Step 6 Real construction from Rational Baby Rudin

Question

Rudin's Books establishes

Multiplication is a little more bothersome than addition in the present context , since products of negative rationals are positive. For this reason we confine ourselves first to $R^+$, the set of all $\alpha\in R$ with $a>0^*$.

If $\alpha\in R^+$ and $\beta\in R^+$ , we define $\alpha\beta$ to be the set of all $p$ such that $p \leq rs$ for some choice of $r\in\alpha$, $s\in\beta$, $r>0$, $s>0$. We define $1^*$ to be the set of all $q<1$.

Then the axioms (M) and (D) of Definition 1.12 hold, with $R^+$ in place of $F$, and with $1^*$ in the role of $1$.

I have proven that $\alpha\beta$ satisfies (M1), (M2), (M3), (M4) but I'm stuck in (M5). I found how construct $1/\alpha$ however I have some questions: Query regarding proof of Existence Theorem (1.19, pg-17, Baby Rudin). It define $1/\alpha$ as:

Fix $\alpha\in R^+$. Let $\beta^+$ be the set of all $p$ with the following property: There exists $r>1$ such that $\frac{1}{pr}\notin\alpha$. Furthermore let $\beta=\beta^+ \cup 0^*\cup 0$ be $1/\alpha$.

Then they prove that $\beta$ is a cut such that $\beta\in R^+$. They also proof that $\alpha\beta=1^*$:

If $r\in\alpha$ and $s\in\beta$, $r>0$, $s>0$, then $1/s\notin\alpha$, $r<1/s$, hence $rs<1$. Thus $\alpha\beta\subset 1^∗$. To prove the opposite inclusion, pick $u\in 1^*$, $u>0$, put $v=\left[(1+u)/2\right]^2$, $w=2/(1+u)$. Then $u<v$, $w>1$, and there is an integer $n$ such that $w^n\in\alpha$ but $w^{n+1}\notin\alpha$. Put $p=1/w^{n+2}$. Then $p\in\beta$, since $(1/p)(1/w)=w^{n+1}\notin\alpha$ and $v=1/w^2=w^np\in\alpha\beta$, so that $u\in\alpha\beta$. Thus $1^∗\subset\alpha\beta$. We conclude that $\alpha\beta=1^*$.

Questions

1. I want to know what is the intuitive or general idea behinde the election of $v$ and $w$ because I think that we could take some different $v$ and $w$.
2. Furthermore why $u<v$?.
3. Why there's some intenger $n$ such that $w^n\in\alpha$ and $w^{n+1}\notin\alpha$.
4. ¿Is there some construction of $1/\alpha$ such that it's different of defined $\beta$ here?

Answer 1

I was looking for some part of the questions and I found another construction of $1/\alpha$. This construction is very similar to $-\alpha$ construction presented in Baby Rudin.

Given $\alpha\in\mathbb{R}^+$. Let us define $ \beta = \{p\in\mathbb{Q}:p\leq 0\}\cap \{p\in\mathbb{Q}:p>0$ and there's $r>0$ such that $\frac{1}{p}-r\notin\alpha \}$. We show that $ \beta $ is a cut and $ \beta\alpha = 1^* $.

proof:

• Let $ \beta^+ $ be the set $ \{p\in\mathbb{Q}:p>0$ and there's $r>0$ such that $\frac{1}{p}-r\notin\alpha\} $. $ \beta^+ $ is not empty because we can found some $ p>0 $, $ r>0 $, $ s>0 $ and $ s\notin\alpha $ such that $ s < \frac{1}{p} - r $. And $\beta$ is not $ \mathbb{Q} $ because there's some $ s\in\alpha $, $ s>0 $, $ p>0 $ such that $ \frac{1}{p}<s $. $ \beta $ satisfies (I).
• Pick $ p\in\beta^+ $ and pick $ r>0 $ so that $ \frac{1}{p}-r\notin\alpha $. If $ 0<q<p $ then $ \frac{1}{p}-r<\frac{1}{q}-r $ hence $ \frac{1}{q}-r\notin\alpha $ and $ q\in\beta^+ $. $ \beta $ satisfies (II).
• Pick $p\in\beta^+$ and $ r>0 $ so that $ \frac{1}{p}-r\notin\alpha $ then every $ s $ with $ 0<s<r $ satisfies $ r-\frac{1}{p}<s $, because $r-\frac{1}{p}<0$, hence $ \frac{1}{q}=\frac{1}{p}-r+s>0 $ and $ \frac{1}{q}<\frac{1}{p} $ wich implies $ p<q $ and $ \frac{1}{q}-s\notin \alpha $ with $ s>0 $, then $ q\in\beta^+ $. $ \beta $ satifies (III).

• It's clear that $ \alpha\beta $ is all $ p \leq rs $ for some $ s\in\alpha $, $ s>0 $, $ r\in\beta^+ $. Pick $ s\in\alpha $, $ s>0 $, $ r\in\beta^+ $, $ t>0 $ so that $ \frac{1}{r}-t\notin\alpha $ then $ s<\frac{1}{r}-t $ and $ sr<1-tr $ hence $ sr<1 $ wich implies that $ \alpha\beta \subset 1^* $.

• It's clear that every $ p\in 1^* $ with $ p\leq 0 $ is in $ \alpha\beta $. Let us take $ 0<p<1 $, then there's some $ n\in\mathbb{N} $ such that $ p<\frac{n}{n+2} $ and every $ m\geq n $ satifies $ p<\frac{m}{m+2} $.
  Pick $ r\in\alpha $, $ r>0 $, $ 0<q<r/n $, then there's some $ k\in\mathbb{N} $ such that $ kq\in\alpha $ and $ (k+1)q\notin\alpha $. If $ k<n $ then $ k=n-l $ for some $ l\in\mathbb{N} $ which implies $(k+1)/n = 1+(1-l)/n \leq 1$ and $ (k+1)q/n\leq q < r/n $ and $ (k+1)q\in\alpha $ which is contradictory, hence $ k\geq n $.
  It's also clear that $ \frac{1}{(k+2)q}\in\beta $ because $ (k+2)q - q\notin\alpha $. This implies $ p<(kq)\cdot\frac{1}{(k+2)q} $ and $ p\in\alpha\beta $. Hence $ 1^*\subset\alpha\beta$ and $\alpha\beta=1^*$.