Rudin Chapter 1, Exercise 20: new identity element for Dedekind cuts

Question

I am working on exercise 20 in Chapter 1 of Baby Rudin on Dedekind cuts.

With reference to the Appendix, suppose that property (III) were omitted from the definition of a cut. Keep the same definitions of order and addition. Show that the resulting ord ered set has the least-upper-bound property, that addition satisfies axiom (A1) to (A4) (with a slightly different zero element!) but that (A5) fails.

Property (III) of cuts is that no cut has a maximal element, so our working definition of cut is a subset of rationals that is non-trivial and closed downwards.

I am trying to understand the intuition for the solution to this problem redefines the $0$ element, $0*$, from the set of negative rationals to the set of non-negative rationals. Surely, omitting property (III) removes the need to exclude $0$ and allows a cut to have a negative element. But why is it crucial to put $0$ back into the set? After working the problem out, I did notice that the existence of the zero element is crucial to invalidating axiom A5, the additive inverse. But wouldn't it be equally valid to include it, in practice? If that were the case, couldn't we define Dedekind cuts without property (III), with the same zero element as before, and still establish a complete ordered field, $\mathbf{R}$? I must be missing something.

Answer 1

We are constructing a field, hence we want the additive identity.

The previous zero element, $0_-$, the set of all negative rational number is no longer the additive identity because we can find another cut, say $A$ such that $A+0_- \ne A$.

For example, we can pick $A$ to be $0^*$, the set of non-positive rational number. This cut is different from $0_-$. We have $0^* + 0_-=0_-$, which shows that $0_-$ is not the additive identity.

In fact, $0^*$ is the new additive identity.