Proof in Rudin. Denseness of $\mathbb{Q}$ in $\mathbb{R}$.

Question

There is a point in Rudin's proof where he says that by the Archemedean property, there exists $m_1, m_2 \in \mathbb{N}$ such that $$nx <m_1$$ and $$-nx<m_2$$ and therefore $$-m_2 <nx <m_1.$$No issues there.

Then he makes the claim that then there exists an $m\in \mathbb{Z}$ (with $m_2\leq m \leq m_1$) such that $$m-1 \leq nx <m,$$ virtually saying that $nx$ lives between two consecutive integers. Now this is certainly $intuitively$ true (and perhaps so intuitively true that Rudin didn't feel as though he had to prove it), but I was wondering how to $prove$ it? I couldn't think of anything that stayed away from saying things like "obviously" or "goes without saying".

Answer 1

It simply uses that any subset of $\mathbf Z$ which is bounded from below (and in particular, any finite subset) is well-ordered, like $\mathbf N$, i.e. has a least element.

SSo here, thet of $m\in\mathbf Z$ such that $nx< m$ is bounded from below by the integer $m_2$, so there is a least such $m$, which by definition means that $m-1\le nx$.

Answer 2

Let $m_0=\max\{m\in\mathbb Z:\ m\leq nx\}$. If $m_0+1\leq nx$, then we would contradict the maximality of $m_0$, so $m_0+1>nx$. Thus $$ m_0\leq nx< m_0+1. $$ If you want to be really careful, you need to prove that on any bounded set of integers, the supremum is a maximum.