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All the statements of the Archimedean property with which I am acquainted fundamentally uses ℕ -- more than as a totally ordered semi-group, really being the 'standard model' of the naturals. It is a fundamental ingredient in showing that the reals are (up to isomorphism) the only complete totally ordered field.

In this particular result, there are two non ingredients which seem to be of a fundamentally different nature: Dedekind completeness (which is about subsets, while all the other axioms are about elements), and the Archimedean property, which pre-supposes the existence of the Naturals. But because being 'Archimedean' already makes sense in (ordered) monoids, that is the one I am most interested in.

Crazy scenario: replace the 'naturals' in the Archimedean property with a non-standard model of the 'naturals' (call this N-Archimedean). Now the reasoning for the argument that all Archimedean totally ordered fields are sub-fields of ℝ readily lifts, but no longer proves quite the same thing.

In other words, it seems that this unique position of ℝ is in part due to ℕ already being baked in to the question. And thus my question: is there a proper generalization of the Archimedean property which is ``properly abstract''?

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If you've got an ordered field, then can't you embed the naturals into it by sending them to an $n$-ary sum of the multiplicative unit? That is, you know by construction that any ordered field must have at least as many elements of the field as the naturals. So there's no extra presupposition in using N, unless I'm missing something very basic to your question. (A similar argument seems to work for a totally ordered monoid with an element that's not equal to the zero.) – Neel Krishnaswami Feb 17 '10 at 18:00
That is certainly part of the answer - in other words, the combination of 'total order' and 'field' already allows (as a conservative extension?) the creation of the naturals. And indeed, for a monoid with a distinguished non-zero element, that is enough to 'create' the naturals. But then, can we factor our definitions so that these derived notions are not directly baked in, but rather become theorems? – Jacques Carette Feb 17 '10 at 18:33
up vote 13 down vote accepted

It is not surprising that some versions of the Archimedean property concern subsets of the order rather than merely elements. The reason is that the Archimedean property is provably not expressible in a first order manner.

This is because the structure of the reals R, as an ordered field, say, (but one can add any structure at all), has elementary extensions to nonstandard models R* which are non-Archimedean. This means that any statement in the language of ordered fields that is true in the reals R will also be true in the nonstandard reals. To prove that such models exist is an elementary application of the Compactness theorem, and one can also construct them directly via the ultrapower construction. One can also control the cofinality of the nonstandard order. For example, one can arrange that every countable subset of R* is bounded. Since all these various nonstandard models R* satisfy exactly the same first order truths as the standard reals R, but are non-Archimedean, it follows that being Archimedean is not first-order expressible.

Being Archimedean is, of course, second-order expressible, and the usual definition is a second order definition. As Neel mentions in the comments, the natural numbers are identifiable as the smallest subset of the ordered field containing 0 and closed under successor n+1.

If one adds the natural numbers N as a predicate to the original model, so that one is looking at R as an ordered field with a unary predicate holding of the natural numbers, then the nonstandard model R* will include a nonstandard version N* of the natural numbers. This new field R*, which is not Archimedean, will nevertheless appear to be Archimedean relative to the nonstandard natural numbers N*. For example, for any x and y in R*, there will be a number n in N* such that nx > y.

Indeed, one can do amazing things along this line. Suppose that V is the entire set-theoretic universe, and let V* be a nonstandard version of it (such as an ultrapower by a nonprincipal ultrafilter on the natural numbers). Inside V*, the structure R* is thought to be the actual real numbers and so V* thinks R* is Archimedean, even though back in V we can see that it is mistaken, precisely because V* is using the wrong set of natural numbers for its conclusion. The model V* simply cannot see the true set of natural numbers sitting inside R*, because it does not have that set.

More generally, one can similarly describe what it should mean for any ordered field F to be Archimedean relative to a subring R. Perhaps this simple idea is the generalization for which you are looking? It is mainly amounting to the question of whether the subring is cofinal in the original order.

Thus, it is very natural to look at the possible cofinalities of the orders that arise in ordered fields (or the other types of structures that you consider). For any infinite regular cardinal κ, one may find an elementary extension of the reals R to a nonstandard ordered field R*, where the order of R* has a cofinal κ sequence. To do this, just perform a series of κ many extensions, each with new elements on top of the previous model. In κ many steps, the union of the structures you built will have an order with cofinality exactly κ.

If one only uses the ultrapower construction to construct the nonstandard models, however, then there are limits on the resulting cofinality of the order. Understanding these limits is a large part of Shelah's deep work on PCF (= possible cofinality) theory.

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Tarski showed the first-order theory of the reals is decidable. But of course the first-order theory of the naturals isn't. So (assuming consistency, I guess) we know you cannot construct the naturals in the reals using only first-order notions. But we can construct the natural numbers if we are allowed to quantify on sets of reals.

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A nonstandard model of the naturals is merely an abstract structure sharing the first order properties of the natural numbers. But it does not share all properties of the natural numbers, and is not isomorphic to the natural numbers. For example, a nonstandard model of the naturals may have any infinite cardinality, so that no bijection between these nonstandard naturals and the naturals exist, let alone an isomorphism.

If you define the naturals as the substructure obtained from a totally ordered field by taking the smallest set including the additive identity $0$ and being closed under addition by the multiplicative identity $1$, you get an isomorphic version of $\mathbb{N}$ instead of a nonstandard version of the naturals, according to an old result of Dedekind.

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I assume that, in your second paragraph, you mean a totally ordered field (which then implies characteristic 0)? This links in with Neel's comment whereby there is enough structure around to 'create' the naturals, so one may as well assume it. – Jacques Carette Feb 17 '10 at 18:39
Yes, you are right. – Michael Greinecker Feb 17 '10 at 19:02

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