$U\subset [0,\infty)$ is open and unbounded $\Rightarrow \exists x$ such that $U\cap \{nx;n\in \mathbb N\}$ is infinite.

I want to show that:

Let $U\subset [0,\infty)$ be open and unbounded. Show that there is a number $x\in (0,\infty)$ such that $U\cap \{nx;n\in \mathbb N\}$ is infinite.

Because of $U$ is open, $U$ is a countable union of open intervals. if $U$ contain an interval $(a,\infty)$, we are done. But if all intervals which contain in $U$ was bounded, what we can do? can somebody give me a hint?

Solutions Collecting From Web of "$U\subset [0,\infty)$ is open and unbounded $\Rightarrow \exists x$ such that $U\cap \{nx;n\in \mathbb N\}$ is infinite."

For $x>0$ let $N_x=\{nx:n\in\Bbb N\}$, and suppose that $U\cap N_x$ is finite for each $x>0$. Then for each $x\ge 0$, there is a $q_x\in\Bbb Q$ such that $U\cap N_x\subseteq[0,q_x)$. For each $q\in\Bbb Q$ let $A_q=\{x\ge 0:q_x=q\}$; evidently $[0,\to)=\bigcup_{q\in\Bbb Q}A_q$. By the Baire category theorem there are a $q\in\Bbb Q$ and $a,b>0$ such that $a<b$, and $(a,b)\subseteq\operatorname{cl}A_q$. Let $D=A_q\cap(a,b)$; $D$ is dense in $(a,b)$, and $U\cap N_x\subseteq[0,q)$ for each $x\in D$.

Now let $d=b-a$, and let $m\in\Bbb Z^+$ be greater than $\max\left\{\frac{q}a,\frac{a}d\right\}$. Suppose that $n\ge m$; then

$$nb-(n+1)a=n(b-a)-a=nd-a\ge md-a>0\;,$$

so $(n+1)a<nb$, and it follows that

$$\bigcup_{n\ge m}(na,nb)=(ma,\to)$$

and hence that $\bigcup_{x\in D}N_x$ is dense in $(ma,\to)$. Let $V=U\cap(ma,\to)$; then $V\cap\bigcup_{x\in D}N_x\ne\varnothing$, but on the other hand

$$V\cap\bigcup_{x\in D}N_x=(ma,\to)\cap\left(U\cap\bigcup_{x\in D}N_x\right)\subseteq(ma,\to)\cap[0,q)\subseteq(q,\to)\cap[0,q)=\varnothing\;,$$

which is absurd.