# Set of all dedekind cuts

I have learned that set of all dedekind cut has properties of real number we know.

1. How do i know whether the set of all cuts is stronger than real number? i.e. set of all dedekind cuts may have a property P, which real number we know doesn't have.

2. Let A be a set of all dedekind cuts of $Q$. Let B be a set of all cuts of A. I heard that B has exactly the same property of A, thus A and B both can be viewed as real number. Please suggest me any book or sites i can study this..

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According to one construction $\mathbb R$ is the set of all Dedekind cuts in $\mathbb Q$, in which case "they" trivially have the same properties -- so if you want an interesting answer you'll need to specify what a "real number" means to you. –  Henning Makholm Jun 13 '12 at 21:10
The set of all Dedekind cuts of $\mathbb{Q}$ is a very specific set (modulo the choice of $\mathbb{Q}$), so it is bound to have set-theoretic properties that other constructions of $\mathbb{R}$ do not have. But the point is that as a complete ordered field, it does not. –  Zhen Lin Jun 13 '12 at 21:12
1. I think this depends on what you define the real numbers to be. 1. is trivial if you define the real numbers to be thing you obtain by the Dedekind Construction. Sometimes the reals are defined to be the unique ordered field with the least upper bound property. However when you say have the same property, in model theoretic terms you should specify the language where your property comes from. By the uniqueness above, there is a order ring isomorphism hence all property state-able in the language of ordered rings hold for one "real number" if and only if it holds for the Dedekind Cut real number. Even considering only linear structure on $\mathbb{R}$, there is a result that states that $(R, <)$ is in the unique complete linearly ordering that has a countable dense subset isomorphic to $(\mathbb{Q}, <)$.
2. Now if $A$ is what you obtained by the standard first Dedekind cut construction. You can prove that $\mathbb{Q}$ is dense in $A$. In general, if you preform the dedekind cut construction on $A$, you obtain another linear ordering which is complete and contains $A$ densely. If $A$ is contains in this new $B$ densely, then $\mathbb{Q}$ is contained in $B$ densely. $B$ is complete, so by the result at the end of 1. it is isomorphic to $\mathbb{R}$, at least as linear structures.
You can find these result stated quickly in Jech's $\textit{Set Theory}$ chapter 4.