# What if one of Dedekind cut's properties were omitted?

1. Sep 18, 2008

### Ronn

Hello.

I have a question about Dedekind' cut.
Problem #20 of Baby rudin's p23 asks: prove why axiom (A5) on page 5 fails if cuts had maximum elements.

(A5): To every x in F( a field) corresponds an element -x in F such that x + (-x) = 0.

I guess Archimedean property is a starting point to prove A5 fails. To do that I need to understand the relation between the existence of largest element and Archimedean Property. In what sense are they related? I am puzzled. Please help me out.

2. Sep 20, 2008

### gunch

You should go through the proof and observe where the maximum element property is used. The first place is in the proof of (A4). This can easily be fixed by letting $$0^*$$ be all non-positive rational numbers instead of all negative (it can easily be shown that it must be defined in this way, because for cuts with a maximum 0 must be included, but no positive number can be included). The proof of (A5) never use the fact that cuts don't have maximum elements, but it uses the old definition of $$0^*$$. It starts to break down when the Archimedean property is used, since it's not necessarily true when w=0, but with the new definition of $$0^*$$ we can have w=0. However we already proved earlier that r+s < 0 if $$r\in \alpha, s \in \beta$$ so it certainly can't work with this definition of the negative. We still need to prove that it can't work with any definition of the negative.

We can easily show that we can define a negative such that there exist a negative for all rational numbers. Simply let $$b = \{p | \forall x\in a \,.\, p < -x \}$$. So to arrive at a contradiction lets consider the cut of an irrational number. Let,
$$a = \{x | x^2 < 2 \textrm{ or x is negative}\}$$
We can easily verify that this is indeed a cut. Assume that there exist a set b such that $$a+b = 0^*$$. We know that a doesn't contain a maximum element. Since $$0^* \subseteq a +b$$, we have x+y = 0, for some x in a and y in b. There exist an element z>x in a, but then $$0 < z+y \in a+b$$ which is a contradiction since no positive number can be in a+b. The same proof works for all irrational numbers.