Show that $\inf \{ m+n\omega: m+n\omega>0,and,m,n\in{Z}\}= 0$, where $\omega>0 $ is irrational. Let $\omega\in\mathbb {R}$ be an irrational positive number. Set
$$A=\{m+n\omega: m+n\omega&gt;0,and,m,n\in{Z}\}.$$

Show that $\inf{A}=0.$

How should I start this problem? I don't get this problem.
Hint

In this context ($\omega$ irrational) it can be shown that the set $\{n\omega-\lfloor n\omega\rfloor\mid n\in\mathbb Z\}$ is a dense subset of $(0,1)$. Also it can be shown to be a subset of $A$.
What $\inf A =0$ means is that, for any $\epsilon &gt;0$ there is some multiple of $\omega$ that is within distance $\epsilon$ of some integer.

Hint 1: Pigeonhole principle. 

For simplicity assume $\omega &lt;1$. Then each interval $[0,1], [1,2], [2,3]$ contains a multiple of $\omega$, call them $n \omega, m \omega$ and $r\omega$. Each of those three intervals is divided into a "first half" and a "second half". For example $[0,1] = [0,1/2] \cup [1/2, 1]$. So two of $n \omega, m \omega$ and $r\omega$ are in the same half. For example $n\omega \in [1/2,1]$ and $r\omega \in [2+1/2,3]$. That implies $r\omega -2 \in [1/2,1]$ and so $r\omega -2$ and $n\omega$ are distance less than $1/2$ apart. Meaning $|(r\omega -2)-n\omega| &lt; 1/2 \implies |(r-n)\omega -2| &lt; 1/2$. So that $\inf A &lt; 1/2$. How does this generalise?

Hint 2: (independent) Compactness of the circle