# Question about the Dedekind completion of a non-archimedean ordered field.

Suppose we have a non-archimedean ordered field $F$. No such field is Dedekind-complete, for that property implies the Archimedean one. But, we can of course fill in the gaps and form the Dedekind completion. Suppose we do this. Denote the resulting structure by $D(F)$. Define the arithmetic operations in the same way as was done for the completion of the rationals to make the reals. What do we get?

It's not a field, that's for sure: consider the sup of the integers of the field, $s$. Now, we can see from the definition of addition of cuts that $s + 1 = s$. This means that addition is not invertible, so it's not a field. What is it? Does it have interesting properties, despite its non-fieldness?

The question: suppose we consider the set of elements $D(F) \backslash F$. Do all these elements "absorb" at least one nonzero element of $D(F)$ -- i.e. given any $x \in D(F) \backslash F$, does there exist at least one $y \in D(F)$ such that $x + y = x$ and $y \ne 0$? If not, what conditions must be met by the field $F$ and element $x$ for it to be "non-absorptive"?

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I don't have the free time to really sink my teeth into this at the moment, but it's a great question. I favorited it and will try to keep an eye on it. –  Pete L. Clark May 13 '13 at 3:30

$\newcommand\Q{\mathbb{Q}}$

It is a great question! In general, the answer is no, for there are fields with elements in the Dedekind closure that do not absorb in this way.

Here is an example. Let $\Q^*$ be a nonstandard version of the rational field, obtained for example as an ultrapower of $\Q$, so that it is an elementary extension of $\Q$. So this is a nonarchimedean ordered field. Furthermore, $\sqrt{2}$ does not exist in $\Q^*$. But $\sqrt{2}$ fills a Dedekind cut $S$ in $\Q^*$ — the cut of elements in $\Q^*$ that are negative or whose square is less than $2$ — and so it is in $D(\Q^*)\setminus \Q^*$. But notice that the cut $S$ has the property that it can be bridged with arbitrarily small elements of $\Q^*$, since this fact is true in $\Q$. In other words, for any $0\lt y$, no matter how small, there is $u<\sqrt{2}$ such that $u+y>\sqrt{2}$. It follows that $\sqrt{2}<\sqrt{2}+y$ for any positive $y$, and so $\sqrt{2}$ is not absorbing.

In any field, one can characterize the absorbing elements exactly by this bridging property. Namely, if $x\in D(F)\setminus F$, then the following are equivalent:

1. $x$ is absorbing; that is, $x+y=x$ for some nonzero positive $y\in F$.
2. $x-u$ is bounded away from $0$, for $u\lt x$.
3. $(x-u)\to 0$ as $u\to x$ from below.
4. The cut determined by $x$ is not "bridgeable", that is, it cannot be jumped over with arbitrarily small elements of $F$.

The $\sqrt{2}$ example worked precisely because that cut is bridgeable, and can be jumped over with arbitrarily small elements of $\Q^*$, since the analogue of this is true in the standard model $\Q$. Meanwhile, the cut in your example of the supremum of the finite elements cannot be jumped over even with any finite elements.

Update. Finally, let me say that this bridging idea points out the right way to form the Dedekind completion of a nonarchimedean ordered field: you should let $D(F)$ fill only the bridgeable cuts, rather than all cuts. If you do it this way, you will be able to define the field structure on the completion, and you will thereby produce an ordered field, in which every bridgeable cut is filled, and in which the original field is dense.

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The bridgeable cuts are known as the "Dedekindian" cuts, and the collection of all Dedekindian cuts in any ordered field forms an ordered field in which every Dedekindian cut is filled, in which the original field is dense. This is known as the Scott completion of the field. –  JDH Oct 18 '13 at 12:54
So is every, or even any, hyperreal field $^{*} \mathbb{R}$ "Dedekindian-complete" (as opposed to "Dedekind-complete"), i.e. every "Dedekindian cut" is filled? Are the surreal numbers $\mathbf{No}$ complete in this sense? –  mike4ty4 Oct 19 '13 at 0:15
I don't believe that every nonstandard model of the theory of the reals $\mathbb{R}^*$ is Scott complete, since there could be Dedekindian cuts that are not part of the nonstandard world. Meanwhile, No is Scott complete for cuts that are sets, since no set cut is bridgeable in No, as any set of infinitesimals is bounded away from zero. –  JDH Oct 19 '13 at 11:32
By the way, I think this question is probably better-placed on MathOverflow.net. –  JDH Oct 19 '13 at 11:32