Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Peano Axioms (partially) formalize our intuitive notion of arithmetic. Partially because they also describe the behaviour of nonstandard models and there are some theorems that they can not prove that might seem, at first sight, to be within their domain e.g. Goodstein's Theorem and some of Harvey Friedman's combinatorial theorems http://arxiv.org/abs/math/9811187. So there is room for some clever soul to find additional axioms which would be independent of PA and also convey a natural intuitive property of the integers. These additional axioms would prove new theorems and also restrict the class of nonstandard models.

Could such a process ever be complete in the sense that all nonstandard models would be excluded?

PS I read the FAQ and am not sure if this is a suitable question. My apologies if it is not.

Larry Wickert, Truth or Consequences, New Mexico

share|improve this question
Not if you restrict to first-order axioms, thanks to Lowenheim-Skolem: en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem . If you don't, then see en.wikipedia.org/wiki/… . –  Qiaochu Yuan Oct 17 '11 at 1:18
I suspect that the OP means something nonstandard by "nonstandard," namely not elementarily equivalent to the standard model. The theory of all first-order propositions that hold true in the standard model would only have models elementarily equivalent to the standard model, but this theory would lack even a recursive axiomatization, by Godel. –  David Feldman Oct 17 '11 at 4:49
add comment

3 Answers

up vote 6 down vote accepted

There is no hope for a first-order theory to eliminate non-standard models. If a first-order theory over a countable language has an infinite model then it has models of all infinite cardinalities (Löwenheim-Skolem Theorem). In the case of a theory $T$ of arithmetic. If the standard model satisfies $T$, then $T$ must have an uncountable model, which has to be non-standard (though there will be countable non-standard models of $T$ as well).

If you formulate PA in second-order logic, then it is already complete with respect to the standard semantics.

share|improve this answer
Alternatively, one can use Los's theorem instead of Lowenheim-Skolem: the ultrapower ${}^* {\bf N}$ of ${\bf N}$ obeys all the first order sentences that the natural numbers do, but is clearly a non-isomorphic model of arithmetic. –  Terry Tao Oct 17 '11 at 2:57
Yes, and a simple compactness trick will also do it. In fact, it seems that it is hard not to prove that there are nonstandard models! –  François G. Dorais Oct 17 '11 at 3:30
add comment

There is a stronger result than the ones stated. As noted above, by compactness (and the L\"owenheim-Skolem theorems) we have a lot of nonstandard models.

However, by G\"odel's First Incompleteness Theorem, we have that no recursive (consistent) extension of PA can be complete. So, no finite set of axioms (or even axiom schemes, like induction) can ever even nail down the complete theory of arithmetic. Not only will there be nonstandard models, but there will be models which satisfy statements which are not "true."

share|improve this answer
add comment

Peano's axioms are categorical if you take the second order formulation, i.e. no non-standard models. If you replace the second order axiom by a first order axiom scheme, they cease being categorical by the compactness theorem, as Francois stated. If you embed the second order version into a first-order model of set they, they remain categorical relative to the set theory, but the set theory itself has non-standard models.

Non-standard models are useful because they help us formalize some features of infinitary processes, i.e. non-standard analysis, geometric group theory, etc. In practice, model theorists would usually have way more structure than simply Peano's axioms, such as a real closed field or hyperbolic group.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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