d$, $(n+1)d\in [x-c,x+c]$.
A: Let's take base $\mathfrak{B}$ of intervals for standart topology in $\mathbb{R}$. Let $I$ be an arbitrary interval (without loss of generality, assume it lies in $\mathbb{R}_{>0}$), so $I\in\mathfrak{B}$, and $|I|=\delta$. Then take $h\in H$ so that $0<h<\delta/2$.
If you take $h,h+h,h+h+h,\ldots$, then one of them will be in $I$, otherwise $\delta/2>(k+1)h-kh>|I|=\delta$, which is impossible. 
As the result, we get that for every element $I$ of base $\mathfrak{B}$ $I\cap H\neq\emptyset$, which means that $H$ is dense in $\mathbb{R}$.
A: It is enough to show that $(a, b)\cap H\neq \phi$ for all $b>a>0$. Suppose not, then $\left(\frac{a}{n}, \frac{b}{n}\right)\cap H =\phi$ for all $n$. (If not, for $x\in \left(\frac{a}{n}, \frac{b}{n}\right)\cap H$, $nx\in (a, b)\cap H$, contradiction.) Now choose $N\in\mathbb{N}$ s.t. $\frac{b}{a}>\frac{N+1}{N}=1+\frac{1}{N}$, then $\frac{b}{n+1}>\frac{a}{n}$ for all $n>N$. Thus
$$ \phi=\bigcup_{n=1}^{\infty}\left(\frac{a}{n}, \frac{b}{n}\right)\cap H\supset \left(0, \frac{b}{N}\right)\cap H$$
and $\left(0, \frac{b}{N}\right)\cap H=\phi$, which contradicts to the condition when $\epsilon=\frac{b}{N}$. 
