In the appendix on Dedekind cuts in Rudin's book, he defines for each $r \in \mathbb{Q}$ the Dedekind cut $r^* = \{q \in \mathbb{Q} \mid q < r\}$. He then asserts that $$r^* s^* = (rs)^*.$$ Rudin omits the proof and states that it is similar for the proof that $(r+s)^* = r^* + s^*$. I do not understand how this is the case, in part because I need to consider three cases: both positive, positive negative, and one positive and one negative.
Suppose both are positive, either $r^*, s^* > 0*$. Then $r^* s^* = \{ab \mid a \in r^*, b \in s^*, ab > 0\}$. To show inclusion in the right-hand side, I need to show that given an $ab \in r^* s^*$, I have $ab < rs$. Since $a \in r^*$, $a < r$. Since $b \in s^*$, $b < s$. So $ab < rs$, giving the first inclusion.
I'm having trouble with the second inclusion. If $q \in (rs)^*$, then $q < rs$. Attempting to follow the proof for addition, I can then say that $rs - q > 0$ and define $rs - q = t$ .Then I want to construct, with $t$, two positive rational numbers whose product is $q$, the first term of which is in $r^*$, the second of which is in $s^*$. I can't figure out how to do this. Subtracting $t$, as in the addition proof, doesn't work, and multiplying it gives an extraneous $t^2$ factor.
I haven't had much success with the remaining cases, but I think I could make more progress if I could get this first one. Any help would be appreciated.
