Closure of this set is $\mathbb R^+$ 
Let $A$ be a subset of $\mathbb R^+$ which is not bounded.
Prove that $\displaystyle \operatorname{cl}(\bigcup_{n\in\mathbb N^*}\frac{1}{n}A)=\mathbb R^+$

where $\frac{1}{n}A$ denotes $\{\frac{a}{n} \; | \; a\in A\}$.
I cannot make any significant progress with this...
Can someone give me a hint?
 A: Let $a_n \in A$ and $a_n \to +\infty$.
For any rational number $\dfrac{p}{q}$, we have
$$\dfrac{p}{q} = \dfrac{a_n}{\dfrac{qa_n}{p}}, \forall n$$
$$\left|\dfrac{a_n}{\dfrac{qa_n}{p}} - \dfrac{a_n}{\lfloor\dfrac{qa_n}{p}\rfloor}\right| \to 0
$$
when $n\to +\infty$.
Or let's remark $\dfrac{a_n}{\dfrac{qa_n}{p}} \leq \dfrac{a_n}{\lfloor\dfrac{qa_n}{p}\rfloor} \leq \dfrac{a_n}{\dfrac{qa_n}{p} - 1}$ and use squeezing theorem.
So all rational numebers are in the closure, so are all irrationals
A: I believe one of the previous answers is correct, but here is my version. Suppose the contrary, that is there is some open interval $(a,b)$ (where $0<a<b$) that does not intersect any of the $A/n$. Use that $1<b/a$ and $(n+1)/n$ approaches $1$ as $n$ approaches $\infty$ to pick $N$ such that $(n+1)/n < b/a$ whenever $n\ge N$. Then $(n+1)a<n b$ hence the intervals $(na,nb)$ and $((n+1)a,(n+1)b)$ overlap. But intervals of this form cover $(Na,\infty)$ and none of them intersects $A$, that is $A$ is disjoint from $(Na,\infty)$, and hence cannot be unbounded, a contradiction. 
