Rudin Dedekind Cuts - Define Negative Multiple - Step 8

Question

I am following the construction of the reals using Dedekind Cuts in Rudin's Chapter 1 Appendix. I would appreciate some review on one of the last steps, Step 8, where he looks at multiplication of the reals.

I am trying to prove property (b), namely that $r^*s^*=(rs)^*$. Here is a good proof for when $r, s>0$. But what about when $r>0, s<0$?

A few relevant definitions: (Note that here, so far, $\mathbb{R}$ is the set of Dedekind cuts and $\alpha, \beta$ are cuts.)

1. Fix $\alpha \in \mathbb{R}$. Let $-\alpha$ be the set of all $p\in\mathbb{Q}$ with the following property: There exists $q\in\mathbb{Q}$ with $q>0$ such that $-p-q\notin \alpha$.
2. $\mathbb{R}^+$ is the set of all $\alpha \in \mathbb{R}$ with $\alpha > 0^*$. ($0^*$ is the set all negative rational numbers.) Then if $\alpha, \beta \in \mathbb{R}^+$, we define $\alpha\beta$ to be the set of all $p$ such that $p\leq rs$ for some choice of $r\in\alpha, s\in\beta$ with $r,s>0$.
3. If $\alpha>0$ and $\beta <0$, $\alpha\beta = -(\alpha(-\beta))$.
4. We associate with each $r \in \mathbb{Q}$ the set $r^*$ which consists of all $p \in \mathbb{Q}$ such that $p<r$. These $r^*$ are cuts so the above apply.

Here is what I have. It relies heavily on manipulating definitions so if someone could check this I would be grateful. Thank you.

First, I show $r^*s^*\subset (rs)^*$:

• Suppose $p\in r^*s^*$.
• By (3), $r^*s^*=(-(r^*(-s)^*))$.
• Thus, by (1), there exists $q>0$ such that $-p-q\notin r^*(-s)^*$.
• By (2), this means that for all $u, v$ with $u\in r^*, v\in (-s)^*$, and $u, v >0$, we have $-p-q>uv$.
• By (4), for all such $u$ and $v$, we know $u<r$ and $v<-s$, so this means $-p-q\geq r(-s)$.
• Since $q>0$, this means $-p> r(-s)$
• This means $p < rs$.
• And by (4), this means $p\in (rs)^*$.

Next I show $(rs)^* \subset r^*s^*$:

• Suppose $p \in (rs)^*$. Then by (4), $p<rs$.
• Since $r>0, s<0$, we have $rs < 0$, and so $-p>r(-s)$.
• Then there exists $q > 0$ such that $-p-q > r(-s)$.
• By (4), for all $u \in r^*, v\in (-s)^*$, we have $u<r$ and $v<-s$. Then for all $u \in r^*$ and $v \in (-s)^*$ with $u, v >0$, we have $-p-q>uv$.
• Therefore, by (2), $-p-q\notin r^*(-s)^*$.
• Therefore, by (1), $p \in -(r^*(-s)^*)$.
• By (3), $-(r^*(-s)^*) = r^*s^*$, and so $p \in r^*s^*$.

EDIT: I have noticed one issue with the above. Twice I wrote that by (3), $-(r^*(-s)^*) = r^*s^*$. However, a strict application of the definition of the product of a positive and negative cut would be that $r^*s^* = -(r^*(-(s^*)))$ (the difference is subtle, but there).

However, it is not hard to show that for any rational cut, $-(r^*) = (-r)^*$.

Suppose $p\in (-r)^*$. Then $p\in\mathbb{Q}$ and $p<-r$. Then there exists $q\in\mathbb{Q}$ with $q>0$ such that $p+q<-r$. Then $-p-q>r$, so $-p-q\notin r^*$, and by definition $p\in -(r^*)$.

Next, suppose $p\in -(r^*)$. Then by definition there exists $q\in\mathbb{Q}$ with $q>0$ such that $-p-q\notin r^*$, i.e. $-p-q\geq r$. Then $-p>r$, so $p<-r$ and $p \in (-r)^*$.

Answer 1

Based on my edit and Brian's comment above, it seems like what is in the post body is correct.