I was studying Rudin's Principle of Mathematical Analysis. In his construction of the reals there is a part I am having trouble comprehending. It concerns the additive inverse of the real.
I understand that for a real number $\alpha$ , which is a cut, the additive inverse is given by the cut $\beta$ - which is the set of all rational numbers $p$ with the property that , there exists a rational $r > 0$ such that $-p-r \notin \alpha $ .
Now we need to show that $\alpha + \beta$ is indeed $0$ .
For this we need to show that the cut given by $\alpha + \beta$ is a subset of the cut given by $0$ and the cut given by $0$ is a subset of $\alpha + \beta$ .
I am having trouble understanding proof of the second part viz. the cut given by $0$ is a subset of $\alpha + \beta$ .
As is customary let us denote the cut representing the real $0$ as $0^*$ .
Here is how Ruding approaches the proof :-
Pick a rational $v \in 0^*$ put $w = -v/2$ . Then , $w>0$ and there is an integer $n$ such that $nw \in \alpha$ but $(n+1)w \notin \alpha$ . This is the statement I am having trouble proving rigorously.
Rudin states that this follows from the fact that $\mathbb Q$ has the Archimedean property. Now I appreciate that $\mathbb Q$ has the Archimedean property and that it can be proven without recourse to a least upper bound property which $\mathbb Q$ conspicuously lacks.
But I am unable to prove the existence of $n$ rigorously. I am having trouble doing so because the set of rationals(cut) $\alpha$ does not have a largest element.
Here is how I could somewhat proceed :- Consider the set of integers(let us call it $I$), $m$ such that $mw \in \alpha$ . This set is non-empty . As we know $\alpha$ is non empty and therefore it has at least one rational say $q$ . Now since $\mathbb Q$ has the Archimedean property there exists an integer $m_1$ such that $m_1 w < q $ . Thus the set of integers $I$ is non empty . Similarly we can show that the set $I$ is bounded above. Because $\alpha$ is bounded above. The largest element of $I$ is the $n$ we are looking for. And $(n+1) w$ won't be in $\alpha$.
Is my proof ok ? I am having second thoughts about it.