Prove Inf A = 0 where set $A = \{m + n\omega: m + n\omega > 0, m, n \in \mathbb{Z}\}$, $\omega$ is a positive irrational number. I found a solution of this question but the solution seemingly showed that there is only $0$ in set $A$ and this is obviously impossible. Where I am wrong? And in the last part of this solution, it says $1 = m_0\alpha$. Is it because we can set $n = 0$, and $m = 1$, then $1$ is in $A$ and every member of $A$ is equal to some multiplication of $\alpha$? 
 A: Given $x\in\mathbb{R}$, we denote the largest integer that is not
greater than $x$ by $[x]$. That is, $[x]=\max\{n\in\mathbb{Z}\mid n\leq x\}$.
Define a function $\theta:\mathbb{R}\rightarrow[0,1)$ by $\theta(x)=x-[x]$.
For each $n\in\mathbb{N}$, define $x_{n}=\theta(n\omega)$. Firstly,
observe that $x_{m}\neq x_{n}$ whenever $m\neq n$. (For, suppose
the contrary that there exist $m\neq n$ such that $x_{m}=x_{n}$.
Then $m\omega=N_{1}+x_{m}$ and $n\omega=N_{2}+x_{n}$ for some integers
$N_{1}$and $N_{2}$. Now we have: $m\omega-N_{1}=x_{m}=x_{n}=n\omega-N_{2}$
which implies $\omega=\frac{N_{1}-N_{2}}{m-n}\in\mathbb{Q}$, which
is a contradiction)
Let $\varepsilon>0$ be given. Since the sequence $(x_{n})_{n}$ is
bounded, by Bolzano-Weierstrass Theorem, it has a convergent subsequence.
In particular, there exist distinct $n_{1},n_{2}\in\mathbb{N}$ such
that $0<|x_{n_{1}}-x_{n_{2}}|<\varepsilon$. Without loss of generality,
we may assume that $x_{n_{1}}>x_{n_{2}}$ (otherwise, swap $x_{n_{1}}$
and $x_{n_{2}}$). Note that $x_{n_{1}}=n_{1}\omega-N_{1}$ and $x_{n_{2}}=n_{2}\omega-N_{2}$
for some integers $N_{1}$, $N_{2}$. Hence $x_{n_{1}}-x_{n_{2}}=(n_{1}-n_{2})\omega-(N_{1}-N_{2})\in A$.
This shows that $\inf A=0$.
