Dedekind Cuts in Rudin' analysis - Step 4 This question refers to the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.
More specifically, in the last paragraph of step 4, Rudin says that for $\alpha$ a fixed cut, and given $v \in 0^*$, setting $w=- v / 2$, there exists an integer $n$ such that $nw \in \alpha$ but $(n+1)w$ is not inside $\alpha$. Rudin says that this depends on the Archimedean property of the rationals, however he has not proved it. 
Could somebody prove the existence of the integer $n$?
 A: By definition there is some rational $q$ that is not in $\alpha$. Let $p=|q|+1$; then $p>q$, so $p\notin\alpha$, and $p>0$. The rational $w$ is also positive, so by the Archimedean property of the rationals there is a positive integer $m$ such that $mw>p$, and therefore $mw\notin\alpha$.
We also know that there is some rational $r\in\alpha$. Let $s=-|r|-1$; then $s<r$, so $s\in\alpha$, and $-s>0$, so again by the Archimedean property there is a positive integer $k$ such that $kw>-s$. But then $(-k)w<s$, so $(-k)w\in\alpha$. We now have positive integers $k$ and $m$ such that $-kw\in\alpha$ and $mw\notin\alpha$.
Let $S=\{i\in\Bbb N:(i-k)w\notin\alpha\}$; $m+k\in S$, so $S\ne\varnothing$. Since $\Bbb N$ is well-ordered, $S$ has a smallest element, $j$. Note that $0\notin S$, since $-kw\in\alpha$, so $j>0$. Let $n=(j-1)-k$. Then $nw\in\alpha$, and $(n+1)w\notin\alpha$.
A: If $v\in 0^*$, then $v<0$, so $w>0$. Let $\gamma=\sup \alpha$.  The Archimedean property says that for any $\gamma$ there is some integer $m$ such that $mw\geq\gamma$. Since $\mathbb{N}$ is well-ordered, there is a smallest $m$ such that $mw\geq\gamma$, call this $n+1$. This means that $nw<\gamma$, and hence $nw\in \alpha$. But also $(n+1)w\notin \alpha$ since it is larger than $\gamma$ (OK, so there might be some subtleties with endpoints you should think about, but essentially this is the idea).
