How can I be more confident that my proof is correct? (Real Analysis) I am going through a textbook to prepare for Real Analysis and I recently tried the problem:
Let $w\in\mathbb{R}$ be an irrational positive number. Set $A = \{ m+nw \mid m+nw > 0,  m,n\in\mathbb{Z} \}$.
Show that $\inf A = 0$.
Attempt:
Define $\alpha = \inf A$.
Clearly $\alpha \geq 0$, for if not, $\alpha < 0$ where $0$ is a greater lower bound than $\alpha$.
We claim that $\alpha = 0$. Suppose not, that is, $\alpha > 0$.
Let $\epsilon = \alpha - m - wn$ for integers $m,n$ s.t. $\alpha > m + wn$.
Then $n\epsilon = n(\alpha - m - wn) \Rightarrow n\epsilon = n\alpha - nm - wn^2 \Rightarrow mn + wn^2 = n\alpha - n\epsilon \Rightarrow mn + wn^2 = n(\alpha - \epsilon).$
$\Rightarrow m + wn = \alpha - \epsilon \in A$, with $\alpha - \epsilon < \alpha$, which contradicts the fact that $\alpha \leq x$ for $\forall x \in A$.
Hence inf $A = 0$.
My problem lies in whether my first let step for $\epsilon$ is correct. How can I be more confident?
 A: We shall show that positive numbers of the form $m+wn$ for $m,n\in\mathbb{Z}$ can get arbitrarily small.  Let $\epsilon>0$ be arbitrary.
Without loss of generality, assume that $w>0$.  Take $N$ to be a positive integer such that $N\geq\frac{1+w}{\epsilon}$.  For $m,n\in\{0,1,2,\ldots,N\}$, we have $0\leq m+wn\leq N(1+w)$.  By the Pigeonhole Principle, there are $m_1,n_1,m_2,n_2\in\{0,1,2,\ldots,N\}$ such that $\left(m_1,n_1\right)\neq \left(m_2,n_2\right)$ and $$\Big|\left(m_1+wn_1\right)-\left(m_2+wn_2\right)\Big|\leq\frac{N(1+w)}{(N+1)^2-1}=\frac{1+w}{2+N}<\frac{1+w}{N}\leq\epsilon\,.$$
Without loss of generality, we assume that $m_1+wn_1\geq m_2+wn_2$.  Define $\mu:=m_1-m_2$ and $\nu:=n_1-n_2$.  Then, $\mu,\nu\in\mathbb{N}$ are such that $\mu+w\nu<\epsilon$.  Clearly, $\mu+w\nu\neq 0$ (as $w\notin\mathbb{Q}$).  Thus, $0<\mu+w\nu<\epsilon$.
If you use continued fractions, you can also show that there exists a sequence $\left\{\left(p_k,q_k\right)\right\}_{k\in\mathbb{N}}$ with $p_k,q_k\in\mathbb{Z}$ for all $k\in\mathbb{N}$ and $0<q_1<q_2<\ldots$ such that $\left|\frac{p_k}{q_k}-w\right|<\frac{1}{q_k^2}$ holds for every $k\in\mathbb{N}$.  Then, when $k$ is sufficiently large, $q_k\geq\frac{1}{\epsilon}$, and so $\left|p_k-wq_k\right|<\frac{1}{q_k}\leq\epsilon$, whence you can then take $\left(\mu,\nu\right)$ to be either $\left(p_k,-q_k\right)$ or $\left(-p_k,q_k\right)$.
A: Consider $G=\{m+nw: m,n\in \mathbb {Z}\}.$ The irrationality of $w$ shows that each element of $G$ has only one representation as $m+nw.$ Now given $n,$ there is a unique $m$ such that $m+nw\in (0,1).$ It follows that there are infinitely many points of $G$ in $(0,1).$ By Bolzano-Weierstrass, $G$ has an accumulation point. So given $\epsilon>0,$ we can find $x,y \in G$ such that $0<|x-y|<\epsilon.$ Note both $x-y,y-x$ are in $G,$ and one of them is positive. This implies $G$ contains arbitrarily small positive numbers, giving the desired conclusion. (Actually $G$ is dense in $\mathbb {R}.$)
