2
$\begingroup$

I am doing a research in foundation of maths, especially in logic area. Right now, I am building a set $\mathbb{L}$ through this five axioms:

  1. Additive axiom: closeness, commutative, associative, 0 as identity, additive inverse.
  2. Multiplicative axiom: closeness, commutative, associative, 1 as identity, multiplicative inverse.
  3. Distributive axiom.
  4. Total partial order axiom: reflexive, transitivity, anti-symmetry, totality.
  5. Existence of positive infinitesimal.

As you might noticed already, axioms 1-4 are just like axioms in reals. The only addition in here is axiom 5, which make everything be much more complicated.

My problem for now is I want to prove that $\mathbb{L}$ is consistent and therefore, semantically, I need to provide a model for it. Can you help me with it? I have been thinking about this for weeks and still cannot come up with a right one. I thought $\mathbb{R}^\omega$ with a lexicographic order would be a right model, but unfortunately, I didn't think it is. Cheers!

$\endgroup$
  • 1
    $\begingroup$ You should read about non-standard analysis. More specific Robinson and Los' work. $\endgroup$ – Asaf Karagila Nov 4 '15 at 11:36
  • $\begingroup$ Hi Asaf. Thank for your reply. I did read it from Rob Glodblatt's book. But I don't think it helps me much in answering my question. :-( $\endgroup$ – Anggha Nugraha Nov 4 '15 at 11:51
  • 1
    $\begingroup$ Note that "existence of positive infinitesimal" is not a first-order statement without further clarification, unlike 1-4 which seem to be axioms for an ordered field. Robinson's approach constructs a model of an ordered field in which infinitesimals exist, so in that respect, Asaf's suggestion is quite relevant. $\endgroup$ – hardmath Nov 4 '15 at 11:55
  • 1
    $\begingroup$ That is a term, hyperreal number, used by some authors, yes. $\endgroup$ – hardmath Nov 4 '15 at 12:04
  • 1
    $\begingroup$ @NoahSchweber One should be a bit careful about where the model begins and ends here. The hyperreals see themselves as Archimedean. This is not inconsistent because to them the natural numbers include the infinite hypernaturals, and every infinitesimal hyperreal is less than the reciprocal of some hypernatural. $\endgroup$ – Ian Nov 4 '15 at 23:29
2
$\begingroup$

What you are describing are precisely the non-Archimedean ordered fields. ("Non-Archimedean" refers to the existence of infinitesimals and infinite elements; the "ordered field" bit covers the rest.)

There are lots of these (e.g., continuum-many non-isomorphic countable examples). One way to construct them is via ultrapowers: fix a nonprincipal ultrafilter $\mathcal{U}$ on a set $I$, and let $$\prod_I\mathbb{R}/\mathcal{U}$$ be the set of maps $I\rightarrow \mathbb{R}$, thought of as $I$-length sequences of reals, modulo the equivalence relation $$(a_i)_{i\in I}\sim (b_i)_{i\in I}\iff \{i: a_i=b_i\}\in\mathcal{U}.$$ This is a set, and we can define an ordering $<$ and operations $+$ and $\times$ on it in the obvious way (we have to check that this is well-defined, but this is not hard and a good exercise). The resulting object is a non-Archimedean (since $\mathcal{U}$ is nonprincipal) ordered field.

In fact, the ordered field $\prod_I\mathbb{R}/\mathcal{U}$ is an elementary extension of the ordered field of real numbers; this is (an instance of) Los' Theorem. So, among other things, it is a real closed field, and its first-order theory is computable.


As far as examples of these things go: there are natural examples of non-Archimedean ordered fields, such as fields of power series (see e.g. https://en.wikipedia.org/wiki/Formal_power_series#Ring_structure).

$\endgroup$
1
$\begingroup$

Ultrapowers and power series are not needed to give an example of a non-archimedean ordered field. Let $\Bbb{Q}[x]$ denote the ring of polynomials in the indeterminate $x$ over the rational numbers $\Bbb{Q}$. Then the field of rational functions $\Bbb{Q}\{x\} = \{ \frac{f}{g} \mathrel{|} f, g \in \Bbb{Q}[x], g \neq 0\}$ can be ordered in such a way that $x < \frac{1}{n}$ for every positive integer $n$. Then $\Bbb{Q}\{x\}$ is a non-archimedean ordered field (and every non-archimedean ordered field has a sub-ordered-field isomorphic to $\Bbb{Q}\{x\}$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.